\(\int \sinh ^3(a+b \log (c x^n)) \, dx\) [268]

Optimal result

Integrand size = 13, antiderivative size = 149 \[ \int \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {6 b^3 n^3 x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}+\frac {6 b^2 n^2 x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}-\frac {3 b n x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac {x \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.81 \[ \int \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (-3 b n \left (-1+9 b^2 n^2\right ) \cosh \left (a+b \log \left (c x^n\right )\right )+3 b n \left (-1+b^2 n^2\right ) \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-2 \left (1-13 b^2 n^2+\left (-1+b^2 n^2\right ) \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )\right )}{4-40 b^2 n^2+36 b^4 n^4} \] Input:

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

Output:

(x*(-3*b*n*(-1 + 9*b^2*n^2)*Cosh[a + b*Log[c*x^n]] + 3*b*n*(-1 + b^2*n^2)* 
Cosh[3*(a + b*Log[c*x^n])] - 2*(1 - 13*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)
 

Rubi [A] (verified)

Time = 0.34 (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 = {6045, 6043}

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.

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

\(\Big \downarrow \) 6045

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

\(\Big \downarrow \) 6043

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 6043

Int[Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> Simp[(- 
x)*(Sinh[d*(a + b*Log[c*x^n])]/(b^2*d^2*n^2 - 1)), x] + Simp[b*d*n*x*(Cosh[ 
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 6045

Int[Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> Si 
mp[(-x)*(Sinh[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])]*(Sinh[d*(a + b*Log[c*x^n])]^(p - 1)/(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[Sinh[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]
 

Maple [A] (verified)

Time = 9.10 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.70

method	result	size
parallelrisch	\(\frac {6 x \left (b^{3} n^{3} {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{6}-2 b^{2} n^{2} {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{5}+\left (-3 b^{3} n^{3}+2 b n \right ) {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}+\frac {4 \left (4 b^{2} n^{2}-1\right ) {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{3}}{3}+\left (-3 b^{3} n^{3}+2 b n \right ) {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}-2 b^{2} n^{2} \tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )+b^{3} n^{3}\right )}{\left (9 b^{4} n^{4}-10 b^{2} n^{2}+1\right ) \left ({\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{6}-3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}+3 {\tanh \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}-1\right )}\)	\(254\)

Input:

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

Output:

6*x*(b^3*n^3*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^6-2*b^2*n^2*tanh(1/2*a+b*ln(( 
c*x^n)^(1/2)))^5+(-3*b^3*n^3+2*b*n)*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^4+4/3* 
(4*b^2*n^2-1)*tanh(1/2*a+b*ln((c*x^n)^(1/2)))^3+(-3*b^3*n^3+2*b*n)*tanh(1/ 
2*a+b*ln((c*x^n)^(1/2)))^2-2*b^2*n^2*tanh(1/2*a+b*ln((c*x^n)^(1/2)))+b^3*n 
^3)/(9*b^4*n^4-10*b^2*n^2+1)/(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)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.34 \[ \int \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3 \, {\left (b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 9 \, {\left (b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - {\left (b^{2} n^{2} - 1\right )} x \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 3 \, {\left (9 \, b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - 3 \, {\left ({\left (b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - {\left (9 \, b^{2} n^{2} - 1\right )} x\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{4 \, {\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Piecewise((Integral(sinh(a - log(c*x**n)/n)**3, x), Eq(b, -1/n)), (Integra 
l(sinh(a - log(c*x**n)/(3*n))**3, x), Eq(b, -1/(3*n))), (Integral(sinh(a + 
 log(c*x**n)/(3*n))**3, x), Eq(b, 1/(3*n))), (Integral(sinh(a + log(c*x**n 
)/n)**3, x), Eq(b, 1/n)), (9*b**3*n**3*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) - 6*b**3*n**3*x*cosh(a 
+ b*log(c*x**n))**3/(9*b**4*n**4 - 10*b**2*n**2 + 1) - 7*b**2*n**2*x*sinh( 
a + b*log(c*x**n))**3/(9*b**4*n**4 - 10*b**2*n**2 + 1) + 6*b**2*n**2*x*sin 
h(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) - 3*b*n*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) + x*sinh(a + b*log(c*x**n))**3/(9*b**4*n**4 - 1 
0*b**2*n**2 + 1), True))
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.77 \[ \int \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{3 \, b} x e^{\left (3 \, b \log \left (x^{n}\right ) + 3 \, a\right )}}{8 \, {\left (3 \, b n + 1\right )}} - \frac {3 \, c^{b} x e^{\left (b \log \left (x^{n}\right ) + a\right )}}{8 \, {\left (b n + 1\right )}} - \frac {3 \, x e^{\left (-b \log \left (x^{n}\right ) - a\right )}}{8 \, {\left (b c^{b} n - c^{b}\right )}} + \frac {x e^{\left (-3 \, b \log \left (x^{n}\right ) - 3 \, a\right )}}{8 \, {\left (3 \, b c^{3 \, b} n - c^{3 \, b}\right )}} \] Input:

integrate(sinh(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* 
b*log(x^n) - 3*a)/(3*b*c^(3*b)*n - c^(3*b))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 665 vs. \(2 (150) = 300\).

Time = 0.16 (sec) , antiderivative size = 665, normalized size of antiderivative = 4.46 \[ \int \sinh ^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(sinh(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))
 

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.62 \[ \int \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{-3\,a}}{{\left (c\,x^n\right )}^{3\,b}\,\left (24\,b\,n-8\right )}-\frac {3\,x\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (8\,b\,n-8\right )}+\frac {x\,{\mathrm {e}}^{3\,a}\,{\left (c\,x^n\right )}^{3\,b}}{24\,b\,n+8}-\frac {3\,x\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{8\,b\,n+8} \] Input:

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

Output:

(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)) + (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)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.17 \[ \int \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (3 x^{6 b n} e^{6 a} c^{6 b} b^{3} n^{3}-x^{6 b n} e^{6 a} c^{6 b} b^{2} n^{2}-3 x^{6 b n} e^{6 a} c^{6 b} b n +x^{6 b n} e^{6 a} c^{6 b}-27 x^{4 b n} e^{4 a} c^{4 b} b^{3} n^{3}+27 x^{4 b n} e^{4 a} c^{4 b} b^{2} n^{2}+3 x^{4 b n} e^{4 a} c^{4 b} b n -3 x^{4 b n} e^{4 a} c^{4 b}-27 x^{2 b n} e^{2 a} c^{2 b} b^{3} n^{3}-27 x^{2 b n} e^{2 a} c^{2 b} b^{2} n^{2}+3 x^{2 b n} e^{2 a} c^{2 b} b n +3 x^{2 b n} e^{2 a} c^{2 b}+3 b^{3} n^{3}+b^{2} n^{2}-3 b n -1\right )}{8 x^{3 b n} e^{3 a} c^{3 b} \left (9 b^{4} n^{4}-10 b^{2} n^{2}+1\right )} \] Input:

int(sinh(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))