3.6.97 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{3/2}} \, dx\) [597]

Optimal. Leaf size=464 \[ -\frac {i d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {4 i b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 b^2 d \left (1+c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b^2 d \left (1+c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b^2 d \left (1+c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.45, antiderivative size = 464, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {5796, 5838, 5787, 5797, 3799, 2221, 2317, 2438, 5798, 5789, 4265} \begin {gather*} \frac {4 i b d \left (c^2 x^2+1\right )^{3/2} \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {i d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {d x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b d \left (c^2 x^2+1\right )^{3/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 b^2 d \left (c^2 x^2+1\right )^{3/2} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b^2 d \left (c^2 x^2+1\right )^{3/2} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b^2 d \left (c^2 x^2+1\right )^{3/2} \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/(Sqrt[d + I*c*d*x]*(f - I*c*f*x)^(3/2)),x]

[Out]

((-I)*d*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (d*x*(1 + c^2*x^2)
*(a + b*ArcSinh[c*x])^2)/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (d*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x
])^2)/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + ((4*I)*b*d*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x])*ArcTan
[E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (2*b*d*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]
)*Log[1 + E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (2*b^2*d*(1 + c^2*x^2)^(3/2)*Poly
Log[2, (-I)*E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (2*b^2*d*(1 + c^2*x^2)^(3/2)*PolyLo
g[2, I*E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (b^2*d*(1 + c^2*x^2)^(3/2)*PolyLog[2, -E
^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))

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

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Mathematica [A]
time = 1.46, size = 511, normalized size = 1.10 \begin {gather*} \frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left ((-1-i) b^2 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\left (-i a^2+a^2 c x+4 i a b \sqrt {1+c^2 x^2} \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-2 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+4 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (1+e^{\sinh ^{-1}(c x)}\right )-a b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )+2 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (-\cos \left (\frac {1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )-4 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right ) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+4 b^2 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,-i e^{-\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 )+b \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \left (-i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (2 a+3 b \pi -4 i b \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )+\left (2 a-3 b \pi +4 i b \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{c d f^2 (-i+c x) (i+c x) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/(Sqrt[d + I*c*d*x]*(f - I*c*f*x)^(3/2)),x]

[Out]

(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*((-1 - I)*b^2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/2] - Si
nh[ArcSinh[c*x]/2]) + ((-I)*a^2 + a^2*c*x + (4*I)*a*b*Sqrt[1 + c^2*x^2]*ArcTan[Tanh[ArcSinh[c*x]/2]] - (2*I)*b
^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 + I/E^ArcSinh[c*x]] + (4*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 + E^ArcSinh[c*x]] - a
*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] + (2*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]]
- (4*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[Cosh[ArcSinh[c*x]/2]])*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2]) +
4*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)/E^ArcSinh[c*x]]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2]) + b*Sq
rt[1 + c^2*x^2]*ArcSinh[c*x]*((-I)*Cosh[ArcSinh[c*x]/2]*(2*a + 3*b*Pi - (4*I)*b*Log[1 + I/E^ArcSinh[c*x]]) + (
2*a - 3*b*Pi + (4*I)*b*Log[1 + I/E^ArcSinh[c*x]])*Sinh[ArcSinh[c*x]/2])))/(c*d*f^2*(-I + c*x)*(I + c*x)*(Cosh[
ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2]))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (-i c f x +f \right )^{\frac {3}{2}} \sqrt {i c d x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(3/2)/(d+I*c*d*x)^(1/2),x)

[Out]

int((a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(3/2)/(d+I*c*d*x)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(3/2)/(d+I*c*d*x)^(1/2),x, algorithm="maxima")

[Out]

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

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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/(f-I*c*f*x)^(3/2)/(d+I*c*d*x)^(1/2),x, algorithm="fricas")

[Out]

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

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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}}{\sqrt {i d \left (c x - i\right )} \left (- i f \left (c x + i\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/(f-I*c*f*x)**(3/2)/(d+I*c*d*x)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))**2/(sqrt(I*d*(c*x - I))*(-I*f*(c*x + I))**(3/2)), 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/(f-I*c*f*x)^(3/2)/(d+I*c*d*x)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/(sqrt(I*c*d*x + d)*(-I*c*f*x + f)^(3/2)), 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}{\sqrt {d+c\,d\,x\,1{}\mathrm {i}}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}} \,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)^(1/2)*(f - c*f*x*1i)^(3/2)),x)

[Out]

int((a + b*asinh(c*x))^2/((d + c*d*x*1i)^(1/2)*(f - c*f*x*1i)^(3/2)), x)

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