Optimal. Leaf size=113 \[ \frac{i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{c \sec (a+b x)}{b}-\frac{c \tanh ^{-1}(\cos (a+b x))}{b}+\frac{d x \sec (a+b x)}{b}-\frac{2 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]
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Rubi [A] time = 0.13045, antiderivative size = 122, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2622, 321, 207, 4420, 6271, 12, 4183, 2279, 2391, 3770} \[ \frac{i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{(c+d x) \sec (a+b x)}{b}-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}-\frac{2 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{d x \tanh ^{-1}(\cos (a+b x))}{b} \]
Antiderivative was successfully verified.
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Rule 2622
Rule 321
Rule 207
Rule 4420
Rule 6271
Rule 12
Rule 4183
Rule 2279
Rule 2391
Rule 3770
Rubi steps
\begin{align*} \int (c+d x) \csc (a+b x) \sec ^2(a+b x) \, dx &=-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}+\frac{(c+d x) \sec (a+b x)}{b}-d \int \left (-\frac{\tanh ^{-1}(\cos (a+b x))}{b}+\frac{\sec (a+b x)}{b}\right ) \, dx\\ &=-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}+\frac{(c+d x) \sec (a+b x)}{b}+\frac{d \int \tanh ^{-1}(\cos (a+b x)) \, dx}{b}-\frac{d \int \sec (a+b x) \, dx}{b}\\ &=\frac{d x \tanh ^{-1}(\cos (a+b x))}{b}-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{(c+d x) \sec (a+b x)}{b}+\frac{d \int b x \csc (a+b x) \, dx}{b}\\ &=\frac{d x \tanh ^{-1}(\cos (a+b x))}{b}-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{(c+d x) \sec (a+b x)}{b}+d \int x \csc (a+b x) \, dx\\ &=-\frac{2 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{d x \tanh ^{-1}(\cos (a+b x))}{b}-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{(c+d x) \sec (a+b x)}{b}-\frac{d \int \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac{d \int \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac{2 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{d x \tanh ^{-1}(\cos (a+b x))}{b}-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{(c+d x) \sec (a+b x)}{b}+\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}\\ &=-\frac{2 d x \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{d x \tanh ^{-1}(\cos (a+b x))}{b}-\frac{(c+d x) \tanh ^{-1}(\cos (a+b x))}{b}-\frac{d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac{i d \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}+\frac{(c+d x) \sec (a+b x)}{b}\\ \end{align*}
Mathematica [A] time = 0.504331, size = 212, normalized size = 1.88 \[ \frac{d \left (i \left (\text{PolyLog}\left (2,-e^{i (a+b x)}\right )-\text{PolyLog}\left (2,e^{i (a+b x)}\right )\right )+(a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )\right )}{b^2}-\frac{a d \log \left (\tan \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}+\frac{d \log \left (\cos \left (\frac{1}{2} (a+b x)\right )-\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}-\frac{d \log \left (\sin \left (\frac{1}{2} (a+b x)\right )+\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}+\frac{c \sec (a+b x)}{b}+\frac{c \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b}-\frac{c \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b}+\frac{d x \sec (a+b x)}{b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.29, size = 160, normalized size = 1.4 \begin{align*} 2\,{\frac{{{\rm e}^{i \left ( bx+a \right ) }} \left ( dx+c \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }}-{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{b}}+{\frac{2\,id\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{id{\it dilog} \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{id{\it dilog} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{da\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.04559, size = 1085, normalized size = 9.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.581003, size = 1046, normalized size = 9.26 \begin{align*} \frac{2 \, b d x - i \, d \cos \left (b x + a\right ){\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right ){\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right ){\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right ){\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) -{\left (b d x + b c\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) -{\left (b d x + b c\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) +{\left (b c - a d\right )} \cos \left (b x + a\right ) \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2} i \, \sin \left (b x + a\right ) + \frac{1}{2}\right ) +{\left (b c - a d\right )} \cos \left (b x + a\right ) \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) - \frac{1}{2} i \, \sin \left (b x + a\right ) + \frac{1}{2}\right ) +{\left (b d x + a d\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) +{\left (b d x + a d\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - d \cos \left (b x + a\right ) \log \left (\sin \left (b x + a\right ) + 1\right ) + d \cos \left (b x + a\right ) \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, b c}{2 \, b^{2} \cos \left (b x + a\right )} \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 ) \csc{\left (a + b x \right )} \sec ^{2}{\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 )} \csc \left (b x + a\right ) \sec \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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