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

Optimal. Leaf size=68 \[ \frac {16 e^{4 a} x \left (c x^n\right )^{4 b} \, _2F_1\left (4,\frac {1}{2} \left (4+\frac {1}{b n}\right );\frac {1}{2} \left (6+\frac {1}{b n}\right );e^{2 a} \left (c x^n\right )^{2 b}\right )}{4 b n+1} \]

[Out]

16*exp(4*a)*x*(c*x^n)^(4*b)*hypergeom([4, 2+1/2/b/n],[3+1/2/b/n],exp(2*a)*(c*x^n)^(2*b))/(4*b*n+1)

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Rubi [A]  time = 0.07, antiderivative size = 68, 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 = {5546, 5548, 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[Csch[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 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 5546

Int[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[
x^(1/n - 1)*Csch[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 5548

Int[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), 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 {csch}^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 {csch}^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*}

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Mathematica [B]  time = 9.02, size = 200, normalized size = 2.94 \[ \frac {x \left (4 \left (4 b^2 n^2-1\right ) \, _2F_1\left (1,\frac {1}{2 b n};1+\frac {1}{2 b n};e^{2 \left (a+b \log \left (c x^n\right )\right )}\right )+\text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \left (\left (1-12 b^2 n^2\right ) \cosh \left (a+b \log \left (c x^n\right )\right )+\left (4 b^2 n^2-1\right ) \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-4 b n \sinh \left (a+b \log \left (c x^n\right )\right )\right )+4 e^{2 a} (2 b n-1) \left (c x^n\right )^{2 b} \, _2F_1\left (1,1+\frac {1}{2 b n};2+\frac {1}{2 b n};e^{2 \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{24 b^3 n^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csch[a + b*Log[c*x^n]]^4,x]

[Out]

(x*(4*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 + b*Log[c*
x^n]))] + 4*(-1 + 4*b^2*n^2)*Hypergeometric2F1[1, 1/(2*b*n), 1 + 1/(2*b*n), E^(2*(a + b*Log[c*x^n]))] + Csch[a
+ b*Log[c*x^n]]^3*((1 - 12*b^2*n^2)*Cosh[a + b*Log[c*x^n]] + (-1 + 4*b^2*n^2)*Cosh[3*(a + b*Log[c*x^n])] - 4*
b*n*Sinh[a + b*Log[c*x^n]])))/(24*b^3*n^3)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(csch(b*log(c*x^n) + a)^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.81, size = 0, normalized size = 0.00 \[ \int \mathrm {csch}\left (a +b \ln \left (c \,x^{n}\right )\right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 16 \, {\left (4 \, b^{2} n^{2} - 1\right )} \int \frac {1}{96 \, {\left (b^{3} c^{b} n^{3} e^{\left (b \log \left (x^{n}\right ) + a\right )} + b^{3} n^{3}\right )}}\,{d x} - 16 \, {\left (4 \, b^{2} n^{2} - 1\right )} \int \frac {1}{96 \, {\left (b^{3} c^{b} n^{3} e^{\left (b \log \left (x^{n}\right ) + a\right )} - b^{3} n^{3}\right )}}\,{d x} - \frac {{\left (2 \, b c^{4 \, b} n + c^{4 \, b}\right )} x e^{\left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right )} + 2 \, {\left (6 \, b^{2} c^{2 \, b} n^{2} - b c^{2 \, b} n - c^{2 \, b}\right )} x e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} - {\left (4 \, b^{2} n^{2} - 1\right )} x}{3 \, {\left (b^{3} c^{6 \, b} n^{3} e^{\left (6 \, b \log \left (x^{n}\right ) + 6 \, a\right )} - 3 \, b^{3} c^{4 \, b} n^{3} e^{\left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right )} + 3 \, b^{3} c^{2 \, b} n^{3} e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} - b^{3} n^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*(4*b^2*n^2 - 1)*integrate(1/96/(b^3*c^b*n^3*e^(b*log(x^n) + a) + b^3*n^3), x) - 16*(4*b^2*n^2 - 1)*integrat
e(1/96/(b^3*c^b*n^3*e^(b*log(x^n) + 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(a + b*log(c*x^n))^4,x)

[Out]

int(1/sinh(a + b*log(c*x^n))^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{4}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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