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

Optimal. Leaf size=470 \[ \frac{5 i d^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{10 i d^5 (1+i c x)^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^5 (1+i c x)^4 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b d^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i b d^5 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 b d^5 \left (c^2 x^2+1\right )^{5/2} \log (c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{5 b d^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

((-I)*b*d^5*x*(1 + c^2*x^2)^(5/2))/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (((8*I)/3)*b*d^5*(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)) - (5*b*d^5*(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^5*(1 + I*c*x)^4*(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)) + (((10*I)/3)*d^5*(1 + I*c*x)^2*(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)) + ((5*I)*d^5*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*x]))/(c*(d + I*
c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (5*d^5*(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)) + (28*b*d^5*(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 = 0.438858, antiderivative size = 470, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {5712, 669, 641, 215, 5819, 627, 43, 5675} \[ \frac{5 i d^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{10 i d^5 (1+i c x)^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^5 (1+i c x)^4 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b d^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i b d^5 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 b d^5 \left (c^2 x^2+1\right )^{5/2} \log (c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{5 b d^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*b*d^5*x*(1 + c^2*x^2)^(5/2))/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (((8*I)/3)*b*d^5*(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)) - (5*b*d^5*(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^5*(1 + I*c*x)^4*(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)) + (((10*I)/3)*d^5*(1 + I*c*x)^2*(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)) + ((5*I)*d^5*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*x]))/(c*(d + I*
c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (5*d^5*(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)) + (28*b*d^5*(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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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 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)^{5/2} \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)^5 \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^5 (1+i c x)^4 \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{10 i d^5 (1+i c x)^2 \left (1+c^2 x^2\right )^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{5 i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \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{5 i d^5}{c}-\frac{2 i d^5 (1+i c x)^4}{3 c \left (1+c^2 x^2\right )^2}+\frac{10 i d^5 (1+i c x)^2}{3 c \left (1+c^2 x^2\right )}+\frac{5 d^5 \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{5 i b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^5 (1+i c x)^4 \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{10 i d^5 (1+i c x)^2 \left (1+c^2 x^2\right )^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{5 i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \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^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{(1+i c x)^4}{\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 (10 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{(1+i c x)^2}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (5 b d^5 \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{5 i b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{5 b d^5 \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^5 (1+i c x)^4 \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{10 i d^5 (1+i c x)^2 \left (1+c^2 x^2\right )^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{5 i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \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^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{(1+i c x)^2}{(1-i c x)^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (10 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1+i c x}{1-i c x} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{5 i b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{5 b d^5 \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^5 (1+i c x)^4 \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{10 i d^5 (1+i c x)^2 \left (1+c^2 x^2\right )^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{5 i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \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^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (1-\frac{4}{(i+c x)^2}-\frac{4 i}{i+c x}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (10 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-1+\frac{2 i}{i+c x}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{i b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i b d^5 \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{5 b d^5 \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^5 (1+i c x)^4 \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{10 i d^5 (1+i c x)^2 \left (1+c^2 x^2\right )^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{5 i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \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{28 b d^5 \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 [B]  time = 8.66659, size = 1331, normalized size = 2.83 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((I*a*d^2)/f^3 + (((8*I)/3)*a*d^2)/(f^3*(I + c*x)^2) - (28*a*d^2)
/(3*f^3*(I + c*x))))/c + (5*a*d^(5/2)*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)
]])/(c*f^(5/2)) - ((I/6)*b*d^2*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*(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*Log[Sqrt[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[ArcSinh[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) + (b*d^2*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]*((14*I - 3*ArcSinh[c*x])*Ar
cSinh[c*x] + (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] - 14*Log[Sqrt[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]] + 42*Log[Sqrt[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]] + 28*Log[Sqrt[1 + c^2*x^2]] +
Sqrt[1 + c^2*x^2]*(ArcSinh[c*x]*(14*I + 3*ArcSinh[c*x]) - (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 14*Log[Sqrt[1
+ c^2*x^2]]))*Sinh[ArcSinh[c*x]/2]))/(6*c*f^3*(1 + I*c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[ArcSin
h[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) - ((I/12)*b*d^2*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]*(9 - (35*I)*
ArcSinh[c*x] + 9*ArcSinh[c*x]^2 + (52*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + 26*Log[Sqrt[1 + c^2*x^2]])) + Cosh[Arc
Sinh[c*x]/2]*(20 + (24*I)*ArcSinh[c*x] + 27*ArcSinh[c*x]^2 + (156*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + 78*Log[Sqr
t[1 + c^2*x^2]]) - I*(3*(-I + ArcSinh[c*x])*Cosh[(5*ArcSinh[c*x])/2] + 2*(13 + (7*I)*ArcSinh[c*x] + 18*ArcSinh
[c*x]^2 + (104*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + (3*I)*(I + ArcSinh[c*x])*Cosh[2*ArcSinh[c*x]] + 52*Log[Sqrt[1
+ c^2*x^2]] + Sqrt[1 + c^2*x^2]*(6 + (38*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 + (52*I)*ArcTan[Coth[ArcSinh[c*x]
/2]] + 26*Log[Sqrt[1 + c^2*x^2]]))*Sinh[ArcSinh[c*x]/2])))/(c*f^3*(-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)

________________________________________________________________________________________

Maple [F]  time = 0.309, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \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((d+I*c*d*x)^(5/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Exception raised: AttributeError