Optimal. Leaf size=306 \[ -\frac{b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-m-1)} (d \sin (e+f x))^{m+1} F_1\left (\frac{1}{2};\frac{1}{2} (-m-1),2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^2}-\frac{a^2 d \cos (e+f x) \sin ^2(e+f x)^{\frac{1-m}{2}} (d \sin (e+f x))^{m-1} F_1\left (\frac{1}{2};\frac{1-m}{2},2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{2 a b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac{1}{2};-\frac{m}{2},2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.420374, antiderivative size = 306, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2824, 3189, 429, 16} \[ -\frac{b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-m-1)} (d \sin (e+f x))^{m+1} F_1\left (\frac{1}{2};\frac{1}{2} (-m-1),2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^2}-\frac{a^2 d \cos (e+f x) \sin ^2(e+f x)^{\frac{1-m}{2}} (d \sin (e+f x))^{m-1} F_1\left (\frac{1}{2};\frac{1-m}{2},2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{2 a b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac{1}{2};-\frac{m}{2},2;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 2824
Rule 3189
Rule 429
Rule 16
Rubi steps
\begin{align*} \int \frac{(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx &=\int \left (\frac{a^2 (d \sin (e+f x))^m}{\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))^m}{\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))^m}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2}\right ) \, dx\\ &=a^2 \int \frac{(d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx-(2 a b) \int \frac{\sin (e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx+b^2 \int \frac{\sin ^2(e+f x) (d \sin (e+f x))^m}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2} \, dx\\ &=\frac{b^2 \int \frac{(d \sin (e+f x))^{2+m}}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2} \, dx}{d^2}-\frac{(2 a b) \int \frac{(d \sin (e+f x))^{1+m}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx}{d}-\frac{\left (a^2 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 ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{1}{2} (-1+m)}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{a^2 d F_1\left (\frac{1}{2};\frac{1-m}{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) (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac{1-m}{2}}}{\left (a^2-b^2\right )^2 f}-\frac{\left (b^2 (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 ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{1+m}{2}}}{\left (-a^2+b^2-b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{d f}+\frac{\left (2 a b (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{m/2}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{b^2 F_1\left (\frac{1}{2};\frac{1}{2} (-1-m),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) (d \sin (e+f x))^{1+m} \sin ^2(e+f x)^{\frac{1}{2} (-1-m)}}{\left (a^2-b^2\right )^2 d f}-\frac{a^2 d F_1\left (\frac{1}{2};\frac{1-m}{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) (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac{1-m}{2}}}{\left (a^2-b^2\right )^2 f}+\frac{2 a b F_1\left (\frac{1}{2};-\frac{m}{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) (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}}{\left (a^2-b^2\right )^2 f}\\ \end{align*}
Mathematica [B] time = 18.8918, size = 1856, normalized size = 6.07 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.66, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\sin \left ( fx+e \right ) \right ) ^{m}}{ \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 (d \sin \left (f x + e\right )\right )^{m}}{{\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 (d \sin \left (f x + e\right )\right )^{m}}{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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 (d \sin \left (f x + e\right )\right )^{m}}{{\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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