Optimal. Leaf size=137 \[ -\frac {\sqrt {a+b \text {ArcCos}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \text {ArcCos}(c x)}-\frac {\sqrt {b} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{8 c^2}-\frac {\sqrt {b} \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{8 c^2} \]
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Rubi [A]
time = 0.24, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4726, 4810,
3393, 3387, 3386, 3432, 3385, 3433} \begin {gather*} -\frac {\sqrt {\pi } \sqrt {b} \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi } \sqrt {b}}\right )}{8 c^2}-\frac {\sqrt {\pi } \sqrt {b} \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{8 c^2}-\frac {\sqrt {a+b \text {ArcCos}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \text {ArcCos}(c x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 3385
Rule 3386
Rule 3387
Rule 3393
Rule 3432
Rule 3433
Rule 4726
Rule 4810
Rubi steps
\begin {align*} \int x \sqrt {a+b \cos ^{-1}(c x)} \, dx &=\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)}+\frac {1}{4} (b c) \int \frac {x^2}{\sqrt {1-c^2 x^2} \sqrt {a+b \cos ^{-1}(c x)}} \, dx\\ &=\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {b \text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{4 c^2}\\ &=\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {b \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {a+b x}}+\frac {\cos (2 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{4 c^2}\\ &=-\frac {\sqrt {a+b \cos ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {b \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{8 c^2}\\ &=-\frac {\sqrt {a+b \cos ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {\left (b \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{8 c^2}-\frac {\left (b \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{8 c^2}\\ &=-\frac {\sqrt {a+b \cos ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \cos \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{4 c^2}-\frac {\sin \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \sin \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{4 c^2}\\ &=-\frac {\sqrt {a+b \cos ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {\sqrt {b} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{8 c^2}-\frac {\sqrt {b} \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{8 c^2}\\ \end {align*}
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Mathematica [A]
time = 0.15, size = 123, normalized size = 0.90 \begin {gather*} -\frac {-2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)} \cos (2 \text {ArcCos}(c x))+\sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right )+\sqrt {\pi } S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{8 \sqrt {\frac {1}{b}} c^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.31, size = 192, normalized size = 1.40
method | result | size |
default | \(\frac {-\sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b +\sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b +4 \arccos \left (c x \right ) \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b +4 \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a}{16 c^{2} \sqrt {a +b \arccos \left (c x \right )}}\) | \(192\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
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 \begin {gather*} \int x \sqrt {a + b \operatorname {acos}{\left (c x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [C] Result contains complex when optimal does not.
time = 0.74, size = 448, normalized size = 3.27 \begin {gather*} -\frac {i \, \sqrt {\pi } a \sqrt {b} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {2 i \, a}{b}\right )}}{4 \, {\left (b + \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} + \frac {\sqrt {\pi } b^{\frac {3}{2}} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {2 i \, a}{b}\right )}}{16 \, {\left (b + \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} + \frac {i \, \sqrt {\pi } a \sqrt {b} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {2 i \, a}{b}\right )}}{4 \, {\left (b - \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} + \frac {\sqrt {\pi } b^{\frac {3}{2}} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {2 i \, a}{b}\right )}}{16 \, {\left (b - \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} - \frac {i \, \sqrt {\pi } a \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {2 i \, a}{b}\right )}}{4 \, c^{2} {\left (\sqrt {b} - \frac {i \, b^{\frac {3}{2}}}{{\left | b \right |}}\right )}} + \frac {i \, \sqrt {\pi } a \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {2 i \, a}{b}\right )}}{4 \, \sqrt {b} c^{2} {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}} + \frac {\sqrt {b \arccos \left (c x\right ) + a} e^{\left (2 i \, \arccos \left (c x\right )\right )}}{8 \, c^{2}} + \frac {\sqrt {b \arccos \left (c x\right ) + a} e^{\left (-2 i \, \arccos \left (c x\right )\right )}}{8 \, c^{2}} \end {gather*}
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 \begin {gather*} \int x\,\sqrt {a+b\,\mathrm {acos}\left (c\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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