Integrand size = 25, antiderivative size = 122 \[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\cos ^2(e+f x),\frac {b \cos ^2(e+f x)}{a+b}\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{f} \]
[Out]
Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3268, 441, 440} \[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\cos ^2(e+f x),\frac {b \cos ^2(e+f x)}{a+b}\right )}{f} \]
[In]
[Out]
Rule 440
Rule 441
Rule 3268
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (d (d \sin (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \sin ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \left (1-x^2\right )^{\frac {1}{2} (-1+m)} \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\left (d \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} (d \sin (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \sin ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \left (1-x^2\right )^{\frac {1}{2} (-1+m)} \left (1-\frac {b x^2}{a+b}\right )^p \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\cos ^2(e+f x),\frac {b \cos ^2(e+f x)}{a+b}\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{f} \\ \end{align*}
Time = 0.65 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},-p,\frac {3+m}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{f (1+m)} \]
[In]
[Out]
\[\int \left (d \sin \left (f x +e \right )\right )^{m} {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
[In]
[Out]
\[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
[In]
[Out]
Timed out. \[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
[In]
[Out]
\[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
[In]
[Out]
Timed out. \[ \int (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]
[In]
[Out]