\(\int x^m \cosh ^3(a+b \log (c x^n)) \, dx\) [245]

Optimal result

Integrand size = 17, antiderivative size = 203 \[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {6 b^2 (1+m) n^2 x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2-9 b^2 n^2\right ) \left ((1+m)^2-b^2 n^2\right )}+\frac {(1+m) x^{1+m} \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-9 b^2 n^2}+\frac {6 b^3 n^3 x^{1+m} \sinh \left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2-9 b^2 n^2\right ) \left ((1+m)^2-b^2 n^2\right )}-\frac {3 b n x^{1+m} \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-9 b^2 n^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.44 \[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} x^{1+m} \left (\frac {3 \sinh (b n \log (x)) \left (-b n \cosh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+(1+m) \sinh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )}{(1+m-b n) (1+m+b n)}+\frac {3 \cosh (b n \log (x)) \left ((1+m) \cosh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )-b n \sinh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )}{(1+m-b n) (1+m+b n)}+\frac {\sinh (3 b n \log (x)) \left (-3 b n \cosh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+(1+m) \sinh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-3 b n) (1+m+3 b n)}+\frac {\cosh (3 b n \log (x)) \left ((1+m) \cosh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )-3 b n \sinh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-3 b n) (1+m+3 b n)}\right ) \] Input:

Integrate[x^m*Cosh[a + b*Log[c*x^n]]^3,x]
 

Output:

(x^(1 + m)*((3*Sinh[b*n*Log[x]]*(-(b*n*Cosh[a - b*n*Log[x] + b*Log[c*x^n]] 
) + (1 + m)*Sinh[a - b*n*Log[x] + b*Log[c*x^n]]))/((1 + m - b*n)*(1 + m + 
b*n)) + (3*Cosh[b*n*Log[x]]*((1 + m)*Cosh[a - b*n*Log[x] + b*Log[c*x^n]] - 
 b*n*Sinh[a - b*n*Log[x] + b*Log[c*x^n]]))/((1 + m - b*n)*(1 + m + b*n)) + 
 (Sinh[3*b*n*Log[x]]*(-3*b*n*Cosh[3*(a - b*n*Log[x] + b*Log[c*x^n])] + (1 
+ m)*Sinh[3*(a - b*n*Log[x] + b*Log[c*x^n])]))/((1 + m - 3*b*n)*(1 + m + 3 
*b*n)) + (Cosh[3*b*n*Log[x]]*((1 + m)*Cosh[3*(a - b*n*Log[x] + b*Log[c*x^n 
])] - 3*b*n*Sinh[3*(a - b*n*Log[x] + b*Log[c*x^n])]))/((1 + m - 3*b*n)*(1 
+ m + 3*b*n))))/4
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6056, 6054}

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.

\(\displaystyle \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 6056

\(\displaystyle -\frac {6 b^2 n^2 \int x^m \cosh \left (a+b \log \left (c x^n\right )\right )dx}{(m+1)^2-9 b^2 n^2}+\frac {(m+1) x^{m+1} \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{-9 b^2 n^2+m^2+2 m+1}-\frac {3 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-9 b^2 n^2}\)

\(\Big \downarrow \) 6054

\(\displaystyle \frac {(m+1) x^{m+1} \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{-9 b^2 n^2+m^2+2 m+1}-\frac {3 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-9 b^2 n^2}-\frac {6 b^2 n^2 \left (\frac {(m+1) x^{m+1} \cosh \left (a+b \log \left (c x^n\right )\right )}{(-b n+m+1) (b n+m+1)}-\frac {b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2}\right )}{(m+1)^2-9 b^2 n^2}\)

Input:

Int[x^m*Cosh[a + b*Log[c*x^n]]^3,x]
 

Output:

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

Defintions of rubi rules used

rule 6054

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

rule 6056

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

Maple [A] (verified)

Time = 66.28 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.41

method	result	size
parallelrisch	\(\frac {7 \left (-\frac {18 \tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right ) \left ({\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}-\frac {2 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}}{3}+1\right ) n^{3} b^{3}}{7}+\left (1+m \right ) \left ({\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}-\frac {10 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}}{7}+1\right ) \left ({\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}+1\right ) n^{2} b^{2}+\frac {6 \tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right ) \left (1+m \right )^{2} {\left ({\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}+1\right )}^{2} n b}{7}-\frac {\left (1+m \right )^{3} {\left ({\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}+1\right )}^{3}}{7}\right ) x^{1+m}}{9 {\left (\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )-1\right )}^{3} \left (b n +\frac {m}{3}+\frac {1}{3}\right ) \left (b n +m +1\right ) \left (b n -m -1\right ) \left (b n -\frac {m}{3}-\frac {1}{3}\right ) {\left (\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )+1\right )}^{3}}\)	\(286\)

Input:

int(x^m*cosh(a+b*ln(c*x^n))^3,x,method=_RETURNVERBOSE)
 

Output:

7/9*(-18/7*tanh(1/2*a+b*ln((c*x^n)^(1/2)))*(tanh(1/2*a+b*ln((c*x^n)^(1/2)) 
)^4-2/3*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^2+1)*n^3*b^3+(1+m)*(tanh(1/2*a+b*l 
n((c*x^n)^(1/2)))^4-10/7*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^2+1)*(tanh(1/2*a+ 
b*ln((c*x^n)^(1/2)))^2+1)*n^2*b^2+6/7*tanh(1/2*a+b*ln((c*x^n)^(1/2)))*(1+m 
)^2*(tanh(1/2*a+b*ln((c*x^n)^(1/2)))^2+1)^2*n*b-1/7*(1+m)^3*(tanh(1/2*a+b* 
ln((c*x^n)^(1/2)))^2+1)^3)*x^(1+m)/(tanh(1/2*a+b*ln((c*x^n)^(1/2)))-1)^3/( 
b*n+1/3*m+1/3)/(b*n+m+1)/(b*n-m-1)/(b*n-1/3*m-1/3)/(tanh(1/2*a+b*ln((c*x^n 
)^(1/2)))+1)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 584 vs. \(2 (214) = 428\).

Time = 0.10 (sec) , antiderivative size = 584, normalized size of antiderivative = 2.88 \[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*((m^3 - (b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x*cosh(b*n*log(x) + b*log 
(c) + a)^3*cosh(m*log(x)) + 3*(m^3 - 9*(b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1 
)*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + 3*((b^3*n^3 - (b*m^2 
+ 2*b*m + b)*n)*x*cosh(m*log(x)) + (b^3*n^3 - (b*m^2 + 2*b*m + b)*n)*x*sin 
h(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a)^3 + 3*((m^3 - (b^2*m + b^2)*n 
^2 + 3*m^2 + 3*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + ( 
m^3 - (b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x*cosh(b*n*log(x) + b*log(c) + 
a)*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a)^2 + 3*(3*(b^3*n^3 - (b* 
m^2 + 2*b*m + b)*n)*x*cosh(b*n*log(x) + b*log(c) + a)^2*cosh(m*log(x)) + ( 
9*b^3*n^3 - (b*m^2 + 2*b*m + b)*n)*x*cosh(m*log(x)) + (3*(b^3*n^3 - (b*m^2 
 + 2*b*m + b)*n)*x*cosh(b*n*log(x) + b*log(c) + a)^2 + (9*b^3*n^3 - (b*m^2 
 + 2*b*m + b)*n)*x)*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a) + ((m^ 
3 - (b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a) 
^3 + 3*(m^3 - 9*(b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x*cosh(b*n*log(x) + b 
*log(c) + a))*sinh(m*log(x)))/(9*b^4*n^4 + m^4 + 4*m^3 - 10*(b^2*m^2 + 2*b 
^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)
 

Sympy [F]

\[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((log(x)*cosh(a)**3, Eq(b, 0) & Eq(m, -1)), (Integral(x**m*cosh(- 
a + m*log(c*x**n)/(3*n) + log(c*x**n)/(3*n))**3, x), Eq(b, (-m - 1)/(3*n)) 
), (Integral(x**m*cosh(-a + m*log(c*x**n)/n + log(c*x**n)/n)**3, x), Eq(b, 
 (-m - 1)/n)), (Integral(x**m*cosh(a + m*log(c*x**n)/(3*n) + log(c*x**n)/( 
3*n))**3, x), Eq(b, (m + 1)/(3*n))), (Integral(x**m*cosh(a + m*log(c*x**n) 
/n + log(c*x**n)/n)**3, x), Eq(b, (m + 1)/n)), (-6*b**3*n**3*x*x**m*sinh(a 
 + b*log(c*x**n))**3/(9*b**4*n**4 - 10*b**2*m**2*n**2 - 20*b**2*m*n**2 - 1 
0*b**2*n**2 + m**4 + 4*m**3 + 6*m**2 + 4*m + 1) + 9*b**3*n**3*x*x**m*sinh( 
a + b*log(c*x**n))*cosh(a + b*log(c*x**n))**2/(9*b**4*n**4 - 10*b**2*m**2* 
n**2 - 20*b**2*m*n**2 - 10*b**2*n**2 + m**4 + 4*m**3 + 6*m**2 + 4*m + 1) + 
 6*b**2*m*n**2*x*x**m*sinh(a + b*log(c*x**n))**2*cosh(a + b*log(c*x**n))/( 
9*b**4*n**4 - 10*b**2*m**2*n**2 - 20*b**2*m*n**2 - 10*b**2*n**2 + m**4 + 4 
*m**3 + 6*m**2 + 4*m + 1) - 7*b**2*m*n**2*x*x**m*cosh(a + b*log(c*x**n))** 
3/(9*b**4*n**4 - 10*b**2*m**2*n**2 - 20*b**2*m*n**2 - 10*b**2*n**2 + m**4 
+ 4*m**3 + 6*m**2 + 4*m + 1) + 6*b**2*n**2*x*x**m*sinh(a + b*log(c*x**n))* 
*2*cosh(a + b*log(c*x**n))/(9*b**4*n**4 - 10*b**2*m**2*n**2 - 20*b**2*m*n* 
*2 - 10*b**2*n**2 + m**4 + 4*m**3 + 6*m**2 + 4*m + 1) - 7*b**2*n**2*x*x**m 
*cosh(a + b*log(c*x**n))**3/(9*b**4*n**4 - 10*b**2*m**2*n**2 - 20*b**2*m*n 
**2 - 10*b**2*n**2 + m**4 + 4*m**3 + 6*m**2 + 4*m + 1) - 3*b*m**2*n*x*x**m 
*sinh(a + b*log(c*x**n))*cosh(a + b*log(c*x**n))**2/(9*b**4*n**4 - 10*b...
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.68 \[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{3 \, b} x e^{\left (3 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) + 3 \, a\right )}}{8 \, {\left (3 \, b n + m + 1\right )}} + \frac {3 \, c^{b} x e^{\left (b \log \left (x^{n}\right ) + m \log \left (x\right ) + a\right )}}{8 \, {\left (b n + m + 1\right )}} - \frac {3 \, x e^{\left (-b \log \left (x^{n}\right ) + m \log \left (x\right ) - a\right )}}{8 \, {\left (b c^{b} n - c^{b} {\left (m + 1\right )}\right )}} - \frac {x e^{\left (-3 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) - 3 \, a\right )}}{8 \, {\left (3 \, b c^{3 \, b} n - c^{3 \, b} {\left (m + 1\right )}\right )}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3225 vs. \(2 (214) = 428\).

Time = 0.17 (sec) , antiderivative size = 3225, normalized size of antiderivative = 15.89 \[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \] Input:

integrate(x^m*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)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 
20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) + 27/8*b^3*c^b* 
n^3*x*x^(b*n)*x^m*e^a/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 1 
0*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 1/8*b^2*c^(3*b)*m*n^2*x*x^(3*b*n)*x 
^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 
 4*m^3 + 6*m^2 + 4*m + 1) - 27/8*b^2*c^b*m*n^2*x*x^(b*n)*x^m*e^a/(9*b^4*n^ 
4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m 
 + 1) - 3/8*b*c^(3*b)*m^2*n*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^ 
2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 1/8*b 
^2*c^(3*b)*n^2*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^ 
2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 3/8*b*c^b*m^2*n*x* 
x^(b*n)*x^m*e^a/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2* 
n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 27/8*b^2*c^b*n^2*x*x^(b*n)*x^m*e^a/(9*b^4 
*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 
4*m + 1) + 1/8*c^(3*b)*m^3*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2 
*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 3/4*b* 
c^(3*b)*m*n*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m 
*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) + 3/8*c^b*m^3*x*x^(b*n) 
*x^m*e^a/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4 
*m^3 + 6*m^2 + 4*m + 1) - 3/4*b*c^b*m*n*x*x^(b*n)*x^m*e^a/(9*b^4*n^4 - ...
 

Mupad [B] (verification not implemented)

Time = 2.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.58 \[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3\,x\,x^m\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (8\,m-8\,b\,n+8\right )}+\frac {x\,x^m\,{\mathrm {e}}^{-3\,a}}{{\left (c\,x^n\right )}^{3\,b}\,\left (8\,m-24\,b\,n+8\right )}+\frac {x\,x^m\,{\mathrm {e}}^{3\,a}\,{\left (c\,x^n\right )}^{3\,b}}{8\,m+24\,b\,n+8}+\frac {3\,x\,x^m\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{8\,m+8\,b\,n+8} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 784, normalized size of antiderivative = 3.86 \[ \int x^m \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{m} x \left (1+3 m +x^{6 b n} e^{6 a} c^{6 b} m^{3}+3 x^{6 b n} e^{6 a} c^{6 b} m^{2}+3 x^{6 b n} e^{6 a} c^{6 b} m +3 x^{4 b n} e^{4 a} c^{4 b} m^{3}-3 b^{3} n^{3}+3 x^{6 b n} e^{6 a} c^{6 b} b^{3} n^{3}-3 x^{6 b n} e^{6 a} c^{6 b} b n +27 x^{4 b n} e^{4 a} c^{4 b} b^{3} n^{3}-27 x^{4 b n} e^{4 a} c^{4 b} b^{2} n^{2}-27 x^{2 b n} e^{2 a} c^{2 b} b^{3} n^{3}+3 x^{2 b n} e^{2 a} c^{2 b} b n +3 b n +9 x^{4 b n} e^{4 a} c^{4 b} m^{2}+9 x^{4 b n} e^{4 a} c^{4 b} m +9 x^{2 b n} e^{2 a} c^{2 b} m^{2}+9 x^{2 b n} e^{2 a} c^{2 b} m +m^{3}-x^{6 b n} e^{6 a} c^{6 b} b^{2} n^{2}-x^{6 b n} e^{6 a} c^{6 b} b^{2} m \,n^{2}-3 x^{6 b n} e^{6 a} c^{6 b} b \,m^{2} n -6 x^{6 b n} e^{6 a} c^{6 b} b m n -27 x^{4 b n} e^{4 a} c^{4 b} b^{2} m \,n^{2}-3 x^{4 b n} e^{4 a} c^{4 b} b \,m^{2} n -27 x^{2 b n} e^{2 a} c^{2 b} b^{2} m \,n^{2}+3 x^{2 b n} e^{2 a} c^{2 b} b \,m^{2} n +6 x^{2 b n} e^{2 a} c^{2 b} b m n -6 x^{4 b n} e^{4 a} c^{4 b} b m n -b^{2} n^{2}+3 m^{2}+6 b m n -b^{2} m \,n^{2}+3 b \,m^{2} n -3 x^{4 b n} e^{4 a} c^{4 b} b n -27 x^{2 b n} e^{2 a} c^{2 b} b^{2} n^{2}+3 x^{2 b n} e^{2 a} c^{2 b}+3 x^{2 b n} e^{2 a} c^{2 b} m^{3}+3 x^{4 b n} e^{4 a} c^{4 b}+x^{6 b n} e^{6 a} c^{6 b}\right )}{8 x^{3 b n} e^{3 a} c^{3 b} \left (9 b^{4} n^{4}-10 b^{2} m^{2} n^{2}-20 b^{2} m \,n^{2}-10 b^{2} n^{2}+m^{4}+4 m^{3}+6 m^{2}+4 m +1\right )} \] Input:

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

Output:

(x**m*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*m*n**2 - 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*m**2*n - 6*x**(6*b*n)*e**(6*a)*c**(6*b)*b*m*n - 3*x** 
(6*b*n)*e**(6*a)*c**(6*b)*b*n + x**(6*b*n)*e**(6*a)*c**(6*b)*m**3 + 3*x**( 
6*b*n)*e**(6*a)*c**(6*b)*m**2 + 3*x**(6*b*n)*e**(6*a)*c**(6*b)*m + 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*m*n**2 - 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*m**2*n - 6*x**(4*b*n)*e**(4*a)* 
c**(4*b)*b*m*n - 3*x**(4*b*n)*e**(4*a)*c**(4*b)*b*n + 3*x**(4*b*n)*e**(4*a 
)*c**(4*b)*m**3 + 9*x**(4*b*n)*e**(4*a)*c**(4*b)*m**2 + 9*x**(4*b*n)*e**(4 
*a)*c**(4*b)*m + 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*m*n**2 - 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*m**2* 
n + 6*x**(2*b*n)*e**(2*a)*c**(2*b)*b*m*n + 3*x**(2*b*n)*e**(2*a)*c**(2*b)* 
b*n + 3*x**(2*b*n)*e**(2*a)*c**(2*b)*m**3 + 9*x**(2*b*n)*e**(2*a)*c**(2*b) 
*m**2 + 9*x**(2*b*n)*e**(2*a)*c**(2*b)*m + 3*x**(2*b*n)*e**(2*a)*c**(2*b) 
- 3*b**3*n**3 - b**2*m*n**2 - b**2*n**2 + 3*b*m**2*n + 6*b*m*n + 3*b*n + m 
**3 + 3*m**2 + 3*m + 1))/(8*x**(3*b*n)*e**(3*a)*c**(3*b)*(9*b**4*n**4 - 10 
*b**2*m**2*n**2 - 20*b**2*m*n**2 - 10*b**2*n**2 + m**4 + 4*m**3 + 6*m**2 + 
 4*m + 1))