Integrand size = 16, antiderivative size = 144 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=-\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 a+\frac {2 b c}{d}} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b} \]
-1/2*(d*x+c)^(1+m)/d/(1+m)+2^(-3-m)*exp(2*a-2*b*c/d)*(d*x+c)^m*GAMMA(1+m,- 2*b*(d*x+c)/d)/b/((-b*(d*x+c)/d)^m)-2^(-3-m)*exp(-2*a+2*b*c/d)*(d*x+c)^m*G AMMA(1+m,2*b*(d*x+c)/d)/b/((b*(d*x+c)/d)^m)
Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\frac {1}{8} (c+d x)^m \left (-\frac {4 (c+d x)}{d (1+m)}+\frac {2^{-m} e^{2 a-\frac {2 b c}{d}} \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m} e^{-2 a+\frac {2 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b}\right ) \]
((c + d*x)^m*((-4*(c + d*x))/(d*(1 + m)) + (E^(2*a - (2*b*c)/d)*Gamma[1 + m, (-2*b*(c + d*x))/d])/(2^m*b*(-((b*(c + d*x))/d))^m) - (E^(-2*a + (2*b*c )/d)*Gamma[1 + m, (2*b*(c + d*x))/d])/(2^m*b*((b*(c + d*x))/d)^m)))/8
Time = 0.42 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.
\(\displaystyle \int \sinh ^2(a+b x) (c+d x)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (i a+i b x)^2 \left (-(c+d x)^m\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int (c+d x)^m \sin (i a+i b x)^2dx\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cosh (2 a+2 b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2^{-m-3} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m-3} e^{\frac {2 b c}{d}-2 a} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 b (c+d x)}{d}\right )}{b}-\frac {(c+d x)^{m+1}}{2 d (m+1)}\) |
-1/2*(c + d*x)^(1 + m)/(d*(1 + m)) + (2^(-3 - m)*E^(2*a - (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (-2*b*(c + d*x))/d])/(b*(-((b*(c + d*x))/d))^m) - (2^( -3 - m)*E^(-2*a + (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (2*b*(c + d*x))/d])/ (b*((b*(c + d*x))/d)^m)
3.1.74.3.1 Defintions of rubi rules used
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]))
\[\int \left (d x +c \right )^{m} \sinh \left (b x +a \right )^{2}d x\]
Time = 0.08 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.67 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=-\frac {{\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) + {\left (d m + d\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) + 4 \, {\left (b d x + b c\right )} \cosh \left (m \log \left (d x + c\right )\right ) + 4 \, {\left (b d x + b c\right )} \sinh \left (m \log \left (d x + c\right )\right )}{8 \, {\left (b d m + b d\right )}} \]
-1/8*((d*m + d)*cosh((d*m*log(2*b/d) - 2*b*c + 2*a*d)/d)*gamma(m + 1, 2*(b *d*x + b*c)/d) - (d*m + d)*cosh((d*m*log(-2*b/d) + 2*b*c - 2*a*d)/d)*gamma (m + 1, -2*(b*d*x + b*c)/d) - (d*m + d)*gamma(m + 1, 2*(b*d*x + b*c)/d)*si nh((d*m*log(2*b/d) - 2*b*c + 2*a*d)/d) + (d*m + d)*gamma(m + 1, -2*(b*d*x + b*c)/d)*sinh((d*m*log(-2*b/d) + 2*b*c - 2*a*d)/d) + 4*(b*d*x + b*c)*cosh (m*log(d*x + c)) + 4*(b*d*x + b*c)*sinh(m*log(d*x + c)))/(b*d*m + b*d)
\[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\int \left (c + d x\right )^{m} \sinh ^{2}{\left (a + b x \right )}\, dx \]
Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.71 \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=-\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} E_{-m}\left (\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {{\left (d x + c\right )}^{m + 1} e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} E_{-m}\left (-\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {{\left (d x + c\right )}^{m + 1}}{2 \, d {\left (m + 1\right )}} \]
-1/4*(d*x + c)^(m + 1)*e^(-2*a + 2*b*c/d)*exp_integral_e(-m, 2*(d*x + c)*b /d)/d - 1/4*(d*x + c)^(m + 1)*e^(2*a - 2*b*c/d)*exp_integral_e(-m, -2*(d*x + c)*b/d)/d - 1/2*(d*x + c)^(m + 1)/(d*(m + 1))
\[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sinh \left (b x + a\right )^{2} \,d x } \]
Timed out. \[ \int (c+d x)^m \sinh ^2(a+b x) \, dx=\int {\mathrm {sinh}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]