3.134 \(\int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx\)

Optimal. Leaf size=444 \[ -\frac {5}{8} a^2 d \sinh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {15}{8} a^2 d \sinh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \sinh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \cosh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {Shi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {15}{8} a^2 d \cosh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {Shi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {5}{8} a^2 d \cosh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {Shi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {4 a^2 \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{x} \]

[Out]

-4*a^2*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^4*(a+I*a*sinh(d*x+c))^(1/2)/x+5/4*a^2*d*cosh(1/2*c+1/4*I*Pi)*sech(1/2*c+1/
4*I*Pi+1/2*d*x)*Shi(1/2*d*x)*(a+I*a*sinh(d*x+c))^(1/2)+15/8*I*a^2*d*sinh(3/2*c+1/4*I*Pi)*sech(1/2*c+1/4*I*Pi+1
/2*d*x)*Shi(3/2*d*x)*(a+I*a*sinh(d*x+c))^(1/2)-5/8*a^2*d*cosh(5/2*c+1/4*I*Pi)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*Shi
(5/2*d*x)*(a+I*a*sinh(d*x+c))^(1/2)-5/8*a^2*d*Chi(5/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*sinh(5/2*c+1/4*I*Pi)*(
a+I*a*sinh(d*x+c))^(1/2)+15/8*I*a^2*d*Chi(3/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*cosh(3/2*c+1/4*I*Pi)*(a+I*a*si
nh(d*x+c))^(1/2)+5/4*a^2*d*Chi(1/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*sinh(1/2*c+1/4*I*Pi)*(a+I*a*sinh(d*x+c))^
(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 444, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3319, 3313, 3303, 3298, 3301} \[ -\frac {5}{8} a^2 d \sinh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {15}{8} a^2 d \sinh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \sinh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \cosh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {Shi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {15}{8} a^2 d \cosh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {Shi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {5}{8} a^2 d \cosh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {Shi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {4 a^2 \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Sinh[c + d*x])^(5/2)/x^2,x]

[Out]

(-4*a^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^4*Sqrt[a + I*a*Sinh[c + d*x]])/x - (5*a^2*d*CoshIntegral[(5*d*x)/2]*Sec
h[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[(5*c)/2 + (I/4)*Pi]*Sqrt[a + I*a*Sinh[c + d*x]])/8 - (15*a^2*d*CoshIntegral[(
3*d*x)/2]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[(6*c - I*Pi)/4]*Sqrt[a + I*a*Sinh[c + d*x]])/8 + (5*a^2*d*CoshIn
tegral[(d*x)/2]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[(2*c + I*Pi)/4]*Sqrt[a + I*a*Sinh[c + d*x]])/4 + (5*a^2*d*
Cosh[(2*c + I*Pi)/4]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]]*SinhIntegral[(d*x)/2])/4 - (15
*a^2*d*Cosh[(6*c - I*Pi)/4]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]]*SinhIntegral[(3*d*x)/2]
)/8 - (5*a^2*d*Cosh[(5*c)/2 + (I/4)*Pi]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]]*SinhIntegra
l[(5*d*x)/2])/8

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx &=\left (4 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\sinh ^5\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right )}{x^2} \, dx\\ &=-\frac {4 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{x}+\left (10 a^2 d \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \left (\frac {\cosh \left (\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}\right )}{8 x}+\frac {3 \cosh \left (\frac {1}{4} (6 c+i \pi )+\frac {3 d x}{2}\right )}{16 x}-\frac {\cosh \left (\frac {1}{4} (10 c-i \pi )+\frac {5 d x}{2}\right )}{16 x}\right ) \, dx\\ &=-\frac {4 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{x}-\frac {1}{8} \left (5 a^2 d \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\cosh \left (\frac {1}{4} (10 c-i \pi )+\frac {5 d x}{2}\right )}{x} \, dx+\frac {1}{4} \left (5 a^2 d \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\cosh \left (\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}\right )}{x} \, dx+\frac {1}{8} \left (15 a^2 d \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\cosh \left (\frac {1}{4} (6 c+i \pi )+\frac {3 d x}{2}\right )}{x} \, dx\\ &=-\frac {4 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{x}+\frac {1}{8} \left (5 i a^2 d \cosh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\sinh \left (\frac {5 d x}{2}\right )}{x} \, dx+\frac {1}{8} \left (15 i a^2 d \cosh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\sinh \left (\frac {3 d x}{2}\right )}{x} \, dx-\frac {1}{4} \left (5 i a^2 d \cosh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\sinh \left (\frac {d x}{2}\right )}{x} \, dx+\frac {1}{8} \left (5 i a^2 d \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\cosh \left (\frac {5 d x}{2}\right )}{x} \, dx+\frac {1}{8} \left (15 i a^2 d \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (6 c-i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\cosh \left (\frac {3 d x}{2}\right )}{x} \, dx-\frac {1}{4} \left (5 i a^2 d \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (2 c+i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \frac {\cosh \left (\frac {d x}{2}\right )}{x} \, dx\\ &=-\frac {4 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{x}-\frac {5}{8} a^2 d \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {15}{8} a^2 d \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (6 c-i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (2 c+i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \cosh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {d x}{2}\right )-\frac {15}{8} a^2 d \cosh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {3 d x}{2}\right )-\frac {5}{8} a^2 d \cosh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {5 d x}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 2.21, size = 347, normalized size = 0.78 \[ \frac {a^2 (\sinh (c+d x)-i)^2 \sqrt {a+i a \sinh (c+d x)} \left (5 d x \sinh \left (\frac {5 c}{2}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )+5 i d x \cosh \left (\frac {5 c}{2}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )-10 i d x \left (\cosh \left (\frac {c}{2}\right )-i \sinh \left (\frac {c}{2}\right )\right ) \text {Chi}\left (\frac {d x}{2}\right )+15 d x \left (\sinh \left (\frac {3 c}{2}\right )-i \cosh \left (\frac {3 c}{2}\right )\right ) \text {Chi}\left (\frac {3 d x}{2}\right )-10 i d x \sinh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )-15 i d x \sinh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+5 i d x \sinh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )-10 d x \cosh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )+15 d x \cosh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+5 d x \cosh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )+20 i \sinh \left (\frac {1}{2} (c+d x)\right )+10 i \sinh \left (\frac {3}{2} (c+d x)\right )-2 i \sinh \left (\frac {5}{2} (c+d x)\right )+20 \cosh \left (\frac {1}{2} (c+d x)\right )-10 \cosh \left (\frac {3}{2} (c+d x)\right )-2 \cosh \left (\frac {5}{2} (c+d x)\right )\right )}{8 x \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Sinh[c + d*x])^(5/2)/x^2,x]

[Out]

(a^2*(-I + Sinh[c + d*x])^2*Sqrt[a + I*a*Sinh[c + d*x]]*(20*Cosh[(c + d*x)/2] - 10*Cosh[(3*(c + d*x))/2] - 2*C
osh[(5*(c + d*x))/2] + (5*I)*d*x*Cosh[(5*c)/2]*CoshIntegral[(5*d*x)/2] - (10*I)*d*x*CoshIntegral[(d*x)/2]*(Cos
h[c/2] - I*Sinh[c/2]) + 15*d*x*CoshIntegral[(3*d*x)/2]*((-I)*Cosh[(3*c)/2] + Sinh[(3*c)/2]) + 5*d*x*CoshIntegr
al[(5*d*x)/2]*Sinh[(5*c)/2] + (20*I)*Sinh[(c + d*x)/2] + (10*I)*Sinh[(3*(c + d*x))/2] - (2*I)*Sinh[(5*(c + d*x
))/2] - 10*d*x*Cosh[c/2]*SinhIntegral[(d*x)/2] - (10*I)*d*x*Sinh[c/2]*SinhIntegral[(d*x)/2] + 15*d*x*Cosh[(3*c
)/2]*SinhIntegral[(3*d*x)/2] - (15*I)*d*x*Sinh[(3*c)/2]*SinhIntegral[(3*d*x)/2] + 5*d*x*Cosh[(5*c)/2]*SinhInte
gral[(5*d*x)/2] + (5*I)*d*x*Sinh[(5*c)/2]*SinhIntegral[(5*d*x)/2]))/(8*x*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)
/2])^5)

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fricas [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((a+I*a*sinh(d*x+c))^(5/2)/x^2,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^2,x, algorithm="giac")

[Out]

integrate((I*a*sinh(d*x + c) + a)^(5/2)/x^2, x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*sinh(d*x+c))^(5/2)/x^2,x)

[Out]

int((a+I*a*sinh(d*x+c))^(5/2)/x^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^2,x, algorithm="maxima")

[Out]

integrate((I*a*sinh(d*x + c) + a)^(5/2)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sinh(c + d*x)*1i)^(5/2)/x^2,x)

[Out]

int((a + a*sinh(c + d*x)*1i)^(5/2)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (i a \left (\sinh {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(d*x+c))**(5/2)/x**2,x)

[Out]

Integral((I*a*(sinh(c + d*x) - I))**(5/2)/x**2, x)

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