3.565 \(\int \frac{(d+i c d x)^{3/2} (a+b \sinh ^{-1}(c x))}{(f-i c f x)^{5/2}} \, dx\)

Optimal. Leaf size=362 \[ \frac{2 i d^4 (1+i c x) \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^4 (1+i c x)^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^4 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{4 i b d^4 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 b d^4 \left (c^2 x^2+1\right )^{5/2} \log (c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^4 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

(((4*I)/3)*b*d^4*(1 + c^2*x^2)^(5/2))/(c*(I + c*x)*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b*d^4*(1 + c^2*
x^2)^(5/2)*ArcSinh[c*x]^2)/(2*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((2*I)/3)*d^4*(1 + I*c*x)^3*(1 + c
^2*x^2)*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((2*I)*d^4*(1 + I*c*x)*(1 + c^2*x^
2)^2*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (d^4*(1 + c^2*x^2)^(5/2)*ArcSinh[c*x]
*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (8*b*d^4*(1 + c^2*x^2)^(5/2)*Log[I + c*x]
)/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.39221, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {5712, 669, 653, 215, 5819, 627, 43, 31, 5675} \[ \frac{2 i d^4 (1+i c x) \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^4 (1+i c x)^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^4 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{4 i b d^4 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 b d^4 \left (c^2 x^2+1\right )^{5/2} \log (c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^4 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*I)/3)*b*d^4*(1 + c^2*x^2)^(5/2))/(c*(I + c*x)*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b*d^4*(1 + c^2*
x^2)^(5/2)*ArcSinh[c*x]^2)/(2*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((2*I)/3)*d^4*(1 + I*c*x)^3*(1 + c
^2*x^2)*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((2*I)*d^4*(1 + I*c*x)*(1 + c^2*x^
2)^2*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (d^4*(1 + c^2*x^2)^(5/2)*ArcSinh[c*x]
*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (8*b*d^4*(1 + c^2*x^2)^(5/2)*Log[I + 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 669

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

Rule 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 215

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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Wit
h[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 +
c^2*x^2], u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[p + 1/2, 0]
&& GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 5675

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

Rubi steps

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

Mathematica [A]  time = 5.63159, size = 706, normalized size = 1.95 \[ \frac{\frac{12 a d^{3/2} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )}{f^{5/2}}-\frac{16 a d (2 c x+i) \sqrt{d+i c d x} \sqrt{f-i c f x}}{f^3 (c x+i)^2}+\frac{b d \sqrt{d+i c d x} \sqrt{f-i c f x} \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right ) \left (-2 i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \left (14 \log \left (c^2 x^2+1\right )+\sqrt{c^2 x^2+1} \left (7 \log \left (c^2 x^2+1\right )+\sinh ^{-1}(c x) \left (3 \sinh ^{-1}(c x)+14 i\right )-28 i \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )+6 \sinh ^{-1}(c x)^2+4 i \sinh ^{-1}(c x)-56 i \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+4\right )+\cosh \left (\frac{3}{2} \sinh ^{-1}(c x)\right ) \left (-7 \log \left (c^2 x^2+1\right )+\left (-3 \sinh ^{-1}(c x)+14 i\right ) \sinh ^{-1}(c x)+28 i \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )+\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \left (21 \log \left (c^2 x^2+1\right )+9 \sinh ^{-1}(c x)^2+6 i \sinh ^{-1}(c x)-84 i \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+8\right )\right )}{f^3 (1+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 )^4}-\frac{2 i b d \sqrt{d+i c d x} \sqrt{f-i c f x} \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right ) \left (2 \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \left (\log \left (c^2 x^2+1\right )+\frac{1}{2} \sqrt{c^2 x^2+1} \left (\log \left (c^2 x^2+1\right )+2 i \sinh ^{-1}(c x)+4 i \tan ^{-1}\left (\coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )+2 i \sinh ^{-1}(c x)+4 i \tan ^{-1}\left (\coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+2\right )-\cosh \left (\frac{3}{2} \sinh ^{-1}(c x)\right ) \left (\frac{1}{2} i \log \left (c^2 x^2+1\right )+\sinh ^{-1}(c x)-2 \tan ^{-1}\left (\coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )+\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \left (\frac{3}{2} i \log \left (c^2 x^2+1\right )+3 \sinh ^{-1}(c x)-6 \tan ^{-1}\left (\coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+4 i\right )\right )}{f^3 (1+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 )^4}}{12 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-16*a*d*(I + 2*c*x)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/(f^3*(I + c*x)^2) + (12*a*d^(3/2)*Log[c*d*f*x + Sqr
t[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]])/f^(5/2) - ((2*I)*b*d*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(C
osh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(-(Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcTan[Coth[ArcSin
h[c*x]/2]] + (I/2)*Log[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)/2)*Log[1 + c^2*x^2]) + 2*(2 + (2*I)*ArcSinh[c*x] + (4*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[1 +
c^2*x^2] + (Sqrt[1 + c^2*x^2]*((2*I)*ArcSinh[c*x] + (4*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[1 + c^2*x^2]))/2
)*Sinh[ArcSinh[c*x]/2]))/(f^3*(1 + I*c*x)*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) + (b*d*Sqrt[d + I
*c*d*x]*Sqrt[f - I*c*f*x]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(Cosh[(3*ArcSinh[c*x])/2]*((14*I - 3
*ArcSinh[c*x])*ArcSinh[c*x] + (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] - 7*Log[1 + c^2*x^2]) + Cosh[ArcSinh[c*x]/2]
*(8 + (6*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 - (84*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 21*Log[1 + c^2*x^2]) - (2*
I)*(4 + (4*I)*ArcSinh[c*x] + 6*ArcSinh[c*x]^2 - (56*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 14*Log[1 + c^2*x^2] + Sq
rt[1 + c^2*x^2]*(ArcSinh[c*x]*(14*I + 3*ArcSinh[c*x]) - (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 7*Log[1 + c^2*x^
2]))*Sinh[ArcSinh[c*x]/2]))/(f^3*(1 + I*c*x)*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4))/(12*c)

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Maple [F]  time = 0.291, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( d+icdx \right ) ^{{\frac{3}{2}}} \left ( f-icfx \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),x)

[Out]

int((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),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((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),x, algorithm="fricas")

[Out]

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

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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((d+I*c*d*x)**(3/2)*(a+b*asinh(c*x))/(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((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: AttributeError