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

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

[Out]

(I*f*(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)) + (f*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)) + (f*(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*f*(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*f*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x])*L
og[1 + E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (2*b^2*f*(1 + c^2*x^2)^(3/2)*PolyLog
[2, (-I)*E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (2*b^2*f*(1 + c^2*x^2)^(3/2)*PolyLog[2
, I*E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (b^2*f*(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))

________________________________________________________________________________________

Rubi [A]  time = 0.753496, antiderivative size = 464, normalized size of antiderivative = 1., 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 = {5712, 5821, 5687, 5714, 3718, 2190, 2279, 2391, 5717, 5693, 4180} \[ -\frac{2 b^2 f \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{2 b^2 f \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{b^2 f \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{f \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{i f \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{f x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{2 b f \left (c^2 x^2+1\right )^{3/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{4 i b f \left (c^2 x^2+1\right )^{3/2} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*f*(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)) + (f*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)) + (f*(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*f*(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*f*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x])*L
og[1 + E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (2*b^2*f*(1 + c^2*x^2)^(3/2)*PolyLog
[2, (-I)*E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (2*b^2*f*(1 + c^2*x^2)^(3/2)*PolyLog[2
, I*E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (b^2*f*(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 5712

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 5821

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]))

Rule 5687

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*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
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 5714

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 3718

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 5693

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 4180

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} \sqrt{f-i c f x}} \, dx &=\frac{\left (1+c^2 x^2\right )^{3/2} \int \frac{(f-i c f 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{f \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{3/2}}-\frac{i c f 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 (f \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 f \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 f \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{f 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 f \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 f \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 f \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{f 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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \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{f 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{f \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 f \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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \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{f 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{f \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 f \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 f \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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \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{f 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{f \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 f \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 f \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 f \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 f \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 f \left (1+c^2 x^2\right )^{3/2}\right ) \operatorname{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 f \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{f 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{f \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 f \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 f \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 f \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 f \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 f \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*}

Mathematica [A]  time = 2.18953, size = 508, normalized size = 1.09 \[ \frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left (4 b^2 \sqrt{c^2 x^2+1} \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right ) \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+\left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right ) \left (a^2 c x+i a^2-a b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )-4 i a b \sqrt{c^2 x^2+1} \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+2 i \pi b^2 \sqrt{c^2 x^2+1} \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+4 i \pi b^2 \sqrt{c^2 x^2+1} \log \left (e^{\sinh ^{-1}(c x)}+1\right )-2 i \pi b^2 \sqrt{c^2 x^2+1} \log \left (\sin \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )-4 i \pi b^2 \sqrt{c^2 x^2+1} \log \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )+b \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \left (\sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \left (2 a-4 i b \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+\pi b\right )+i \cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \left (2 a+4 i b \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\pi b\right )\right )+(-1+i) b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)^2 \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-\sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{c d^2 f (c x-i) (c x+i) \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/((d + I*c*d*x)^(3/2)*Sqrt[f - I*c*f*x]),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] - (4*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[Cosh[ArcSinh[c*x]/2]] - (2*I)*b^2*Pi*S
qrt[1 + c^2*x^2]*Log[Sin[(Pi + (2*I)*ArcSinh[c*x])/4]])*(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*Sqrt[1 +
c^2*x^2]*ArcSinh[c*x]*(I*Cosh[ArcSinh[c*x]/2]*(2*a - b*Pi + (4*I)*b*Log[1 - I/E^ArcSinh[c*x]]) + (2*a + b*Pi -
(4*I)*b*Log[1 - I/E^ArcSinh[c*x]])*Sinh[ArcSinh[c*x]/2])))/(c*d^2*f*(-I + c*x)*(I + c*x)*(Cosh[ArcSinh[c*x]/2
] + I*Sinh[ArcSinh[c*x]/2]))

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Maple [F]  time = 0.286, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2} \left ( d+icdx \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{f-icfx}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} +{\left (c^{2} d^{2} f x - i \, c d^{2} f\right )}{\rm integral}\left (\frac{-i \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a^{2} - 2 \,{\left (\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\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{3} d^{2} f x^{3} - i \, c^{2} d^{2} f x^{2} + c d^{2} f x - i \, d^{2} f}, x\right )}{c^{2} d^{2} f x - i \, c d^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out