Optimal. Leaf size=96 \[ -\frac {\csc (c+d x)}{a d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3 d}-\frac {a \sin (c+d x)}{b^2 d}+\frac {\sin ^2(c+d x)}{2 b d} \]
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Rubi [A]
time = 0.10, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 908}
\begin {gather*} \frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3 d}-\frac {b \log (\sin (c+d x))}{a^2 d}-\frac {a \sin (c+d x)}{b^2 d}-\frac {\csc (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 908
Rule 2916
Rubi steps
\begin {align*} \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {b^2 \left (b^2-x^2\right )^2}{x^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{x^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {\text {Subst}\left (\int \left (-a+\frac {b^4}{a x^2}-\frac {b^4}{a^2 x}+x+\frac {\left (a^2-b^2\right )^2}{a^2 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac {\csc (c+d x)}{a d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3 d}-\frac {a \sin (c+d x)}{b^2 d}+\frac {\sin ^2(c+d x)}{2 b d}\\ \end {align*}
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Mathematica [A]
time = 0.13, size = 86, normalized size = 0.90 \begin {gather*} \frac {-\frac {2 \csc (c+d x)}{a}-\frac {2 b \log (\sin (c+d x))}{a^2}+\frac {2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3}-\frac {2 a \sin (c+d x)}{b^2}+\frac {\sin ^2(c+d x)}{b}}{2 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.15, size = 90, normalized size = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{b^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} a^{2}}}{d}\) | \(90\) |
default | \(\frac {-\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{b^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} a^{2}}}{d}\) | \(90\) |
risch | \(-\frac {i x \,a^{2}}{b^{3}}+\frac {2 i x}{b}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}-\frac {2 i a^{2} c}{b^{3} d}+\frac {4 i c}{b d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{b^{3} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a^{2} d}\) | \(276\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 91, normalized size = 0.95 \begin {gather*} -\frac {\frac {2 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} - \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}} + \frac {2}{a \sin \left (d x + c\right )} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b^{3}}}{2 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.43, size = 133, normalized size = 1.39 \begin {gather*} \frac {4 \, a^{3} b \cos \left (d x + c\right )^{2} - 4 \, b^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 4 \, a^{3} b - 4 \, a b^{3} + 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - a^{2} b^{2}\right )} \sin \left (d x + c\right )}{4 \, a^{2} b^{3} d \sin \left (d x + c\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 4.71, size = 105, normalized size = 1.09 \begin {gather*} -\frac {\frac {2 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3}}}{2 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.05, size = 233, normalized size = 2.43 \begin {gather*} \frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,{\left (a^2-b^2\right )}^2}{a^2\,b^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2-2\,b^2\right )}{b^3\,d}-\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+b^2\right )}{b^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^2+b^2\right )}{b^2}-\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{b}+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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