Optimal. Leaf size=275 \[ \frac{\tan ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{f}-\frac{(3 a+8 b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}-\frac{a (5 a+8 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
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Rubi [A] time = 0.36881, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3196, 467, 577, 582, 524, 426, 424, 421, 419} \[ \frac{\tan ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{f}-\frac{(3 a+8 b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}-\frac{a (5 a+8 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
Antiderivative was successfully verified.
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Rule 3196
Rule 467
Rule 577
Rule 582
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^2\right )^{3/2}}{\left (1-x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2 \sqrt{a+b x^2} \left (3 a+6 b x^2\right )}{\left (1-x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 f}\\ &=-\frac{(a+2 b) \sin ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (-6 a (a+3 b)-3 b (3 a+8 b) x^2\right )}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 f}\\ &=-\frac{(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}-\frac{(a+2 b) \sin ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{-3 a b (3 a+8 b)-24 b^2 (a+2 b) x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{9 b f}\\ &=-\frac{(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}-\frac{(a+2 b) \sin ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}+\frac{\left (8 (a+2 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 f}-\frac{\left (a (5 a+8 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 f}\\ &=-\frac{(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}-\frac{(a+2 b) \sin ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}+\frac{\left (8 (a+2 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 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{\left (a (5 a+8 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 f \sqrt{a+b \sin ^2(e+f x)}}\\ &=-\frac{(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}+\frac{8 (a+2 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 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{a (5 a+8 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 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(a+2 b) \sin ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}\\ \end{align*}
Mathematica [A] time = 2.87417, size = 211, normalized size = 0.77 \[ \frac{-\frac{\tan (e+f x) \sec ^2(e+f x) \left (\left (64 a^2+160 a b+17 b^2\right ) \cos (2 (e+f x))+32 a^2-2 b (6 a+17 b) \cos (4 (e+f x))+108 a b-b^2 \cos (6 (e+f x))+18 b^2\right )}{4 \sqrt{2}}-4 a (5 a+8 b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )+32 a (a+2 b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{12 f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 2.241, size = 419, normalized size = 1.5 \begin{align*} -{\frac{1}{ \left ( -3+3\,\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( \sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{b}^{2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{6}+\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}b \left ( 3\,a+7\,b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}\sin \left ( fx+e \right ) -\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 4\,{a}^{2}+13\,ab+9\,{b}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \sin \left ( fx+e \right ) -\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}a \left ( 5\,{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) a+8\,{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b-8\,{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) a-16\,{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{- \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 1+\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{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{4}\,{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 (-{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )^{4}, 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{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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