Optimal. Leaf size=156 \[ -\frac{(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac{(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.498398, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4019, 4008, 3787, 3770, 3767, 8} \[ -\frac{(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac{(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4019
Rule 4008
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac{\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^3(c+d x) (3 a (B-C)-a (B-6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^2(c+d x) \left (2 a^2 (4 B-9 C)-a^2 (7 B-27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\int \sec (c+d x) \left (-15 a^3 (B-3 C)+a^3 (7 B-27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(7 B-27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac{(B-3 C) \int \sec (c+d x) \, dx}{a^3}\\ &=\frac{(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(7 B-27 C) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac{(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 1.64531, size = 294, normalized size = 1.88 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left ((22 B-57 C) \tan ^3\left (\frac{1}{2} (c+d x)\right )+(87 C-22 B) \tan \left (\frac{1}{2} (c+d x)\right )+96 (B-C) \sin ^{10}\left (\frac{1}{2} (c+d x)\right ) \csc ^7(c+d x)-4 (7 B-12 C) \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-15 (B-3 C) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\frac{1}{4} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) ((7 B-12 C) \cos (c+d x)-4 B+9 C)+15 (B-3 C) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{15 a^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.057, size = 245, normalized size = 1.6 \begin{align*} -{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) C}{d{a}^{3}}}-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) C}{d{a}^{3}}}-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.96901, size = 386, normalized size = 2.47 \begin{align*} \frac{3 \, C{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \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{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B{\left (\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.521296, size = 668, normalized size = 4.28 \begin{align*} \frac{15 \,{\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (11 \, B - 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (17 \, B - 57 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (32 \, B - 117 \, C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\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 ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.21224, size = 251, normalized size = 1.61 \begin{align*} \frac{\frac{60 \,{\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \,{\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac{120 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac{3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 20 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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