\(\int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx\) [602]

Optimal result

Integrand size = 37, antiderivative size = 522 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

1/3*d^3*(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)+4/3*b*d^3*(c^2*x^2+1)^(5/
2)*(a+b*arcsinh(c*x))*ln(1+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)-4/3*b^2*d^3*(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*d^3*(c^2*x^2+1)^
(5/2)*(a+b*arcsinh(c*x))*sec(1/4*Pi+1/2*I*arcsinh(c*x))^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+4/3*I*b^2*d^3*
(c^2*x^2+1)^(5/2)*tan(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*I*d^3*(c^2*x^2+1)^(
5/2)*(a+b*arcsinh(c*x))^2*tan(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*I*d^3*(c^2*
x^2+1)^(5/2)*(a+b*arcsinh(c*x))^2*sec(1/4*Pi+1/2*I*arcsinh(c*x))^2*tan(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x
)^(5/2)/(f-I*c*f*x)^(5/2)

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 522, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {5796, 5844, 5843, 3399, 4271, 3852, 8, 4269, 3797, 2221, 2317, 2438} \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {d^3 \left (c^2 x^2+1\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^3 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 b^2 d^3 \left (c^2 x^2+1\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b^2 d^3 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[In]

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

[Out]

(d^3*(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*b*d^3*(1 +
c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
- (4*b^2*d^3*(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)
) + (2*b*d^3*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])*Sec[Pi/4 + (I/2)*ArcSinh[c*x]]^2)/(3*c*(d + I*c*d*x)^(5/
2)*(f - I*c*f*x)^(5/2)) + (((4*I)/3)*b^2*d^3*(1 + c^2*x^2)^(5/2)*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/(c*(d + I*c*d
*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((I/3)*d^3*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2*Tan[Pi/4 + (I/2)*ArcSin
h[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - ((I/3)*d^3*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2*S
ec[Pi/4 + (I/2)*ArcSinh[c*x]]^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3399

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

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4271

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-b^2)*(c + d*x)^m*Cot[e
+ f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))), Int[(c +
d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)^m*(b*Csc[e + f*x])^
(n - 2), x], x] - Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

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 5843

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

Rule 5844

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(d+i c d x)^3 (a+b \text {arcsinh}(c x))^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 {2 d^3 (a+b \text {arcsinh}(c x))^2}{(i+c x)^2 \sqrt {1+c^2 x^2}}-\frac {i d^3 (a+b \text {arcsinh}(c x))^2}{(i+c x) \sqrt {1+c^2 x^2}}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = -\frac {\left (i d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {(a+b \text {arcsinh}(c x))^2}{(i+c x) \sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {(a+b \text {arcsinh}(c x))^2}{(i+c x)^2 \sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = -\frac {\left (i d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {(a+b x)^2}{i c+c \sinh (x)} \, dx,x,\text {arcsinh}(c x)\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 c d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {(a+b x)^2}{(i c+c \sinh (x))^2} \, dx,x,\text {arcsinh}(c x)\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = -\frac {\left (d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x)^2 \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x)^2 \csc ^4\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = \frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x)^2 \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x) \cot \left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 b^2 d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = \frac {d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (4 i b d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x) \cot \left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (4 i b d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {e^{-x} (a+b x)}{1+i e^{-x}} \, dx,x,\text {arcsinh}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (4 i b^2 d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int 1 \, dx,x,\cot \left (\frac {\pi }{4}-\frac {1}{2} i \text {arcsinh}(c x)\right )\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = \frac {d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (8 i b d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {e^{-x} (a+b x)}{1+i e^{-x}} \, dx,x,\text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (4 b^2 d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1+i e^{-x}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = \frac {d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (8 b^2 d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1+i e^{-x}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (4 b^2 d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{-\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = \frac {d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 b^2 d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = \frac {d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 10.07 (sec) , antiderivative size = 788, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {\sqrt {i d (-i+c x)} \sqrt {-i f (i+c x)} \left (\frac {2 i a^2}{3 f^3 (i+c x)^2}-\frac {a^2}{3 f^3 (i+c x)}\right )}{c}-\frac {i a b \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {-d f \left (1+c^2 x^2\right )} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right ) \left (-\cosh \left (\frac {3}{2} \text {arcsinh}(c x)\right ) \left (\text {arcsinh}(c x)-2 \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+i \log \left (\sqrt {1+c^2 x^2}\right )\right )+\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right ) \left (4 i+3 \text {arcsinh}(c x)-6 \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+3 i \log \left (\sqrt {1+c^2 x^2}\right )\right )+2 \left (\sqrt {1+c^2 x^2} \left (i \text {arcsinh}(c x)+2 i \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\log \left (\sqrt {1+c^2 x^2}\right )\right )+2 \left (1+i \text {arcsinh}(c x)+2 i \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\log \left (\sqrt {1+c^2 x^2}\right )\right )\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )}{3 c f^3 (1+i c x) \sqrt {-((-i d+c d x) (i f+c f x))} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^4}-\frac {i b^2 (-i+c x) \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {-d f \left (1+c^2 x^2\right )} \left ((-1-i) \text {arcsinh}(c x)^2-\frac {2 \text {arcsinh}(c x) (2 i+\text {arcsinh}(c x))}{i+c x}-2 i (\pi -2 i \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )-i \pi \left (3 \text {arcsinh}(c x)-4 \log \left (1+e^{\text {arcsinh}(c x)}\right )-2 \log \left (-\cos \left (\frac {1}{4} (\pi +2 i \text {arcsinh}(c x))\right )\right )+4 \log \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )\right )+4 \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )-\frac {4 \text {arcsinh}(c x)^2 \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}{\left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^3}+\frac {2 \left (4+\text {arcsinh}(c x)^2\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}{\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}\right )}{3 c f^3 \sqrt {-((-i d+c d x) (i f+c f x))} \sqrt {1+c^2 x^2} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^2} \]

[In]

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

[Out]

(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((((2*I)/3)*a^2)/(f^3*(I + c*x)^2) - a^2/(3*f^3*(I + c*x))))/c -
((I/3)*a*b*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*(Cosh[ArcSinh[c*x]/2]
+ I*Sinh[ArcSinh[c*x]/2])*(-(Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcTan[Coth[ArcSinh[c*x]/2]] + I*Log[S
qrt[1 + c^2*x^2]])) + Cosh[ArcSinh[c*x]/2]*(4*I + 3*ArcSinh[c*x] - 6*ArcTan[Coth[ArcSinh[c*x]/2]] + (3*I)*Log[
Sqrt[1 + c^2*x^2]]) + 2*(Sqrt[1 + c^2*x^2]*(I*ArcSinh[c*x] + (2*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[Sqrt[1 +
c^2*x^2]]) + 2*(1 + I*ArcSinh[c*x] + (2*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[Sqrt[1 + c^2*x^2]]))*Sinh[ArcSi
nh[c*x]/2]))/(c*f^3*(1 + I*c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh
[c*x]/2])^4) - ((I/3)*b^2*(-I + c*x)*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2
))]*((-1 - I)*ArcSinh[c*x]^2 - (2*ArcSinh[c*x]*(2*I + ArcSinh[c*x]))/(I + c*x) - (2*I)*(Pi - (2*I)*ArcSinh[c*x
])*Log[1 + I/E^ArcSinh[c*x]] - I*Pi*(3*ArcSinh[c*x] - 4*Log[1 + E^ArcSinh[c*x]] - 2*Log[-Cos[(Pi + (2*I)*ArcSi
nh[c*x])/4]] + 4*Log[Cosh[ArcSinh[c*x]/2]]) + 4*PolyLog[2, (-I)/E^ArcSinh[c*x]] - (4*ArcSinh[c*x]^2*Sinh[ArcSi
nh[c*x]/2])/(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^3 + (2*(4 + ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/
(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/(c*f^3*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[1 + c^2*
x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^2)

Maple [F]

\[\int \frac {\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2} \sqrt {i c d x +d}}{\left (-i c f x +f \right )^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int \frac {\sqrt {i d \left (c x - i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (- i f \left (c x + i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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