3.6.93 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{(d+i c d x)^{5/2} \sqrt {f-i c f x}} \, dx\) [593]

Optimal. Leaf size=942 \[ -\frac {2 i b^2 f^2 \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b f^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 x^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

-2/3*I*b^2*f^2*(c^2*x^2+1)^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-2/3*b^2*f^2*x*(c^2*x^2+1)^2/(d+I*c*d*x)^(5/
2)/(f-I*c*f*x)^(5/2)+1/3*b^2*f^2*(c^2*x^2+1)^(5/2)*arcsinh(c*x)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*b*f^
2*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-2/3*I*b*f^2*x*(c^2*x^2+1)^(3/2)*(
a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*b*c*f^2*x^2*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))/(d+
I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-4/3*I*b*f^2*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2)
)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*f^2*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*
x)^(5/2)-1/3*c^2*f^2*x^3*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+2/3*f^2*x*(c^2*x
^2+1)^2*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*f^2*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^
2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+2/3*I*f^2*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*
f*x)^(5/2)-2/3*b*f^2*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)/c/(d+I*c*d*x)^(5/2)/
(f-I*c*f*x)^(5/2)-2/3*b^2*f^2*(c^2*x^2+1)^(5/2)*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/c/(d+I*c*d*x)^(5/2)/(f-I
*c*f*x)^(5/2)+2/3*b^2*f^2*(c^2*x^2+1)^(5/2)*polylog(2,I*(c*x+(c^2*x^2+1)^(1/2)))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*
x)^(5/2)-1/3*b^2*f^2*(c^2*x^2+1)^(5/2)*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(
5/2)

________________________________________________________________________________________

Rubi [A]
time = 0.87, antiderivative size = 942, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 18, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {5796, 5838, 5788, 5787, 5797, 3799, 2221, 2317, 2438, 5798, 197, 5789, 4265, 267, 5800, 5810, 294, 221} \begin {gather*} -\frac {c^2 f^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 x^3}{3 (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) x^2}{3 (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 \left (c^2 x^2+1\right )^2 x}{3 (i c x d+d)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2 x}{3 (i c x d+d)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 x}{3 (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) x}{3 (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 f^2 \left (c^2 x^2+1\right )^2}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 f^2 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}+\frac {b f^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i b f^2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b f^2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 \left (c^2 x^2+1\right )^{5/2} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b^2 f^2 \left (c^2 x^2+1\right )^{5/2} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}}-\frac {b^2 f^2 \left (c^2 x^2+1\right )^{5/2} \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (i c x d+d)^{5/2} (f-i c f x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/((d + I*c*d*x)^(5/2)*Sqrt[f - I*c*f*x]),x]

[Out]

(((-2*I)/3)*b^2*f^2*(1 + c^2*x^2)^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (2*b^2*f^2*x*(1 + c^2*x^2)^
2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (b^2*f^2*(1 + c^2*x^2)^(5/2)*ArcSinh[c*x])/(3*c*(d + I*c*d*x)
^(5/2)*(f - I*c*f*x)^(5/2)) + (b*f^2*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c*(d + I*c*d*x)^(5/2)*(f - I
*c*f*x)^(5/2)) - (((2*I)/3)*b*f^2*x*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/((d + I*c*d*x)^(5/2)*(f - I*c*f*
x)^(5/2)) - (b*c*f^2*x^2*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
+ (((2*I)/3)*f^2*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (f^2*x*(
1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (c^2*f^2*x^3*(1 + c^2*x^2)*
(a + b*ArcSinh[c*x])^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*f^2*x*(1 + c^2*x^2)^2*(a + b*ArcSinh[
c*x])^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (f^2*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/(3*c*(
d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((4*I)/3)*b*f^2*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])*ArcTan[E^A
rcSinh[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (2*b*f^2*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])*
Log[1 + E^(2*ArcSinh[c*x])])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (2*b^2*f^2*(1 + c^2*x^2)^(5/2)*Po
lyLog[2, (-I)*E^ArcSinh[c*x]])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*b^2*f^2*(1 + c^2*x^2)^(5/2)*
PolyLog[2, I*E^ArcSinh[c*x]])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b^2*f^2*(1 + c^2*x^2)^(5/2)*Pol
yLog[2, -E^(2*ArcSinh[c*x])])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((c + d*x)^(m
+ 1)/(d*(m + 1))), x] + Dist[2*I, Int[(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x]
, x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5787

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh
[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Dist[b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[x*((a + b*ArcS
inh[c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5788

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(
p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*(p + 1))), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a
+ b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2
)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &
& LtQ[p, -1] && NeQ[p, -3/2]

Rule 5789

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sech[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rule 5796

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :>
Dist[(d + e*x)^q*((f + g*x)^q/(1 + c^2*x^2)^q), Int[(d + e*x)^(p - q)*(1 + c^2*x^2)^q*(a + b*ArcSinh[c*x])^n,
x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 + e^2, 0] && HalfIntegerQ[p,
q] && GeQ[p - q, 0]

Rule 5797

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/e, Subst[Int[(
a + b*x)^n*Tanh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rule 5798

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^
(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^p], Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e
, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5800

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] - Dist[b*c*(n/(f*(m + 1)))*Simp[(
d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]
/; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 5810

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] + (-Dist[f^2*((m - 1)/(2*e*(p +
1))), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(2*c*(p + 1)))*Simp[
(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]
) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 1]

Rule 5838

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g
}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && ((EqQ[n, 1] && GtQ[p,
-1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] && LtQ[p, -2]))

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{5/2} \sqrt {f-i c f x}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(f-i c f x)^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \left (\frac {f^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}}-\frac {2 i c f^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}}-\frac {c^2 f^2 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {\left (f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i c f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (c^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (4 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 b c f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 b c^3 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {b f^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 x^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 b c f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (4 b c f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 i b^2 c f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (b^2 c^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i b^2 f^2 \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b f^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 x^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (4 b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i b^2 f^2 \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b f^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 x^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (4 b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i b^2 f^2 \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b f^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 x^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (4 b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i b^2 f^2 \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b f^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 x^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i b^2 f^2 \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b f^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f^2 x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c f^2 x^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 f^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b f^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b^2 f^2 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 4.62, size = 524, normalized size = 0.56 \begin {gather*} \frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left (\frac {a^2 (-2 i+c x)}{(-i+c x)^2}-\frac {a b \left (-i \cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right ) \left (\sinh ^{-1}(c x)-2 \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-\frac {1}{2} i \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (-2-3 i \sinh ^{-1}(c x)-6 i \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\frac {3}{2} \log \left (1+c^2 x^2\right )\right )+2 \left (\left (-1+\sqrt {1+c^2 x^2}\right ) \sinh ^{-1}(c x)+2 \left (2+\sqrt {1+c^2 x^2}\right ) \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\frac {1}{2} i \left (-2+\left (2+\sqrt {1+c^2 x^2}\right ) \log \left (1+c^2 x^2\right )\right )\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{\sqrt {1+c^2 x^2} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^3}-\frac {b^2 \left ((1-i) \sinh ^{-1}(c x)^2-\frac {\sinh ^{-1}(c x) \left (-2 i+\sinh ^{-1}(c x)\right )}{-i+c x}+2 \left (-i \pi +2 \sinh ^{-1}(c x)\right ) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+i \pi \left (\sinh ^{-1}(c x)-4 \log \left (1+e^{\sinh ^{-1}(c x)}\right )+4 \log \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+2 \log \left (\sin \left (\frac {1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )\right )-4 \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )-\frac {2 \sinh ^{-1}(c x)^2 \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )}{\left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^3}-\frac {2 \left (-2+\sinh ^{-1}(c x)^2\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )}{\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )}\right )}{\sqrt {1+c^2 x^2}}\right )}{3 c d^3 f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/((d + I*c*d*x)^(5/2)*Sqrt[f - I*c*f*x]),x]

[Out]

(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*((a^2*(-2*I + c*x))/(-I + c*x)^2 - (a*b*((-I)*Cosh[(3*ArcSinh[c*x])/2]*(A
rcSinh[c*x] - 2*ArcTan[Coth[ArcSinh[c*x]/2]] - (I/2)*Log[1 + c^2*x^2]) + Cosh[ArcSinh[c*x]/2]*(-2 - (3*I)*ArcS
inh[c*x] - (6*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + (3*Log[1 + c^2*x^2])/2) + 2*((-1 + Sqrt[1 + c^2*x^2])*ArcSinh[
c*x] + 2*(2 + Sqrt[1 + c^2*x^2])*ArcTan[Coth[ArcSinh[c*x]/2]] + (I/2)*(-2 + (2 + Sqrt[1 + c^2*x^2])*Log[1 + c^
2*x^2]))*Sinh[ArcSinh[c*x]/2]))/(Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^3) - (b^2*(
(1 - I)*ArcSinh[c*x]^2 - (ArcSinh[c*x]*(-2*I + ArcSinh[c*x]))/(-I + c*x) + 2*((-I)*Pi + 2*ArcSinh[c*x])*Log[1
- I/E^ArcSinh[c*x]] + I*Pi*(ArcSinh[c*x] - 4*Log[1 + E^ArcSinh[c*x]] + 4*Log[Cosh[ArcSinh[c*x]/2]] + 2*Log[Sin
[(Pi + (2*I)*ArcSinh[c*x])/4]]) - 4*PolyLog[2, I/E^ArcSinh[c*x]] - (2*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Co
sh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^3 - (2*(-2 + ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[
c*x]/2] + I*Sinh[ArcSinh[c*x]/2])))/Sqrt[1 + c^2*x^2]))/(3*c*d^3*f)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (i c d x +d \right )^{\frac {5}{2}} \sqrt {-i c f x +f}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x)

[Out]

int((a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x)

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x, algorithm="fricas")

[Out]

1/3*((b^2*c*x - 2*I*b^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 3*(c^3*d^3*f*x^
2 - 2*I*c^2*d^3*f*x - c*d^3*f)*integral(-1/3*(3*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a^2 + 2*(3*sqrt(I*c*d*x +
d)*sqrt(-I*c*f*x + f)*a*b + (b^2*c*x - 2*I*b^2)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f))*log(c
*x + sqrt(c^2*x^2 + 1)))/(c^4*d^3*f*x^4 - 2*I*c^3*d^3*f*x^3 - 2*I*c*d^3*f*x - d^3*f), x))/(c^3*d^3*f*x^2 - 2*I
*c^2*d^3*f*x - c*d^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (i d \left (c x - i\right )\right )^{\frac {5}{2}} \sqrt {- i f \left (c x + i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/(d+I*c*d*x)**(5/2)/(f-I*c*f*x)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))**2/((I*d*(c*x - I))**(5/2)*sqrt(-I*f*(c*x + I))), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/((I*c*d*x + d)^(5/2)*sqrt(-I*c*f*x + f)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^2/((d + c*d*x*1i)^(5/2)*(f - c*f*x*1i)^(1/2)),x)

[Out]

int((a + b*asinh(c*x))^2/((d + c*d*x*1i)^(5/2)*(f - c*f*x*1i)^(1/2)), x)

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