Optimal. Leaf size=68 \[ -\frac{i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a+b x)}\right )}{2 d}-\frac{i \cos ^{-1}(a+b x)^2}{2 d}+\frac{\cos ^{-1}(a+b x) \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )}{d} \]
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Rubi [A] time = 0.0813552, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {4806, 12, 4626, 3719, 2190, 2279, 2391} \[ -\frac{i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a+b x)}\right )}{2 d}-\frac{i \cos ^{-1}(a+b x)^2}{2 d}+\frac{\cos ^{-1}(a+b x) \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 4806
Rule 12
Rule 4626
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\cos ^{-1}(a+b x)}{\frac{a d}{b}+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b \cos ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cos ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}(a+b x)\right )}{d}\\ &=-\frac{i \cos ^{-1}(a+b x)^2}{2 d}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}(a+b x)\right )}{d}\\ &=-\frac{i \cos ^{-1}(a+b x)^2}{2 d}+\frac{\cos ^{-1}(a+b x) \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )}{d}-\frac{\operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}(a+b x)\right )}{d}\\ &=-\frac{i \cos ^{-1}(a+b x)^2}{2 d}+\frac{\cos ^{-1}(a+b x) \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )}{d}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}(a+b x)}\right )}{2 d}\\ &=-\frac{i \cos ^{-1}(a+b x)^2}{2 d}+\frac{\cos ^{-1}(a+b x) \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )}{d}-\frac{i \text{Li}_2\left (-e^{2 i \cos ^{-1}(a+b x)}\right )}{2 d}\\ \end{align*}
Mathematica [A] time = 0.0414374, size = 59, normalized size = 0.87 \[ -\frac{i \left (\text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a+b x)}\right )+\cos ^{-1}(a+b x) \left (\cos ^{-1}(a+b x)+2 i \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )\right )\right )}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.089, size = 85, normalized size = 1.3 \begin{align*}{\frac{-{\frac{i}{2}} \left ( \arccos \left ( bx+a \right ) \right ) ^{2}}{d}}+{\frac{\arccos \left ( bx+a \right ) }{d}\ln \left ( 1+ \left ( bx+a+i\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) ^{2} \right ) }-{\frac{{\frac{i}{2}}}{d}{\it polylog} \left ( 2,- \left ( bx+a+i\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) ^{2} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arccos \left (b x + a\right )}{b d x + a 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*} \frac{b \int \frac{\operatorname{acos}{\left (a + b x \right )}}{a + b x}\, dx}{d} \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{\arccos \left (b x + a\right )}{d x + \frac{a d}{b}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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