Optimal. Leaf size=88 \[ \frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-3 b) \log (\sin (c+d x)+1)}{4 d}+\frac {\tan ^2(c+d x) (a+b \sin (c+d x))}{2 d}+\frac {3 b \sin (c+d x)}{2 d} \]
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Rubi [A] time = 0.08, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2721, 819, 774, 633, 31} \[ \frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-3 b) \log (\sin (c+d x)+1)}{4 d}+\frac {\tan ^2(c+d x) (a+b \sin (c+d x))}{2 d}+\frac {3 b \sin (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 31
Rule 633
Rule 774
Rule 819
Rule 2721
Rubi steps
\begin {align*} \int (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx &=\frac {\operatorname {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 {(a+b \sin (c+d x)) \tan ^2(c+d x)}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {x \left (2 a b^2+3 b^2 x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {3 b \sin (c+d x)}{2 d}+\frac {(a+b \sin (c+d x)) \tan ^2(c+d x)}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {-3 b^4-2 a b^2 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {3 b \sin (c+d x)}{2 d}+\frac {(a+b \sin (c+d x)) \tan ^2(c+d x)}{2 d}-\frac {(2 a-3 b) \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}-\frac {(2 a+3 b) \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-3 b) \log (1+\sin (c+d x))}{4 d}+\frac {3 b \sin (c+d x)}{2 d}+\frac {(a+b \sin (c+d x)) \tan ^2(c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 77, normalized size = 0.88 \[ \frac {a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d}-\frac {b \sin (c+d x) \tan ^2(c+d x)}{d}-\frac {3 b \left (\tanh ^{-1}(\sin (c+d x))-\tan (c+d x) \sec (c+d x)\right )}{2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 90, normalized size = 1.02 \[ \frac {{\left (2 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right ) + 2 \, a}{4 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 96, normalized size = 1.09 \[ \frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b \left (\sin ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 b \sin \left (d x +c \right )}{2 d}-\frac {3 b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.62, size = 73, normalized size = 0.83 \[ \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 4 \, b \sin \left (d x + c\right ) - \frac {2 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.73, size = 176, normalized size = 2.00 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+\frac {3\,b}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-\frac {3\,b}{2}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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 \[ \int \left (a + b \sin {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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