3.284 \(\int \frac {1}{x \sinh ^{\frac {5}{2}}(a+b \log (c x^n))} \, dx\)

Optimal. Leaf size=111 \[ -\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {2 i \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )} F\left (\left .\frac {1}{2} \left (i a+i b \log \left (c x^n\right )-\frac {\pi }{2}\right )\right |2\right )}{3 b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \]

[Out]

-2/3*cosh(a+b*ln(c*x^n))/b/n/sinh(a+b*ln(c*x^n))^(3/2)-2/3*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n))^2)^(1/2)/s
in(1/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n))*EllipticF(cos(1/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n)),2^(1/2))*(I*sinh(a+b*ln(c
*x^n)))^(1/2)/b/n/sinh(a+b*ln(c*x^n))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2636, 2642, 2641} \[ -\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {2 i \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )} F\left (\left .\frac {1}{2} \left (i a+i b \log \left (c x^n\right )-\frac {\pi }{2}\right )\right |2\right )}{3 b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sinh[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(-2*Cosh[a + b*Log[c*x^n]])/(3*b*n*Sinh[a + b*Log[c*x^n]]^(3/2)) + (((2*I)/3)*EllipticF[(I*a - Pi/2 + I*b*Log[
c*x^n])/2, 2]*Sqrt[I*Sinh[a + b*Log[c*x^n]]])/(b*n*Sqrt[Sinh[a + b*Log[c*x^n]]])

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{x \sinh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx,x,\log \left (c x^n\right )\right )}{3 n}\\ &=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac {\sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )} \operatorname {Subst}\left (\int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx,x,\log \left (c x^n\right )\right )}{3 n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}\\ &=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {2 i F\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b \log \left (c x^n\right )\right )\right |2\right ) \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}}{3 b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 122, normalized size = 1.10 \[ -\frac {2 \left (\sinh \left (a+b \log \left (c x^n\right )\right ) \sqrt {-\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right )+\cosh \left (a+b \log \left (c x^n\right )\right )\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sinh[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(-2*(Cosh[a + b*Log[c*x^n]] + Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(a + b*Log[c*x^n])] + Sinh[2*(a + b*Log[
c*x^n])]]*Sinh[a + b*Log[c*x^n]]*Sqrt[1 - Cosh[2*(a + b*Log[c*x^n])] - Sinh[2*(a + b*Log[c*x^n])]]))/(3*b*n*Si
nh[a + b*Log[c*x^n]]^(3/2))

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{x \sinh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sinh(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")

[Out]

integral(1/(x*sinh(b*log(c*x^n) + a)^(5/2)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sinh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sinh(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

integrate(1/(x*sinh(b*log(c*x^n) + a)^(5/2)), x)

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maple [A]  time = 0.08, size = 144, normalized size = 1.30 \[ -\frac {i \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+2 \left (\cosh ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{3 n \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )^{\frac {3}{2}} \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/sinh(a+b*ln(c*x^n))^(5/2),x)

[Out]

-1/3/n/sinh(a+b*ln(c*x^n))^(3/2)*(I*(1-I*sinh(a+b*ln(c*x^n)))^(1/2)*2^(1/2)*(I*sinh(a+b*ln(c*x^n))+1)^(1/2)*(I
*sinh(a+b*ln(c*x^n)))^(1/2)*EllipticF((1-I*sinh(a+b*ln(c*x^n)))^(1/2),1/2*2^(1/2))*sinh(a+b*ln(c*x^n))+2*cosh(
a+b*ln(c*x^n))^2)/cosh(a+b*ln(c*x^n))/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sinh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sinh(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(x*sinh(b*log(c*x^n) + a)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x\,{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*sinh(a + b*log(c*x^n))^(5/2)),x)

[Out]

int(1/(x*sinh(a + b*log(c*x^n))^(5/2)), 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(1/x/sinh(a+b*ln(c*x**n))**(5/2),x)

[Out]

Timed out

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