\(\int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx\) [264]

Optimal result

Integrand size = 18, antiderivative size = 141 \[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\frac {2^{-2 (3+m)} e^{4 a} x^m (-b x)^{-m} \Gamma (1+m,-4 b x)}{b}-\frac {2^{-4-m} e^{2 a} x^m (-b x)^{-m} \Gamma (1+m,-2 b x)}{b}-\frac {2^{-4-m} e^{-2 a} x^m (b x)^{-m} \Gamma (1+m,2 b x)}{b}+\frac {2^{-2 (3+m)} e^{-4 a} x^m (b x)^{-m} \Gamma (1+m,4 b x)}{b} \] Output:

exp(4*a)*x^m*GAMMA(1+m,-4*b*x)/(2^(6+2*m))/b/((-b*x)^m)-2^(-4-m)*exp(2*a)* 
x^m*GAMMA(1+m,-2*b*x)/b/((-b*x)^m)-2^(-4-m)*x^m*GAMMA(1+m,2*b*x)/b/exp(2*a 
)/((b*x)^m)+x^m*GAMMA(1+m,4*b*x)/(2^(6+2*m))/b/exp(4*a)/((b*x)^m)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79 \[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\frac {4^{-3-m} e^{-4 a} x^m \left (-b^2 x^2\right )^{-m} \left (e^{8 a} (b x)^m \Gamma (1+m,-4 b x)-2^{2+m} e^{6 a} (b x)^m \Gamma (1+m,-2 b x)+(-b x)^m \left (-2^{2+m} e^{2 a} \Gamma (1+m,2 b x)+\Gamma (1+m,4 b x)\right )\right )}{b} \] Input:

Integrate[x^m*Cosh[a + b*x]*Sinh[a + b*x]^3,x]
 

Output:

(4^(-3 - m)*x^m*(E^(8*a)*(b*x)^m*Gamma[1 + m, -4*b*x] - 2^(2 + m)*E^(6*a)* 
(b*x)^m*Gamma[1 + m, -2*b*x] + (-(b*x))^m*(-(2^(2 + m)*E^(2*a)*Gamma[1 + m 
, 2*b*x]) + Gamma[1 + m, 4*b*x])))/(b*E^(4*a)*(-(b^2*x^2))^m)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5971, 2009}

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.

\(\displaystyle \int x^m \sinh ^3(a+b x) \cosh (a+b x) \, dx\)

\(\Big \downarrow \) 5971

\(\displaystyle \int \left (\frac {1}{8} x^m \sinh (4 a+4 b x)-\frac {1}{4} x^m \sinh (2 a+2 b x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^{4 a} 2^{-2 (m+3)} x^m (-b x)^{-m} \Gamma (m+1,-4 b x)}{b}-\frac {e^{2 a} 2^{-m-4} x^m (-b x)^{-m} \Gamma (m+1,-2 b x)}{b}-\frac {e^{-2 a} 2^{-m-4} x^m (b x)^{-m} \Gamma (m+1,2 b x)}{b}+\frac {e^{-4 a} 2^{-2 (m+3)} x^m (b x)^{-m} \Gamma (m+1,4 b x)}{b}\)

Input:

Int[x^m*Cosh[a + b*x]*Sinh[a + b*x]^3,x]
 

Output:

(E^(4*a)*x^m*Gamma[1 + m, -4*b*x])/(2^(2*(3 + m))*b*(-(b*x))^m) - (2^(-4 - 
 m)*E^(2*a)*x^m*Gamma[1 + m, -2*b*x])/(b*(-(b*x))^m) - (2^(-4 - m)*x^m*Gam 
ma[1 + m, 2*b*x])/(b*E^(2*a)*(b*x)^m) + (x^m*Gamma[1 + m, 4*b*x])/(2^(2*(3 
 + m))*b*E^(4*a)*(b*x)^m)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5971

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

Maple [F]

\[\int x^{m} \cosh \left (b x +a \right ) \sinh \left (b x +a \right )^{3}d x\]

Input:

int(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x)
 

Output:

int(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.22 \[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\frac {\cosh \left (m \log \left (4 \, b\right ) + 4 \, a\right ) \Gamma \left (m + 1, 4 \, b x\right ) - 4 \, \cosh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 1, 2 \, b x\right ) - 4 \, \cosh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 1, -2 \, b x\right ) + \cosh \left (m \log \left (-4 \, b\right ) - 4 \, a\right ) \Gamma \left (m + 1, -4 \, b x\right ) - \Gamma \left (m + 1, 4 \, b x\right ) \sinh \left (m \log \left (4 \, b\right ) + 4 \, a\right ) + 4 \, \Gamma \left (m + 1, 2 \, b x\right ) \sinh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) + 4 \, \Gamma \left (m + 1, -2 \, b x\right ) \sinh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) - \Gamma \left (m + 1, -4 \, b x\right ) \sinh \left (m \log \left (-4 \, b\right ) - 4 \, a\right )}{64 \, b} \] Input:

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

Output:

1/64*(cosh(m*log(4*b) + 4*a)*gamma(m + 1, 4*b*x) - 4*cosh(m*log(2*b) + 2*a 
)*gamma(m + 1, 2*b*x) - 4*cosh(m*log(-2*b) - 2*a)*gamma(m + 1, -2*b*x) + c 
osh(m*log(-4*b) - 4*a)*gamma(m + 1, -4*b*x) - gamma(m + 1, 4*b*x)*sinh(m*l 
og(4*b) + 4*a) + 4*gamma(m + 1, 2*b*x)*sinh(m*log(2*b) + 2*a) + 4*gamma(m 
+ 1, -2*b*x)*sinh(m*log(-2*b) - 2*a) - gamma(m + 1, -4*b*x)*sinh(m*log(-4* 
b) - 4*a))/b
 

Sympy [F]

\[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\int x^{m} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}\, dx \] Input:

integrate(x**m*cosh(b*x+a)*sinh(b*x+a)**3,x)
 

Output:

Integral(x**m*sinh(a + b*x)**3*cosh(a + b*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.83 \[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\frac {1}{16} \, \left (4 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (-4 \, a\right )} \Gamma \left (m + 1, 4 \, b x\right ) - \frac {1}{8} \, \left (2 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (-2 \, a\right )} \Gamma \left (m + 1, 2 \, b x\right ) + \frac {1}{8} \, \left (-2 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (2 \, a\right )} \Gamma \left (m + 1, -2 \, b x\right ) - \frac {1}{16} \, \left (-4 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (4 \, a\right )} \Gamma \left (m + 1, -4 \, b x\right ) \] Input:

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

Output:

1/16*(4*b*x)^(-m - 1)*x^(m + 1)*e^(-4*a)*gamma(m + 1, 4*b*x) - 1/8*(2*b*x) 
^(-m - 1)*x^(m + 1)*e^(-2*a)*gamma(m + 1, 2*b*x) + 1/8*(-2*b*x)^(-m - 1)*x 
^(m + 1)*e^(2*a)*gamma(m + 1, -2*b*x) - 1/16*(-4*b*x)^(-m - 1)*x^(m + 1)*e 
^(4*a)*gamma(m + 1, -4*b*x)
 

Giac [F]

\[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\int { x^{m} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} \,d x } \] Input:

integrate(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate(x^m*cosh(b*x + a)*sinh(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\int x^m\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^3 \,d x \] Input:

int(x^m*cosh(a + b*x)*sinh(a + b*x)^3,x)
 

Output:

int(x^m*cosh(a + b*x)*sinh(a + b*x)^3, x)
 

Reduce [F]

\[ \int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx=\frac {x^{m} e^{8 b x +8 a}-4 x^{m} e^{6 b x +6 a}-e^{4 b x +8 a} \left (\int \frac {x^{m} e^{4 b x}}{x}d x \right ) m +4 e^{4 b x +6 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m +4 e^{4 b x +2 a} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m -e^{4 b x} \left (\int \frac {x^{m}}{e^{4 b x} x}d x \right ) m -4 x^{m} e^{2 b x +2 a}+x^{m}}{64 e^{4 b x +4 a} b} \] Input:

int(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x)
 

Output:

(x**m*e**(8*a + 8*b*x) - 4*x**m*e**(6*a + 6*b*x) - e**(8*a + 4*b*x)*int((x 
**m*e**(4*b*x))/x,x)*m + 4*e**(6*a + 4*b*x)*int((x**m*e**(2*b*x))/x,x)*m + 
 4*e**(2*a + 4*b*x)*int(x**m/(e**(2*b*x)*x),x)*m - e**(4*b*x)*int(x**m/(e* 
*(4*b*x)*x),x)*m - 4*x**m*e**(2*a + 2*b*x) + x**m)/(64*e**(4*a + 4*b*x)*b)