3.241 \(\int \cosh ^3(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=149 \[ \frac {x \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {3 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {6 b^2 n^2 x \cosh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1}+\frac {6 b^3 n^3 x \sinh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1} \]

[Out]

-6*b^2*n^2*x*cosh(a+b*ln(c*x^n))/(9*b^4*n^4-10*b^2*n^2+1)+x*cosh(a+b*ln(c*x^n))^3/(-9*b^2*n^2+1)+6*b^3*n^3*x*s
inh(a+b*ln(c*x^n))/(9*b^4*n^4-10*b^2*n^2+1)-3*b*n*x*cosh(a+b*ln(c*x^n))^2*sinh(a+b*ln(c*x^n))/(-9*b^2*n^2+1)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5520, 5518} \[ \frac {6 b^3 n^3 x \sinh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1}+\frac {x \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {6 b^2 n^2 x \cosh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1}-\frac {3 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*Log[c*x^n]]^3,x]

[Out]

(-6*b^2*n^2*x*Cosh[a + b*Log[c*x^n]])/(1 - 10*b^2*n^2 + 9*b^4*n^4) + (x*Cosh[a + b*Log[c*x^n]]^3)/(1 - 9*b^2*n
^2) + (6*b^3*n^3*x*Sinh[a + b*Log[c*x^n]])/(1 - 10*b^2*n^2 + 9*b^4*n^4) - (3*b*n*x*Cosh[a + b*Log[c*x^n]]^2*Si
nh[a + b*Log[c*x^n]])/(1 - 9*b^2*n^2)

Rule 5518

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

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 ^3\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {x \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {3 b n x \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {\left (6 b^2 n^2\right ) \int \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx}{1-9 b^2 n^2}\\ &=-\frac {6 b^2 n^2 x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}+\frac {x \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac {6 b^3 n^3 x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}-\frac {3 b n x \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.56, size = 117, normalized size = 0.79 \[ \frac {x \left (\left (3-27 b^2 n^2\right ) \cosh \left (a+b \log \left (c x^n\right )\right )+\left (1-b^2 n^2\right ) \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )+6 b n \sinh \left (a+b \log \left (c x^n\right )\right ) \left (\left (b^2 n^2-1\right ) \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+5 b^2 n^2-1\right )\right )}{36 b^4 n^4-40 b^2 n^2+4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*Log[c*x^n]]^3,x]

[Out]

(x*((3 - 27*b^2*n^2)*Cosh[a + b*Log[c*x^n]] + (1 - b^2*n^2)*Cosh[3*(a + b*Log[c*x^n])] + 6*b*n*(-1 + 5*b^2*n^2
+ (-1 + b^2*n^2)*Cosh[2*(a + b*Log[c*x^n])])*Sinh[a + b*Log[c*x^n]]))/(4 - 40*b^2*n^2 + 36*b^4*n^4)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 199, normalized size = 1.34 \[ -\frac {{\left (b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{3} + 3 \, {\left (b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - 3 \, {\left (b^{3} n^{3} - b n\right )} x \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{3} + 3 \, {\left (9 \, b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) - 3 \, {\left (3 \, {\left (b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + {\left (9 \, b^{3} n^{3} - b n\right )} x\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{4 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((b^2*n^2 - 1)*x*cosh(b*n*log(x) + b*log(c) + a)^3 + 3*(b^2*n^2 - 1)*x*cosh(b*n*log(x) + b*log(c) + a)*si
nh(b*n*log(x) + b*log(c) + a)^2 - 3*(b^3*n^3 - b*n)*x*sinh(b*n*log(x) + b*log(c) + a)^3 + 3*(9*b^2*n^2 - 1)*x*
cosh(b*n*log(x) + b*log(c) + a) - 3*(3*(b^3*n^3 - b*n)*x*cosh(b*n*log(x) + b*log(c) + a)^2 + (9*b^3*n^3 - b*n)
*x)*sinh(b*n*log(x) + b*log(c) + a))/(9*b^4*n^4 - 10*b^2*n^2 + 1)

________________________________________________________________________________________

giac [B]  time = 0.22, size = 665, normalized size = 4.46 \[ \frac {3 \, b^{3} c^{3 \, b} n^{3} x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} + \frac {27 \, b^{3} c^{b} n^{3} x x^{b n} e^{a}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac {b^{2} c^{3 \, b} n^{2} x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac {27 \, b^{2} c^{b} n^{2} x x^{b n} e^{a}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac {3 \, b c^{3 \, b} n x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac {27 \, b^{3} n^{3} x e^{\left (-a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} - \frac {3 \, b^{3} n^{3} x e^{\left (-3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} - \frac {3 \, b c^{b} n x x^{b n} e^{a}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} + \frac {c^{3 \, b} x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac {27 \, b^{2} n^{2} x e^{\left (-a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} - \frac {b^{2} n^{2} x e^{\left (-3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} + \frac {3 \, c^{b} x x^{b n} e^{a}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} + \frac {3 \, b n x e^{\left (-a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} + \frac {3 \, b n x e^{\left (-3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} + \frac {3 \, x e^{\left (-a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} + \frac {x e^{\left (-3 \, a\right )}}{8 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*b^3*c^(3*b)*n^3*x*x^(3*b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1) + 27/8*b^3*c^b*n^3*x*x^(b*n)*e^a/(9*b^4*n
^4 - 10*b^2*n^2 + 1) - 1/8*b^2*c^(3*b)*n^2*x*x^(3*b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1) - 27/8*b^2*c^b*n^2
*x*x^(b*n)*e^a/(9*b^4*n^4 - 10*b^2*n^2 + 1) - 3/8*b*c^(3*b)*n*x*x^(3*b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1)
- 27/8*b^3*n^3*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) - 3/8*b^3*n^3*x*e^(-3*a)/((9*b^4*n^4 - 10*
b^2*n^2 + 1)*c^(3*b)*x^(3*b*n)) - 3/8*b*c^b*n*x*x^(b*n)*e^a/(9*b^4*n^4 - 10*b^2*n^2 + 1) + 1/8*c^(3*b)*x*x^(3*
b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1) - 27/8*b^2*n^2*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) -
1/8*b^2*n^2*x*e^(-3*a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^(3*b)*x^(3*b*n)) + 3/8*c^b*x*x^(b*n)*e^a/(9*b^4*n^4 -
10*b^2*n^2 + 1) + 3/8*b*n*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) + 3/8*b*n*x*e^(-3*a)/((9*b^4*n^4
- 10*b^2*n^2 + 1)*c^(3*b)*x^(3*b*n)) + 3/8*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) + 1/8*x*e^(-3*
a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^(3*b)*x^(3*b*n))

________________________________________________________________________________________

maple [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \cosh ^{3}\left (a +b \ln \left (c \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 115, normalized size = 0.77 \[ \frac {c^{3 \, b} x e^{\left (3 \, b \log \left (x^{n}\right ) + 3 \, a\right )}}{8 \, {\left (3 \, b n + 1\right )}} + \frac {3 \, c^{b} x e^{\left (b \log \left (x^{n}\right ) + a\right )}}{8 \, {\left (b n + 1\right )}} - \frac {3 \, x e^{\left (-b \log \left (x^{n}\right ) - a\right )}}{8 \, {\left (b c^{b} n - c^{b}\right )}} - \frac {x e^{\left (-3 \, a\right )}}{8 \, {\left (3 \, b c^{3 \, b} n - c^{3 \, b}\right )} {\left (x^{n}\right )}^{3 \, b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.05, size = 94, normalized size = 0.63 \[ \frac {x\,{\mathrm {e}}^{3\,a}\,{\left (c\,x^n\right )}^{3\,b}}{24\,b\,n+8}-\frac {x\,{\mathrm {e}}^{-3\,a}}{{\left (c\,x^n\right )}^{3\,b}\,\left (24\,b\,n-8\right )}-\frac {3\,x\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (8\,b\,n-8\right )}+\frac {3\,x\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{8\,b\,n+8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*log(c*x^n))^3,x)

[Out]

(x*exp(3*a)*(c*x^n)^(3*b))/(24*b*n + 8) - (x*exp(-3*a))/((c*x^n)^(3*b)*(24*b*n - 8)) - (3*x*exp(-a))/((c*x^n)^
b*(8*b*n - 8)) + (3*x*exp(a)*(c*x^n)^b)/(8*b*n + 8)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \int \cosh ^{3}{\left (a - \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {1}{n} \\\int \cosh ^{3}{\left (a - \frac {\log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = - \frac {1}{3 n} \\\int \cosh ^{3}{\left (a + \frac {\log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = \frac {1}{3 n} \\\int \cosh ^{3}{\left (a + \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {1}{n} \\- \frac {6 b^{3} n^{3} x \sinh ^{3}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} + \frac {9 b^{3} n^{3} x \sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} + \frac {6 b^{2} n^{2} x \sinh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} - \frac {7 b^{2} n^{2} x \cosh ^{3}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} - \frac {3 b n x \sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} + \frac {x \cosh ^{3}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((Integral(cosh(a - log(c*x**n)/n)**3, x), Eq(b, -1/n)), (Integral(cosh(a - log(c*x**n)/(3*n))**3, x)
, Eq(b, -1/(3*n))), (Integral(cosh(a + log(c*x**n)/(3*n))**3, x), Eq(b, 1/(3*n))), (Integral(cosh(a + log(c*x*
*n)/n)**3, x), Eq(b, 1/n)), (-6*b**3*n**3*x*sinh(a + b*n*log(x) + b*log(c))**3/(9*b**4*n**4 - 10*b**2*n**2 + 1
) + 9*b**3*n**3*x*sinh(a + b*n*log(x) + b*log(c))*cosh(a + b*n*log(x) + b*log(c))**2/(9*b**4*n**4 - 10*b**2*n*
*2 + 1) + 6*b**2*n**2*x*sinh(a + b*n*log(x) + b*log(c))**2*cosh(a + b*n*log(x) + b*log(c))/(9*b**4*n**4 - 10*b
**2*n**2 + 1) - 7*b**2*n**2*x*cosh(a + b*n*log(x) + b*log(c))**3/(9*b**4*n**4 - 10*b**2*n**2 + 1) - 3*b*n*x*si
nh(a + b*n*log(x) + b*log(c))*cosh(a + b*n*log(x) + b*log(c))**2/(9*b**4*n**4 - 10*b**2*n**2 + 1) + x*cosh(a +
b*n*log(x) + b*log(c))**3/(9*b**4*n**4 - 10*b**2*n**2 + 1), True))

________________________________________________________________________________________