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

Optimal. Leaf size=67 \[ \frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{3 b n \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac {2 i F\left (\left .\frac {1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{3 b n} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2636, 2641} \[ \frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{3 b n \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac {2 i F\left (\left .\frac {1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{3 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 122, normalized size = 1.82 \[ \frac {2 \left (\cosh \left (a+b \log \left (c x^n\right )\right ) \sqrt {\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right )+\sinh \left (a+b \log \left (c x^n\right )\right )\right )}{3 b n \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Sinh[a + b*Log[c*x^n]] + 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])]]*Sqrt[1 + Cosh[2*(a + b*Log[c*x^n])] + Sinh[2*(a + b*Log[c*x^n])]]))/(3*b*n*Co
sh[a + b*Log[c*x^n]]^(3/2))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{x \cosh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.40, size = 295, normalized size = 4.40 \[ \frac {2 \left (2 \sqrt {-\left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )}\, \sqrt {-2 \left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-1}\, \EllipticF \left (\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ), \sqrt {2}\right ) \left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )+\sqrt {-\left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )}\, \sqrt {-2 \left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-1}\, \EllipticF \left (\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ), \sqrt {2}\right )+2 \cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )\right ) \sqrt {\left (2 \left (\cosh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-1\right ) \left (\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )}}{3 n \sqrt {2 \left (\sinh ^{4}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )+\sinh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}\, \left (2 \left (\cosh ^{2}\left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )\right )-1\right )^{\frac {3}{2}} \sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/cosh(a+b*ln(c*x^n))^(5/2),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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \cosh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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