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

Optimal. Leaf size=403 \[ -\frac {1}{4} i a^2 \sinh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{2} i a^2 \sinh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} i a^2 \sinh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{2} i a^2 \cosh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Shi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} i a^2 \cosh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Shi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {1}{4} i a^2 \cosh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {Shi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \]

[Out]

5/2*a^2*sinh(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)+5/4*I*a^2*cos
h(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)-1/4*a^2*sinh(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)-1/4*a^2*Chi(5/2*d*x)*sech(1/2*c+1/4*I
*Pi+1/2*d*x)*cosh(5/2*c+1/4*I*Pi)*(a+I*a*sinh(d*x+c))^(1/2)+5/2*a^2*Chi(1/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*
cosh(1/2*c+1/4*I*Pi)*(a+I*a*sinh(d*x+c))^(1/2)+5/4*I*a^2*Chi(3/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*sinh(3/2*c+
1/4*I*Pi)*(a+I*a*sinh(d*x+c))^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 403, 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, 3312, 3303, 3298, 3301} \[ -\frac {1}{4} i a^2 \sinh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{2} i a^2 \sinh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} i a^2 \sinh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{2} i a^2 \cosh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Shi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} i a^2 \cosh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Shi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {1}{4} i a^2 \cosh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {Shi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-I/4)*a^2*CoshIntegral[(5*d*x)/2]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[(5*c)/2 - (I/4)*Pi]*Sqrt[a + I*a*Sinh[c
+ d*x]] + ((5*I)/2)*a^2*CoshIntegral[(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]] + ((5*I)/4)*a^2*CoshIntegral[(3*d*x)/2]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[(6*c + I*Pi)/4]*S
qrt[a + I*a*Sinh[c + d*x]] + ((5*I)/2)*a^2*Cosh[(2*c - I*Pi)/4]*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Si
nh[c + d*x]]*SinhIntegral[(d*x)/2] + ((5*I)/4)*a^2*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] - (I/4)*a^2*Cosh[(5*c)/2 - (I/4)*Pi]*Sech[c/2 + (I/4)*Pi + (d*x)/
2]*Sqrt[a + I*a*Sinh[c + d*x]]*SinhIntegral[(5*d*x)/2]

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && 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} \, 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} \, dx\\ &=-\left (\left (4 i 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 \left (\frac {5 i \sinh \left (\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}\right )}{8 x}+\frac {5 i \sinh \left (\frac {1}{4} (6 c+i \pi )+\frac {3 d x}{2}\right )}{16 x}-\frac {i \sinh \left (\frac {1}{4} (10 c-i \pi )+\frac {5 d x}{2}\right )}{16 x}\right ) \, dx\right )\\ &=-\left (\frac {1}{4} \left (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 \left (\frac {1}{4} (10 c-i \pi )+\frac {5 d x}{2}\right )}{x} \, dx\right )+\frac {1}{4} \left (5 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 \left (\frac {1}{4} (6 c+i \pi )+\frac {3 d x}{2}\right )}{x} \, dx+\frac {1}{2} \left (5 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 \left (\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}\right )}{x} \, dx\\ &=-\left (\frac {1}{4} \left (a^2 \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\right )+\frac {1}{2} \left (5 a^2 \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}{4} \left (5 a^2 \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 (a^2 \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}{2} \left (5 a^2 \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 {1}{4} \left (5 a^2 \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} i a^2 \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 {5}{2} i a^2 \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} i a^2 \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}{2} i a^2 \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 {5}{4} i a^2 \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 {1}{4} i a^2 \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 = 1.37, size = 242, normalized size = 0.60 \[ \frac {a^2 (\sinh (c+d x)-i)^2 \sqrt {a+i a \sinh (c+d x)} \left (i \sinh \left (\frac {5 c}{2}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )+\cosh \left (\frac {5 c}{2}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )-10 \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \text {Chi}\left (\frac {d x}{2}\right )+5 \left (\cosh \left (\frac {3 c}{2}\right )-i \sinh \left (\frac {3 c}{2}\right )\right ) \text {Chi}\left (\frac {3 d x}{2}\right )-10 \sinh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )+5 \sinh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+\sinh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )-10 i \cosh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )-5 i \cosh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+i \cosh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )\right )}{4 \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-I + Sinh[c + d*x])^2*Sqrt[a + I*a*Sinh[c + d*x]]*(Cosh[(5*c)/2]*CoshIntegral[(5*d*x)/2] - 10*CoshIntegr
al[(d*x)/2]*(Cosh[c/2] + I*Sinh[c/2]) + 5*CoshIntegral[(3*d*x)/2]*(Cosh[(3*c)/2] - I*Sinh[(3*c)/2]) + I*CoshIn
tegral[(5*d*x)/2]*Sinh[(5*c)/2] - (10*I)*Cosh[c/2]*SinhIntegral[(d*x)/2] - 10*Sinh[c/2]*SinhIntegral[(d*x)/2]
- (5*I)*Cosh[(3*c)/2]*SinhIntegral[(3*d*x)/2] + 5*Sinh[(3*c)/2]*SinhIntegral[(3*d*x)/2] + I*Cosh[(5*c)/2]*Sinh
Integral[(5*d*x)/2] + Sinh[(5*c)/2]*SinhIntegral[(5*d*x)/2]))/(4*(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,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 \[ \int \frac {{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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