Optimal. Leaf size=161 \[ \frac{(2 A-7 B+27 C) \tan (c+d x)}{15 a^3 d}+\frac{(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.45006, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4084, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac{(2 A-7 B+27 C) \tan (c+d x)}{15 a^3 d}+\frac{(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4084
Rule 4019
Rule 4008
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac{(A-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) (a (2 A+3 B-3 C)+a (A-B+6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A+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 (A+4 B-9 C)+a^2 (2 A-7 B+27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A+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 (2 A-7 B+27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A+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{(B-3 C) \int \sec (c+d x) \, dx}{a^3}+\frac{(2 A-7 B+27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=\frac{(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A+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{(2 A-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{(2 A-7 B+27 C) \tan (c+d x)}{15 a^3 d}-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A+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 [B] time = 6.3667, size = 839, normalized size = 5.21 \[ \frac{16 (3 C-B) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^3}-\frac{16 (3 C-B) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^3}+\frac{\sec \left (\frac{c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-20 A \sin \left (\frac{d x}{2}\right )+160 B \sin \left (\frac{d x}{2}\right )-255 C \sin \left (\frac{d x}{2}\right )+22 A \sin \left (\frac{3 d x}{2}\right )-167 B \sin \left (\frac{3 d x}{2}\right )+567 C \sin \left (\frac{3 d x}{2}\right )-10 A \sin \left (c-\frac{d x}{2}\right )+170 B \sin \left (c-\frac{d x}{2}\right )-600 C \sin \left (c-\frac{d x}{2}\right )+10 A \sin \left (c+\frac{d x}{2}\right )-170 B \sin \left (c+\frac{d x}{2}\right )+375 C \sin \left (c+\frac{d x}{2}\right )-20 A \sin \left (2 c+\frac{d x}{2}\right )+160 B \sin \left (2 c+\frac{d x}{2}\right )-480 C \sin \left (2 c+\frac{d x}{2}\right )+75 B \sin \left (c+\frac{3 d x}{2}\right )-60 C \sin \left (c+\frac{3 d x}{2}\right )+22 A \sin \left (2 c+\frac{3 d x}{2}\right )-167 B \sin \left (2 c+\frac{3 d x}{2}\right )+402 C \sin \left (2 c+\frac{3 d x}{2}\right )+75 B \sin \left (3 c+\frac{3 d x}{2}\right )-225 C \sin \left (3 c+\frac{3 d x}{2}\right )+10 A \sin \left (c+\frac{5 d x}{2}\right )-95 B \sin \left (c+\frac{5 d x}{2}\right )+315 C \sin \left (c+\frac{5 d x}{2}\right )+15 B \sin \left (2 c+\frac{5 d x}{2}\right )+30 C \sin \left (2 c+\frac{5 d x}{2}\right )+10 A \sin \left (3 c+\frac{5 d x}{2}\right )-95 B \sin \left (3 c+\frac{5 d x}{2}\right )+240 C \sin \left (3 c+\frac{5 d x}{2}\right )+15 B \sin \left (4 c+\frac{5 d x}{2}\right )-45 C \sin \left (4 c+\frac{5 d x}{2}\right )+2 A \sin \left (2 c+\frac{7 d x}{2}\right )-22 B \sin \left (2 c+\frac{7 d x}{2}\right )+72 C \sin \left (2 c+\frac{7 d x}{2}\right )+15 C \sin \left (3 c+\frac{7 d x}{2}\right )+2 A \sin \left (4 c+\frac{7 d x}{2}\right )-22 B \sin \left (4 c+\frac{7 d x}{2}\right )+57 C \sin \left (4 c+\frac{7 d x}{2}\right )\right ) \cos \left (\frac{c}{2}+\frac{d x}{2}\right )}{60 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.072, size = 303, normalized size = 1.9 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\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{A}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\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{A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\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 [B] time = 0.981516, size = 473, normalized size = 2.94 \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 )} + \frac{A{\left (\frac{15 \, \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{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.521837, size = 690, normalized size = 4.29 \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 (A - 11 \, B + 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, A - 17 \, B + 57 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (7 \, A - 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{A \sec ^{3}{\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{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 = 1.2796, size = 316, normalized size = 1.96 \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 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 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} + 10 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 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} + 15 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 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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