Optimal. Leaf size=202 \[ -\frac{2 a \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{b^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(2 a+b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{b^2 f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{a \sin (e+f x) \cos (e+f x)}{b f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]
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Rubi [A] time = 0.199586, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3188, 470, 524, 426, 424, 421, 419} \[ -\frac{2 a \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{b^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(2 a+b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{b^2 f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{a \sin (e+f x) \cos (e+f x)}{b f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3188
Rule 470
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \frac{\sin ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{b (a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{a+(-2 a-b) x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{b (a+b) f}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{b (a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (2 a \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 )}{b^2 f}-\frac{\left ((-2 a-b) \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 )}{b^2 (a+b) f}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{b (a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left ((-2 a-b) \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 )}{b^2 (a+b) f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{\left (2 a \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 )}{b^2 f \sqrt{a+b \sin ^2(e+f x)}}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{b (a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(2 a+b) \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)}}{b^2 (a+b) f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{2 a \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}}}{b^2 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.67261, size = 136, normalized size = 0.67 \[ \frac{a \left (-4 (a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )+2 (2 a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )+\sqrt{2} b \sin (2 (e+f x))\right )}{2 b^2 f (a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.25, size = 241, normalized size = 1.2 \begin{align*} -{\frac{a}{{b}^{2} \left ( a+b \right ) \cos \left ( fx+e \right ) f} \left ( 2\,a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) +2\,\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b-2\,a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) -\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b+b \left ( \sin \left ( fx+e \right ) \right ) ^{3}-b\sin \left ( fx+e \right ) \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \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{\sin \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{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 (\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}}{b^{2} \cos \left (f x + e\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + 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{\sin \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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