Optimal. Leaf size=216 \[ -\frac {2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {(11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(2 A-6 B+13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.52, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4084, 4019, 3787, 3767, 8, 3768, 3770} \[ -\frac {2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {(11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(2 A-6 B+13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3768
Rule 3770
Rule 3787
Rule 4019
Rule 4084
Rubi steps
\begin {align*} \int \frac {\sec ^4(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 ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^4(c+d x) (a (A+4 B-4 C)+a (2 A-2 B+7 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^3(c+d x) \left (-3 a^2 (A-6 B+11 C)+a^2 (8 A-18 B+43 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sec ^2(c+d x) \left (-2 a^3 (11 A-36 B+76 C)+15 a^3 (2 A-6 B+13 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 A-6 B+13 C) \int \sec ^3(c+d x) \, dx}{a^3}-\frac {(2 (11 A-36 B+76 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=\frac {(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 A-6 B+13 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac {(2 (11 A-36 B+76 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 6.47, size = 1081, normalized size = 5.00 \[ \text {result too large to display} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 328, normalized size = 1.52 \[ \frac {15 \, {\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (11 \, A - 36 \, B + 76 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A - 114 \, B + 239 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (64 \, A - 234 \, B + 479 \, C\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 288, normalized size = 1.33 \[ \frac {\frac {30 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 465 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.79, size = 433, normalized size = 2.00 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{20 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3 d \,a^{3}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {31 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {7 C}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A}{d \,a^{3}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{3}}-\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{2 d \,a^{3}}+\frac {C}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7 C}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A}{d \,a^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{3}}+\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{2 d \,a^{3}}-\frac {C}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 493, normalized size = 2.28 \[ -\frac {C {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \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 {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, B {\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 )} + A {\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} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.40, size = 227, normalized size = 1.05 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-3\,B+\frac {13\,C}{2}\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-3\,B+5\,C\right )}{4\,a^3}-\frac {2\,A+2\,B-10\,C}{4\,a^3}+\frac {3\,\left (A-B+C\right )}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-7\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-5\,C\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-3\,B+5\,C}{12\,a^3}+\frac {A-B+C}{4\,a^3}\right )}{d} \]
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 \[ \frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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