Optimal. Leaf size=145 \[ -\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Rubi [A]
time = 0.29, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4170, 4104,
4093, 3872, 3855, 3852, 8} \begin {gather*} \frac {(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {3 C \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(A-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3852
Rule 3855
Rule 3872
Rule 4093
Rule 4104
Rule 4170
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A+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 C)-a (A+6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-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-9 C)-a^2 (2 A+27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sec (c+d x) \left (-45 a^3 C+a^3 (2 A+27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(3 C) \int \sec (c+d x) \, dx}{a^3}+\frac {(2 A+27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=-\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(2 A+27 C) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{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. \(457\) vs. \(2(145)=290\).
time = 3.16, size = 457, normalized size = 3.15 \begin {gather*} \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (2880 C \cos ^5\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 )+\sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-5 (4 A+51 C) \sin \left (\frac {d x}{2}\right )+(22 A+567 C) \sin \left (\frac {3 d x}{2}\right )-10 A \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 )+375 C \sin \left (c+\frac {d x}{2}\right )-20 A \sin \left (2 c+\frac {d x}{2}\right )-480 C \sin \left (2 c+\frac {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 )+402 C \sin \left (2 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 )+315 C \sin \left (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 )+240 C \sin \left (3 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 )+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 )+57 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )\right )}{60 a^3 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.38, size = 151, normalized size = 1.04
method | result | size |
derivativedivides | \(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(151\) |
default | \(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(151\) |
risch | \(\frac {2 i \left (45 C \,{\mathrm e}^{6 i \left (d x +c \right )}+225 C \,{\mathrm e}^{5 i \left (d x +c \right )}+20 A \,{\mathrm e}^{4 i \left (d x +c \right )}+480 C \,{\mathrm e}^{4 i \left (d x +c \right )}+10 A \,{\mathrm e}^{3 i \left (d x +c \right )}+600 C \,{\mathrm e}^{3 i \left (d x +c \right )}+22 A \,{\mathrm e}^{2 i \left (d x +c \right )}+567 C \,{\mathrm e}^{2 i \left (d x +c \right )}+10 \,{\mathrm e}^{i \left (d x +c \right )} A +315 C \,{\mathrm e}^{i \left (d x +c \right )}+2 A +72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(208\) |
norman | \(\frac {-\frac {\left (A -9 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}+\frac {\left (A +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (A +81 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {\left (7 A -153 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (25 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {5 \left (27 C +A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (1773 C +53 A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} a^{2}}+\frac {3 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}-\frac {3 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\) | \(232\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 233, normalized size = 1.61 \begin {gather*} \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 )} + \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 = 3.30, size = 222, normalized size = 1.53 \begin {gather*} -\frac {45 \, {\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (A + 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + 57 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A + 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 {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 {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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.51, size = 178, normalized size = 1.23 \begin {gather*} -\frac {\frac {180 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {180 \, C \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 \, 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} + 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 ) + 255 \, 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 = 2.64, size = 150, normalized size = 1.03 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{6\,a^3}+\frac {C}{3\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{4\,a^3}+\frac {2\,C}{a^3}-\frac {2\,A-6\,C}{4\,a^3}\right )}{d}-\frac {6\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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