Integrand size = 12, antiderivative size = 85 \[ \int x^m \cosh ^2(a+b x) \, dx=\frac {x^{1+m}}{2 (1+m)}+\frac {2^{-3-m} e^{2 a} x^m (-b x)^{-m} \Gamma (1+m,-2 b x)}{b}-\frac {2^{-3-m} e^{-2 a} x^m (b x)^{-m} \Gamma (1+m,2 b x)}{b} \] Output:
x^(1+m)/(2+2*m)+2^(-3-m)*exp(2*a)*x^m*GAMMA(1+m,-2*b*x)/b/((-b*x)^m)-2^(-3 -m)*x^m*GAMMA(1+m,2*b*x)/b/exp(2*a)/((b*x)^m)
Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int x^m \cosh ^2(a+b x) \, dx=\frac {1}{8} x^m \left (\frac {4 x}{1+m}+\frac {2^{-m} e^{2 a} (-b x)^{-m} \Gamma (1+m,-2 b x)}{b}-\frac {2^{-m} e^{-2 a} (b x)^{-m} \Gamma (1+m,2 b x)}{b}\right ) \] Input:
Integrate[x^m*Cosh[a + b*x]^2,x]
Output:
(x^m*((4*x)/(1 + m) + (E^(2*a)*Gamma[1 + m, -2*b*x])/(2^m*b*(-(b*x))^m) - Gamma[1 + m, 2*b*x]/(2^m*b*E^(2*a)*(b*x)^m)))/8
Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 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.
\(\displaystyle \int x^m \cosh ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^m \sin \left (i a+i b x+\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \int \left (\frac {1}{2} x^m \cosh (2 a+2 b x)+\frac {x^m}{2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{2 a} 2^{-m-3} x^m (-b x)^{-m} \Gamma (m+1,-2 b x)}{b}-\frac {e^{-2 a} 2^{-m-3} x^m (b x)^{-m} \Gamma (m+1,2 b x)}{b}+\frac {x^{m+1}}{2 (m+1)}\) |
Input:
Int[x^m*Cosh[a + b*x]^2,x]
Output:
x^(1 + m)/(2*(1 + m)) + (2^(-3 - m)*E^(2*a)*x^m*Gamma[1 + m, -2*b*x])/(b*( -(b*x))^m) - (2^(-3 - m)*x^m*Gamma[1 + m, 2*b*x])/(b*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]))
\[\int x^{m} \cosh \left (b x +a \right )^{2}d x\]
Input:
int(x^m*cosh(b*x+a)^2,x)
Output:
int(x^m*cosh(b*x+a)^2,x)
Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.44 \[ \int x^m \cosh ^2(a+b x) \, dx=\frac {4 \, b x \cosh \left (m \log \left (x\right )\right ) - {\left (m + 1\right )} \cosh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 1, 2 \, b x\right ) + {\left (m + 1\right )} \cosh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 1, -2 \, b x\right ) + {\left (m + 1\right )} \Gamma \left (m + 1, 2 \, b x\right ) \sinh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) - {\left (m + 1\right )} \Gamma \left (m + 1, -2 \, b x\right ) \sinh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left (m \log \left (x\right )\right )}{8 \, {\left (b m + b\right )}} \] Input:
integrate(x^m*cosh(b*x+a)^2,x, algorithm="fricas")
Output:
1/8*(4*b*x*cosh(m*log(x)) - (m + 1)*cosh(m*log(2*b) + 2*a)*gamma(m + 1, 2* b*x) + (m + 1)*cosh(m*log(-2*b) - 2*a)*gamma(m + 1, -2*b*x) + (m + 1)*gamm a(m + 1, 2*b*x)*sinh(m*log(2*b) + 2*a) - (m + 1)*gamma(m + 1, -2*b*x)*sinh (m*log(-2*b) - 2*a) + 4*b*x*sinh(m*log(x)))/(b*m + b)
\[ \int x^m \cosh ^2(a+b x) \, dx=\int x^{m} \cosh ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(x**m*cosh(b*x+a)**2,x)
Output:
Integral(x**m*cosh(a + b*x)**2, x)
Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int x^m \cosh ^2(a+b x) \, dx=-\frac {1}{4} \, \left (2 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (-2 \, a\right )} \Gamma \left (m + 1, 2 \, b x\right ) - \frac {1}{4} \, \left (-2 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (2 \, a\right )} \Gamma \left (m + 1, -2 \, b x\right ) + \frac {x^{m + 1}}{2 \, {\left (m + 1\right )}} \] Input:
integrate(x^m*cosh(b*x+a)^2,x, algorithm="maxima")
Output:
-1/4*(2*b*x)^(-m - 1)*x^(m + 1)*e^(-2*a)*gamma(m + 1, 2*b*x) - 1/4*(-2*b*x )^(-m - 1)*x^(m + 1)*e^(2*a)*gamma(m + 1, -2*b*x) + 1/2*x^(m + 1)/(m + 1)
\[ \int x^m \cosh ^2(a+b x) \, dx=\int { x^{m} \cosh \left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^m*cosh(b*x+a)^2,x, algorithm="giac")
Output:
integrate(x^m*cosh(b*x + a)^2, x)
Timed out. \[ \int x^m \cosh ^2(a+b x) \, dx=\int x^m\,{\mathrm {cosh}\left (a+b\,x\right )}^2 \,d x \] Input:
int(x^m*cosh(a + b*x)^2,x)
Output:
int(x^m*cosh(a + b*x)^2, x)
\[ \int x^m \cosh ^2(a+b x) \, dx=\frac {x^{m} e^{4 b x +4 a} m +x^{m} e^{4 b x +4 a}-e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m^{2}-e^{2 b x +4 a} \left (\int \frac {x^{m} e^{2 b x}}{x}d x \right ) m +4 x^{m} e^{2 b x +2 a} b x +e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m^{2}+e^{2 b x} \left (\int \frac {x^{m}}{e^{2 b x} x}d x \right ) m -x^{m} m -x^{m}}{8 e^{2 b x +2 a} b \left (m +1\right )} \] Input:
int(x^m*cosh(b*x+a)^2,x)
Output:
(x**m*e**(4*a + 4*b*x)*m + x**m*e**(4*a + 4*b*x) - e**(4*a + 2*b*x)*int((x **m*e**(2*b*x))/x,x)*m**2 - e**(4*a + 2*b*x)*int((x**m*e**(2*b*x))/x,x)*m + 4*x**m*e**(2*a + 2*b*x)*b*x + e**(2*b*x)*int(x**m/(e**(2*b*x)*x),x)*m**2 + e**(2*b*x)*int(x**m/(e**(2*b*x)*x),x)*m - x**m*m - x**m)/(8*e**(2*a + 2 *b*x)*b*(m + 1))