Optimal. Leaf size=259 \[ -\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{15 b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(a+4 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\sin ^3(e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f} \]
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Rubi [A] time = 0.32, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3188, 478, 582, 524, 426, 424, 421, 419} \[ -\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{15 b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\sin ^3(e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}-\frac {(a+4 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f} \]
Antiderivative was successfully verified.
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Rule 419
Rule 421
Rule 424
Rule 426
Rule 478
Rule 524
Rule 582
Rule 3188
Rubi steps
\begin {align*} \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^4 \sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (3 a+(a+4 b) x^2\right )}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{5 f}\\ &=-\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {a (a+4 b)+\left (-2 a^2+3 a b+8 b^2\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f}\\ &=-\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f}+\frac {\left (\left (-2 a^2+3 a b+8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f}\\ &=-\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (\left (-2 a^2+3 a b+8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}
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Mathematica [A] time = 1.43, size = 199, normalized size = 0.77 \[ \frac {-\sqrt {2} b \sin (2 (e+f x)) \left (8 a^2-4 b (4 a+7 b) \cos (2 (e+f x))+48 a b+3 b^2 \cos (4 (e+f x))+25 b^2\right )+32 a \left (a^2-a b-2 b^2\right ) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )-16 a \left (2 a^2-3 a b-8 b^2\right ) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{240 b^2 f \sqrt {2 a-b \cos (2 (e+f x))+b}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}, x\right ) \]
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 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.55, size = 413, normalized size = 1.59 \[ \frac {3 b^{3} \left (\sin ^{7}\left (f x +e \right )\right )+4 a \,b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+b^{3} \left (\sin ^{5}\left (f x +e \right )\right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-2 a^{2} \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -4 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +8 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}+a^{2} b \left (\sin ^{3}\left (f x +e \right )\right )-4 b^{3} \left (\sin ^{3}\left (f x +e \right )\right )-a^{2} b \sin \left (f x +e \right )-4 a \,b^{2} \sin \left (f x +e \right )}{15 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]
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 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{4}\,{d x} \]
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 \[ \int {\sin \left (e+f\,x\right )}^4\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]
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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