Optimal. Leaf size=117 \[ \frac {i d \text {Li}_2\left (-i e^{i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}-\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]
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Rubi [A] time = 0.07, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4185, 4181, 2279, 2391} \[ \frac {i d \text {Li}_2\left (-i e^{i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}-\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 2279
Rule 2391
Rule 4181
Rule 4185
Rubi steps
\begin {align*} \int (c+d x) \sec ^3(a+b x) \, dx &=-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}+\frac {1}{2} \int (c+d x) \sec (a+b x) \, dx\\ &=-\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}-\frac {d \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{2 b}+\frac {d \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{2 b}\\ &=-\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}+\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}\\ &=-\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {i d \text {Li}_2\left (-i e^{i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}\\ \end {align*}
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Mathematica [B] time = 3.68, size = 389, normalized size = 3.32 \[ \frac {d \left (i \left (\text {Li}_2\left (-e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )-\text {Li}_2\left (e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )\right )+\left (-a-b x+\frac {\pi }{2}\right ) \left (\log \left (1-e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )-\log \left (1+e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )\right )-\left (\frac {\pi }{2}-a\right ) \log \left (\tan \left (\frac {1}{2} \left (-a-b x+\frac {\pi }{2}\right )\right )\right )\right )}{2 b^2}-\frac {d \sin \left (\frac {b x}{2}\right )}{2 b^2 \left (\cos \left (\frac {a}{2}\right )-\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )-\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}+\frac {d \sin \left (\frac {b x}{2}\right )}{2 b^2 \left (\sin \left (\frac {a}{2}\right )+\cos \left (\frac {a}{2}\right )\right ) \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right )+\cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}+\frac {c \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac {c \tan (a+b x) \sec (a+b x)}{2 b}+\frac {d x}{4 b \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )-\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )^2}-\frac {d x}{4 b \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right )+\cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.96, size = 435, normalized size = 3.72 \[ \frac {-i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 2 \, d \cos \left (b x + a\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (b x + a\right )}{4 \, b^{2} \cos \left (b x + a\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \sec \left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.12, size = 267, normalized size = 2.28 \[ -\frac {i \left (d x b \,{\mathrm e}^{3 i \left (b x +a \right )}+c b \,{\mathrm e}^{3 i \left (b x +a \right )}-d x b \,{\mathrm e}^{i \left (b x +a \right )}-c b \,{\mathrm e}^{i \left (b x +a \right )}-i d \,{\mathrm e}^{3 i \left (b x +a \right )}-i d \,{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {i c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{2 b}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{2 b^{2}}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{2 b}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{2 b^{2}}+\frac {i d \dilog \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {i d \dilog \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {i d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]
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 \[ \int \left (c + d x\right ) \sec ^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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