Optimal. Leaf size=200 \[ -i \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-i \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.30912, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5258, 4551, 4530, 3719, 2190, 2279, 2391, 4520} \[ -i \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-i \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 5258
Rule 4551
Rule 4530
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rule 4520
Rubi steps
\begin{align*} \int \frac{\sec ^{-1}(a+b x)}{x} \, dx &=\operatorname{Subst}\left (\int \frac{x \sec (x) \tan (x)}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\operatorname{Subst}\left (\int \frac{x \tan (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=a \operatorname{Subst}\left (\int \frac{x \sin (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )+\operatorname{Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\left (2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\right )-(i a) \operatorname{Subst}\left (\int \frac{e^{i x} x}{1-\sqrt{1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )-(i a) \operatorname{Subst}\left (\int \frac{e^{i x} x}{1+\sqrt{1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-\operatorname{Subst}\left (\int \log \left (1-\frac{a e^{i x}}{1-\sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )-\operatorname{Subst}\left (\int \log \left (1-\frac{a e^{i x}}{1+\sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+\operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )+i \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{a x}{1-\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )+i \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{a x}{1+\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )\\ &=\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-i \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-i \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )+\frac{1}{2} i \text{Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )\\ \end{align*}
Mathematica [A] time = 0.318753, size = 284, normalized size = 1.42 \[ -i \left (\text{PolyLog}\left (2,-\frac{\left (\sqrt{1-a^2}-1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+\text{PolyLog}\left (2,\frac{\left (\sqrt{1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )\right )+\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+\log \left (1+\frac{\left (\sqrt{1-a^2}-1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \left (\sec ^{-1}(a+b x)-2 \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right )\right )+\log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \left (\sec ^{-1}(a+b x)+2 \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right )\right )-4 i \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(a+1) \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.404, size = 374, normalized size = 1.9 \begin{align*}{\rm arcsec} \left (bx+a\right )\ln \left ({ \left ( -a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}+1 \right ) \left ( 1+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) +{\rm arcsec} \left (bx+a\right )\ln \left ({ \left ( a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}-1 \right ) \left ( -1+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) -{\rm arcsec} \left (bx+a\right )\ln \left ( 1+i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) -{\rm arcsec} \left (bx+a\right )\ln \left ( 1-i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) -i{\it dilog} \left ({ \left ( -a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}+1 \right ) \left ( 1+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) -i{\it dilog} \left ({ \left ( a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}-1 \right ) \left ( -1+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) +i{\it dilog} \left ( 1+i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) +i{\it dilog} \left ( 1-i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsec}\left (b x + a\right )}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcsec}\left (b x + a\right )}{x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asec}{\left (a + b x \right )}}{x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsec}\left (b x + a\right )}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]