Optimal. Leaf size=63 \[ \frac {\sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{2 b c^2}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{2 b c^2} \]
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Rubi [A] time = 0.12, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4636, 4406, 12, 3303, 3299, 3302} \[ \frac {\sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3299
Rule 3302
Rule 3303
Rule 4406
Rule 4636
Rubi steps
\begin {align*} \int \frac {x}{a+b \cos ^{-1}(c x)} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{c^2}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 (a+b x)} \, dx,x,\cos ^{-1}(c x)\right )}{c^2}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{2 c^2}\\ &=-\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{2 c^2}+\frac {\sin \left (\frac {2 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{2 c^2}\\ &=\frac {\text {Ci}\left (\frac {2 a}{b}+2 \cos ^{-1}(c x)\right ) \sin \left (\frac {2 a}{b}\right )}{2 b c^2}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 56, normalized size = 0.89 \[ -\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \cos ^{-1}(c x)\right )-\sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{b \arccos \left (c x\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 86, normalized size = 1.37 \[ \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{2}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac {\operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{2 \, b c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 58, normalized size = 0.92 \[ \frac {-\frac {\Si \left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{2 b}+\frac {\Ci \left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{2 b}}{c^{2}} \]
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 \[ \int \frac {x}{b \arccos \left (c x\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{a+b\,\mathrm {acos}\left (c\,x\right )} \,d x \]
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 \[ \int \frac {x}{a + b \operatorname {acos}{\left (c x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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