Optimal. Leaf size=153 \[ \frac{a^3 \left (c^2-3 c d+3 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{d^3 f}-\frac{a^3 (c-3 d) \tan (e+f x)}{d^2 f}-\frac{2 a^3 (c-d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{d^3 f \sqrt{c+d}}+\frac{a^3 \tanh ^{-1}(\sin (e+f x))}{2 d f}+\frac{a^3 \tan (e+f x) \sec (e+f x)}{2 d f} \]
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Rubi [A] time = 0.324612, antiderivative size = 257, normalized size of antiderivative = 1.68, number of steps used = 9, number of rules used = 9, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.29, Rules used = {3987, 102, 154, 157, 63, 217, 203, 93, 205} \[ \frac{a^4 \left (2 c^2-6 c d+7 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{a^3 (2 c-5 d) \tan (e+f x)}{2 d^2 f}+\frac{2 a^4 (c-d)^{5/2} \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{d^3 f \sqrt{c+d} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{\tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{2 d f} \]
Antiderivative was successfully verified.
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Rule 3987
Rule 102
Rule 154
Rule 157
Rule 63
Rule 217
Rule 203
Rule 93
Rule 205
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{5/2}}{\sqrt{a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{2 d f}+\frac{(a \tan (e+f x)) \operatorname{Subst}\left (\int \frac{\sqrt{a+a x} \left (-a^3 (c+2 d)+a^3 (2 c-5 d) x\right )}{\sqrt{a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{2 d f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{a^3 (2 c-5 d) \tan (e+f x)}{2 d^2 f}+\frac{\left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{2 d f}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{a^5 d (c+2 d)+a^5 \left (2 c^2-6 c d+7 d^2\right ) x}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{2 d^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{a^3 (2 c-5 d) \tan (e+f x)}{2 d^2 f}+\frac{\left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{2 d f}+\frac{\left (a^5 (c-d)^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^5 \left (2 c^2-6 c d+7 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{2 d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{a^3 (2 c-5 d) \tan (e+f x)}{2 d^2 f}+\frac{\left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{2 d f}+\frac{\left (2 a^5 (c-d)^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{a c-a d-(-a c-a d) x^2} \, dx,x,\frac{\sqrt{a+a \sec (e+f x)}}{\sqrt{a-a \sec (e+f x)}}\right )}{d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (a^4 \left (2 c^2-6 c d+7 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a-x^2}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{a^3 (2 c-5 d) \tan (e+f x)}{2 d^2 f}+\frac{2 a^4 (c-d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right ) \tan (e+f x)}{d^3 \sqrt{c+d} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{2 d f}+\frac{\left (a^4 \left (2 c^2-6 c d+7 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right )}{d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{a^3 (2 c-5 d) \tan (e+f x)}{2 d^2 f}+\frac{a^4 \left (2 c^2-6 c d+7 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right ) \tan (e+f x)}{d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{2 a^4 (c-d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right ) \tan (e+f x)}{d^3 \sqrt{c+d} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{2 d f}\\ \end{align*}
Mathematica [C] time = 2.28753, size = 419, normalized size = 2.74 \[ \frac{a^3 \cos ^2(e+f x) \sec ^6\left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^3 (c \cos (e+f x)+d) \left (-2 \left (2 c^2-6 c d+7 d^2\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+2 \left (2 c^2-6 c d+7 d^2\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\frac{8 (c-d)^3 (\sin (e)+i \cos (e)) \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}-\frac{4 d (c-3 d) \sin \left (\frac{f x}{2}\right )}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}-\frac{4 d (c-3 d) \sin \left (\frac{f x}{2}\right )}{\left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{d^2}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{d^2}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}\right )}{32 d^3 f (c+d \sec (e+f x))} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.097, size = 491, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.52238, size = 1235, normalized size = 8.07 \begin{align*} \left [\frac{2 \,{\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt{\frac{c - d}{c + d}} \cos \left (f x + e\right )^{2} \log \left (\frac{2 \, c d \cos \left (f x + e\right ) -{\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (c^{2} + c d +{\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{c - d}{c + d}} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) +{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (a^{3} d^{2} - 2 \,{\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, d^{3} f \cos \left (f x + e\right )^{2}}, -\frac{4 \,{\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt{-\frac{c - d}{c + d}} \arctan \left (-\frac{{\left (d \cos \left (f x + e\right ) + c\right )} \sqrt{-\frac{c - d}{c + d}}}{{\left (c - d\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{2} -{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) +{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (a^{3} d^{2} - 2 \,{\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, d^{3} f \cos \left (f x + e\right )^{2}}\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*} a^{3} \left (\int \frac{\sec{\left (e + f x \right )}}{c + d \sec{\left (e + f x \right )}}\, dx + \int \frac{3 \sec ^{2}{\left (e + f x \right )}}{c + d \sec{\left (e + f x \right )}}\, dx + \int \frac{3 \sec ^{3}{\left (e + f x \right )}}{c + d \sec{\left (e + f x \right )}}\, dx + \int \frac{\sec ^{4}{\left (e + f x \right )}}{c + d \sec{\left (e + f x \right )}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.30457, size = 398, normalized size = 2.6 \begin{align*} \frac{\frac{{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{d^{3}} - \frac{{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{d^{3}} + \frac{4 \,{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )}}{\sqrt{-c^{2} + d^{2}} d^{3}} + \frac{2 \,{\left (2 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 5 \, a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{2} d^{2}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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