Optimal. Leaf size=139 \[ \frac{2 \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 d^2 f}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}+\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}} \]
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Rubi [A] time = 0.162471, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3508, 3486, 3769, 3771, 2641} \[ \frac{2 \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 d^2 f}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}+\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3508
Rule 3486
Rule 3769
Rule 3771
Rule 2641
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx &=-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}-2 \int \frac{-\frac{a^2}{2}-b^2+\frac{1}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}-\left (-a^2-2 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{3/2}} \, dx\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}+\frac{\left (a^2+2 b^2\right ) \int \sqrt{d \sec (e+f x)} \, dx}{3 d^2}\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}+\frac{\left (\left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx}{3 d^2}\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 d^2 f}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.519407, size = 101, normalized size = 0.73 \[ \frac{\sec ^2(e+f x) \left (2 \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )+a^2 \sin (2 (e+f x))-2 a b \cos (2 (e+f x))-2 a b-b^2 \sin (2 (e+f x))\right )}{3 f (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.277, size = 320, normalized size = 2.3 \begin{align*}{\frac{2}{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{a}^{2}+2\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{b}^{2}+i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{a}^{2}+2\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{b}^{2}-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab+\cos \left ( fx+e \right ) \sin \left ( fx+e \right ){a}^{2}-\cos \left ( fx+e \right ) \sin \left ( fx+e \right ){b}^{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{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{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\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 (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \sqrt{d \sec \left (f x + e\right )}}{d^{2} \sec \left (f x + e\right )^{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{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}{\left (d \sec{\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{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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