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

Optimal. Leaf size=517 \[ -\frac {5 i d^4 (1+i c x) \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {15 i d^4 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 i d^4 (1+i c x)^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {15 d^4 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {b c d^4 x^2 \left (c^2 x^2+1\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b d^4 (1+i c x)^2 \left (c^2 x^2+1\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {3 i b d^4 x \left (c^2 x^2+1\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {8 b d^4 \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 b d^4 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)^2}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

3/2*I*b*d^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*d^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*d^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*d
^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*d^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*d^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*d^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*d^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*d^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 = 0.42, antiderivative size = 517, 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 = {5712, 669, 671, 641, 215, 5819, 627, 43, 5675} \[ -\frac {5 i d^4 (1+i c x) \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {15 i d^4 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 i d^4 (1+i c x)^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {15 d^4 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {b c d^4 x^2 \left (c^2 x^2+1\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {5 b d^4 (1+i c x)^2 \left (c^2 x^2+1\right )^{3/2}}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {3 i b d^4 x \left (c^2 x^2+1\right )^{3/2}}{2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {8 b d^4 \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {15 b d^4 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)^2}{4 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/2)*b*d^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*d^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*d^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*d^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)*d^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)*d^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)*d^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*d^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*d^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 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 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 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 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 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 671

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 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
]

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 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])

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)^{3/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{3/2} \int \frac {(d+i c d 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 d^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 d^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 d^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 d^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 d^4}{2 c}-\frac {5 i d^4 (1+i c x)}{2 c}-\frac {2 i d^4 (1+i c x)^3}{c \left (1+c^2 x^2\right )}-\frac {15 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 d^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 = 4.02, size = 781, normalized size = 1.51 \[ \frac {\frac {4 a d^2 \left (c^2 x^2-7 i c x+24\right ) \sqrt {d+i c d x} \sqrt {f-i c f x}}{f^2 (c x+i)}-\frac {60 a d^{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 )}{f^{3/2}}+\frac {4 b d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (2 \left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (4 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+i \log \left (c^2 x^2+1\right )\right )-\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 )\right )+4 \sinh ^{-1}(c x) \left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{f^2 \sqrt {c^2 x^2+1} \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 d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (2 \sinh ^{-1}(c x) \left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (-8 \sqrt {c^2 x^2+1}-i \sinh \left (2 \sinh ^{-1}(c x)\right )+8\right )+\left (\sinh \left (2 \sinh ^{-1}(c x)\right )-8 i \left (\sqrt {c^2 x^2+1}+1\right )\right ) \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (8 i \log \left (c^2 x^2+1\right )+16 c x+i \cosh \left (2 \sinh ^{-1}(c x)\right )+32 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )-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 )\right )}{f^2 \sqrt {c^2 x^2+1} \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 d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\sinh ^{-1}(c x) \left (-\left (\sqrt {c^2 x^2+1}-2\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \left (\sqrt {c^2 x^2+1}+2\right ) \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (i \log \left (c^2 x^2+1\right )+c x-4 \tan ^{-1}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )-\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 )\right )\right )}{f^2 \sqrt {c^2 x^2+1} \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} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*d^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(24 - (7*I)*c*x + c^2*x^2))/(f^2*(I + c*x)) - (60*a*d^(5/2)*Log[
c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]])/f^(3/2) + (4*b*d^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])) + 4*ArcSinh[c*x]*((-I)*Cosh[ArcSin
h[c*x]/2] + Sinh[ArcSinh[c*x]/2]) + 2*(4*ArcTan[Tanh[ArcSinh[c*x]/2]] + I*Log[1 + c^2*x^2])*(I*Cosh[ArcSinh[c*
x]/2] + Sinh[ArcSinh[c*x]/2])))/(f^2*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])) + (16*
b*d^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])))/(f^2*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])) + (b*d^2*Sqrt[d + I*c*d*x]*Sqrt[
f - I*c*f*x]*(-10*ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2]) + (16*c*x + 32*ArcTan[Tanh[Ar
cSinh[c*x]/2]] + I*Cosh[2*ArcSinh[c*x]] + (8*I)*Log[1 + c^2*x^2])*(I*Cosh[ArcSinh[c*x]/2] + 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[ArcSinh[c
*x]/2]*((-8*I)*(1 + Sqrt[1 + c^2*x^2]) + Sinh[2*ArcSinh[c*x]]))))/(f^2*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2]
- I*Sinh[ArcSinh[c*x]/2])))/(8*c)

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b c^{2} d^{2} x^{2} - 2 i \, b c d^{2} x - b d^{2}\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (a c^{2} d^{2} x^{2} - 2 i \, a c d^{2} x - a d^{2}\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{c^{2} f^{2} x^{2} + 2 i \, c f^{2} x - f^{2}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (i c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )}{\left (-i c f x +f \right )^{\frac {3}{2}}}\, dx \]

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)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (\frac {c^{2} d^{3} x^{3}}{\sqrt {c^{2} d f x^{2} + d f} f} - \frac {8 i \, c d^{3} x^{2}}{\sqrt {c^{2} d f x^{2} + d f} f} + \frac {17 \, d^{3} x}{\sqrt {c^{2} d f x^{2} + d f} f} - \frac {15 \, d^{3} \operatorname {arsinh}\left (c x\right )}{\sqrt {d f} c f} - \frac {24 i \, d^{3}}{\sqrt {c^{2} d f x^{2} + d f} c f}\right )} a + b \int \frac {{\left (i \, c d x + d\right )}^{\frac {5}{2}} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (-i \, c f x + f\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(3/2),x)

[Out]

Timed out

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