Optimal. Leaf size=132 \[ \frac {C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A+4 B-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Rubi [A]
time = 0.24, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4169, 4093,
4083, 3855, 3879} \begin {gather*} \frac {(6 A+4 B-29 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(3 A+2 B-7 C) \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 4169
Rubi steps
\begin {align*} \int \frac {\sec ^2(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 ^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) (a (3 A+2 B-2 C)+5 a C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A+2 B-7 C) \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 (3 A+2 B-7 C)-15 a^2 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A+4 B-29 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}+\frac {C \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A+2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A+4 B-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(132)=264\).
time = 1.73, size = 277, normalized size = 2.10 \begin {gather*} -\frac {\left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \left (240 C \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 (3 A+4 B-29 C) \sin \left (\frac {d x}{2}\right )-15 (A-5 C) \sin \left (c+\frac {d x}{2}\right )+15 A \sin \left (c+\frac {3 d x}{2}\right )+10 B \sin \left (c+\frac {3 d x}{2}\right )-95 C \sin \left (c+\frac {3 d x}{2}\right )+15 C \sin \left (2 c+\frac {3 d x}{2}\right )+3 A \sin \left (2 c+\frac {5 d x}{2}\right )+2 B \sin \left (2 c+\frac {5 d x}{2}\right )-22 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )\right )}{15 a^3 d (1+\cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.58, size = 144, normalized size = 1.09
method | result | size |
derivativedivides | \(\frac {-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {4 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(144\) |
default | \(\frac {-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {4 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(144\) |
risch | \(-\frac {2 i \left (15 C \,{\mathrm e}^{4 i \left (d x +c \right )}-15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+75 C \,{\mathrm e}^{3 i \left (d x +c \right )}-15 A \,{\mathrm e}^{2 i \left (d x +c \right )}-20 B \,{\mathrm e}^{2 i \left (d x +c \right )}+145 C \,{\mathrm e}^{2 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )} A -10 B \,{\mathrm e}^{i \left (d x +c \right )}+95 C \,{\mathrm e}^{i \left (d x +c \right )}-3 A -2 B +22 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(185\) |
norman | \(\frac {\frac {\left (A -B -9 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}-\frac {\left (A -B +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (A +B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (7 A +3 B -43 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}+\frac {\left (9 A +B -11 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (-59 C +9 A +7 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(223\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 232, normalized size = 1.76 \begin {gather*} -\frac {C {\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 {B {\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}} - \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\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 = 1.99, size = 193, normalized size = 1.46 \begin {gather*} \frac {15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (3 \, A + 2 \, B - 22 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 2 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 7 \, B - 32 \, C\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 ^{2}{\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 ^{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 {C \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 = 0.52, size = 180, normalized size = 1.36 \begin {gather*} \frac {\frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, C \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} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, 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 ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C 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 = 3.30, size = 134, normalized size = 1.02 \begin {gather*} \frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{12\,a^3}-\frac {A+B-3\,C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-A+3\,C}{4\,a^3}+\frac {A-B+C}{4\,a^3}-\frac {A+B-3\,C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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