Integrand size = 16, antiderivative size = 162 \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {i 2^{-3-m} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-3-m} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b} \] Output:
1/2*(d*x+c)^(1+m)/d/(1+m)+I*2^(-3-m)*exp(2*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+ m,-2*I*b*(d*x+c)/d)/b/((-I*b*(d*x+c)/d)^m)-I*2^(-3-m)*(d*x+c)^m*GAMMA(1+m, 2*I*b*(d*x+c)/d)/b/exp(2*I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)
Time = 0.80 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93 \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\frac {1}{8} (c+d x)^m \left (\frac {4 c+4 d x}{d+d m}+\frac {i 2^{-m} e^{2 i \left (a-\frac {b c}{d}\right )} \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-m} e^{-2 i \left (a-\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b}\right ) \] Input:
Integrate[(c + d*x)^m*Sin[a + b*x]^2,x]
Output:
((c + d*x)^m*((4*c + 4*d*x)/(d + d*m) + (I*E^((2*I)*(a - (b*c)/d))*Gamma[1 + m, ((-2*I)*b*(c + d*x))/d])/(2^m*b*(((-I)*b*(c + d*x))/d)^m) - (I*Gamma [1 + m, ((2*I)*b*(c + d*x))/d])/(2^m*b*E^((2*I)*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m)))/8
Time = 0.43 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, 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.
\(\displaystyle \int \sin ^2(a+b x) (c+d x)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^2 (c+d x)^mdx\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} \cos (2 a+2 b x) (c+d x)^m\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i 2^{-m-3} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-m-3} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {(c+d x)^{m+1}}{2 d (m+1)}\) |
Input:
Int[(c + d*x)^m*Sin[a + b*x]^2,x]
Output:
(c + d*x)^(1 + m)/(2*d*(1 + m)) + (I*2^(-3 - m)*E^((2*I)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-2*I)*b*(c + d*x))/d])/(b*(((-I)*b*(c + d*x))/d)^ m) - (I*2^(-3 - m)*(c + d*x)^m*Gamma[1 + m, ((2*I)*b*(c + d*x))/d])/(b*E^( (2*I)*(a - (b*c)/d))*((I*b*(c + d*x))/d)^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]))
\[\int \left (d x +c \right )^{m} \sin \left (b x +a \right )^{2}d x\]
Input:
int((d*x+c)^m*sin(b*x+a)^2,x)
Output:
int((d*x+c)^m*sin(b*x+a)^2,x)
Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.84 \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\frac {{\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + 4 \, {\left (b d x + b c\right )} {\left (d x + c\right )}^{m}}{8 \, {\left (b d m + b d\right )}} \] Input:
integrate((d*x+c)^m*sin(b*x+a)^2,x, algorithm="fricas")
Output:
1/8*((I*d*m + I*d)*e^(-(d*m*log(-2*I*b/d) + 2*I*b*c - 2*I*a*d)/d)*gamma(m + 1, -2*(I*b*d*x + I*b*c)/d) + (-I*d*m - I*d)*e^(-(d*m*log(2*I*b/d) - 2*I* b*c + 2*I*a*d)/d)*gamma(m + 1, -2*(-I*b*d*x - I*b*c)/d) + 4*(b*d*x + b*c)* (d*x + c)^m)/(b*d*m + b*d)
\[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int \left (c + d x\right )^{m} \sin ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate((d*x+c)**m*sin(b*x+a)**2,x)
Output:
Integral((c + d*x)**m*sin(a + b*x)**2, x)
\[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*x+c)^m*sin(b*x+a)^2,x, algorithm="maxima")
Output:
-1/2*((d*m + d)*integrate((d*x + c)^m*cos(2*b*x + 2*a), x) - e^(m*log(d*x + c) + log(d*x + c)))/(d*m + d)
\[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*x+c)^m*sin(b*x+a)^2,x, algorithm="giac")
Output:
integrate((d*x + c)^m*sin(b*x + a)^2, x)
Timed out. \[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^m \,d x \] Input:
int(sin(a + b*x)^2*(c + d*x)^m,x)
Output:
int(sin(a + b*x)^2*(c + d*x)^m, x)
\[ \int (c+d x)^m \sin ^2(a+b x) \, dx=\int \left (d x +c \right )^{m} \sin \left (b x +a \right )^{2}d x \] Input:
int((d*x+c)^m*sin(b*x+a)^2,x)
Output:
int((c + d*x)**m*sin(a + b*x)**2,x)