Optimal. Leaf size=206 \[ \frac{a (2 a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{\sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 b f} \]
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Rubi [A] time = 0.198534, antiderivative size = 206, 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, 479, 524, 426, 424, 421, 419} \[ \frac{a (2 a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{\sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 b f} \]
Antiderivative was successfully verified.
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Rule 3188
Rule 479
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \frac{\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{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 b f}+\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 )}{3 b f}\\ &=-\frac{\cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 b 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 )}{3 b^2 f}+\frac{\left (a (2 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 )}{3 b^2 f}\\ &=-\frac{\cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 b f}-\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 )}{3 b^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\left (a (2 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 )}{3 b^2 f \sqrt{a+b \sin ^2(e+f x)}}\\ &=-\frac{\cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 b f}-\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)}}{3 b^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{a (2 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}}}{3 b^2 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.977425, size = 163, normalized size = 0.79 \[ \frac{b \sin (2 (e+f x)) (-2 a+b \cos (2 (e+f x))-b)+2 \sqrt{2} a (2 a-b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-4 \sqrt{2} a (a-b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{6 \sqrt{2} b^2 f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.089, size = 268, normalized size = 1.3 \begin{align*}{\frac{1}{3\,{b}^{2}\cos \left ( fx+e \right ) f} \left ({b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5}+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 ){a}^{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 ) b-2\,\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 ){a}^{2}+2\,\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 ) ab+ab \left ( \sin \left ( fx+e \right ) \right ) ^{3}-{b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}-\sin \left ( fx+e \right ) ab \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}}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{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 \cos \left (f x + e\right )^{2} - a - b}, 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}}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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