Integrand size = 22, antiderivative size = 267 \[ \int (c+d x)^m \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {3^{-1-m} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )}{8 b}-\frac {3^{-1-m} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )}{8 b} \]
-1/8*exp(I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-I*b*(d*x+c)/d)/b/((-I*b*(d*x+c) /d)^m)-1/8*(d*x+c)^m*GAMMA(1+m,I*b*(d*x+c)/d)/b/exp(I*(a-b*c/d))/((I*b*(d* x+c)/d)^m)-1/8*3^(-1-m)*exp(3*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-3*I*b*(d*x +c)/d)/b/((-I*b*(d*x+c)/d)^m)-1/8*3^(-1-m)*(d*x+c)^m*GAMMA(1+m,3*I*b*(d*x+ c)/d)/b/exp(3*I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)
Time = 0.77 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.94 \[ \int (c+d x)^m \cos ^2(a+b x) \sin (a+b x) \, dx=\frac {e^{-\frac {3 i (b c+a d)}{d}} (c+d x)^m \left (3 e^{\frac {2 i (b c+a d)}{d}} \left (-e^{2 i a} \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )-e^{\frac {2 i b c}{d}} \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )\right )-3^{-m} \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (e^{6 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )+e^{\frac {6 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )\right )\right )}{24 b} \]
((c + d*x)^m*(3*E^(((2*I)*(b*c + a*d))/d)*(-((E^((2*I)*a)*Gamma[1 + m, ((- I)*b*(c + d*x))/d])/(((-I)*b*(c + d*x))/d)^m) - (E^(((2*I)*b*c)/d)*Gamma[1 + m, (I*b*(c + d*x))/d])/((I*b*(c + d*x))/d)^m) - (E^((6*I)*a)*((I*b*(c + d*x))/d)^m*Gamma[1 + m, ((-3*I)*b*(c + d*x))/d] + E^(((6*I)*b*c)/d)*(((-I )*b*(c + d*x))/d)^m*Gamma[1 + m, ((3*I)*b*(c + d*x))/d])/(3^m*((b^2*(c + d *x)^2)/d^2)^m)))/(24*b*E^(((3*I)*(b*c + a*d))/d))
Time = 0.48 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4906, 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.
\(\displaystyle \int \sin (a+b x) \cos ^2(a+b x) (c+d x)^m \, dx\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle \int \left (\frac {1}{4} \sin (a+b x) (c+d x)^m+\frac {1}{4} \sin (3 a+3 b x) (c+d x)^m\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {3^{-m-1} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i b (c+d x)}{d}\right )}{8 b}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {3^{-m-1} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i b (c+d x)}{d}\right )}{8 b}\) |
-1/8*(E^(I*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-I)*b*(c + d*x))/d])/ (b*(((-I)*b*(c + d*x))/d)^m) - ((c + d*x)^m*Gamma[1 + m, (I*b*(c + d*x))/d ])/(8*b*E^(I*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m) - (3^(-1 - m)*E^((3*I)* (a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-3*I)*b*(c + d*x))/d])/(8*b*(((- I)*b*(c + d*x))/d)^m) - (3^(-1 - m)*(c + d*x)^m*Gamma[1 + m, ((3*I)*b*(c + d*x))/d])/(8*b*E^((3*I)*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m)
3.1.70.3.1 Defintions of rubi rules used
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
\[\int \left (d x +c \right )^{m} \cos \left (x b +a \right )^{2} \sin \left (x b +a \right )d x\]
Time = 0.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.70 \[ \int (c+d x)^m \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {3 \, e^{\left (-\frac {d m \log \left (\frac {i \, b}{d}\right ) - i \, b c + i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {i \, b d x + i \, b c}{d}\right ) + e^{\left (-\frac {d m \log \left (-\frac {3 i \, b}{d}\right ) + 3 i \, b c - 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + 3 \, e^{\left (-\frac {d m \log \left (-\frac {i \, b}{d}\right ) + i \, b c - i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-i \, b d x - i \, b c}{d}\right ) + e^{\left (-\frac {d m \log \left (\frac {3 i \, b}{d}\right ) - 3 i \, b c + 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{24 \, b} \]
-1/24*(3*e^(-(d*m*log(I*b/d) - I*b*c + I*a*d)/d)*gamma(m + 1, (I*b*d*x + I *b*c)/d) + e^(-(d*m*log(-3*I*b/d) + 3*I*b*c - 3*I*a*d)/d)*gamma(m + 1, -3* (I*b*d*x + I*b*c)/d) + 3*e^(-(d*m*log(-I*b/d) + I*b*c - I*a*d)/d)*gamma(m + 1, (-I*b*d*x - I*b*c)/d) + e^(-(d*m*log(3*I*b/d) - 3*I*b*c + 3*I*a*d)/d) *gamma(m + 1, -3*(-I*b*d*x - I*b*c)/d))/b
\[ \int (c+d x)^m \cos ^2(a+b x) \sin (a+b x) \, dx=\int \left (c + d x\right )^{m} \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}\, dx \]
\[ \int (c+d x)^m \cos ^2(a+b x) \sin (a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) \,d x } \]
\[ \int (c+d x)^m \cos ^2(a+b x) \sin (a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) \,d x } \]
Timed out. \[ \int (c+d x)^m \cos ^2(a+b x) \sin (a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^m \,d x \]