Optimal. Leaf size=108 \[ \frac{2 (B-C) \tan (c+d x)}{a d}-\frac{(2 B-3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{(B-C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(2 B-3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]
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Rubi [A] time = 0.239867, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4019, 3787, 3767, 8, 3768, 3770} \[ \frac{2 (B-C) \tan (c+d x)}{a d}-\frac{(2 B-3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{(B-C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(2 B-3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4019
Rule 3787
Rule 3767
Rule 8
Rule 3768
Rule 3770
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx &=\int \frac{\sec ^3(c+d x) (B+C \sec (c+d x))}{a+a \sec (c+d x)} \, dx\\ &=\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac{\int \sec ^2(c+d x) (2 a (B-C)-a (2 B-3 C) \sec (c+d x)) \, dx}{a^2}\\ &=\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{(2 B-3 C) \int \sec ^3(c+d x) \, dx}{a}+\frac{(2 (B-C)) \int \sec ^2(c+d x) \, dx}{a}\\ &=-\frac{(2 B-3 C) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{(2 B-3 C) \int \sec (c+d x) \, dx}{2 a}-\frac{(2 (B-C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d}\\ &=-\frac{(2 B-3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{2 (B-C) \tan (c+d x)}{a d}-\frac{(2 B-3 C) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}\\ \end{align*}
Mathematica [B] time = 0.664843, size = 383, normalized size = 3.55 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (2 (2 B-3 C) \cos \left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+(2 B-3 C) \cos \left (\frac{3}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+4 B \sin \left (\frac{1}{2} (c+d x)\right )+4 B \sin \left (\frac{5}{2} (c+d x)\right )+2 B \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 B \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-2 C \sin \left (\frac{1}{2} (c+d x)\right )+2 C \sin \left (\frac{3}{2} (c+d x)\right )-4 C \sin \left (\frac{5}{2} (c+d x)\right )-3 C \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 C \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 a d (\cos (c+d x)+1)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.054, size = 252, normalized size = 2.3 \begin{align*}{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{B}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{B}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 0.949285, size = 381, normalized size = 3.53 \begin{align*} -\frac{C{\left (\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{2 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, B{\left (\frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (d x + c\right )}{{\left (a - \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.513692, size = 386, normalized size = 3.57 \begin{align*} -\frac{{\left ({\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (B - C\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, B - C\right )} \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{4 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\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*} \frac{\int \frac{B \sec ^{3}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.15507, size = 211, normalized size = 1.95 \begin{align*} -\frac{\frac{{\left (2 \, B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{{\left (2 \, B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{2 \,{\left (B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} + \frac{2 \,{\left (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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