Optimal. Leaf size=247 \[ -\frac{i b \text{PolyLog}\left (2,-\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{i b \text{PolyLog}\left (2,-\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e} \]
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Rubi [A] time = 0.37276, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5224, 2518} \[ -\frac{i b \text{PolyLog}\left (2,-\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{i b \text{PolyLog}\left (2,-\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e} \]
Antiderivative was successfully verified.
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Rule 5224
Rule 2518
Rubi steps
\begin{align*} \int \frac{a+b \sec ^{-1}(c x)}{d+e x} \, dx &=\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e-\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e+\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e}-\frac{b \int \frac{\log \left (1+\frac{\left (e-\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{\sqrt{1-\frac{1}{c^2 x^2}} x^2} \, dx}{c e}-\frac{b \int \frac{\log \left (1+\frac{\left (e+\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{\sqrt{1-\frac{1}{c^2 x^2}} x^2} \, dx}{c e}+\frac{b \int \frac{\log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{\sqrt{1-\frac{1}{c^2 x^2}} x^2} \, dx}{c e}\\ &=\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e-\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e+\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e}-\frac{i b \text{Li}_2\left (-\frac{\left (e-\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{i b \text{Li}_2\left (-\frac{\left (e+\sqrt{-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{i b \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e}\\ \end{align*}
Mathematica [A] time = 0.587821, size = 333, normalized size = 1.35 \[ \frac{a \log (d+e x)}{e}+\frac{b \left (-i \left (\text{PolyLog}\left (2,\frac{\left (\sqrt{e^2-c^2 d^2}-e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )+\text{PolyLog}\left (2,-\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )\right )+\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+\log \left (1+\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right ) \left (2 \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right )+\sec ^{-1}(c x)\right )+\log \left (1+\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right ) \left (\sec ^{-1}(c x)-2 \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right )\right )+4 i \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(e-c d) \tan \left (\frac{1}{2} \sec ^{-1}(c x)\right )}{\sqrt{e^2-c^2 d^2}}\right )-\sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )}{e} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.374, size = 456, normalized size = 1.9 \begin{align*}{\frac{a\ln \left ( cex+dc \right ) }{e}}+{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ({ \left ( dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}+e \right ) \left ( e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }+{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ({ \left ( -dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}-e \right ) \left ( -e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{ib}{e}{\it dilog} \left ({ \left ( dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}+e \right ) \left ( e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{ib}{e}{\it dilog} \left ({ \left ( -dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}-e \right ) \left ( -e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ( 1+i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }-{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ( 1-i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }+{\frac{ib}{e}{\it dilog} \left ( 1+i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }+{\frac{ib}{e}{\it dilog} \left ( 1-i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )}{e x + d}\,{d x} + \frac{a \log \left (e x + d\right )}{e} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arcsec}\left (c x\right ) + a}{e x + d}, x\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 \frac{a + b \operatorname{asec}{\left (c x \right )}}{d + e x}\, 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 \frac{b \operatorname{arcsec}\left (c x\right ) + a}{e x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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