Optimal. Leaf size=126 \[ \frac {(2 a+b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(2 a-b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {a^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) d} \]
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Rubi [A]
time = 0.14, 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 = {2800, 1661,
815} \begin {gather*} \frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac {a^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}+\frac {(2 a+b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(2 a-b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 815
Rule 1661
Rule 2800
Rubi steps
\begin {align*} \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {x^3}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) d}+\frac {\text {Subst}\left (\int \frac {\frac {a b^4}{a^2-b^2}-\frac {b^2 \left (2 a^2-b^2\right ) x}{a^2-b^2}}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) d}+\frac {\text {Subst}\left (\int \left (-\frac {b^2 (2 a+b)}{2 (a+b)^2 (b-x)}-\frac {2 a^3 b^2}{(a-b)^2 (a+b)^2 (a+x)}+\frac {(2 a-b) b^2}{2 (a-b)^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {(2 a+b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(2 a-b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {a^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {\sec ^2(c+d x) (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [A]
time = 0.35, size = 117, normalized size = 0.93 \begin {gather*} \frac {\frac {(2 a+b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac {(2 a-b) \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 a^3 \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}-\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}}{4 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.34, size = 121, normalized size = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(121\) |
default | \(\frac {-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(121\) |
norman | \(\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{3} \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (2 a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {\left (2 a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\) | \(226\) |
risch | \(-\frac {i a x}{a^{2}-2 a b +b^{2}}-\frac {i b x}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {i b x}{2 a^{2}-4 a b +2 b^{2}}-\frac {i a c}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {i a x}{a^{2}+2 a b +b^{2}}+\frac {i b c}{2 \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {2 i a^{3} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {i b c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {i a c}{\left (a^{2}-2 a b +b^{2}\right ) d}+\frac {2 i a^{3} x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {i \left (-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{\left (a^{2}-2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(455\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 142, normalized size = 1.13 \begin {gather*} -\frac {\frac {4 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (2 \, a + b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 157, normalized size = 1.25 \begin {gather*} -\frac {4 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{3} + 2 \, a b^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \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 = 0.48, size = 177, normalized size = 1.40 \begin {gather*} -\frac {\frac {4 \, a^{3} b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {{\left (2 \, a - b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (2 \, a + b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a^{3} \sin \left (d x + c\right )^{2} - a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) - a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 12.29, size = 217, normalized size = 1.72 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (2\,a-b\right )}{2\,d\,{\left (a-b\right )}^2}-\frac {\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2-b^2}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^3\,\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 )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (2\,a+b\right )}{2\,d\,{\left (a+b\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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