3.184 \(\int \text {sech}^3(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=70 \[ \frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \, _2F_1\left (3,\frac {3 b+\frac {1}{n}}{2 b};\frac {1}{2} \left (5+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{3 b n+1} \]

[Out]

8*exp(3*a)*x*(c*x^n)^(3*b)*hypergeom([3, 1/2*(3*b+1/n)/b],[5/2+1/2/b/n],-exp(2*a)*(c*x^n)^(2*b))/(3*b*n+1)

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Rubi [A]  time = 0.07, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5545, 5547, 263, 364} \[ \frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \, _2F_1\left (3,\frac {3 b+\frac {1}{n}}{2 b};\frac {1}{2} \left (5+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{3 b n+1} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*Log[c*x^n]]^3,x]

[Out]

(8*E^(3*a)*x*(c*x^n)^(3*b)*Hypergeometric2F1[3, (3*b + n^(-1))/(2*b), (5 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b)
)])/(1 + 3*b*n)

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 5545

Int[Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[
x^(1/n - 1)*Sech[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n
, 1])

Rule 5547

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[2^p/E^(a*d*p), Int[(e*x)^m
/(x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p), x], x] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {\left (x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int x^{-1+\frac {1}{n}} \text {sech}^3(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (8 e^{-3 a} x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1-3 b+\frac {1}{n}}}{\left (1+e^{-2 a} x^{-2 b}\right )^3} \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (8 e^{-3 a} x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+3 b+\frac {1}{n}}}{\left (e^{-2 a}+x^{2 b}\right )^3} \, dx,x,c x^n\right )}{n}\\ &=\frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \, _2F_1\left (3,\frac {3 b+\frac {1}{n}}{2 b};\frac {1}{2} \left (5+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+3 b n}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 101, normalized size = 1.44 \[ \frac {x \left (2 e^a (b n-1) \left (c x^n\right )^b \, _2F_1\left (1,\frac {1}{2} \left (1+\frac {1}{b n}\right );\frac {1}{2} \left (3+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )+\left (b n \tanh \left (a+b \log \left (c x^n\right )\right )+1\right ) \text {sech}\left (a+b \log \left (c x^n\right )\right )\right )}{2 b^2 n^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sech[a + b*Log[c*x^n]]^3,x]

[Out]

(x*(2*E^a*(-1 + b*n)*(c*x^n)^b*Hypergeometric2F1[1, (1 + 1/(b*n))/2, (3 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b))
] + Sech[a + b*Log[c*x^n]]*(1 + b*n*Tanh[a + b*Log[c*x^n]])))/(2*b^2*n^2)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

integral(sech(b*log(c*x^n) + a)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

integrate(sech(b*log(c*x^n) + a)^3, x)

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maple [F]  time = 2.00, size = 0, normalized size = 0.00 \[ \int \mathrm {sech}\left (a +b \ln \left (c \,x^{n}\right )\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(a+b*ln(c*x^n))^3,x)

[Out]

int(sech(a+b*ln(c*x^n))^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 8 \, {\left (b^{2} c^{b} n^{2} - c^{b}\right )} \int \frac {e^{\left (b \log \left (x^{n}\right ) + a\right )}}{8 \, {\left (b^{2} c^{2 \, b} n^{2} e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + b^{2} n^{2}\right )}}\,{d x} + \frac {{\left (b c^{3 \, b} n + c^{3 \, b}\right )} x e^{\left (3 \, b \log \left (x^{n}\right ) + 3 \, a\right )} - {\left (b c^{b} n - c^{b}\right )} x e^{\left (b \log \left (x^{n}\right ) + a\right )}}{b^{2} c^{4 \, b} n^{2} e^{\left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right )} + 2 \, b^{2} c^{2 \, b} n^{2} e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + b^{2} n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*log(c*x^n))^3,x, algorithm="maxima")

[Out]

8*(b^2*c^b*n^2 - c^b)*integrate(1/8*e^(b*log(x^n) + a)/(b^2*c^(2*b)*n^2*e^(2*b*log(x^n) + 2*a) + b^2*n^2), x)
+ ((b*c^(3*b)*n + c^(3*b))*x*e^(3*b*log(x^n) + 3*a) - (b*c^b*n - c^b)*x*e^(b*log(x^n) + a))/(b^2*c^(4*b)*n^2*e
^(4*b*log(x^n) + 4*a) + 2*b^2*c^(2*b)*n^2*e^(2*b*log(x^n) + 2*a) + b^2*n^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cosh(a + b*log(c*x^n))^3,x)

[Out]

int(1/cosh(a + b*log(c*x^n))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {sech}^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*ln(c*x**n))**3,x)

[Out]

Integral(sech(a + b*log(c*x**n))**3, x)

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