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

Optimal. Leaf size=386 \[ -\frac{2 b^2 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b \left (c^2 x^2+1\right )^{5/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 x \left (c^2 x^2+1\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

-(b^2*x*(1 + c^2*x^2)^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (b*(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)) + (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)) + (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)) + (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*
b*(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*(1 + c^2*x^2)^(5/2)*PolyLog[2, -E^(2*ArcSinh[c*x])])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(
5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.531256, antiderivative size = 386, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.27, Rules used = {5712, 5690, 5687, 5714, 3718, 2190, 2279, 2391, 5717, 191} \[ -\frac{2 b^2 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b \left (c^2 x^2+1\right )^{5/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 x \left (c^2 x^2+1\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*x*(1 + c^2*x^2)^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (b*(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)) + (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)) + (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)) + (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*
b*(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*(1 + c^2*x^2)^(5/2)*PolyLog[2, -E^(2*ArcSinh[c*x])])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(
5/2))

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 5690

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[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 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 191

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx &=\frac{\left (1+c^2 x^2\right )^{5/2} \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{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{\left (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 (2 b c \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{b \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{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{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 \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 (4 b c \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{b^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 \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{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{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 (4 b \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{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{b^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 \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{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{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{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{\left (8 b \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{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{b^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 \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{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{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{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 b \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 (4 b^2 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{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{b^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 \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{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{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{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 b \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 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{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{b^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 \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{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{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{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 b \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 \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*}

Mathematica [A]  time = 7.82093, size = 642, normalized size = 1.66 \[ \frac{-b^2 \left (-16 \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-16 \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+2 \sqrt{c^2 x^2+1} \left (\left (6 \sinh ^{-1}(c x)-3 i \pi \right ) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+i \left (-3 i \sinh ^{-1}(c x)^2+6 \pi \sinh ^{-1}(c x)+2 i \sinh ^{-1}(c x)+3 \left (\pi -2 i \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-12 \pi \log \left (e^{\sinh ^{-1}(c x)}+1\right )+3 \pi \log \left (\sin \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )-3 \pi \log \left (-\cos \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )+12 \pi \log \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )\right )+c x-6 c x \sinh ^{-1}(c x)^2-2 \sinh \left (3 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)^2+\sinh \left (3 \sinh ^{-1}(c x)\right )+2 \sinh ^{-1}(c x)^2 \cosh \left (3 \sinh ^{-1}(c x)\right )+4 i \pi \sinh ^{-1}(c x) \cosh \left (3 \sinh ^{-1}(c x)\right )+4 \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+4 \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )-2 i \pi \log \left (1-i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+2 i \pi \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )-8 i \pi \log \left (e^{\sinh ^{-1}(c x)}+1\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+8 i \pi \cosh \left (3 \sinh ^{-1}(c x)\right ) \log \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )-2 i \pi \cosh \left (3 \sinh ^{-1}(c x)\right ) \log \left (-\cos \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )+2 i \pi \cosh \left (3 \sinh ^{-1}(c x)\right ) \log \left (\sin \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )\right )+4 a^2 c x \left (2 c^2 x^2+3\right )+2 a b \left (\sqrt{c^2 x^2+1} \left (2-3 \log \left (c^2 x^2+1\right )\right )-\log \left (c^2 x^2+1\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+2 \sinh ^{-1}(c x) \left (3 c x+\sinh \left (3 \sinh ^{-1}(c x)\right )\right )\right )}{12 d^2 f^2 \left (c^3 x^2+c\right ) \sqrt{d+i c d x} \sqrt{f-i c f x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*a^2*c*x*(3 + 2*c^2*x^2) - b^2*(c*x - 6*c*x*ArcSinh[c*x]^2 + (4*I)*Pi*ArcSinh[c*x]*Cosh[3*ArcSinh[c*x]] + 2*
ArcSinh[c*x]^2*Cosh[3*ArcSinh[c*x]] - (2*I)*Pi*Cosh[3*ArcSinh[c*x]]*Log[1 - I/E^ArcSinh[c*x]] + 4*ArcSinh[c*x]
*Cosh[3*ArcSinh[c*x]]*Log[1 - I/E^ArcSinh[c*x]] + (2*I)*Pi*Cosh[3*ArcSinh[c*x]]*Log[1 + I/E^ArcSinh[c*x]] + 4*
ArcSinh[c*x]*Cosh[3*ArcSinh[c*x]]*Log[1 + I/E^ArcSinh[c*x]] - (8*I)*Pi*Cosh[3*ArcSinh[c*x]]*Log[1 + E^ArcSinh[
c*x]] - (2*I)*Pi*Cosh[3*ArcSinh[c*x]]*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]] + (8*I)*Pi*Cosh[3*ArcSinh[c*x]]*L
og[Cosh[ArcSinh[c*x]/2]] + (2*I)*Pi*Cosh[3*ArcSinh[c*x]]*Log[Sin[(Pi + (2*I)*ArcSinh[c*x])/4]] + 2*Sqrt[1 + c^
2*x^2]*(((-3*I)*Pi + 6*ArcSinh[c*x])*Log[1 - I/E^ArcSinh[c*x]] + I*((2*I)*ArcSinh[c*x] + 6*Pi*ArcSinh[c*x] - (
3*I)*ArcSinh[c*x]^2 + 3*(Pi - (2*I)*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] - 12*Pi*Log[1 + E^ArcSinh[c*x]] -
3*Pi*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]] + 12*Pi*Log[Cosh[ArcSinh[c*x]/2]] + 3*Pi*Log[Sin[(Pi + (2*I)*ArcSi
nh[c*x])/4]])) - 16*(1 + c^2*x^2)^(3/2)*PolyLog[2, (-I)/E^ArcSinh[c*x]] - 16*(1 + c^2*x^2)^(3/2)*PolyLog[2, I/
E^ArcSinh[c*x]] + Sinh[3*ArcSinh[c*x]] - 2*ArcSinh[c*x]^2*Sinh[3*ArcSinh[c*x]]) + 2*a*b*(Sqrt[1 + c^2*x^2]*(2
- 3*Log[1 + c^2*x^2]) - Cosh[3*ArcSinh[c*x]]*Log[1 + c^2*x^2] + 2*ArcSinh[c*x]*(3*c*x + Sinh[3*ArcSinh[c*x]]))
)/(12*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(c + c^3*x^2))

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Maple [F]  time = 0.278, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2} \left ( d+icdx \right ) ^{-{\frac{5}{2}}} \left ( f-icfx \right ) ^{-{\frac{5}{2}}}}\, 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)^(5/2)/(f-I*c*f*x)^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a b c{\left (\frac{1}{c^{4} d^{\frac{5}{2}} f^{\frac{5}{2}} x^{2} + c^{2} d^{\frac{5}{2}} f^{\frac{5}{2}}} - \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{\frac{5}{2}} f^{\frac{5}{2}}}\right )} + \frac{2}{3} \, a b{\left (\frac{x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac{3}{2}} d f} + \frac{2 \, x}{\sqrt{c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \, a^{2}{\left (\frac{x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac{3}{2}} d f} + \frac{2 \, x}{\sqrt{c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} + b^{2} \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{{\left (i \, c d x + d\right )}^{\frac{5}{2}}{\left (-i \, c f x + f\right )}^{\frac{5}{2}}}\,{d x} \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)^(5/2)/(f-I*c*f*x)^(5/2),x, algorithm="maxima")

[Out]

1/3*a*b*c*(1/(c^4*d^(5/2)*f^(5/2)*x^2 + c^2*d^(5/2)*f^(5/2)) - 2*log(c^2*x^2 + 1)/(c^2*d^(5/2)*f^(5/2))) + 2/3
*a*b*(x/((c^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d^2*f^2))*arcsinh(c*x) + 1/3*a^2*(x/((c
^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d^2*f^2)) + b^2*integrate(log(c*x + sqrt(c^2*x^2 +
1))^2/((I*c*d*x + d)^(5/2)*(-I*c*f*x + f)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2 \, b^{2} c^{2} x^{3} + 3 \, b^{2} x\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 3 \,{\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )}{\rm integral}\left (\frac{3 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a^{2} + 2 \,{\left (3 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a b -{\left (2 \, b^{2} c^{3} x^{3} + 3 \, b^{2} c x\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{3 \,{\left (c^{6} d^{3} f^{3} x^{6} + 3 \, c^{4} d^{3} f^{3} x^{4} + 3 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )}}, x\right )}{3 \,{\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )}} \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)^(5/2)/(f-I*c*f*x)^(5/2),x, algorithm="fricas")

[Out]

1/3*((2*b^2*c^2*x^3 + 3*b^2*x)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 3*(c^4*d^
3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*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 - (2*b^2*c^3*x^3 + 3*b^2*c*x)*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^6*d^3*f^3*x^6 + 3*c^4*d^3*f^3*x^4 + 3*c^2*d^3*f^3*x^2 + d^3*f^3),
x))/(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)

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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)**(5/2)/(f-I*c*f*x)**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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)^(5/2)/(f-I*c*f*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: AttributeError