3.185 \(\int \text{sech}^4(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=69 \[ \frac{16 e^{4 a} x \left (c x^n\right )^{4 b} \, _2F_1\left (4,\frac{1}{2} \left (4+\frac{1}{b n}\right );\frac{1}{2} \left (6+\frac{1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{4 b n+1} \]

[Out]

(16*E^(4*a)*x*(c*x^n)^(4*b)*Hypergeometric2F1[4, (4 + 1/(b*n))/2, (6 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b))])/
(1 + 4*b*n)

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Rubi [A]  time = 0.0730036, antiderivative size = 69, normalized size of antiderivative = 1., 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} \[ \frac{16 e^{4 a} x \left (c x^n\right )^{4 b} \, _2F_1\left (4,\frac{1}{2} \left (4+\frac{1}{b n}\right );\frac{1}{2} \left (6+\frac{1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{4 b n+1} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*Log[c*x^n]]^4,x]

[Out]

(16*E^(4*a)*x*(c*x^n)^(4*b)*Hypergeometric2F1[4, (4 + 1/(b*n))/2, (6 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b))])/
(1 + 4*b*n)

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]

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])

Rubi steps

\begin{align*} \int \text{sech}^4\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}^4(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac{\left (16 e^{-4 a} x \left (c x^n\right )^{-1/n}\right ) \operatorname{Subst}\left (\int \frac{x^{-1-4 b+\frac{1}{n}}}{\left (1+e^{-2 a} x^{-2 b}\right )^4} \, dx,x,c x^n\right )}{n}\\ &=\frac{\left (16 e^{-4 a} x \left (c x^n\right )^{-1/n}\right ) \operatorname{Subst}\left (\int \frac{x^{-1+4 b+\frac{1}{n}}}{\left (e^{-2 a}+x^{2 b}\right )^4} \, dx,x,c x^n\right )}{n}\\ &=\frac{16 e^{4 a} x \left (c x^n\right )^{4 b} \, _2F_1\left (4,\frac{1}{2} \left (4+\frac{1}{b n}\right );\frac{1}{2} \left (6+\frac{1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+4 b n}\\ \end{align*}

Mathematica [B]  time = 13.4016, size = 192, normalized size = 2.78 \[ \frac{x \left (\left (8 b^2 n^2-2\right ) \, _2F_1\left (1,\frac{1}{2 b n};1+\frac{1}{2 b n};-e^{2 a} \left (c x^n\right )^{2 b}\right )+\text{sech}^2\left (a+b \log \left (c x^n\right )\right ) \left (\tanh \left (a+b \log \left (c x^n\right )\right ) \left (\left (4 b^2 n^2-1\right ) \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+8 b^2 n^2-1\right )+2 b n\right )-2 e^{2 a} (2 b n-1) \left (c x^n\right )^{2 b} \, _2F_1\left (1,1+\frac{1}{2 b n};2+\frac{1}{2 b n};-e^{2 a} \left (c x^n\right )^{2 b}\right )\right )}{12 b^3 n^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sech[a + b*Log[c*x^n]]^4,x]

[Out]

(x*(-2*E^(2*a)*(-1 + 2*b*n)*(c*x^n)^(2*b)*Hypergeometric2F1[1, 1 + 1/(2*b*n), 2 + 1/(2*b*n), -(E^(2*a)*(c*x^n)
^(2*b))] + (-2 + 8*b^2*n^2)*Hypergeometric2F1[1, 1/(2*b*n), 1 + 1/(2*b*n), -(E^(2*a)*(c*x^n)^(2*b))] + Sech[a
+ b*Log[c*x^n]]^2*(2*b*n + (-1 + 8*b^2*n^2 + (-1 + 4*b^2*n^2)*Cosh[2*(a + b*Log[c*x^n])])*Tanh[a + b*Log[c*x^n
]])))/(12*b^3*n^3)

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Maple [F]  time = 0.105, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm sech} \left (a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(a+b*ln(c*x^n))^4,x)

[Out]

int(sech(a+b*ln(c*x^n))^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 16 \,{\left (4 \, b^{2} n^{2} - 1\right )} \int \frac{1}{48 \,{\left (b^{3} c^{2 \, b} n^{3} e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + b^{3} n^{3}\right )}}\,{d x} + \frac{{\left (2 \, b c^{4 \, b} n + c^{4 \, b}\right )} x e^{\left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right )} - 2 \,{\left (6 \, b^{2} c^{2 \, b} n^{2} - b c^{2 \, b} n - c^{2 \, b}\right )} x e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} -{\left (4 \, b^{2} n^{2} - 1\right )} x}{3 \,{\left (b^{3} c^{6 \, b} n^{3} e^{\left (6 \, b \log \left (x^{n}\right ) + 6 \, a\right )} + 3 \, b^{3} c^{4 \, b} n^{3} e^{\left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right )} + 3 \, b^{3} c^{2 \, b} n^{3} e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + b^{3} n^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*log(c*x^n))^4,x, algorithm="maxima")

[Out]

16*(4*b^2*n^2 - 1)*integrate(1/48/(b^3*c^(2*b)*n^3*e^(2*b*log(x^n) + 2*a) + b^3*n^3), x) + 1/3*((2*b*c^(4*b)*n
+ c^(4*b))*x*e^(4*b*log(x^n) + 4*a) - 2*(6*b^2*c^(2*b)*n^2 - b*c^(2*b)*n - c^(2*b))*x*e^(2*b*log(x^n) + 2*a)
- (4*b^2*n^2 - 1)*x)/(b^3*c^(6*b)*n^3*e^(6*b*log(x^n) + 6*a) + 3*b^3*c^(4*b)*n^3*e^(4*b*log(x^n) + 4*a) + 3*b^
3*c^(2*b)*n^3*e^(2*b*log(x^n) + 2*a) + b^3*n^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\operatorname{sech}\left (b \log \left (c x^{n}\right ) + a\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*log(c*x^n))^4,x, algorithm="fricas")

[Out]

integral(sech(b*log(c*x^n) + a)^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}^{4}{\left (a + b \log{\left (c x^{n} \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*ln(c*x**n))**4,x)

[Out]

Integral(sech(a + b*log(c*x**n))**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}\left (b \log \left (c x^{n}\right ) + a\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(a+b*log(c*x^n))^4,x, algorithm="giac")

[Out]

integrate(sech(b*log(c*x^n) + a)^4, x)