Optimal. Leaf size=125 \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 A-7 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(4 A-29 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Rubi [A]
time = 0.22, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4104, 4093,
4083, 3855, 3879} \begin {gather*} \frac {(4 A-29 B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(2 A-7 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 3855
Rule 3879
Rule 4083
Rule 4093
Rule 4104
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^2(c+d x) (2 a (A-B)+5 a B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 A-7 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (-2 a^2 (2 A-7 B)-15 a^2 B \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 A-7 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(4 A-29 B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}+\frac {B \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {B \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 A-7 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(4 A-29 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.94, size = 197, normalized size = 1.58 \begin {gather*} \frac {-240 B \cos ^6\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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (5 (4 A-29 B) \sin \left (\frac {d x}{2}\right )+75 B \sin \left (c+\frac {d x}{2}\right )+10 A \sin \left (c+\frac {3 d x}{2}\right )-95 B \sin \left (c+\frac {3 d x}{2}\right )+15 B \sin \left (2 c+\frac {3 d x}{2}\right )+2 A \sin \left (2 c+\frac {5 d x}{2}\right )-22 B \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d (1+\cos (c+d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.23, size = 119, normalized size = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {4 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {2 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-4 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(119\) |
default | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {4 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {2 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-4 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(119\) |
risch | \(-\frac {2 i \left (15 B \,{\mathrm e}^{4 i \left (d x +c \right )}+75 B \,{\mathrm e}^{3 i \left (d x +c \right )}-20 A \,{\mathrm e}^{2 i \left (d x +c \right )}+145 B \,{\mathrm e}^{2 i \left (d x +c \right )}-10 \,{\mathrm e}^{i \left (d x +c \right )} A +95 B \,{\mathrm e}^{i \left (d x +c \right )}-2 A +22 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}\) | \(146\) |
norman | \(\frac {\frac {\left (A -11 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (A -B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (A +9 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}-\frac {\left (3 A -43 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}+\frac {\left (-59 B +7 A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a^{2}}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(209\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, size = 187, normalized size = 1.50 \begin {gather*} -\frac {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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 4.24, size = 183, normalized size = 1.46 \begin {gather*} \frac {15 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 32 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}{a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 147, normalized size = 1.18 \begin {gather*} \frac {\frac {60 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\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} + 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} + 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 )}{a^{15}}}{60 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.98, size = 124, normalized size = 0.99 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{12\,a^3}+\frac {A-3\,B}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B}{4\,a^3}+\frac {A-3\,B}{4\,a^3}-\frac {A+3\,B}{4\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}+\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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