3.6.50 \(\int \frac {(f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x))}{(d+i c d x)^{3/2}} \, dx\) [550]

Optimal. Leaf size=518 \[ -\frac {3 i b f^4 x \left (1+c^2 x^2\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {b c f^4 x^2 \left (1+c^2 x^2\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b f^4 (1-i c x)^2 \left (1+c^2 x^2\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 b f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x)^2}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i f^4 (1-i c x)^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 i f^4 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 i f^4 (1-i c x) \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {15 f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {8 b f^4 \left (1+c^2 x^2\right )^{3/2} \log (i-c x)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

-3/2*I*b*f^4*x*(c^2*x^2+1)^(3/2)/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+b*c*f^4*x^2*(c^2*x^2+1)^(3/2)/(d+I*c*d*x)
^(3/2)/(f-I*c*f*x)^(3/2)+5/4*b*f^4*(1-I*c*x)^2*(c^2*x^2+1)^(3/2)/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+15/4*b*
f^4*(c^2*x^2+1)^(3/2)*arcsinh(c*x)^2/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+2*I*f^4*(1-I*c*x)^3*(c^2*x^2+1)*(a+
b*arcsinh(c*x))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+15/2*I*f^4*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))/c/(d+I*c*d*x
)^(3/2)/(f-I*c*f*x)^(3/2)+5/2*I*f^4*(1-I*c*x)*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)
^(3/2)-15/2*f^4*(c^2*x^2+1)^(3/2)*arcsinh(c*x)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-8*b*f^
4*(c^2*x^2+1)^(3/2)*ln(I-c*x)/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)

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Rubi [A]
time = 0.28, antiderivative size = 518, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {5796, 683, 685, 655, 221, 5837, 641, 45, 5783} \begin {gather*} \frac {5 i f^4 (1-i c x) \left (c^2 x^2+1\right )^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 {15 i f^4 \left (c^2 x^2+1\right )^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 {2 i f^4 (1-i c x)^3 \left (c^2 x^2+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 {15 f^4 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \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 {b c f^4 x^2 \left (c^2 x^2+1\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b f^4 (1-i c x)^2 \left (c^2 x^2+1\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {3 i b f^4 x \left (c^2 x^2+1\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {8 b f^4 \left (c^2 x^2+1\right )^{3/2} \log (-c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 b f^4 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)^2}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/2)*b*f^4*x*(1 + c^2*x^2)^(3/2))/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (b*c*f^4*x^2*(1 + c^2*x^2
)^(3/2))/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (5*b*f^4*(1 - I*c*x)^2*(1 + c^2*x^2)^(3/2))/(4*c*(d + I*c
*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + (15*b*f^4*(1 + c^2*x^2)^(3/2)*ArcSinh[c*x]^2)/(4*c*(d + I*c*d*x)^(3/2)*(f -
I*c*f*x)^(3/2)) + ((2*I)*f^4*(1 - I*c*x)^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(3/2)*(f - I*
c*f*x)^(3/2)) + (((15*I)/2)*f^4*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/
2)) + (((5*I)/2)*f^4*(1 - I*c*x)*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3
/2)) - (15*f^4*(1 + c^2*x^2)^(3/2)*ArcSinh[c*x]*(a + b*ArcSinh[c*x]))/(2*c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(
3/2)) - (8*b*f^4*(1 + c^2*x^2)^(3/2)*Log[I - c*x])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))

Rule 45

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 221

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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 655

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 683

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 685

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*(m + 2*p + 1))), x] + Dist[2*c*d*((m + p)/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rule 5783

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

Rule 5796

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 5837

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])

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)^{3/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{3/2} \int \frac {(f-i c f x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{\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 {2 i f^4 (1-i c x)^3 \left (1+c^2 x^2\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 {15 i f^4 \left (1+c^2 x^2\right )^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 {5 i f^4 (1-i c x) \left (1+c^2 x^2\right )^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 {15 f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \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 {\left (b c \left (1+c^2 x^2\right )^{3/2}\right ) \int \left (\frac {15 i f^4}{2 c}+\frac {5 i f^4 (1-i c x)}{2 c}+\frac {2 i f^4 (1-i c x)^3}{c \left (1+c^2 x^2\right )}-\frac {15 f^4 \sinh ^{-1}(c x)}{2 c \sqrt {1+c^2 x^2}}\right ) \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=-\frac {15 i b f^4 x \left (1+c^2 x^2\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b f^4 (1-i c x)^2 \left (1+c^2 x^2\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i f^4 (1-i c x)^3 \left (1+c^2 x^2\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 {15 i f^4 \left (1+c^2 x^2\right )^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 {5 i f^4 (1-i c x) \left (1+c^2 x^2\right )^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 {15 f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \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 {\left (2 i b f^4 \left (1+c^2 x^2\right )^{3/2}\right ) \int \frac {(1-i c x)^3}{1+c^2 x^2} \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (15 b f^4 \left (1+c^2 x^2\right )^{3/2}\right ) \int \frac {\sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=-\frac {15 i b f^4 x \left (1+c^2 x^2\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b f^4 (1-i c x)^2 \left (1+c^2 x^2\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 b f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x)^2}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i f^4 (1-i c x)^3 \left (1+c^2 x^2\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 {15 i f^4 \left (1+c^2 x^2\right )^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 {5 i f^4 (1-i c x) \left (1+c^2 x^2\right )^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 {15 f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \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 {\left (2 i b f^4 \left (1+c^2 x^2\right )^{3/2}\right ) \int \frac {(1-i c x)^2}{1+i c x} \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=-\frac {15 i b f^4 x \left (1+c^2 x^2\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b f^4 (1-i c x)^2 \left (1+c^2 x^2\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 b f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x)^2}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i f^4 (1-i c x)^3 \left (1+c^2 x^2\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 {15 i f^4 \left (1+c^2 x^2\right )^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 {5 i f^4 (1-i c x) \left (1+c^2 x^2\right )^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 {15 f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \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 {\left (2 i b f^4 \left (1+c^2 x^2\right )^{3/2}\right ) \int \left (-3+i c x+\frac {4}{1+i c x}\right ) \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=-\frac {3 i b f^4 x \left (1+c^2 x^2\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {b c f^4 x^2 \left (1+c^2 x^2\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b f^4 (1-i c x)^2 \left (1+c^2 x^2\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 b f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x)^2}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i f^4 (1-i c x)^3 \left (1+c^2 x^2\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 {15 i f^4 \left (1+c^2 x^2\right )^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 {5 i f^4 (1-i c x) \left (1+c^2 x^2\right )^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 {15 f^4 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \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 {8 b f^4 \left (1+c^2 x^2\right )^{3/2} \log (i-c x)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 2.52, size = 779, normalized size = 1.50 \begin {gather*} \frac {\frac {4 a f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (24+7 i c x+c^2 x^2\right )}{d^2 (-i+c x)}-\frac {60 a f^{5/2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )}{d^{3/2}}-\frac {4 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\sinh ^{-1}(c x) \left (-4 i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-4 \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+2 \left (4 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\log \left (1+c^2 x^2\right )\right ) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{d^2 \sqrt {1+c^2 x^2} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}+\frac {16 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (-\sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\left (c x-4 \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-i \log \left (1+c^2 x^2\right )\right ) \left (-i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\sinh ^{-1}(c x) \left (i \left (2+\sqrt {1+c^2 x^2}\right ) \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\left (-2+\sqrt {1+c^2 x^2}\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{d^2 \sqrt {1+c^2 x^2} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}+\frac {b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (-10 \sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-\left (\cosh \left (2 \sinh ^{-1}(c x)\right )+8 \left (2 i c x+4 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\log \left (1+c^2 x^2\right )\right )\right ) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+2 \sinh ^{-1}(c x) \left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (8-8 \sqrt {1+c^2 x^2}+i \sinh \left (2 \sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (8 i \left (1+\sqrt {1+c^2 x^2}\right )+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )\right )}{d^2 \sqrt {1+c^2 x^2} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(24 + (7*I)*c*x + c^2*x^2))/(d^2*(-I + c*x)) - (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^(3/2) - (4*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f -
I*c*f*x]*(ArcSinh[c*x]*((-4*I)*Cosh[ArcSinh[c*x]/2] - 4*Sinh[ArcSinh[c*x]/2]) + ArcSinh[c*x]^2*(Cosh[ArcSinh[
c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) + 2*((4*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + Log[1 + c^2*x^2])*(Cosh[ArcSinh[c*
x]/2] + I*Sinh[ArcSinh[c*x]/2])))/(d^2*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])) + (1
6*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-(ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]))
+ (c*x - 4*ArcTan[Coth[ArcSinh[c*x]/2]] - I*Log[1 + c^2*x^2])*((-I)*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/
2]) + ArcSinh[c*x]*(I*(2 + Sqrt[1 + c^2*x^2])*Cosh[ArcSinh[c*x]/2] - (-2 + Sqrt[1 + c^2*x^2])*Sinh[ArcSinh[c*x
]/2])))/(d^2*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])) + (b*f^2*Sqrt[d + I*c*d*x]*Sqr
t[f - I*c*f*x]*(-10*ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) - (Cosh[2*ArcSinh[c*x]] + 8
*((2*I)*c*x + (4*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + Log[1 + c^2*x^2]))*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c
*x]/2]) + 2*ArcSinh[c*x]*(Sinh[ArcSinh[c*x]/2]*(8 - 8*Sqrt[1 + c^2*x^2] + I*Sinh[2*ArcSinh[c*x]]) + Cosh[ArcSi
nh[c*x]/2]*((8*I)*(1 + Sqrt[1 + c^2*x^2]) + Sinh[2*ArcSinh[c*x]]))))/(d^2*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]
/2] + I*Sinh[ArcSinh[c*x]/2])))/(8*c)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (-i c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )}{\left (i c d x +d \right )^{\frac {3}{2}}}\, dx\]

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)^(3/2),x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

1/2*(c^2*f^3*x^3/(sqrt(c^2*d*f*x^2 + d*f)*d) + 8*I*c*f^3*x^2/(sqrt(c^2*d*f*x^2 + d*f)*d) + 17*f^3*x/(sqrt(c^2*
d*f*x^2 + d*f)*d) - 15*f^3*arcsinh(c*x)/(sqrt(d*f)*c*d) + 24*I*f^3/(sqrt(c^2*d*f*x^2 + d*f)*c*d))*a + b*integr
ate((-I*c*f*x + f)^(5/2)*log(c*x + sqrt(c^2*x^2 + 1))/(I*c*d*x + d)^(3/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))*(f - c*f*x*1i)^(5/2))/(d + c*d*x*1i)^(3/2),x)

[Out]

int(((a + b*asinh(c*x))*(f - c*f*x*1i)^(5/2))/(d + c*d*x*1i)^(3/2), x)

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