Optimal. Leaf size=322 \[ \frac{2 a b \cos (e+f x) \sin ^2(e+f x)^{-\frac{n p}{2}} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac{1}{2};-\frac{n p}{2},2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{b^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-n p-1)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (-n p-1),2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{a^2 \cot (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (1-n p),2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.508049, antiderivative size = 322, normalized size of antiderivative = 1., 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 = {2826, 2824, 3189, 429, 16} \[ \frac{2 a b \cos (e+f x) \sin ^2(e+f x)^{-\frac{n p}{2}} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac{1}{2};-\frac{n p}{2},2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{b^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-n p-1)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (-n p-1),2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{a^2 \cot (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (1-n p),2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 2826
Rule 2824
Rule 3189
Rule 429
Rule 16
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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}
Mathematica [B] time = 19.0039, size = 2036, normalized size = 6.32 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.394, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c \left ( d\sin \left ( fx+e \right ) \right ) ^{p} \right ) ^{n}}{ \left ( a+b\sin \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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