Integrand size = 13, antiderivative size = 149 \[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {6 b^2 n^2 x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}+\frac {x \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac {6 b^3 n^3 x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}-\frac {3 b n x \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2} \] Output:
-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*sinh(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)
Time = 0.34 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (\left (3-27 b^2 n^2\right ) \cosh \left (a+b \log \left (c x^n\right )\right )+\left (1-b^2 n^2\right ) \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )+6 b n \left (-1+5 b^2 n^2+\left (-1+b^2 n^2\right ) \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )\right )}{4-40 b^2 n^2+36 b^4 n^4} \] Input:
Integrate[Cosh[a + b*Log[c*x^n]]^3,x]
Output:
(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)
Time = 0.37 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6046, 6044}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 6046 |
| \(\displaystyle -\frac {6 b^2 n^2 \int \cosh \left (a+b \log \left (c x^n\right )\right )dx}{1-9 b^2 n^2}+\frac {x \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {3 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}\) |
| \(\Big \downarrow \) 6044 |
| \(\displaystyle \frac {x \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {3 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}-\frac {6 b^2 n^2 \left (\frac {x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2}-\frac {b n x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2}\right )}{1-9 b^2 n^2}\) |
Input:
Int[Cosh[a + b*Log[c*x^n]]^3,x]
Output:
(x*Cosh[a + b*Log[c*x^n]]^3)/(1 - 9*b^2*n^2) - (3*b*n*x*Cosh[a + b*Log[c*x ^n]]^2*Sinh[a + b*Log[c*x^n]])/(1 - 9*b^2*n^2) - (6*b^2*n^2*((x*Cosh[a + b *Log[c*x^n]])/(1 - b^2*n^2) - (b*n*x*Sinh[a + b*Log[c*x^n]])/(1 - b^2*n^2) ))/(1 - 9*b^2*n^2)
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]
Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> Si mp[(-x)*(Cosh[d*(a + b*Log[c*x^n])]^p/(b^2*d^2*n^2*p^2 - 1)), 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] + Simp[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]) /; FreeQ[{a, b, c, d, n} , x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 - 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(356\) vs. \(2(149)=298\).
Time = 5.25 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.40
| method | result | size |
| parallelrisch | \(-\frac {x \left (1+3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}-6 b n \tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )-12 b n {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{3}+3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4} b^{2} n^{2}+3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2} b^{2} n^{2}-7 b^{2} n^{2}+3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}+{\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{6}+18 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{5} b^{3} n^{3}-7 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{6} b^{2} n^{2}-12 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{3} b^{3} n^{3}+18 \tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right ) b^{3} n^{3}-6 b n {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{5}\right )}{\left (9 b^{4} n^{4}-10 b^{2} n^{2}+1\right ) \left ({\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{6}-3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}+3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}-1\right )}\) | \(357\) |
Input:
int(cosh(a+b*ln(c*x^n))^3,x,method=_RETURNVERBOSE)
Output:
-x/(9*b^4*n^4-10*b^2*n^2+1)*(1+3*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^2-6*b*n*t anh(1/2*a+b*ln((c*x^n)^(1/2)))-12*b*n*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^3+3* tanh(1/2*a+b*ln((c*x^n)^(1/2)))^4*b^2*n^2+3*tanh(1/2*a+b*ln((c*x^n)^(1/2)) )^2*b^2*n^2-7*b^2*n^2+3*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^4+tanh(1/2*a+b*ln( (c*x^n)^(1/2)))^6+18*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^5*b^3*n^3-7*tanh(1/2* a+b*ln((c*x^n)^(1/2)))^6*b^2*n^2-12*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^3*b^3* n^3+18*tanh(1/2*a+b*ln((c*x^n)^(1/2)))*b^3*n^3-6*b*n*tanh(1/2*a+b*ln((c*x^ n)^(1/2)))^5)/(tanh(1/2*a+b*ln((c*x^n)^(1/2)))^6-3*tanh(1/2*a+b*ln((c*x^n) ^(1/2)))^4+3*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^2-1)
Time = 0.10 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.34 \[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {{\left (b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 3 \, {\left (b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 3 \, {\left (b^{3} n^{3} - b n\right )} x \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 3 \, {\left (9 \, b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - 3 \, {\left (3 \, {\left (b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + {\left (9 \, b^{3} n^{3} - b n\right )} x\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{4 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} \] Input:
integrate(cosh(a+b*log(c*x^n))^3,x, algorithm="fricas")
Output:
-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)*sinh(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*co sh(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)
\[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \int \cosh ^{3}{\left (a - \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {1}{n} \\\int \cosh ^{3}{\left (a - \frac {\log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = - \frac {1}{3 n} \\\int \cosh ^{3}{\left (a + \frac {\log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = \frac {1}{3 n} \\\int \cosh ^{3}{\left (a + \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {1}{n} \\- \frac {6 b^{3} n^{3} x \sinh ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} + \frac {9 b^{3} n^{3} x \sinh {\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} + \frac {6 b^{2} n^{2} x \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} - \frac {7 b^{2} n^{2} x \cosh ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} - \frac {3 b n x \sinh {\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} + \frac {x \cosh ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} - 10 b^{2} n^{2} + 1} & \text {otherwise} \end {cases} \] Input:
integrate(cosh(a+b*ln(c*x**n))**3,x)
Output:
Piecewise((Integral(cosh(a - log(c*x**n)/n)**3, x), Eq(b, -1/n)), (Integra l(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*log(c*x**n))**3/(9*b* *4*n**4 - 10*b**2*n**2 + 1) + 9*b**3*n**3*x*sinh(a + b*log(c*x**n))*cosh(a + b*log(c*x**n))**2/(9*b**4*n**4 - 10*b**2*n**2 + 1) + 6*b**2*n**2*x*sinh (a + b*log(c*x**n))**2*cosh(a + b*log(c*x**n))/(9*b**4*n**4 - 10*b**2*n**2 + 1) - 7*b**2*n**2*x*cosh(a + b*log(c*x**n))**3/(9*b**4*n**4 - 10*b**2*n* *2 + 1) - 3*b*n*x*sinh(a + b*log(c*x**n))*cosh(a + b*log(c*x**n))**2/(9*b* *4*n**4 - 10*b**2*n**2 + 1) + x*cosh(a + b*log(c*x**n))**3/(9*b**4*n**4 - 10*b**2*n**2 + 1), True))
Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.77 \[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{3 \, b} x e^{\left (3 \, b \log \left (x^{n}\right ) + 3 \, a\right )}}{8 \, {\left (3 \, b n + 1\right )}} + \frac {3 \, c^{b} x e^{\left (b \log \left (x^{n}\right ) + a\right )}}{8 \, {\left (b n + 1\right )}} - \frac {3 \, x e^{\left (-b \log \left (x^{n}\right ) - a\right )}}{8 \, {\left (b c^{b} n - c^{b}\right )}} - \frac {x e^{\left (-3 \, a\right )}}{8 \, {\left (3 \, b c^{3 \, b} n - c^{3 \, b}\right )} {\left (x^{n}\right )}^{3 \, b}} \] Input:
integrate(cosh(a+b*log(c*x^n))^3,x, algorithm="maxima")
Output:
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))
Leaf count of result is larger than twice the leaf count of optimal. 665 vs. \(2 (150) = 300\).
Time = 0.14 (sec) , antiderivative size = 665, normalized size of antiderivative = 4.46 \[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\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}} \] Input:
integrate(cosh(a+b*log(c*x^n))^3,x, algorithm="giac")
Output:
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 - 1 0*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))
Time = 2.01 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.63 \[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{3\,a}\,{\left (c\,x^n\right )}^{3\,b}}{24\,b\,n+8}-\frac {x\,{\mathrm {e}}^{-3\,a}}{{\left (c\,x^n\right )}^{3\,b}\,\left (24\,b\,n-8\right )}-\frac {3\,x\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (8\,b\,n-8\right )}+\frac {3\,x\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{8\,b\,n+8} \] Input:
int(cosh(a + b*log(c*x^n))^3,x)
Output:
(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)
Time = 0.26 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.17 \[ \int \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (3 x^{6 b n} e^{6 a} c^{6 b} b^{3} n^{3}-x^{6 b n} e^{6 a} c^{6 b} b^{2} n^{2}-3 x^{6 b n} e^{6 a} c^{6 b} b n +x^{6 b n} e^{6 a} c^{6 b}+27 x^{4 b n} e^{4 a} c^{4 b} b^{3} n^{3}-27 x^{4 b n} e^{4 a} c^{4 b} b^{2} n^{2}-3 x^{4 b n} e^{4 a} c^{4 b} b n +3 x^{4 b n} e^{4 a} c^{4 b}-27 x^{2 b n} e^{2 a} c^{2 b} b^{3} n^{3}-27 x^{2 b n} e^{2 a} c^{2 b} b^{2} n^{2}+3 x^{2 b n} e^{2 a} c^{2 b} b n +3 x^{2 b n} e^{2 a} c^{2 b}-3 b^{3} n^{3}-b^{2} n^{2}+3 b n +1\right )}{8 x^{3 b n} e^{3 a} c^{3 b} \left (9 b^{4} n^{4}-10 b^{2} n^{2}+1\right )} \] Input:
int(cosh(a+b*log(c*x^n))^3,x)
Output:
(x*(3*x**(6*b*n)*e**(6*a)*c**(6*b)*b**3*n**3 - x**(6*b*n)*e**(6*a)*c**(6*b )*b**2*n**2 - 3*x**(6*b*n)*e**(6*a)*c**(6*b)*b*n + x**(6*b*n)*e**(6*a)*c** (6*b) + 27*x**(4*b*n)*e**(4*a)*c**(4*b)*b**3*n**3 - 27*x**(4*b*n)*e**(4*a) *c**(4*b)*b**2*n**2 - 3*x**(4*b*n)*e**(4*a)*c**(4*b)*b*n + 3*x**(4*b*n)*e* *(4*a)*c**(4*b) - 27*x**(2*b*n)*e**(2*a)*c**(2*b)*b**3*n**3 - 27*x**(2*b*n )*e**(2*a)*c**(2*b)*b**2*n**2 + 3*x**(2*b*n)*e**(2*a)*c**(2*b)*b*n + 3*x** (2*b*n)*e**(2*a)*c**(2*b) - 3*b**3*n**3 - b**2*n**2 + 3*b*n + 1))/(8*x**(3 *b*n)*e**(3*a)*c**(3*b)*(9*b**4*n**4 - 10*b**2*n**2 + 1))