3.196 \(\int \frac{\text{sech}^{\frac{5}{2}}(a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=97 \[ \frac{2 \sinh \left (a+b \log \left (c x^n\right )\right ) \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac{2 i \sqrt{\text{sech}\left (a+b \log \left (c x^n\right )\right )} \sqrt{\cosh \left (a+b \log \left (c x^n\right )\right )} \text{EllipticF}\left (\frac{1}{2} i \left (a+b \log \left (c x^n\right )\right ),2\right )}{3 b n} \]

[Out]

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

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Rubi [A]  time = 0.0614106, antiderivative size = 97, normalized size of antiderivative = 1., 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 = {3768, 3771, 2641} \[ \frac{2 \sinh \left (a+b \log \left (c x^n\right )\right ) \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac{2 i \sqrt{\text{sech}\left (a+b \log \left (c x^n\right )\right )} \sqrt{\cosh \left (a+b \log \left (c x^n\right )\right )} F\left (\left .\frac{1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{3 b n} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*Log[c*x^n]]^(5/2)/x,x]

[Out]

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

Rule 3768

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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]

Rubi steps

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

Mathematica [A]  time = 0.161581, size = 74, normalized size = 0.76 \[ \frac{2 \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right ) \left (\sinh \left (a+b \log \left (c x^n\right )\right )-i \cosh ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right ) \text{EllipticF}\left (\frac{1}{2} i \left (a+b \log \left (c x^n\right )\right ),2\right )\right )}{3 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sech[a + b*Log[c*x^n]]^(3/2)*((-I)*Cosh[a + b*Log[c*x^n]]^(3/2)*EllipticF[(I/2)*(a + b*Log[c*x^n]), 2] + Si
nh[a + b*Log[c*x^n]]))/(3*b*n)

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Maple [B]  time = 0.47, size = 295, normalized size = 3. \begin{align*}{\frac{2}{3\,bn} \left ( 2\,\sqrt{- \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}}\sqrt{-2\, \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) ,\sqrt{2} \right ) \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}+\sqrt{- \left ( \sinh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) \right ) ^{2}}\sqrt{-2\, \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cosh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) ,\sqrt{2} \right ) +2\,\cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2} \right ) \sqrt{ \left ( 2\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}-1 \right ) \left ( \sinh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) \right ) ^{2}}{\frac{1}{\sqrt{2\, \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{4}+ \left ( \sinh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) \right ) ^{2}}}} \left ( 2\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}-1 \right ) ^{-{\frac{3}{2}}} \left ( \sinh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/n*(2*(-sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)*(-2*sinh(1/2*a+1/2*b*ln(c*x^n))^2-1)^(1/2)*EllipticF(cosh(1/2*
a+1/2*b*ln(c*x^n)),2^(1/2))*sinh(1/2*a+1/2*b*ln(c*x^n))^2+(-sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)*(-2*sinh(1/2*
a+1/2*b*ln(c*x^n))^2-1)^(1/2)*EllipticF(cosh(1/2*a+1/2*b*ln(c*x^n)),2^(1/2))+2*cosh(1/2*a+1/2*b*ln(c*x^n))*sin
h(1/2*a+1/2*b*ln(c*x^n))^2)*((2*cosh(1/2*a+1/2*b*ln(c*x^n))^2-1)*sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)/(2*sinh(
1/2*a+1/2*b*ln(c*x^n))^4+sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)/(2*cosh(1/2*a+1/2*b*ln(c*x^n))^2-1)^(3/2)/sinh(1
/2*a+1/2*b*ln(c*x^n))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}\left (b \log \left (c x^{n}\right ) + a\right )^{\frac{5}{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{sech}\left (b \log \left (c x^{n}\right ) + a\right )^{\frac{5}{2}}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*ln(c*x**n))**(5/2)/x,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out