Optimal. Leaf size=406 \[ -\frac {3 a b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-m-1)} (d \sin (e+f x))^{m+1} F_1\left (\frac {1}{2};\frac {1}{2} (-m-1),3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^3}+\frac {3 a^2 b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac {1}{2};-\frac {m}{2},3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}+\frac {b^3 \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac {1}{2};\frac {1}{2} (-m-2),3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}-\frac {a^3 d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} F_1\left (\frac {1}{2};\frac {1-m}{2},3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3} \]
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Rubi [A] time = 0.55, antiderivative size = 406, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2824, 3189, 429, 16} \[ -\frac {3 a b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-m-1)} (d \sin (e+f x))^{m+1} F_1\left (\frac {1}{2};\frac {1}{2} (-m-1),3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^3}-\frac {a^3 d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} F_1\left (\frac {1}{2};\frac {1-m}{2},3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}+\frac {b^3 \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac {1}{2};\frac {1}{2} (-m-2),3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}+\frac {3 a^2 b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac {1}{2};-\frac {m}{2},3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3} \]
Antiderivative was successfully verified.
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Rule 16
Rule 429
Rule 2824
Rule 3189
Rubi steps
\begin {align*} \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx &=\int \left (\frac {a^3 (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}-\frac {3 a^2 b \sin (e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac {3 a b^2 \sin ^2(e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac {b^3 \sin ^3(e+f x) (d \sin (e+f x))^m}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^3}\right ) \, dx\\ &=a^3 \int \frac {(d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx-\left (3 a^2 b\right ) \int \frac {\sin (e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx+\left (3 a b^2\right ) \int \frac {\sin ^2(e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx+b^3 \int \frac {\sin ^3(e+f x) (d \sin (e+f x))^m}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^3} \, dx\\ &=\frac {b^3 \int \frac {(d \sin (e+f x))^{3+m}}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^3} \, dx}{d^3}+\frac {\left (3 a b^2\right ) \int \frac {(d \sin (e+f x))^{2+m}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx}{d^2}-\frac {\left (3 a^2 b\right ) \int \frac {(d \sin (e+f x))^{1+m}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx}{d}-\frac {\left (a^3 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 )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {a^3 d F_1\left (\frac {1}{2};\frac {1-m}{2},3;\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 )^3 f}-\frac {\left (3 a 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 )^3} \, dx,x,\cos (e+f x)\right )}{d f}+\frac {\left (3 a^2 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 )^3} \, dx,x,\cos (e+f x)\right )}{f}-\frac {\left (b^3 (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {2+m}{2}}}{\left (-a^2+b^2-b^2 x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {3 a b^2 F_1\left (\frac {1}{2};\frac {1}{2} (-1-m),3;\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 )^3 d f}-\frac {a^3 d F_1\left (\frac {1}{2};\frac {1-m}{2},3;\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 )^3 f}+\frac {b^3 F_1\left (\frac {1}{2};\frac {1}{2} (-2-m),3;\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 )^3 f}+\frac {3 a^2 b F_1\left (\frac {1}{2};-\frac {m}{2},3;\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 )^3 f}\\ \end {align*}
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Mathematica [B] time = 18.69, size = 2298, normalized size = 5.66 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (d \sin \left (f x + e\right )\right )^{m}}{3 \, a b^{2} \cos \left (f x + e\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]
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 \[ \int \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.77, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x +e \right )\right )^{m}}{\left (a +b \sin \left (f x +e \right )\right )^{3}}\, dx \]
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 \[ \int \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]
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 \[ \int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^m}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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