\(\int x^{2+m} \sinh ^2(a+b x) \, dx\) [86]

Optimal result

Integrand size = 14, antiderivative size = 85 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=-\frac {x^{3+m}}{2 (3+m)}+\frac {2^{-5-m} e^{2 a} x^m (-b x)^{-m} \Gamma (3+m,-2 b x)}{b^3}-\frac {2^{-5-m} e^{-2 a} x^m (b x)^{-m} \Gamma (3+m,2 b x)}{b^3} \] Output:

-1/2*x^(3+m)/(3+m)+2^(-5-m)*exp(2*a)*x^m*GAMMA(3+m,-2*b*x)/b^3/((-b*x)^m)- 
2^(-5-m)*x^m*GAMMA(3+m,2*b*x)/b^3/exp(2*a)/((b*x)^m)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\frac {1}{32} x^m \left (-\frac {16 x^3}{3+m}+\frac {2^{-m} e^{2 a} (-b x)^{-m} \Gamma (3+m,-2 b x)}{b^3}-\frac {2^{-m} e^{-2 a} (b x)^{-m} \Gamma (3+m,2 b x)}{b^3}\right ) \] Input:

Integrate[x^(2 + m)*Sinh[a + b*x]^2,x]
 

Output:

(x^m*((-16*x^3)/(3 + m) + (E^(2*a)*Gamma[3 + m, -2*b*x])/(2^m*b^3*(-(b*x)) 
^m) - Gamma[3 + m, 2*b*x]/(2^m*b^3*E^(2*a)*(b*x)^m)))/32
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 3793, 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.

\(\displaystyle \int x^{m+2} \sinh ^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -x^{m+2} \sin (i a+i b x)^2dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int x^{m+2} \sin (i a+i b x)^2dx\)

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^{2 a} 2^{-m-5} x^m (-b x)^{-m} \Gamma (m+3,-2 b x)}{b^3}-\frac {e^{-2 a} 2^{-m-5} x^m (b x)^{-m} \Gamma (m+3,2 b x)}{b^3}-\frac {x^{m+3}}{2 (m+3)}\)

Input:

Int[x^(2 + m)*Sinh[a + b*x]^2,x]
 

Output:

-1/2*x^(3 + m)/(3 + m) + (2^(-5 - m)*E^(2*a)*x^m*Gamma[3 + m, -2*b*x])/(b^ 
3*(-(b*x))^m) - (2^(-5 - m)*x^m*Gamma[3 + m, 2*b*x])/(b^3*E^(2*a)*(b*x)^m)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.60 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=-\frac {4 \, b x \cosh \left ({\left (m + 2\right )} \log \left (x\right )\right ) + {\left (m + 3\right )} \cosh \left ({\left (m + 2\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 3, 2 \, b x\right ) - {\left (m + 3\right )} \cosh \left ({\left (m + 2\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 3, -2 \, b x\right ) - {\left (m + 3\right )} \Gamma \left (m + 3, 2 \, b x\right ) \sinh \left ({\left (m + 2\right )} \log \left (2 \, b\right ) + 2 \, a\right ) + {\left (m + 3\right )} \Gamma \left (m + 3, -2 \, b x\right ) \sinh \left ({\left (m + 2\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m + 2\right )} \log \left (x\right )\right )}{8 \, {\left (b m + 3 \, b\right )}} \] Input:

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

Output:

-1/8*(4*b*x*cosh((m + 2)*log(x)) + (m + 3)*cosh((m + 2)*log(2*b) + 2*a)*ga 
mma(m + 3, 2*b*x) - (m + 3)*cosh((m + 2)*log(-2*b) - 2*a)*gamma(m + 3, -2* 
b*x) - (m + 3)*gamma(m + 3, 2*b*x)*sinh((m + 2)*log(2*b) + 2*a) + (m + 3)* 
gamma(m + 3, -2*b*x)*sinh((m + 2)*log(-2*b) - 2*a) + 4*b*x*sinh((m + 2)*lo 
g(x)))/(b*m + 3*b)
 

Sympy [F]

\[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\int x^{m + 2} \sinh ^{2}{\left (a + b x \right )}\, dx \] Input:

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

Output:

Integral(x**(m + 2)*sinh(a + b*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=-\frac {1}{4} \, \left (2 \, b x\right )^{-m - 3} x^{m + 3} e^{\left (-2 \, a\right )} \Gamma \left (m + 3, 2 \, b x\right ) - \frac {1}{4} \, \left (-2 \, b x\right )^{-m - 3} x^{m + 3} e^{\left (2 \, a\right )} \Gamma \left (m + 3, -2 \, b x\right ) - \frac {x^{m + 3}}{2 \, {\left (m + 3\right )}} \] Input:

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

Output:

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

Giac [F]

\[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\int { x^{m + 2} \sinh \left (b x + a\right )^{2} \,d x } \] Input:

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

Output:

integrate(x^(m + 2)*sinh(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\int x^{m+2}\,{\mathrm {sinh}\left (a+b\,x\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^{2+m} \sinh ^2(a+b x) \, dx=\frac {-6 e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m +e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m^{4}+6 e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m^{3}+11 e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m^{2}-e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m^{4}-6 e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m^{3}-11 e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m^{2}+6 e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m -10 x^{m} b m x -12 x^{m} b^{2} x^{2}+x^{m} e^{4 b x +4 a} m^{3}+6 x^{m} e^{4 b x +4 a} m^{2}+11 x^{m} e^{4 b x +4 a} m +4 x^{m} e^{4 b x +4 a} b^{2} m \,x^{2}-2 x^{m} e^{4 b x +4 a} b \,m^{2} x -10 x^{m} e^{4 b x +4 a} b m x -16 x^{m} e^{2 b x +2 a} b^{3} x^{3}+12 x^{m} e^{4 b x +4 a} b^{2} x^{2}-12 x^{m} e^{4 b x +4 a} b x -4 x^{m} b^{2} m \,x^{2}-2 x^{m} b \,m^{2} x -x^{m} m^{3}+6 x^{m} e^{4 b x +4 a}-6 x^{m} m^{2}-11 x^{m} m -6 x^{m}-12 x^{m} b x}{32 e^{2 b x +2 a} b^{3} \left (m +3\right )} \] Input:

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

Output:

(4*x**m*e**(4*a + 4*b*x)*b**2*m*x**2 + 12*x**m*e**(4*a + 4*b*x)*b**2*x**2 
- 2*x**m*e**(4*a + 4*b*x)*b*m**2*x - 10*x**m*e**(4*a + 4*b*x)*b*m*x - 12*x 
**m*e**(4*a + 4*b*x)*b*x + x**m*e**(4*a + 4*b*x)*m**3 + 6*x**m*e**(4*a + 4 
*b*x)*m**2 + 11*x**m*e**(4*a + 4*b*x)*m + 6*x**m*e**(4*a + 4*b*x) - e**(4* 
a + 2*b*x)*int((x**m*e**(2*b*x))/x,x)*m**4 - 6*e**(4*a + 2*b*x)*int((x**m* 
e**(2*b*x))/x,x)*m**3 - 11*e**(4*a + 2*b*x)*int((x**m*e**(2*b*x))/x,x)*m** 
2 - 6*e**(4*a + 2*b*x)*int((x**m*e**(2*b*x))/x,x)*m - 16*x**m*e**(2*a + 2* 
b*x)*b**3*x**3 + e**(2*b*x)*int(x**m/(e**(2*b*x)*x),x)*m**4 + 6*e**(2*b*x) 
*int(x**m/(e**(2*b*x)*x),x)*m**3 + 11*e**(2*b*x)*int(x**m/(e**(2*b*x)*x),x 
)*m**2 + 6*e**(2*b*x)*int(x**m/(e**(2*b*x)*x),x)*m - 4*x**m*b**2*m*x**2 - 
12*x**m*b**2*x**2 - 2*x**m*b*m**2*x - 10*x**m*b*m*x - 12*x**m*b*x - x**m*m 
**3 - 6*x**m*m**2 - 11*x**m*m - 6*x**m)/(32*e**(2*a + 2*b*x)*b**3*(m + 3))