\(\int x^m \sinh ^2(a+b \log (c x^n)) \, dx\) [271]

Optimal result

Integrand size = 17, antiderivative size = 120 \[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b^2 n^2 x^{1+m}}{(1+m) \left ((1+m)^2-4 b^2 n^2\right )}-\frac {2 b n x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-4 b^2 n^2}+\frac {(1+m) x^{1+m} \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-4 b^2 n^2} \] Output:

2*b^2*n^2*x^(1+m)/(1+m)/((1+m)^2-4*b^2*n^2)-2*b*n*x^(1+m)*cosh(a+b*ln(c*x^ 
n))*sinh(a+b*ln(c*x^n))/((1+m)^2-4*b^2*n^2)+(1+m)*x^(1+m)*sinh(a+b*ln(c*x^ 
n))^2/((1+m)^2-4*b^2*n^2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{1+m} \left (-1-2 m-m^2+4 b^2 n^2+(1+m)^2 \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-2 b (1+m) n \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right )}{2 (1+m) (1+m-2 b n) (1+m+2 b n)} \] Input:

Integrate[x^m*Sinh[a + b*Log[c*x^n]]^2,x]
 

Output:

(x^(1 + m)*(-1 - 2*m - m^2 + 4*b^2*n^2 + (1 + m)^2*Cosh[2*(a + b*Log[c*x^n 
])] - 2*b*(1 + m)*n*Sinh[2*(a + b*Log[c*x^n])]))/(2*(1 + m)*(1 + m - 2*b*n 
)*(1 + m + 2*b*n))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6055, 15}

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.

\(\displaystyle \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 6055

\(\displaystyle \frac {2 b^2 n^2 \int x^mdx}{(m+1)^2-4 b^2 n^2}+\frac {(m+1) x^{m+1} \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{-4 b^2 n^2+m^2+2 m+1}-\frac {2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {(m+1) x^{m+1} \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{-4 b^2 n^2+m^2+2 m+1}-\frac {2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}+\frac {2 b^2 n^2 x^{m+1}}{(m+1) \left ((m+1)^2-4 b^2 n^2\right )}\)

Input:

Int[x^m*Sinh[a + b*Log[c*x^n]]^2,x]
 

Output:

(2*b^2*n^2*x^(1 + m))/((1 + m)*((1 + m)^2 - 4*b^2*n^2)) - (2*b*n*x^(1 + m) 
*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]])/((1 + m)^2 - 4*b^2*n^2) + 
((1 + m)*x^(1 + m)*Sinh[a + b*Log[c*x^n]]^2)/(1 + 2*m + m^2 - 4*b^2*n^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 6055

Int[((e_.)*(x_))^(m_.)*Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p 
_), x_Symbol] :> Simp[(-(m + 1))*(e*x)^(m + 1)*(Sinh[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)*Cosh 
[d*(a + b*Log[c*x^n])]*(Sinh[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*Sinh[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 [F]

\[\int x^{m} {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]

Input:

int(x^m*sinh(a+b*ln(c*x^n))^2,x)
 

Output:

int(x^m*sinh(a+b*ln(c*x^n))^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.07 \[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {{\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} \cosh \left (m \log \left (x\right )\right ) + {\left (4 \, b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} x \cosh \left (m \log \left (x\right )\right ) + {\left ({\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (m \log \left (x\right )\right ) + {\left (m^{2} + 2 \, m + 1\right )} x \sinh \left (m \log \left (x\right )\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 4 \, {\left ({\left (b m + b\right )} n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \cosh \left (m \log \left (x\right )\right ) + {\left (b m + b\right )} n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (m \log \left (x\right )\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + {\left ({\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + {\left (4 \, b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} x\right )} \sinh \left (m \log \left (x\right )\right )}{2 \, {\left (m^{3} - 4 \, {\left (b^{2} m + b^{2}\right )} n^{2} + 3 \, m^{2} + 3 \, m + 1\right )}} \] Input:

integrate(x^m*sinh(a+b*log(c*x^n))^2,x, algorithm="fricas")
 

Output:

1/2*((m^2 + 2*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2*cosh(m*log(x)) + 
(4*b^2*n^2 - m^2 - 2*m - 1)*x*cosh(m*log(x)) + ((m^2 + 2*m + 1)*x*cosh(m*l 
og(x)) + (m^2 + 2*m + 1)*x*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a) 
^2 - 4*((b*m + b)*n*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + (b* 
m + b)*n*x*cosh(b*n*log(x) + b*log(c) + a)*sinh(m*log(x)))*sinh(b*n*log(x) 
 + b*log(c) + a) + ((m^2 + 2*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2 + 
(4*b^2*n^2 - m^2 - 2*m - 1)*x)*sinh(m*log(x)))/(m^3 - 4*(b^2*m + b^2)*n^2 
+ 3*m^2 + 3*m + 1)
 

Sympy [F]

\[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \log {\left (x \right )} \sinh ^{2}{\left (a \right )} & \text {for}\: b = 0 \wedge m = -1 \\\int x^{m} \sinh ^{2}{\left (- a + \frac {m \log {\left (c x^{n} \right )}}{2 n} + \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {m + 1}{2 n} \\\int x^{m} \sinh ^{2}{\left (a + \frac {m \log {\left (c x^{n} \right )}}{2 n} + \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {m + 1}{2 n} \\\int \frac {\sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{x}\, dx & \text {for}\: m = -1 \\\frac {2 b^{2} n^{2} x x^{m} \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} - \frac {2 b^{2} n^{2} x x^{m} \cosh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} + \frac {2 b m n x x^{m} \sinh {\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} + \frac {2 b n x x^{m} \sinh {\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} - \frac {m^{2} x x^{m} \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} - \frac {2 m x x^{m} \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} - \frac {x x^{m} \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} & \text {otherwise} \end {cases} \] Input:

integrate(x**m*sinh(a+b*ln(c*x**n))**2,x)
 

Output:

Piecewise((log(x)*sinh(a)**2, Eq(b, 0) & Eq(m, -1)), (Integral(x**m*sinh(- 
a + m*log(c*x**n)/(2*n) + log(c*x**n)/(2*n))**2, x), Eq(b, -(m + 1)/(2*n)) 
), (Integral(x**m*sinh(a + m*log(c*x**n)/(2*n) + log(c*x**n)/(2*n))**2, x) 
, Eq(b, (m + 1)/(2*n))), (Integral(sinh(a + b*log(c*x**n))**2/x, x), Eq(m, 
 -1)), (2*b**2*n**2*x*x**m*sinh(a + b*log(c*x**n))**2/(4*b**2*m*n**2 + 4*b 
**2*n**2 - m**3 - 3*m**2 - 3*m - 1) - 2*b**2*n**2*x*x**m*cosh(a + b*log(c* 
x**n))**2/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 3*m - 1) + 2*b*m* 
n*x*x**m*sinh(a + b*log(c*x**n))*cosh(a + b*log(c*x**n))/(4*b**2*m*n**2 + 
4*b**2*n**2 - m**3 - 3*m**2 - 3*m - 1) + 2*b*n*x*x**m*sinh(a + b*log(c*x** 
n))*cosh(a + b*log(c*x**n))/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 
 3*m - 1) - m**2*x*x**m*sinh(a + b*log(c*x**n))**2/(4*b**2*m*n**2 + 4*b**2 
*n**2 - m**3 - 3*m**2 - 3*m - 1) - 2*m*x*x**m*sinh(a + b*log(c*x**n))**2/( 
4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 3*m - 1) - x*x**m*sinh(a + b 
*log(c*x**n))**2/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 3*m - 1), 
True))
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{2 \, b} x e^{\left (2 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) + 2 \, a\right )}}{4 \, {\left (2 \, b n + m + 1\right )}} - \frac {x e^{\left (-2 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) - 2 \, a\right )}}{4 \, {\left (2 \, b c^{2 \, b} n - c^{2 \, b} {\left (m + 1\right )}\right )}} - \frac {x^{m + 1}}{2 \, {\left (m + 1\right )}} \] Input:

integrate(x^m*sinh(a+b*log(c*x^n))^2,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 758 vs. \(2 (127) = 254\).

Time = 0.17 (sec) , antiderivative size = 758, normalized size of antiderivative = 6.32 \[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx =\text {Too large to display} \] Input:

integrate(x^m*sinh(a+b*log(c*x^n))^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.62 \[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m\,{\mathrm {e}}^{-2\,a}}{{\left (c\,x^n\right )}^{2\,b}\,\left (4\,m-8\,b\,n+4\right )}-\frac {x\,x^m}{2\,m+2}+\frac {x\,x^m\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}}{4\,m+8\,b\,n+4} \] Input:

int(x^m*sinh(a + b*log(c*x^n))^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.18 \[ \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{m} x \left (2 x^{4 b n} e^{4 a} c^{4 b} b m n +2 x^{4 b n} e^{4 a} c^{4 b} b n -x^{4 b n} e^{4 a} c^{4 b} m^{2}-2 x^{4 b n} e^{4 a} c^{4 b} m -x^{4 b n} e^{4 a} c^{4 b}-8 x^{2 b n} e^{2 a} c^{2 b} b^{2} n^{2}+2 x^{2 b n} e^{2 a} c^{2 b} m^{2}+4 x^{2 b n} e^{2 a} c^{2 b} m +2 x^{2 b n} e^{2 a} c^{2 b}-2 b m n -2 b n -m^{2}-2 m -1\right )}{4 x^{2 b n} e^{2 a} c^{2 b} \left (4 b^{2} m \,n^{2}+4 b^{2} n^{2}-m^{3}-3 m^{2}-3 m -1\right )} \] Input:

int(x^m*sinh(a+b*log(c*x^n))^2,x)
 

Output:

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