Optimal. Leaf size=224 \[ \frac{\tan (e+f x)}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 b \sin (e+f x) \cos (e+f x)}{f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
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Rubi [A] time = 0.225916, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3196, 471, 527, 524, 426, 424, 421, 419} \[ \frac{\tan (e+f x)}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 b \sin (e+f x) \cos (e+f x)}{f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
Antiderivative was successfully verified.
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Rule 3196
Rule 471
Rule 527
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \frac{\tan ^2(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^2}{\left (1-x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{(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-b x^2}{\sqrt{1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=-\frac{2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\tan (e+f x)}{(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 (a-b)-2 a b x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{a (a+b)^2 f}\\ &=-\frac{2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\tan (e+f x)}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (2 \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 )}{(a+b)^2 f}+\frac{\left (\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 )}{(a+b) f}\\ &=-\frac{2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\tan (e+f x)}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (2 \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 )}{(a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\left (\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 )}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ &=-\frac{2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 \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)}}{(a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\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}}}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\tan (e+f x)}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.885533, size = 145, normalized size = 0.65 \[ \frac{2 \tan (e+f x) (a-b \cos (2 (e+f x)))+\sqrt{2} (a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-2 \sqrt{2} a \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{\sqrt{2} f (a+b)^2 \sqrt{2 a-b \cos (2 (e+f x))+b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 2.174, size = 278, normalized size = 1.2 \begin{align*}{\frac{1}{ \left ( a+b \right ) ^{2}\cos \left ( fx+e \right ) f}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) +b\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) -2\,a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) -2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) b+a\sin \left ( fx+e \right ) +b\sin \left ( fx+e \right ) \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(-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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )^{2}}{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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \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{\tan \left (f x + e\right )^{2}}{{\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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