Optimal. Leaf size=181 \[ -i \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )-i \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )-\frac{1}{2} i \sin ^{-1}(a+b x)^2 \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.278964, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4805, 4741, 4521, 2190, 2279, 2391} \[ -i \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )-i \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )-\frac{1}{2} i \sin ^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4805
Rule 4741
Rule 4521
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\sin ^{-1}(a+b x)}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{-\frac{a}{b}+\frac{x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x \cos (x)}{-\frac{a}{b}+\frac{\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{1}{2} i \sin ^{-1}(a+b x)^2+\frac{i \operatorname{Subst}\left (\int \frac{e^{i x} x}{-\frac{i a}{b}-\frac{\sqrt{1-a^2}}{b}+\frac{e^{i x}}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}+\frac{i \operatorname{Subst}\left (\int \frac{e^{i x} x}{-\frac{i a}{b}+\frac{\sqrt{1-a^2}}{b}+\frac{e^{i x}}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{1}{2} i \sin ^{-1}(a+b x)^2+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )-\operatorname{Subst}\left (\int \log \left (1+\frac{e^{i x}}{\left (-\frac{i a}{b}-\frac{\sqrt{1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )-\operatorname{Subst}\left (\int \log \left (1+\frac{e^{i x}}{\left (-\frac{i a}{b}+\frac{\sqrt{1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac{1}{2} i \sin ^{-1}(a+b x)^2+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )+i \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{\left (-\frac{i a}{b}-\frac{\sqrt{1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )+i \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{\left (-\frac{i a}{b}+\frac{\sqrt{1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )\\ &=-\frac{1}{2} i \sin ^{-1}(a+b x)^2+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )-i \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )-i \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )\\ \end{align*}
Mathematica [A] time = 0.0159664, size = 197, normalized size = 1.09 \[ -i \text{PolyLog}\left (2,-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}-i a}\right )-i \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1+\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )+\sin ^{-1}(a+b x) \log \left (1+\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )-\frac{1}{2} i \sin ^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.128, size = 579, normalized size = 3.2 \begin{align*} -{\frac{i}{2}} \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}+{\frac{\arcsin \left ( bx+a \right ){a}^{2}}{{a}^{2}-1}\ln \left ({ \left ( ia+\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }+{\frac{\arcsin \left ( bx+a \right ){a}^{2}}{{a}^{2}-1}\ln \left ({ \left ( ia-\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia-\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }-{\frac{i{a}^{2}}{{a}^{2}-1}{\it dilog} \left ({ \left ( ia-\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia-\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }-{\frac{i{a}^{2}}{{a}^{2}-1}{\it dilog} \left ({ \left ( ia+\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }+{\frac{i}{{a}^{2}-1}{\it dilog} \left ({ \left ( ia-\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia-\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }+{\frac{i}{{a}^{2}-1}{\it dilog} \left ({ \left ( ia+\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }-{\frac{\arcsin \left ( bx+a \right ) }{{a}^{2}-1}\ln \left ({ \left ( ia-\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia-\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }-{\frac{\arcsin \left ( bx+a \right ) }{{a}^{2}-1}\ln \left ({ \left ( ia+\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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{\arcsin \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{asin}{\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{\arcsin \left (b x + a\right )}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]