Integrand size = 14, antiderivative size = 86 \[ \int x^{1+m} \sinh ^2(a+b x) \, dx=-\frac {x^{2+m}}{2 (2+m)}-\frac {2^{-4-m} e^{2 a} x^m (-b x)^{-m} \Gamma (2+m,-2 b x)}{b^2}-\frac {2^{-4-m} e^{-2 a} x^m (b x)^{-m} \Gamma (2+m,2 b x)}{b^2} \] Output:
-1/2*x^(2+m)/(2+m)-2^(-4-m)*exp(2*a)*x^m*GAMMA(2+m,-2*b*x)/b^2/((-b*x)^m)- 2^(-4-m)*x^m*GAMMA(2+m,2*b*x)/b^2/exp(2*a)/((b*x)^m)
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int x^{1+m} \sinh ^2(a+b x) \, dx=\frac {1}{16} x^m \left (-\frac {8 x^2}{2+m}-\frac {2^{-m} e^{2 a} (-b x)^{-m} \Gamma (2+m,-2 b x)}{b^2}-\frac {2^{-m} e^{-2 a} (b x)^{-m} \Gamma (2+m,2 b x)}{b^2}\right ) \] Input:
Integrate[x^(1 + m)*Sinh[a + b*x]^2,x]
Output:
(x^m*((-8*x^2)/(2 + m) - (E^(2*a)*Gamma[2 + m, -2*b*x])/(2^m*b^2*(-(b*x))^ m) - Gamma[2 + m, 2*b*x]/(2^m*b^2*E^(2*a)*(b*x)^m)))/16
Time = 0.35 (sec) , antiderivative size = 86, 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.
\(\displaystyle \int x^{m+1} \sinh ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -x^{m+1} \sin (i a+i b x)^2dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int x^{m+1} \sin (i a+i b x)^2dx\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\int \left (\frac {x^{m+1}}{2}-\frac {1}{2} x^{m+1} \cosh (2 a+2 b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 a} 2^{-m-4} x^m (-b x)^{-m} \Gamma (m+2,-2 b x)}{b^2}-\frac {e^{-2 a} 2^{-m-4} x^m (b x)^{-m} \Gamma (m+2,2 b x)}{b^2}-\frac {x^{m+2}}{2 (m+2)}\) |
Input:
Int[x^(1 + m)*Sinh[a + b*x]^2,x]
Output:
-1/2*x^(2 + m)/(2 + m) - (2^(-4 - m)*E^(2*a)*x^m*Gamma[2 + m, -2*b*x])/(b^ 2*(-(b*x))^m) - (2^(-4 - m)*x^m*Gamma[2 + m, 2*b*x])/(b^2*E^(2*a)*(b*x)^m)
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]))
\[\int x^{1+m} \sinh \left (b x +a \right )^{2}d x\]
Input:
int(x^(1+m)*sinh(b*x+a)^2,x)
Output:
int(x^(1+m)*sinh(b*x+a)^2,x)
Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.58 \[ \int x^{1+m} \sinh ^2(a+b x) \, dx=-\frac {4 \, b x \cosh \left ({\left (m + 1\right )} \log \left (x\right )\right ) + {\left (m + 2\right )} \cosh \left ({\left (m + 1\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 2, 2 \, b x\right ) - {\left (m + 2\right )} \cosh \left ({\left (m + 1\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 2, -2 \, b x\right ) - {\left (m + 2\right )} \Gamma \left (m + 2, 2 \, b x\right ) \sinh \left ({\left (m + 1\right )} \log \left (2 \, b\right ) + 2 \, a\right ) + {\left (m + 2\right )} \Gamma \left (m + 2, -2 \, b x\right ) \sinh \left ({\left (m + 1\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m + 1\right )} \log \left (x\right )\right )}{8 \, {\left (b m + 2 \, b\right )}} \] Input:
integrate(x^(1+m)*sinh(b*x+a)^2,x, algorithm="fricas")
Output:
-1/8*(4*b*x*cosh((m + 1)*log(x)) + (m + 2)*cosh((m + 1)*log(2*b) + 2*a)*ga mma(m + 2, 2*b*x) - (m + 2)*cosh((m + 1)*log(-2*b) - 2*a)*gamma(m + 2, -2* b*x) - (m + 2)*gamma(m + 2, 2*b*x)*sinh((m + 1)*log(2*b) + 2*a) + (m + 2)* gamma(m + 2, -2*b*x)*sinh((m + 1)*log(-2*b) - 2*a) + 4*b*x*sinh((m + 1)*lo g(x)))/(b*m + 2*b)
\[ \int x^{1+m} \sinh ^2(a+b x) \, dx=\int x^{m + 1} \sinh ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(x**(1+m)*sinh(b*x+a)**2,x)
Output:
Integral(x**(m + 1)*sinh(a + b*x)**2, x)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int x^{1+m} \sinh ^2(a+b x) \, dx=-\frac {1}{4} \, \left (2 \, b x\right )^{-m - 2} x^{m + 2} e^{\left (-2 \, a\right )} \Gamma \left (m + 2, 2 \, b x\right ) - \frac {1}{4} \, \left (-2 \, b x\right )^{-m - 2} x^{m + 2} e^{\left (2 \, a\right )} \Gamma \left (m + 2, -2 \, b x\right ) - \frac {x^{m + 2}}{2 \, {\left (m + 2\right )}} \] Input:
integrate(x^(1+m)*sinh(b*x+a)^2,x, algorithm="maxima")
Output:
-1/4*(2*b*x)^(-m - 2)*x^(m + 2)*e^(-2*a)*gamma(m + 2, 2*b*x) - 1/4*(-2*b*x )^(-m - 2)*x^(m + 2)*e^(2*a)*gamma(m + 2, -2*b*x) - 1/2*x^(m + 2)/(m + 2)
\[ \int x^{1+m} \sinh ^2(a+b x) \, dx=\int { x^{m + 1} \sinh \left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^(1+m)*sinh(b*x+a)^2,x, algorithm="giac")
Output:
integrate(x^(m + 1)*sinh(b*x + a)^2, x)
Timed out. \[ \int x^{1+m} \sinh ^2(a+b x) \, dx=\int x^{m+1}\,{\mathrm {sinh}\left (a+b\,x\right )}^2 \,d x \] Input:
int(x^(m + 1)*sinh(a + b*x)^2,x)
Output:
int(x^(m + 1)*sinh(a + b*x)^2, x)
\[ \int x^{1+m} \sinh ^2(a+b x) \, dx=\frac {2 x^{m} e^{4 b x +4 a} b m x +4 x^{m} e^{4 b x +4 a} b x -x^{m} e^{4 b x +4 a} m^{2}-3 x^{m} e^{4 b x +4 a} m -2 x^{m} e^{4 b x +4 a}+e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m^{3}+3 e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m^{2}+2 e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m -8 x^{m} e^{2 b x +2 a} b^{2} x^{2}+e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m^{3}+3 e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m^{2}+2 e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m -2 x^{m} b m x -4 x^{m} b x -x^{m} m^{2}-3 x^{m} m -2 x^{m}}{16 e^{2 b x +2 a} b^{2} \left (m +2\right )} \] Input:
int(x^(1+m)*sinh(b*x+a)^2,x)
Output:
(2*x**m*e**(4*a + 4*b*x)*b*m*x + 4*x**m*e**(4*a + 4*b*x)*b*x - x**m*e**(4* a + 4*b*x)*m**2 - 3*x**m*e**(4*a + 4*b*x)*m - 2*x**m*e**(4*a + 4*b*x) + e* *(4*a + 2*b*x)*int((x**m*e**(2*b*x))/x,x)*m**3 + 3*e**(4*a + 2*b*x)*int((x **m*e**(2*b*x))/x,x)*m**2 + 2*e**(4*a + 2*b*x)*int((x**m*e**(2*b*x))/x,x)* m - 8*x**m*e**(2*a + 2*b*x)*b**2*x**2 + e**(2*b*x)*int(x**m/(e**(2*b*x)*x) ,x)*m**3 + 3*e**(2*b*x)*int(x**m/(e**(2*b*x)*x),x)*m**2 + 2*e**(2*b*x)*int (x**m/(e**(2*b*x)*x),x)*m - 2*x**m*b*m*x - 4*x**m*b*x - x**m*m**2 - 3*x**m *m - 2*x**m)/(16*e**(2*a + 2*b*x)*b**2*(m + 2))