3.242 \(\int \cosh ^4(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=191 \[ \frac {x \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}-\frac {4 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}-\frac {12 b^2 n^2 x \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{64 b^4 n^4-20 b^2 n^2+1}+\frac {24 b^3 n^3 x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{64 b^4 n^4-20 b^2 n^2+1}+\frac {24 b^4 n^4 x}{64 b^4 n^4-20 b^2 n^2+1} \]

[Out]

24*b^4*n^4*x/(64*b^4*n^4-20*b^2*n^2+1)-12*b^2*n^2*x*cosh(a+b*ln(c*x^n))^2/(64*b^4*n^4-20*b^2*n^2+1)+x*cosh(a+b
*ln(c*x^n))^4/(-16*b^2*n^2+1)+24*b^3*n^3*x*cosh(a+b*ln(c*x^n))*sinh(a+b*ln(c*x^n))/(64*b^4*n^4-20*b^2*n^2+1)-4
*b*n*x*cosh(a+b*ln(c*x^n))^3*sinh(a+b*ln(c*x^n))/(-16*b^2*n^2+1)

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Rubi [A]  time = 0.05, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5520, 8} \[ -\frac {12 b^2 n^2 x \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{64 b^4 n^4-20 b^2 n^2+1}+\frac {x \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}-\frac {4 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}+\frac {24 b^3 n^3 x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{64 b^4 n^4-20 b^2 n^2+1}+\frac {24 b^4 n^4 x}{64 b^4 n^4-20 b^2 n^2+1} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*Log[c*x^n]]^4,x]

[Out]

(24*b^4*n^4*x)/(1 - 20*b^2*n^2 + 64*b^4*n^4) - (12*b^2*n^2*x*Cosh[a + b*Log[c*x^n]]^2)/(1 - 20*b^2*n^2 + 64*b^
4*n^4) + (x*Cosh[a + b*Log[c*x^n]]^4)/(1 - 16*b^2*n^2) + (24*b^3*n^3*x*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c
*x^n]])/(1 - 20*b^2*n^2 + 64*b^4*n^4) - (4*b*n*x*Cosh[a + b*Log[c*x^n]]^3*Sinh[a + b*Log[c*x^n]])/(1 - 16*b^2*
n^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5520

Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> -Simp[(x*Cosh[d*(a + b*Log[c*x^n])]^p
)/(b^2*d^2*n^2*p^2 - 1), x] + (Dist[(b^2*d^2*n^2*p*(p - 1))/(b^2*d^2*n^2*p^2 - 1), Int[Cosh[d*(a + b*Log[c*x^n
])]^(p - 2), x], x] + Simp[(b*d*n*p*x*Cosh[d*(a + b*Log[c*x^n])]^(p - 1)*Sinh[d*(a + b*Log[c*x^n])])/(b^2*d^2*
n^2*p^2 - 1), x]) /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 - 1, 0]

Rubi steps

\begin {align*} \int \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {x \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}-\frac {4 b n x \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}-\frac {\left (12 b^2 n^2\right ) \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx}{1-16 b^2 n^2}\\ &=-\frac {12 b^2 n^2 x \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-20 b^2 n^2+64 b^4 n^4}+\frac {x \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}+\frac {24 b^3 n^3 x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-20 b^2 n^2+64 b^4 n^4}-\frac {4 b n x \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}+\frac {\left (24 b^4 n^4\right ) \int 1 \, dx}{1-20 b^2 n^2+64 b^4 n^4}\\ &=\frac {24 b^4 n^4 x}{1-20 b^2 n^2+64 b^4 n^4}-\frac {12 b^2 n^2 x \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-20 b^2 n^2+64 b^4 n^4}+\frac {x \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}+\frac {24 b^3 n^3 x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-20 b^2 n^2+64 b^4 n^4}-\frac {4 b n x \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-16 b^2 n^2}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 167, normalized size = 0.87 \[ \frac {x \left (128 b^3 n^3 \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+16 b^3 n^3 \sinh \left (4 \left (a+b \log \left (c x^n\right )\right )\right )+\left (4-64 b^2 n^2\right ) \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\left (1-4 b^2 n^2\right ) \cosh \left (4 \left (a+b \log \left (c x^n\right )\right )\right )-8 b n \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-4 b n \sinh \left (4 \left (a+b \log \left (c x^n\right )\right )\right )+192 b^4 n^4-60 b^2 n^2+3\right )}{8 \left (64 b^4 n^4-20 b^2 n^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*Log[c*x^n]]^4,x]

[Out]

(x*(3 - 60*b^2*n^2 + 192*b^4*n^4 + (4 - 64*b^2*n^2)*Cosh[2*(a + b*Log[c*x^n])] + (1 - 4*b^2*n^2)*Cosh[4*(a + b
*Log[c*x^n])] - 8*b*n*Sinh[2*(a + b*Log[c*x^n])] + 128*b^3*n^3*Sinh[2*(a + b*Log[c*x^n])] - 4*b*n*Sinh[4*(a +
b*Log[c*x^n])] + 16*b^3*n^3*Sinh[4*(a + b*Log[c*x^n])]))/(8*(1 - 20*b^2*n^2 + 64*b^4*n^4))

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fricas [A]  time = 0.56, size = 293, normalized size = 1.53 \[ -\frac {{\left (4 \, b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{4} - 16 \, {\left (4 \, b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{3} + {\left (4 \, b^{2} n^{2} - 1\right )} x \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{4} + 4 \, {\left (16 \, b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 2 \, {\left (3 \, {\left (4 \, b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 2 \, {\left (16 \, b^{2} n^{2} - 1\right )} x\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - 3 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} x - 16 \, {\left ({\left (4 \, b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{3} + {\left (16 \, b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{8 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((4*b^2*n^2 - 1)*x*cosh(b*n*log(x) + b*log(c) + a)^4 - 16*(4*b^3*n^3 - b*n)*x*cosh(b*n*log(x) + b*log(c)
+ a)*sinh(b*n*log(x) + b*log(c) + a)^3 + (4*b^2*n^2 - 1)*x*sinh(b*n*log(x) + b*log(c) + a)^4 + 4*(16*b^2*n^2 -
1)*x*cosh(b*n*log(x) + b*log(c) + a)^2 + 2*(3*(4*b^2*n^2 - 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2 + 2*(16*b^2
*n^2 - 1)*x)*sinh(b*n*log(x) + b*log(c) + a)^2 - 3*(64*b^4*n^4 - 20*b^2*n^2 + 1)*x - 16*((4*b^3*n^3 - b*n)*x*c
osh(b*n*log(x) + b*log(c) + a)^3 + (16*b^3*n^3 - b*n)*x*cosh(b*n*log(x) + b*log(c) + a))*sinh(b*n*log(x) + b*l
og(c) + a))/(64*b^4*n^4 - 20*b^2*n^2 + 1)

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giac [B]  time = 0.24, size = 777, normalized size = 4.07 \[ \frac {b^{3} c^{4 \, b} n^{3} x x^{4 \, b n} e^{\left (4 \, a\right )}}{64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1} + \frac {8 \, b^{3} c^{2 \, b} n^{3} x x^{2 \, b n} e^{\left (2 \, a\right )}}{64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1} + \frac {24 \, b^{4} n^{4} x}{64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1} - \frac {b^{2} c^{4 \, b} n^{2} x x^{4 \, b n} e^{\left (4 \, a\right )}}{4 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} - \frac {4 \, b^{2} c^{2 \, b} n^{2} x x^{2 \, b n} e^{\left (2 \, a\right )}}{64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1} - \frac {b c^{4 \, b} n x x^{4 \, b n} e^{\left (4 \, a\right )}}{4 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} - \frac {b c^{2 \, b} n x x^{2 \, b n} e^{\left (2 \, a\right )}}{2 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} - \frac {8 \, b^{3} n^{3} x e^{\left (-2 \, a\right )}}{{\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{2 \, b} x^{2 \, b n}} - \frac {b^{3} n^{3} x e^{\left (-4 \, a\right )}}{{\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{4 \, b} x^{4 \, b n}} - \frac {15 \, b^{2} n^{2} x}{2 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} + \frac {c^{4 \, b} x x^{4 \, b n} e^{\left (4 \, a\right )}}{16 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} + \frac {c^{2 \, b} x x^{2 \, b n} e^{\left (2 \, a\right )}}{4 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} - \frac {4 \, b^{2} n^{2} x e^{\left (-2 \, a\right )}}{{\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{2 \, b} x^{2 \, b n}} - \frac {b^{2} n^{2} x e^{\left (-4 \, a\right )}}{4 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{4 \, b} x^{4 \, b n}} + \frac {b n x e^{\left (-2 \, a\right )}}{2 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{2 \, b} x^{2 \, b n}} + \frac {b n x e^{\left (-4 \, a\right )}}{4 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{4 \, b} x^{4 \, b n}} + \frac {3 \, x}{8 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )}} + \frac {x e^{\left (-2 \, a\right )}}{4 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{2 \, b} x^{2 \, b n}} + \frac {x e^{\left (-4 \, a\right )}}{16 \, {\left (64 \, b^{4} n^{4} - 20 \, b^{2} n^{2} + 1\right )} c^{4 \, b} x^{4 \, b n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*c^(4*b)*n^3*x*x^(4*b*n)*e^(4*a)/(64*b^4*n^4 - 20*b^2*n^2 + 1) + 8*b^3*c^(2*b)*n^3*x*x^(2*b*n)*e^(2*a)/(64*
b^4*n^4 - 20*b^2*n^2 + 1) + 24*b^4*n^4*x/(64*b^4*n^4 - 20*b^2*n^2 + 1) - 1/4*b^2*c^(4*b)*n^2*x*x^(4*b*n)*e^(4*
a)/(64*b^4*n^4 - 20*b^2*n^2 + 1) - 4*b^2*c^(2*b)*n^2*x*x^(2*b*n)*e^(2*a)/(64*b^4*n^4 - 20*b^2*n^2 + 1) - 1/4*b
*c^(4*b)*n*x*x^(4*b*n)*e^(4*a)/(64*b^4*n^4 - 20*b^2*n^2 + 1) - 1/2*b*c^(2*b)*n*x*x^(2*b*n)*e^(2*a)/(64*b^4*n^4
- 20*b^2*n^2 + 1) - 8*b^3*n^3*x*e^(-2*a)/((64*b^4*n^4 - 20*b^2*n^2 + 1)*c^(2*b)*x^(2*b*n)) - b^3*n^3*x*e^(-4*
a)/((64*b^4*n^4 - 20*b^2*n^2 + 1)*c^(4*b)*x^(4*b*n)) - 15/2*b^2*n^2*x/(64*b^4*n^4 - 20*b^2*n^2 + 1) + 1/16*c^(
4*b)*x*x^(4*b*n)*e^(4*a)/(64*b^4*n^4 - 20*b^2*n^2 + 1) + 1/4*c^(2*b)*x*x^(2*b*n)*e^(2*a)/(64*b^4*n^4 - 20*b^2*
n^2 + 1) - 4*b^2*n^2*x*e^(-2*a)/((64*b^4*n^4 - 20*b^2*n^2 + 1)*c^(2*b)*x^(2*b*n)) - 1/4*b^2*n^2*x*e^(-4*a)/((6
4*b^4*n^4 - 20*b^2*n^2 + 1)*c^(4*b)*x^(4*b*n)) + 1/2*b*n*x*e^(-2*a)/((64*b^4*n^4 - 20*b^2*n^2 + 1)*c^(2*b)*x^(
2*b*n)) + 1/4*b*n*x*e^(-4*a)/((64*b^4*n^4 - 20*b^2*n^2 + 1)*c^(4*b)*x^(4*b*n)) + 3/8*x/(64*b^4*n^4 - 20*b^2*n^
2 + 1) + 1/4*x*e^(-2*a)/((64*b^4*n^4 - 20*b^2*n^2 + 1)*c^(2*b)*x^(2*b*n)) + 1/16*x*e^(-4*a)/((64*b^4*n^4 - 20*
b^2*n^2 + 1)*c^(4*b)*x^(4*b*n))

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maple [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \cosh ^{4}\left (a +b \ln \left (c \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 129, normalized size = 0.68 \[ \frac {c^{4 \, b} x e^{\left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right )}}{16 \, {\left (4 \, b n + 1\right )}} + \frac {c^{2 \, b} x e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}}{4 \, {\left (2 \, b n + 1\right )}} + \frac {3}{8} \, x - \frac {x e^{\left (-2 \, b \log \left (x^{n}\right ) - 2 \, a\right )}}{4 \, {\left (2 \, b c^{2 \, b} n - c^{2 \, b}\right )}} - \frac {x e^{\left (-4 \, a\right )}}{16 \, {\left (4 \, b c^{4 \, b} n - c^{4 \, b}\right )} {\left (x^{n}\right )}^{4 \, b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*c^(4*b)*x*e^(4*b*log(x^n) + 4*a)/(4*b*n + 1) + 1/4*c^(2*b)*x*e^(2*b*log(x^n) + 2*a)/(2*b*n + 1) + 3/8*x -
1/4*x*e^(-2*b*log(x^n) - 2*a)/(2*b*c^(2*b)*n - c^(2*b)) - 1/16*x*e^(-4*a)/((4*b*c^(4*b)*n - c^(4*b))*(x^n)^(4
*b))

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mupad [B]  time = 1.04, size = 102, normalized size = 0.53 \[ \frac {3\,x}{8}-\frac {x\,{\mathrm {e}}^{-2\,a}}{{\left (c\,x^n\right )}^{2\,b}\,\left (8\,b\,n-4\right )}+\frac {x\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}}{8\,b\,n+4}-\frac {x\,{\mathrm {e}}^{-4\,a}}{{\left (c\,x^n\right )}^{4\,b}\,\left (64\,b\,n-16\right )}+\frac {x\,{\mathrm {e}}^{4\,a}\,{\left (c\,x^n\right )}^{4\,b}}{64\,b\,n+16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*log(c*x^n))^4,x)

[Out]

(3*x)/8 - (x*exp(-2*a))/((c*x^n)^(2*b)*(8*b*n - 4)) + (x*exp(2*a)*(c*x^n)^(2*b))/(8*b*n + 4) - (x*exp(-4*a))/(
(c*x^n)^(4*b)*(64*b*n - 16)) + (x*exp(4*a)*(c*x^n)^(4*b))/(64*b*n + 16)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \int \cosh ^{4}{\left (a - \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {1}{2 n} \\\int \cosh ^{4}{\left (a - \frac {\log {\left (c x^{n} \right )}}{4 n} \right )}\, dx & \text {for}\: b = - \frac {1}{4 n} \\\int \cosh ^{4}{\left (a + \frac {\log {\left (c x^{n} \right )}}{4 n} \right )}\, dx & \text {for}\: b = \frac {1}{4 n} \\\int \cosh ^{4}{\left (a + \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {1}{2 n} \\\frac {24 b^{4} n^{4} x \sinh ^{4}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} - \frac {48 b^{4} n^{4} x \sinh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} + \frac {24 b^{4} n^{4} x \cosh ^{4}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} - \frac {24 b^{3} n^{3} x \sinh ^{3}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} + \frac {40 b^{3} n^{3} x \sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{3}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} + \frac {12 b^{2} n^{2} x \sinh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} - \frac {16 b^{2} n^{2} x \cosh ^{4}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} - \frac {4 b n x \sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{3}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} + \frac {x \cosh ^{4}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{64 b^{4} n^{4} - 20 b^{2} n^{2} + 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((Integral(cosh(a - log(c*x**n)/(2*n))**4, x), Eq(b, -1/(2*n))), (Integral(cosh(a - log(c*x**n)/(4*n)
)**4, x), Eq(b, -1/(4*n))), (Integral(cosh(a + log(c*x**n)/(4*n))**4, x), Eq(b, 1/(4*n))), (Integral(cosh(a +
log(c*x**n)/(2*n))**4, x), Eq(b, 1/(2*n))), (24*b**4*n**4*x*sinh(a + b*n*log(x) + b*log(c))**4/(64*b**4*n**4 -
20*b**2*n**2 + 1) - 48*b**4*n**4*x*sinh(a + b*n*log(x) + b*log(c))**2*cosh(a + b*n*log(x) + b*log(c))**2/(64*
b**4*n**4 - 20*b**2*n**2 + 1) + 24*b**4*n**4*x*cosh(a + b*n*log(x) + b*log(c))**4/(64*b**4*n**4 - 20*b**2*n**2
+ 1) - 24*b**3*n**3*x*sinh(a + b*n*log(x) + b*log(c))**3*cosh(a + b*n*log(x) + b*log(c))/(64*b**4*n**4 - 20*b
**2*n**2 + 1) + 40*b**3*n**3*x*sinh(a + b*n*log(x) + b*log(c))*cosh(a + b*n*log(x) + b*log(c))**3/(64*b**4*n**
4 - 20*b**2*n**2 + 1) + 12*b**2*n**2*x*sinh(a + b*n*log(x) + b*log(c))**2*cosh(a + b*n*log(x) + b*log(c))**2/(
64*b**4*n**4 - 20*b**2*n**2 + 1) - 16*b**2*n**2*x*cosh(a + b*n*log(x) + b*log(c))**4/(64*b**4*n**4 - 20*b**2*n
**2 + 1) - 4*b*n*x*sinh(a + b*n*log(x) + b*log(c))*cosh(a + b*n*log(x) + b*log(c))**3/(64*b**4*n**4 - 20*b**2*
n**2 + 1) + x*cosh(a + b*n*log(x) + b*log(c))**4/(64*b**4*n**4 - 20*b**2*n**2 + 1), True))

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