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

Optimal. Leaf size=472 \[ -\frac{5 i f^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{10 i f^5 (1-i c x)^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 i f^5 (1-i c x)^4 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 f^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b f^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i b f^5 \left (c^2 x^2+1\right )^{5/2}}{3 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 b f^5 \left (c^2 x^2+1\right )^{5/2} \log (-c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{5 b f^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

(I*b*f^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*f^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*f^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)*f^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)*f^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)*f^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*f^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*f^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 = 0.444465, antiderivative size = 472, 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} \[ -\frac{5 i f^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{10 i f^5 (1-i c x)^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 i f^5 (1-i c x)^4 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 f^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b f^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i b f^5 \left (c^2 x^2+1\right )^{5/2}}{3 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 b f^5 \left (c^2 x^2+1\right )^{5/2} \log (-c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{5 b f^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*b*f^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*f^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*f^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)*f^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)*f^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)*f^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*f^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*f^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{(f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{(d+i c d x)^{5/2}} \, dx &=\frac{\left (1+c^2 x^2\right )^{5/2} \int \frac{(f-i c f 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 f^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 f^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 f^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 f^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 f^5}{c}+\frac{2 i f^5 (1-i c x)^4}{3 c \left (1+c^2 x^2\right )^2}-\frac{10 i f^5 (1-i c x)^2}{3 c \left (1+c^2 x^2\right )}+\frac{5 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 f^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 = 7.29367, size = 1005, normalized size = 2.13 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-4*I)*a*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-23 - (34*I)*c*x + 3*c^2*x^2))/(d^3*(-I + c*x)^2) + (60*a*
f^(5/2)*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]])/d^(5/2) - (2*b*f^2*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 + (3*I)*A
rcSinh[c*x])*ArcSinh[c*x] - 28*ArcTan[Tanh[ArcSinh[c*x]/2]] + (7*I)*Log[1 + c^2*x^2]) + Cosh[ArcSinh[c*x]/2]*(
84*ArcTan[Tanh[ArcSinh[c*x]/2]] - I*(8 - (6*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 + 21*Log[1 + c^2*x^2])) + 2*(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] + Sqrt[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]))/(d^3*(I + c*x)*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^4) + ((2*I)*b*f^2*Sqrt[d
+ I*c*d*x]*Sqrt[f - I*c*f*x]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])*((-I)*Cosh[(3*ArcSinh[c*x])/2]*(
ArcSinh[c*x] - 2*ArcTan[Coth[ArcSinh[c*x]/2]] - (I/2)*Log[1 + c^2*x^2]) + Cosh[ArcSinh[c*x]/2]*(4 + (3*I)*ArcS
inh[c*x] - (6*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + (3*Log[1 + c^2*x^2])/2) + 2*((2 + Sqrt[1 + c^2*x^2])*ArcSinh[c
*x] + 2*(2 + Sqrt[1 + c^2*x^2])*ArcTan[Coth[ArcSinh[c*x]/2]] + (I/2)*(4 + (2 + Sqrt[1 + c^2*x^2])*Log[1 + c^2*
x^2]))*Sinh[ArcSinh[c*x]/2]))/(d^3*(I + c*x)*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^4) + (b*f^2*Sqrt[
d + I*c*d*x]*Sqrt[f - I*c*f*x]*(I*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/2])*(2*(4 + (6*I)*c*x - 6*c^2*x^2 +
52*(-I + c*x)*ArcTan[Coth[ArcSinh[c*x]/2]] + 13*(1 + I*c*x)*Log[1 + c^2*x^2])*(Cosh[ArcSinh[c*x]/2] + I*Sinh[
ArcSinh[c*x]/2]) + 18*ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^3 + ArcSinh[c*x]*((-24*I)
*Cosh[ArcSinh[c*x]/2] - (35*I)*Cosh[(3*ArcSinh[c*x])/2] + (3*I)*Cosh[(5*ArcSinh[c*x])/2] - 24*Sinh[ArcSinh[c*x
]/2] + 35*Sinh[(3*ArcSinh[c*x])/2] + 3*Sinh[(5*ArcSinh[c*x])/2])))/(d^3*(I + c*x)*(Cosh[ArcSinh[c*x]/2] + I*Si
nh[ArcSinh[c*x]/2])^4))/(12*c)

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Maple [F]  time = 0.296, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( f-icfx \right ) ^{{\frac{5}{2}}} \left ( d+icdx \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))/(d+I*c*d*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((f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))/(d+I*c*d*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 (-i \, b c^{2} f^{2} x^{2} + 2 \, b c f^{2} x + i \, b f^{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} f^{2} x^{2} + 2 \, a c f^{2} x + i \, a f^{2}\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}}{c^{3} d^{3} x^{3} - 3 i \, c^{2} d^{3} x^{2} - 3 \, c d^{3} x + i \, d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: AttributeError