Optimal. Leaf size=120 \[ \frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d}+\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))} \]
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Rubi [A]
time = 0.08, antiderivative size = 120, 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 {a^2}{24 d (a \sin (c+d x)+a)^3}+\frac {a}{32 d (a-a \sin (c+d x))^2}-\frac {3 a}{32 d (a \sin (c+d x)+a)^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {3}{16 d (a \sin (c+d x)+a)}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d} \end {gather*}
Antiderivative was successfully verified.
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Rule 46
Rule 212
Rule 2746
Rubi steps
\begin {align*} \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {a^5 \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \text {Subst}\left (\int \left (\frac {1}{16 a^4 (a-x)^3}+\frac {1}{8 a^5 (a-x)^2}+\frac {1}{8 a^3 (a+x)^4}+\frac {3}{16 a^4 (a+x)^3}+\frac {3}{16 a^5 (a+x)^2}+\frac {5}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))}+\frac {5 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d}+\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 0.10, size = 97, normalized size = 0.81 \begin {gather*} \frac {\sec ^4(c+d x) \left (-8+25 \sin (c+d x)+25 \sin ^2(c+d x)-15 \sin ^3(c+d x)-15 \sin ^4(c+d x)+15 \tanh ^{-1}(\sin (c+d x)) (-1+\sin (c+d x))^2 (1+\sin (c+d x))^3\right )}{48 a d (1+\sin (c+d x))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.23, size = 91, normalized size = 0.76
method | result | size |
derivativedivides | \(\frac {\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}}{d a}\) | \(91\) |
default | \(\frac {\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}}{d a}\) | \(91\) |
risch | \(-\frac {i \left (30 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+110 i {\mathrm e}^{6 i \left (d x +c \right )}+40 \,{\mathrm e}^{7 i \left (d x +c \right )}-110 i {\mathrm e}^{4 i \left (d x +c \right )}+18 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}+40 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d a}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 a d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{16 a d}\) | \(185\) |
norman | \(\frac {\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {11 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}+\frac {13 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 a d}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 a d}\) | \(239\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 130, normalized size = 1.08 \begin {gather*} -\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 15 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 25 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 147, normalized size = 1.22 \begin {gather*} -\frac {30 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 4}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\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 ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 6.79, size = 116, normalized size = 0.97 \begin {gather*} \frac {\frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {3 \, {\left (15 \, \sin \left (d x + c\right )^{2} - 38 \, \sin \left (d x + c\right ) + 25\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {55 \, \sin \left (d x + c\right )^{3} + 201 \, \sin \left (d x + c\right )^{2} + 255 \, \sin \left (d x + c\right ) + 117}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.14, size = 115, normalized size = 0.96 \begin {gather*} \frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{16\,a\,d}-\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{16}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{16}-\frac {25\,{\sin \left (c+d\,x\right )}^2}{48}-\frac {25\,\sin \left (c+d\,x\right )}{48}+\frac {1}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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