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

Optimal. Leaf size=97 \[ \frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text {sech}^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac {6 i \sqrt {\text {sech}\left (a+b \log \left (c x^n\right )\right )} \sqrt {\cosh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{5 b n} \]

[Out]

2/5*sinh(a+b*ln(c*x^n))/b/n/sech(a+b*ln(c*x^n))^(3/2)-6/5*I*(cosh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)/cosh(1/2*a+1
/2*b*ln(c*x^n))*EllipticE(I*sinh(1/2*a+1/2*b*ln(c*x^n)),2^(1/2))*cosh(a+b*ln(c*x^n))^(1/2)*sech(a+b*ln(c*x^n))
^(1/2)/b/n

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Rubi [A]  time = 0.07, antiderivative size = 97, 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 = {3769, 3771, 2639} \[ \frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text {sech}^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac {6 i \sqrt {\text {sech}\left (a+b \log \left (c x^n\right )\right )} \sqrt {\cosh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{5 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-6*I)/5)*Sqrt[Cosh[a + b*Log[c*x^n]]]*EllipticE[(I/2)*(a + b*Log[c*x^n]), 2]*Sqrt[Sech[a + b*Log[c*x^n]]])/
(b*n) + (2*Sinh[a + b*Log[c*x^n]])/(5*b*n*Sech[a + b*Log[c*x^n]]^(3/2))

Rule 2639

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

Rule 3769

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

Rubi steps

\begin {align*} \int \frac {1}{x \text {sech}^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\text {sech}^{\frac {5}{2}}(a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=\frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text {sech}^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {\text {sech}(a+b x)}} \, dx,x,\log \left (c x^n\right )\right )}{5 n}\\ &=\frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text {sech}^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {\left (3 \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 \sqrt {\cosh (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{5 n}\\ &=-\frac {6 i \sqrt {\cosh \left (a+b \log \left (c x^n\right )\right )} E\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 )}}{5 b n}+\frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text {sech}^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 87, normalized size = 0.90 \[ \frac {\sqrt {\text {sech}\left (a+b \log \left (c x^n\right )\right )} \left (\sinh \left (a+b \log \left (c x^n\right )\right )+\sinh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-12 i \sqrt {\cosh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )\right )}{10 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sech[a + b*Log[c*x^n]]]*((-12*I)*Sqrt[Cosh[a + b*Log[c*x^n]]]*EllipticE[(I/2)*(a + b*Log[c*x^n]), 2] + S
inh[a + b*Log[c*x^n]] + Sinh[3*(a + b*Log[c*x^n])]))/(10*b*n)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{x \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.67, size = 256, normalized size = 2.64 \[ \frac {2 \sqrt {\left (2 \left (\cosh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-1\right ) \left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )}\, \left (8 \left (\cosh ^{7}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-16 \left (\cosh ^{5}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )+10 \left (\cosh ^{3}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-3 \sqrt {-\left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )}\, \sqrt {-2 \left (\cosh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )+1}\, \EllipticE \left (\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ), \sqrt {2}\right )-2 \cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )}{5 n \sqrt {2 \left (\sinh ^{4}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )+\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}\, \sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \sqrt {2 \left (\cosh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-1}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/n*((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)*(8*cosh(1/2*a+1/2*b*ln(c*x^n))
^7-16*cosh(1/2*a+1/2*b*ln(c*x^n))^5+10*cosh(1/2*a+1/2*b*ln(c*x^n))^3-3*(-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)^(1/2)*EllipticE(cosh(1/2*a+1/2*b*ln(c*x^n)),2^(1/2))-2*cosh(1/2*a+1/2*b*l
n(c*x^n)))/(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)/sinh(1/2*a+1/2*b*ln(c*x^n))/(
2*cosh(1/2*a+1/2*b*ln(c*x^n))^2-1)^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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