Optimal. Leaf size=126 \[ \frac {5 \tanh ^{-1}(\sin (c+d x))}{32 a^3 d}-\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{12 d (a+a \sin (c+d x))^3}-\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )} \]
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Rubi [A]
time = 0.07, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2746, 46, 212}
\begin {gather*} \frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{32 a^3 d}-\frac {a}{16 d (a \sin (c+d x)+a)^4}-\frac {1}{12 d (a \sin (c+d x)+a)^3}-\frac {3}{32 a d (a \sin (c+d x)+a)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 46
Rule 212
Rule 2746
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {a^3 \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \text {Subst}\left (\int \left (\frac {1}{32 a^5 (a-x)^2}+\frac {1}{4 a^2 (a+x)^5}+\frac {1}{4 a^3 (a+x)^4}+\frac {3}{16 a^4 (a+x)^3}+\frac {1}{8 a^5 (a+x)^2}+\frac {5}{32 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{12 d (a+a \sin (c+d x))^3}-\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {5 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{32 a^2 d}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{32 a^3 d}-\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{12 d (a+a \sin (c+d x))^3}-\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.11, size = 95, normalized size = 0.75 \begin {gather*} -\frac {\sec ^2(c+d x) \left (32+15 \sin (c+d x)-35 \sin ^2(c+d x)-45 \sin ^3(c+d x)-15 \sin ^4(c+d x)+15 \tanh ^{-1}(\sin (c+d x)) (-1+\sin (c+d x)) (1+\sin (c+d x))^4\right )}{96 a^3 d (1+\sin (c+d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.32, size = 91, normalized size = 0.72
method | result | size |
derivativedivides | \(\frac {-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{64}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{64}}{d \,a^{3}}\) | \(91\) |
default | \(\frac {-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{64}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{64}}{d \,a^{3}}\) | \(91\) |
risch | \(-\frac {i \left (90 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}-150 i {\mathrm e}^{6 i \left (d x +c \right )}-200 \,{\mathrm e}^{7 i \left (d x +c \right )}+150 i {\mathrm e}^{4 i \left (d x +c \right )}-142 \,{\mathrm e}^{5 i \left (d x +c \right )}-90 i {\mathrm e}^{2 i \left (d x +c \right )}-200 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{48 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d \,a^{3}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{32 a^{3} d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{32 a^{3} d}\) | \(185\) |
norman | \(\frac {\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {27 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {31 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {31 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {33 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {33 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {9 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {9 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {89 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32 a^{3} d}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32 a^{3} d}\) | \(242\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 146, normalized size = 1.16 \begin {gather*} -\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 45 \, \sin \left (d x + c\right )^{3} + 35 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) - 32\right )}}{a^{3} \sin \left (d x + c\right )^{5} + 3 \, a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - a^{3}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{192 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 228, normalized size = 1.81 \begin {gather*} -\frac {30 \, \cos \left (d x + c\right )^{4} - 130 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 30 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 36}{192 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2}\right )} \sin \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 {\sec ^{3}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\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 = 8.38, size = 116, normalized size = 0.92 \begin {gather*} \frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac {12 \, {\left (5 \, \sin \left (d x + c\right ) - 7\right )}}{a^{3} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 596 \, \sin \left (d x + c\right )^{3} + 1110 \, \sin \left (d x + c\right )^{2} + 996 \, \sin \left (d x + c\right ) + 405}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{768 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.15, size = 129, normalized size = 1.02 \begin {gather*} \frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{32\,a^3\,d}+\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{32}+\frac {15\,{\sin \left (c+d\,x\right )}^3}{32}+\frac {35\,{\sin \left (c+d\,x\right )}^2}{96}-\frac {5\,\sin \left (c+d\,x\right )}{32}-\frac {1}{3}}{d\,\left (-a^3\,{\sin \left (c+d\,x\right )}^5-3\,a^3\,{\sin \left (c+d\,x\right )}^4-2\,a^3\,{\sin \left (c+d\,x\right )}^3+2\,a^3\,{\sin \left (c+d\,x\right )}^2+3\,a^3\,\sin \left (c+d\,x\right )+a^3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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