Optimal. Leaf size=84 \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \tanh ^{-1}(\sin (c+d x))}{8 d} \]
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Rubi [A] time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2836, 12, 88, 206} \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \tanh ^{-1}(\sin (c+d x))}{8 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 88
Rule 206
Rule 2836
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {x^2}{a^2 (a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \operatorname {Subst}\left (\int \frac {x^2}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \operatorname {Subst}\left (\int \left (\frac {1}{4 (a-x)^3}-\frac {1}{4 a (a-x)^2}+\frac {1}{8 a (a+x)^2}-\frac {1}{8 a \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=-\frac {a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 74, normalized size = 0.88 \[ \frac {a \tan ^4(c+d x)}{4 d}-\frac {a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {a \tan (c+d x) \sec (c+d x)}{8 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 135, normalized size = 1.61 \[ \frac {2 \, a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a \sin \left (d x + c\right ) + 2 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 91, normalized size = 1.08 \[ -\frac {2 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a \sin \left (d x + c\right ) - a\right )}}{\sin \left (d x + c\right ) + 1} + \frac {3 \, a \sin \left (d x + c\right )^{2} - 14 \, a \sin \left (d x + c\right ) + 7 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 100, normalized size = 1.19 \[ \frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a \sin \left (d x +c \right )}{8 d}-\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 84, normalized size = 1.00 \[ -\frac {a \log \left (\sin \left (d x + c\right ) + 1\right ) - a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (a \sin \left (d x + c\right )^{2} + 3 \, a \sin \left (d x + c\right ) - 2 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 14.46, size = 167, normalized size = 1.99 \[ -\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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