Integrand size = 28, antiderivative size = 154 \[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {(e+f x)^{1+m}}{a f (1+m)}+\frac {e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )}{2 a d} \]
(f*x+e)^(1+m)/a/f/(1+m)+1/2*exp(I*(c-d*e/f))*(f*x+e)^m*GAMMA(1+m,-I*d*(f*x +e)/f)/a/d/((-I*d*(f*x+e)/f)^m)+1/2*(f*x+e)^m*GAMMA(1+m,I*d*(f*x+e)/f)/a/d /exp(I*(c-d*e/f))/((I*d*(f*x+e)/f)^m)
Time = 0.80 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.43 \[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {d^2 (e+f x)^2}{f^2}\right )^{-m} \left (2 d e^{-i \left (c-\frac {d e}{f}\right )} (e+f x) \left (\frac {d^2 (e+f x)^2}{f^2}\right )^m+f (1+m) \left (\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )+e^{-2 i \left (c-\frac {d e}{f}\right )} f (1+m) \left (-\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{2 a d f (1+m) (1+\sin (c+d x))} \]
(E^(I*(c - (d*e)/f))*(e + f*x)^m*((2*d*(e + f*x)*((d^2*(e + f*x)^2)/f^2)^m )/E^(I*(c - (d*e)/f)) + f*(1 + m)*((I*d*(e + f*x))/f)^m*Gamma[1 + m, ((-I) *d*(e + f*x))/f] + (f*(1 + m)*(((-I)*d*(e + f*x))/f)^m*Gamma[1 + m, (I*d*( e + f*x))/f])/E^((2*I)*(c - (d*e)/f)))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2 ])^2)/(2*a*d*f*(1 + m)*((d^2*(e + f*x)^2)/f^2)^m*(1 + Sin[c + d*x]))
Time = 0.42 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5034, 17, 3042, 3789, 2612}
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.
\(\displaystyle \int \frac {\cos ^2(c+d x) (e+f x)^m}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 5034 |
\(\displaystyle \frac {\int (e+f x)^mdx}{a}-\frac {\int (e+f x)^m \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {(e+f x)^{m+1}}{a f (m+1)}-\frac {\int (e+f x)^m \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(e+f x)^{m+1}}{a f (m+1)}-\frac {\int (e+f x)^m \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3789 |
\(\displaystyle \frac {(e+f x)^{m+1}}{a f (m+1)}-\frac {\frac {1}{2} i \int e^{-i (c+d x)} (e+f x)^mdx-\frac {1}{2} i \int e^{i (c+d x)} (e+f x)^mdx}{a}\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle \frac {(e+f x)^{m+1}}{a f (m+1)}-\frac {-\frac {e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (m+1,-\frac {i d (e+f x)}{f}\right )}{2 d}-\frac {e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (m+1,\frac {i d (e+f x)}{f}\right )}{2 d}}{a}\) |
(e + f*x)^(1 + m)/(a*f*(1 + m)) - (-1/2*(E^(I*(c - (d*e)/f))*(e + f*x)^m*G amma[1 + m, ((-I)*d*(e + f*x))/f])/(d*(((-I)*d*(e + f*x))/f)^m) - ((e + f* x)^m*Gamma[1 + m, (I*d*(e + f*x))/f])/(2*d*E^(I*(c - (d*e)/f))*((I*d*(e + f*x))/f)^m))/a
3.3.89.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c + d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d) )^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m + 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] && !IntegerQ[m]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I /2 Int[(c + d*x)^m/E^(I*(e + f*x)), x], x] - Simp[I/2 Int[(c + d*x)^m*E ^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]
Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.) *Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] - Simp[1/b Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)*Sin[ c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 1] && EqQ[a^ 2 - b^2, 0]
\[\int \frac {\left (f x +e \right )^{m} \left (\cos ^{2}\left (d x +c \right )\right )}{a +a \sin \left (d x +c \right )}d x\]
Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\left (f m + f\right )} e^{\left (-\frac {f m \log \left (\frac {i \, d}{f}\right ) - i \, d e + i \, c f}{f}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, d e}{f}\right ) + {\left (f m + f\right )} e^{\left (-\frac {f m \log \left (-\frac {i \, d}{f}\right ) + i \, d e - i \, c f}{f}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, d e}{f}\right ) + 2 \, {\left (d f x + d e\right )} {\left (f x + e\right )}^{m}}{2 \, {\left (a d f m + a d f\right )}} \]
1/2*((f*m + f)*e^(-(f*m*log(I*d/f) - I*d*e + I*c*f)/f)*gamma(m + 1, (I*d*f *x + I*d*e)/f) + (f*m + f)*e^(-(f*m*log(-I*d/f) + I*d*e - I*c*f)/f)*gamma( m + 1, (-I*d*f*x - I*d*e)/f) + 2*(d*f*x + d*e)*(f*x + e)^m)/(a*d*f*m + a*d *f)
\[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\left (e + f x\right )^{m} \cos ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
\[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{m} \cos \left (d x + c\right )^{2}}{a \sin \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{m} \cos \left (d x + c\right )^{2}}{a \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e+f\,x\right )}^m}{a+a\,\sin \left (c+d\,x\right )} \,d x \]