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

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

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 69, 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} \[ \frac {4 e^{2 a} x \left (c x^n\right )^{2 b} \, _2F_1\left (2,\frac {1}{2} \left (2+\frac {1}{b n}\right );\frac {1}{2} \left (4+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{2 b n+1} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*Log[c*x^n]]^2,x]

[Out]

(4*E^(2*a)*x*(c*x^n)^(2*b)*Hypergeometric2F1[2, (2 + 1/(b*n))/2, (4 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b))])/(
1 + 2*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}^2\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}^2(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (4 e^{-2 a} x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1-2 b+\frac {1}{n}}}{\left (1+e^{-2 a} x^{-2 b}\right )^2} \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (4 e^{-2 a} x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+2 b+\frac {1}{n}}}{\left (e^{-2 a}+x^{2 b}\right )^2} \, dx,x,c x^n\right )}{n}\\ &=\frac {4 e^{2 a} x \left (c x^n\right )^{2 b} \, _2F_1\left (2,\frac {1}{2} \left (2+\frac {1}{b n}\right );\frac {1}{2} \left (4+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+2 b n}\\ \end {align*}

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Mathematica [A]  time = 5.55, size = 126, normalized size = 1.83 \[ \frac {x \left (-\frac {e^{2 a} \left (c x^n\right )^{2 b} \, _2F_1\left (1,1+\frac {1}{2 b n};2+\frac {1}{2 b n};-e^{2 a} \left (c x^n\right )^{2 b}\right )}{2 b n+1}+\, _2F_1\left (1,\frac {1}{2 b n};1+\frac {1}{2 b n};-e^{2 a} \left (c x^n\right )^{2 b}\right )+\tanh \left (a+b \log \left (c x^n\right )\right )\right )}{b n} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sech[a + b*Log[c*x^n]]^2,x]

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sech(b*log(c*x^n) + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.69, size = 0, normalized size = 0.00 \[ \int \mathrm {sech}\left (a +b \ln \left (c \,x^{n}\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, x}{b c^{2 \, b} n e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + b n} + 4 \, \int \frac {1}{2 \, {\left (b c^{2 \, b} n e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + b n\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cosh(a + b*log(c*x^n))^2,x)

[Out]

int(1/cosh(a + b*log(c*x^n))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(a + b*log(c*x**n))**2, x)

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