Optimal. Leaf size=156 \[ \frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Rubi [A]
time = 0.34, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4104,
4093, 3872, 3855, 3852, 8} \begin {gather*} -\frac {(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 {(B-C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(4 B-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 4157
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=\frac {(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) (3 a (B-C)-a (B-6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(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 (4 B-9 C)-a^2 (7 B-27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(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 (7 B-27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(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 {(7 B-27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {(B-3 C) \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(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 {(7 B-27 C) \text {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 {(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(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*}
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Mathematica [A]
time = 1.77, size = 294, normalized size = 1.88 \begin {gather*} \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (-15 (B-3 C) \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 )-4 (7 B-12 C) \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+96 (B-C) \csc ^7(c+d x) \sin ^{10}\left (\frac {1}{2} (c+d x)\right )+(-22 B+87 C) \tan \left (\frac {1}{2} (c+d x)\right )-\frac {1}{4} (-4 B+9 C+(7 B-12 C) \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )+15 (B-3 C) \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 ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+(22 B-57 C) \tan ^3\left (\frac {1}{2} (c+d x)\right )\right )}{15 a^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.57, size = 162, normalized size = 1.04
method | result | size |
derivativedivides | \(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {4 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (12 C -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 C +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(162\) |
default | \(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {4 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (12 C -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 C +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(162\) |
norman | \(\frac {-\frac {\left (B -C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (4 B -9 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}-\frac {\left (7 B -25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {5 \left (8 B -27 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (26 B -81 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {\left (43 B -153 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (553 B -1773 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} a^{2}}+\frac {\left (B -3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {\left (B -3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(249\) |
risch | \(-\frac {2 i \left (15 B \,{\mathrm e}^{6 i \left (d x +c \right )}-45 C \,{\mathrm e}^{6 i \left (d x +c \right )}+75 B \,{\mathrm e}^{5 i \left (d x +c \right )}-225 C \,{\mathrm e}^{5 i \left (d x +c \right )}+160 B \,{\mathrm e}^{4 i \left (d x +c \right )}-480 C \,{\mathrm e}^{4 i \left (d x +c \right )}+170 B \,{\mathrm e}^{3 i \left (d x +c \right )}-600 C \,{\mathrm e}^{3 i \left (d x +c \right )}+167 B \,{\mathrm e}^{2 i \left (d x +c \right )}-567 C \,{\mathrm e}^{2 i \left (d x +c \right )}+95 B \,{\mathrm e}^{i \left (d x +c \right )}-315 C \,{\mathrm e}^{i \left (d x +c \right )}+22 B -72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}\) | \(275\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 286, normalized size = 1.83 \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 )} - 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 )}}{60 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.37, size = 256, normalized size = 1.64 \begin {gather*} \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 (11 \, B - 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (17 \, B - 57 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (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 {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 {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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.52, size = 186, normalized size = 1.19 \begin {gather*} \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 \, 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 \, 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} + 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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.86, size = 168, normalized size = 1.08 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-3\,C\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{4\,a^3}-\frac {3\,C}{2\,a^3}+\frac {2\,B-4\,C}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{6\,a^3}+\frac {2\,B-4\,C}{12\,a^3}\right )}{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 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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