Optimal. Leaf size=117 \[ -\frac{i d \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac{i d \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac{d \sec (a+b x)}{2 b^2}+\frac{i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]
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Rubi [A] time = 0.12925, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4413, 4181, 2279, 2391, 4185} \[ -\frac{i d \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac{i d \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac{d \sec (a+b x)}{2 b^2}+\frac{i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 4413
Rule 4181
Rule 2279
Rule 2391
Rule 4185
Rubi steps
\begin{align*} \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx &=-\int (c+d x) \sec (a+b x) \, dx+\int (c+d x) \sec ^3(a+b x) \, dx\\ &=\frac{2 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{1}{2} \int (c+d x) \sec (a+b x) \, dx+\frac{d \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b}-\frac{d \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{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 )}{b^2}+\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}-\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{i d \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}+\frac{i d \text{Li}_2\left (i e^{i (a+b x)}\right )}{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}+\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*}
Mathematica [B] time = 6.5224, size = 555, normalized size = 4.74 \[ \frac{d x \left (-i \text{PolyLog}\left (2,\frac{1}{2} \left ((1+i)-(1-i) \tan \left (\frac{1}{2} (a+b x)\right )\right )\right )+i \text{PolyLog}\left (2,\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )+i\right )\right )-i \text{PolyLog}\left (2,\frac{1}{2} \left ((1-i) \tan \left (\frac{1}{2} (a+b x)\right )+(1+i)\right )\right )+i \text{PolyLog}\left (2,\frac{1}{2} \left ((1+i) \tan \left (\frac{1}{2} (a+b x)\right )+(1-i)\right )\right )+a \log \left (1-\tan \left (\frac{1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )-1\right )\right )-i \log \left (1-i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac{1}{2}+\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )-1\right )\right )-i \log \left (1+i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )+1\right )\right )+i \log \left (1-i \tan \left (\frac{1}{2} (a+b x)\right )\right ) \log \left (\left (\frac{1}{2}+\frac{i}{2}\right ) \left (\tan \left (\frac{1}{2} (a+b x)\right )+1\right )\right )-a \log \left (\tan \left (\frac{1}{2} (a+b x)\right )+1\right )\right )}{2 b \left (-i \log \left (1-i \tan \left (\frac{1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac{1}{2} (a+b x)\right )\right )+a\right )}-\frac{d \sin \left (\frac{1}{2} (a+b x)\right )}{2 b^2 \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )}+\frac{d \sin \left (\frac{1}{2} (a+b x)\right )}{2 b^2 \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\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{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )^2}-\frac{d x}{4 b \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\right )\right )^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.162, size = 267, normalized size = 2.3 \begin{align*}{\frac{-i \left ( dxb{{\rm e}^{3\,i \left ( bx+a \right ) }}-id{{\rm e}^{3\,i \left ( bx+a \right ) }}+bc{{\rm e}^{3\,i \left ( bx+a \right ) }}-dxb{{\rm e}^{i \left ( bx+a \right ) }}-id{{\rm e}^{i \left ( bx+a \right ) }}-bc{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) ^{2}}}+{\frac{ic\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{2\,b}}+{\frac{d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{2\,{b}^{2}}}-{\frac{d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{2\,b}}-{\frac{d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{2\,{b}^{2}}}-{\frac{{\frac{i}{2}}d{\it dilog} \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{{\frac{i}{2}}d{\it dilog} \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{ida\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.624285, size = 1169, normalized size = 9.99 \begin{align*} \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right ) \tan ^{2}{\left (a + b x \right )} \sec{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} \sec \left (b x + a\right ) \tan \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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