Optimal. Leaf size=177 \[ -i \text {Li}_2\left (\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-\frac {1}{2} i \cos ^{-1}(a+b x)^2 \]
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Rubi [A] time = 0.28, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4806, 4742, 4522, 2190, 2279, 2391} \[ -i \text {PolyLog}\left (2,\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )-i \text {PolyLog}\left (2,\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-\frac {1}{2} i \cos ^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 4522
Rule 4742
Rule 4806
Rubi steps
\begin {align*} \int \frac {\cos ^{-1}(a+b x)}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cos ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x \sin (x)}{-\frac {a}{b}+\frac {\cos (x)}{b}} \, dx,x,\cos ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int \frac {e^{i x} x}{-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}+\frac {i e^{i x}}{b}} \, dx,x,\cos ^{-1}(a+b x)\right )}{b}-\frac {\operatorname {Subst}\left (\int \frac {e^{i x} x}{-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}+\frac {i e^{i x}}{b}} \, dx,x,\cos ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-\operatorname {Subst}\left (\int \log \left (1+\frac {i e^{i x}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\cos ^{-1}(a+b x)\right )-\operatorname {Subst}\left (\int \log \left (1+\frac {i e^{i x}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\cos ^{-1}(a+b x)\right )\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )+i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i x}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \cos ^{-1}(a+b x)}\right )+i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i x}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \cos ^{-1}(a+b x)}\right )\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )\\ \end {align*}
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Mathematica [A] time = 0.21, size = 228, normalized size = 1.29 \[ -i \left (\text {Li}_2\left (\left (a-\sqrt {a^2-1}\right ) e^{i \cos ^{-1}(a+b x)}\right )+\text {Li}_2\left (\left (a+\sqrt {a^2-1}\right ) e^{i \cos ^{-1}(a+b x)}\right )\right )+\log \left (1+\left (\sqrt {a^2-1}-a\right ) e^{i \cos ^{-1}(a+b x)}\right ) \left (\cos ^{-1}(a+b x)-2 \sin ^{-1}\left (\frac {\sqrt {1-a}}{\sqrt {2}}\right )\right )+\log \left (1-\left (\sqrt {a^2-1}+a\right ) e^{i \cos ^{-1}(a+b x)}\right ) \left (\cos ^{-1}(a+b x)+2 \sin ^{-1}\left (\frac {\sqrt {1-a}}{\sqrt {2}}\right )\right )-4 i \sin ^{-1}\left (\frac {\sqrt {1-a}}{\sqrt {2}}\right ) \tan ^{-1}\left (\frac {(a+1) \tan \left (\frac {1}{2} \cos ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )-\frac {1}{2} i \cos ^{-1}(a+b x)^2 \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arccos \left (b x + a\right )}{x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arccos \left (b x + a\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.22, size = 199, normalized size = 1.12 \[ -\frac {i \arccos \left (b x +a \right )^{2}}{2}+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )-i \dilog \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )-i \dilog \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arccos \left (b x + a\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (a+b\,x\right )}{x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acos}{\left (a + b x \right )}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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