3.6.51 \(\int \frac {(f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x))}{(d+i c d x)^{5/2}} \, dx\) [551]

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

[Out]

I*b*f^5*x*(c^2*x^2+1)^(5/2)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+8/3*I*b*f^5*(c^2*x^2+1)^(5/2)/c/(I-c*x)/(d+I*c
*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-5/2*b*f^5*(c^2*x^2+1)^(5/2)*arcsinh(c*x)^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)
+2/3*I*f^5*(1-I*c*x)^4*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-10/3*I*f^5*(1-I*c*
x)^2*(c^2*x^2+1)^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*(c^2*x^2+1)^3*(a+b*arcsinh
(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+5*f^5*(c^2*x^2+1)^(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)+28/3*b*f^5*(c^2*x^2+1)^(5/2)*ln(I-c*x)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)

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Rubi [A]
time = 0.30, antiderivative size = 472, normalized size of antiderivative = 1.00, 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 = {5796, 683, 655, 221, 5837, 641, 45, 5783} \begin {gather*} -\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}} \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)^(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 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 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)^{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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1005\) vs. \(2(472)=944\).
time = 4.91, size = 1005, normalized size = 2.13 \begin {gather*} \frac {-\frac {4 i a f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (-23-34 i c x+3 c^2 x^2\right )}{d^3 (-i+c x)^2}+\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^{5/2}}-\frac {2 b f^2 \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 (\cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right ) \left (\left (-14+3 i \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)-28 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+7 i \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (84 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-i \left (8-6 i \sinh ^{-1}(c x)+9 \sinh ^{-1}(c x)^2+21 \log \left (1+c^2 x^2\right )\right )\right )+2 \left (4-4 i \sinh ^{-1}(c x)+6 \sinh ^{-1}(c x)^2+56 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+14 \log \left (1+c^2 x^2\right )+\sqrt {1+c^2 x^2} \left (\sinh ^{-1}(c x) \left (-14 i+3 \sinh ^{-1}(c x)\right )+28 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+7 \log \left (1+c^2 x^2\right )\right )\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{d^3 (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 f^2 \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 (-i \cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right ) \left (\sinh ^{-1}(c x)-2 \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-\frac {1}{2} i \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (4+3 i \sinh ^{-1}(c x)-6 i \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\frac {3}{2} \log \left (1+c^2 x^2\right )\right )+2 \left (\left (2+\sqrt {1+c^2 x^2}\right ) \sinh ^{-1}(c x)+2 \left (2+\sqrt {1+c^2 x^2}\right ) \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\frac {1}{2} i \left (4+\left (2+\sqrt {1+c^2 x^2}\right ) \log \left (1+c^2 x^2\right )\right )\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{d^3 (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 {b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (2 \left (4+6 i c x-6 c^2 x^2+52 (-i+c x) \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+13 (1+i c x) \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 )+18 \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 )^3+\sinh ^{-1}(c x) \left (-24 i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-35 i \cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right )+3 i \cosh \left (\frac {5}{2} \sinh ^{-1}(c x)\right )-24 \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+35 \sinh \left (\frac {3}{2} \sinh ^{-1}(c x)\right )+3 \sinh \left (\frac {5}{2} \sinh ^{-1}(c x)\right )\right )\right )}{d^3 (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} \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)^(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 = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (-i c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )}{\left (i c d x +d \right )^{\frac {5}{2}}}\, dx\]

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)^(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]
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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(3*I*(c^2*d*f*x^2 + d*f)^(5/2)/(c^5*d^5*x^4 - 4*I*c^4*d^5*x^3 - 6*c^3*d^5*x^2 + 4*I*c^2*d^5*x + c*d^5) -
15*I*(c^2*d*f*x^2 + d*f)^(3/2)*f/(-3*I*c^4*d^4*x^3 - 9*c^3*d^4*x^2 + 9*I*c^2*d^4*x + 3*c*d^4) + 10*I*sqrt(c^2*
d*f*x^2 + d*f)*f^2/(c^3*d^3*x^2 - 2*I*c^2*d^3*x - c*d^3) + 105*I*sqrt(c^2*d*f*x^2 + d*f)*f^2/(3*I*c^2*d^3*x +
3*c*d^3) - 15*f^3*arcsinh(c*x)/(c*d^3*sqrt(f/d)))*a + b*integrate((-I*c*f*x + f)^(5/2)*log(c*x + sqrt(c^2*x^2
+ 1))/(I*c*d*x + d)^(5/2), 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)^(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(-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)**(5/2),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)^(5/2),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 )}^{5/2}} \,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)^(5/2),x)

[Out]

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

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