Optimal. Leaf size=175 \[ \frac {2^{\frac {1}{2}-\frac {p}{2}} \sec (e+f x) (1-\sin (e+f x))^{\frac {p+1}{2}} (a \sin (e+f x)+a)^{m+1} (g \sec (e+f x))^p (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (\frac {1}{2} (2 m-p+1);\frac {p+1}{2},-n;\frac {1}{2} (2 m-p+3);\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m-p+1)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.43, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2926, 2921, 140, 139, 138} \[ \frac {2^{\frac {1}{2}-\frac {p}{2}} \sec (e+f x) (1-\sin (e+f x))^{\frac {p+1}{2}} (a \sin (e+f x)+a)^{m+1} (g \sec (e+f x))^p (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (\frac {1}{2} (2 m-p+1);\frac {p+1}{2},-n;\frac {1}{2} (2 m-p+3);\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m-p+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 138
Rule 139
Rule 140
Rule 2921
Rule 2926
Rubi steps
\begin {align*} \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx &=\left ((g \cos (e+f x))^p (g \sec (e+f x))^p\right ) \int (g \cos (e+f x))^{-p} (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx\\ &=\frac {\left (\sec (e+f x) (g \sec (e+f x))^p (a-a \sin (e+f x))^{\frac {1+p}{2}} (a+a \sin (e+f x))^{\frac {1+p}{2}}\right ) \operatorname {Subst}\left (\int (a-a x)^{\frac {1}{2} (-1-p)} (a+a x)^{m+\frac {1}{2} (-1-p)} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (2^{-\frac {1}{2}-\frac {p}{2}} \sec (e+f x) (g \sec (e+f x))^p (a-a \sin (e+f x))^{-\frac {1}{2}-\frac {p}{2}+\frac {1+p}{2}} \left (\frac {a-a \sin (e+f x)}{a}\right )^{\frac {1}{2}+\frac {p}{2}} (a+a \sin (e+f x))^{\frac {1+p}{2}}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{\frac {1}{2} (-1-p)} (a+a x)^{m+\frac {1}{2} (-1-p)} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (2^{-\frac {1}{2}-\frac {p}{2}} \sec (e+f x) (g \sec (e+f x))^p (a-a \sin (e+f x))^{-\frac {1}{2}-\frac {p}{2}+\frac {1+p}{2}} \left (\frac {a-a \sin (e+f x)}{a}\right )^{\frac {1}{2}+\frac {p}{2}} (a+a \sin (e+f x))^{\frac {1+p}{2}} (c+d \sin (e+f x))^n \left (\frac {a (c+d \sin (e+f x))}{a c-a d}\right )^{-n}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{\frac {1}{2} (-1-p)} (a+a x)^{m+\frac {1}{2} (-1-p)} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^n \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {2^{\frac {1}{2}-\frac {p}{2}} F_1\left (\frac {1}{2} (1+2 m-p);\frac {1+p}{2},-n;\frac {1}{2} (3+2 m-p);\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1+p}{2}} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{a f (1+2 m-p)}\\ \end {align*}
________________________________________________________________________________________
Mathematica [B] time = 11.13, size = 893, normalized size = 5.10 \[ \frac {2 F_1\left (\frac {1-p}{2};m+n-p+1,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),\frac {(d-c) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right ) \cos \left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) (g \sec (e+f x))^p \sin \left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^n}{f \left (\frac {d n F_1\left (\frac {1-p}{2};m+n-p+1,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right ) \cos ^2(e+f x)}{c+d \sin (e+f x)}-2 (n-p) F_1\left (\frac {1-p}{2};m+n-p+1,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),\frac {(d-c) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right ) \sin ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )+\frac {2 (1-p) \left ((m+n-p+1) F_1\left (\frac {3-p}{2};m+n-p+2,-n;\frac {5-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),\frac {(d-c) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right )-\frac {(c-d) n F_1\left (\frac {3-p}{2};m+n-p+1,1-n;\frac {5-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),\frac {(d-c) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right )}{c+d}\right ) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{3-p}-F_1\left (\frac {1-p}{2};m+n-p+1,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),\frac {(d-c) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right )+p F_1\left (\frac {1-p}{2};m+n-p+1,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),\frac {(d-c) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right ) \sin (e+f x)+\frac {2 (c-d) n F_1\left (\frac {1-p}{2};m+n-p+1,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ),\frac {(d-c) \tan ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d}\right ) \sin ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )}{c+d \sin (e+f x)}\right )} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 6.29, size = 0, normalized size = 0.00 \[ \int \left (g \sec \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^p\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________