Optimal. Leaf size=214 \[ \frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^2(e+f x)}{a}\right )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}-\frac {\sin (e+f x) \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b f (2 p+5)} \]
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Rubi [A] time = 0.21, antiderivative size = 214, 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 = {3190, 416, 388, 246, 245} \[ \frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^2(e+f x)}{a}\right )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}-\frac {\sin (e+f x) \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b f (2 p+5)} \]
Antiderivative was successfully verified.
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Rule 245
Rule 246
Rule 388
Rule 416
Rule 3190
Rubi steps
\begin {align*} \int \cos ^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 x^2\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\cos ^2(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\operatorname {Subst}\left (\int \left (a+b x^2\right )^p \left (a+b (5+2 p)-(3 a+b (7+2 p)) x^2\right ) \, dx,x,\sin (e+f x)\right )}{b f (5+2 p)}\\ &=-\frac {(3 a+b (7+2 p)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos ^2(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2+2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\right ) \operatorname {Subst}\left (\int \left (a+b x^2\right )^p \, dx,x,\sin (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)}\\ &=-\frac {(3 a+b (7+2 p)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos ^2(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (\left (3 a^2+2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\right ) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \left (1+\frac {b x^2}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)}\\ &=-\frac {(3 a+b (7+2 p)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos ^2(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2+2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^2(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{b^2 f (3+2 p) (5+2 p)}\\ \end {align*}
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Mathematica [C] time = 0.40, size = 191, normalized size = 0.89 \[ \frac {3 a \sin (e+f x) \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p F_1\left (\frac {1}{2};-2,-p;\frac {3}{2};\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{f \left (2 \sin ^2(e+f x) \left (b p F_1\left (\frac {3}{2};-2,1-p;\frac {5}{2};\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-2 a F_1\left (\frac {3}{2};-1,-p;\frac {5}{2};\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right )+3 a F_1\left (\frac {1}{2};-2,-p;\frac {3}{2};\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right )} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \cos \left (f x + e\right )^{5}, x\right ) \]
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 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.93, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{5}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]
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 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5}\,{d x} \]
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 \[ \int {\cos \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]
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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