Optimal. Leaf size=322 \[ \frac {2 a b F_1\left (\frac {1}{2};-\frac {n p}{2},2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^2 f}-\frac {b^2 F_1\left (\frac {1}{2};\frac {1}{2} (-1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^2 f}-\frac {a^2 F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^2 f} \]
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Rubi [A]
time = 0.35, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2905, 2903,
3268, 440, 16} \begin {gather*} \frac {2 a b \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};-\frac {n p}{2},2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac {b^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-n p-1)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};\frac {1}{2} (-n p-1),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac {a^2 \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 16
Rule 440
Rule 2903
Rule 2905
Rule 3268
Rubi steps
\begin {align*} \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^2} \, dx &=\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {(d \sin (e+f x))^{n p}}{(a+b \sin (e+f x))^2} \, dx\\ &=\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \left (\frac {a^2 (d \sin (e+f x))^{n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2}-\frac {2 a b \sin (e+f x) (d \sin (e+f x))^{n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2}+\frac {b^2 \sin ^2(e+f x) (d \sin (e+f x))^{n p}}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2}\right ) \, dx\\ &=\left (a^2 (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {(d \sin (e+f x))^{n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx-\left (2 a b (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {\sin (e+f x) (d \sin (e+f x))^{n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx+\left (b^2 (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {\sin ^2(e+f x) (d \sin (e+f x))^{n p}}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2} \, dx\\ &=\frac {\left (b^2 (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {(d \sin (e+f x))^{2+n p}}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2} \, dx}{d^2}-\frac {\left (2 a b (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {(d \sin (e+f x))^{1+n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx}{d}-\frac {\left (a^2 d (d \sin (e+f x))^{-n p+2 \left (-\frac {1}{2}+\frac {n p}{2}\right )} \sin ^2(e+f x)^{\frac {1}{2}-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (-1+n p)}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {a^2 F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^2 f}+\frac {\left (2 a b \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {n p}{2}}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}-\frac {\left (b^2 (d \sin (e+f x))^{-n p+2 \left (\frac {1}{2}+\frac {n p}{2}\right )} \sin ^2(e+f x)^{-\frac {1}{2}-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (1+n p)}}{\left (-a^2+b^2-b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{d f}\\ &=\frac {2 a b F_1\left (\frac {1}{2};-\frac {n p}{2},2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^2 f}-\frac {b^2 F_1\left (\frac {1}{2};\frac {1}{2} (-1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^2 f}-\frac {a^2 F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^2 f}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1970\) vs. \(2(322)=644\).
time = 16.89, size = 1970, normalized size = 6.12 \begin {gather*} -\frac {\sec ^2(e+f x)^{\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (-a (2+n p) \left (\left (a^2+b^2\right ) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-2 b^2 F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},2;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )\right )+2 b \left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),2;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )}{a^3 \left (a^2-b^2\right ) f (1+n p) (2+n p) (a+b \sin (e+f x))^2 \left (-\frac {\sec ^2(e+f x)^{1+\frac {n p}{2}} \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (-a (2+n p) \left (\left (a^2+b^2\right ) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-2 b^2 F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},2;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )\right )+2 b \left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),2;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )}{a^3 \left (a^2-b^2\right ) (1+n p) (2+n p)}-\frac {n p \sec ^2(e+f x)^{\frac {n p}{2}} \tan ^2(e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (-a (2+n p) \left (\left (a^2+b^2\right ) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-2 b^2 F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},2;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )\right )+2 b \left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),2;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )}{a^3 \left (a^2-b^2\right ) (1+n p) (2+n p)}-\frac {n p \sec ^2(e+f x)^{\frac {n p}{2}} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{-1+n p} \left (-a (2+n p) \left (\left (a^2+b^2\right ) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-2 b^2 F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},2;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )\right )+2 b \left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),2;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right ) \left (\sqrt {\sec ^2(e+f x)}-\frac {\tan ^2(e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )}{a^3 \left (a^2-b^2\right ) (1+n p) (2+n p)}-\frac {\sec ^2(e+f x)^{\frac {n p}{2}} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (2 b \left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),2;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x)+2 b \left (a^2-b^2\right ) (1+n p) \tan (e+f x) \left (\frac {4 \left (-1+\frac {b^2}{a^2}\right ) \left (1+\frac {n p}{2}\right ) F_1\left (2+\frac {n p}{2};\frac {1}{2} (-1+n p),3;3+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{2+\frac {n p}{2}}-\frac {\left (1+\frac {n p}{2}\right ) (-1+n p) F_1\left (2+\frac {n p}{2};1+\frac {1}{2} (-1+n p),2;3+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{2+\frac {n p}{2}}\right )-a (2+n p) \left (\left (a^2+b^2\right ) \left (\frac {2 \left (-1+\frac {b^2}{a^2}\right ) (1+n p) F_1\left (1+\frac {1}{2} (1+n p);\frac {n p}{2},2;1+\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+n p}-\frac {n p (1+n p) F_1\left (1+\frac {1}{2} (1+n p);1+\frac {n p}{2},1;1+\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+n p}\right )-2 b^2 \left (\frac {4 \left (-1+\frac {b^2}{a^2}\right ) (1+n p) F_1\left (1+\frac {1}{2} (1+n p);\frac {n p}{2},3;1+\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+n p}-\frac {n p (1+n p) F_1\left (1+\frac {1}{2} (1+n p);1+\frac {n p}{2},2;1+\frac {1}{2} (3+n p);-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+n p}\right )\right )\right )}{a^3 \left (a^2-b^2\right ) (1+n p) (2+n p)}\right )} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.55, size = 0, normalized size = 0.00 \[\int \frac {\left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n}}{\left (a +b \sin \left (f x +e \right )\right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n}}{\left (a + b \sin {\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,{\left (d\,\sin \left (e+f\,x\right )\right )}^p\right )}^n}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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