Optimal. Leaf size=220 \[ -\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}+\frac {(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}-\frac {\sin ^2(e+f x) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]
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Rubi [A] time = 0.22, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3186, 416, 388, 246, 245} \[ -\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}+\frac {(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}-\frac {\sin ^2(e+f x) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]
Antiderivative was successfully verified.
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Rule 245
Rule 246
Rule 388
Rule 416
Rule 3186
Rubi steps
\begin {align*} \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx &=-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right )^2 \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)}+\frac {\operatorname {Subst}\left (\int \left (a+b-b x^2\right )^p \left (a-2 b (2+p)-(3 a-2 b (2+p)) x^2\right ) \, dx,x,\cos (e+f x)\right )}{b f (5+2 p)}\\ &=\frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)}-\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname {Subst}\left (\int \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)}\\ &=\frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)}-\frac {\left (\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname {Subst}\left (\int \left (1-\frac {b x^2}{a+b}\right )^p \, dx,x,\cos (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)}\\ &=\frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)}\\ \end {align*}
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Mathematica [C] time = 0.55, size = 98, normalized size = 0.45 \[ \frac {\sin ^5(e+f x) \sqrt {\cos ^2(e+f x)} \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {a+b \sin ^2(e+f x)}{a}\right )^{-p} F_1\left (3;\frac {1}{2},-p;4;\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{6 f} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \sin \left (f x + e\right ), x\right ) \]
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 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 4.56, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{5}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]
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 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \]
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 \[ \int {\sin \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]
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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