Optimal. Leaf size=119 \[ \frac {15 \sqrt {\pi } C\left (\frac {2 \sqrt {\cos ^{-1}(a x)}}{\sqrt {\pi }}\right )}{128 a^2}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {15 \sqrt {\cos ^{-1}(a x)}}{64 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)} \]
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Rubi [A] time = 0.29, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4630, 4708, 4642, 4724, 3312, 3304, 3352} \[ \frac {15 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\cos ^{-1}(a x)}}{\sqrt {\pi }}\right )}{128 a^2}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {15 \sqrt {\cos ^{-1}(a x)}}{64 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)} \]
Antiderivative was successfully verified.
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Rule 3304
Rule 3312
Rule 3352
Rule 4630
Rule 4642
Rule 4708
Rule 4724
Rubi steps
\begin {align*} \int x \cos ^{-1}(a x)^{5/2} \, dx &=\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac {1}{4} (5 a) \int \frac {x^2 \cos ^{-1}(a x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac {15}{16} \int x \sqrt {\cos ^{-1}(a x)} \, dx+\frac {5 \int \frac {\cos ^{-1}(a x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx}{8 a}\\ &=-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac {1}{64} (15 a) \int \frac {x^2}{\sqrt {1-a^2 x^2} \sqrt {\cos ^{-1}(a x)}} \, dx\\ &=-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac {15 \operatorname {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{64 a^2}\\ &=-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac {15 \operatorname {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{64 a^2}\\ &=\frac {15 \sqrt {\cos ^{-1}(a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac {15 \operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{128 a^2}\\ &=\frac {15 \sqrt {\cos ^{-1}(a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac {15 \operatorname {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{64 a^2}\\ &=\frac {15 \sqrt {\cos ^{-1}(a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {5 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac {\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac {15 \sqrt {\pi } C\left (\frac {2 \sqrt {\cos ^{-1}(a x)}}{\sqrt {\pi }}\right )}{128 a^2}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 73, normalized size = 0.61 \[ \frac {15 \sqrt {\pi } C\left (\frac {2 \sqrt {\cos ^{-1}(a x)}}{\sqrt {\pi }}\right )-2 \sqrt {\cos ^{-1}(a x)} \left (\left (15-16 \cos ^{-1}(a x)^2\right ) \cos \left (2 \cos ^{-1}(a x)\right )+20 \cos ^{-1}(a x) \sin \left (2 \cos ^{-1}(a x)\right )\right )}{128 a^2} \]
Antiderivative was successfully verified.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 167, normalized size = 1.40 \[ \frac {5 \, i \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (2 \, i \arccos \left (a x\right )\right )}}{32 \, a^{2}} + \frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (2 \, i \arccos \left (a x\right )\right )}}{8 \, a^{2}} - \frac {5 \, i \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (-2 \, i \arccos \left (a x\right )\right )}}{32 \, a^{2}} + \frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (-2 \, i \arccos \left (a x\right )\right )}}{8 \, a^{2}} - \frac {15 \, \sqrt {\pi } i \operatorname {erf}\left (-{\left (i + 1\right )} \sqrt {\arccos \left (a x\right )}\right )}{256 \, a^{2} {\left (i - 1\right )}} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (2 \, i \arccos \left (a x\right )\right )}}{128 \, a^{2}} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (-2 \, i \arccos \left (a x\right )\right )}}{128 \, a^{2}} + \frac {15 \, \sqrt {\pi } \operatorname {erf}\left ({\left (i - 1\right )} \sqrt {\arccos \left (a x\right )}\right )}{256 \, a^{2} {\left (i - 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 79, normalized size = 0.66 \[ \frac {32 \arccos \left (a x \right )^{\frac {5}{2}} \sqrt {\pi }\, \cos \left (2 \arccos \left (a x \right )\right )-40 \arccos \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \sin \left (2 \arccos \left (a x \right )\right )-30 \cos \left (2 \arccos \left (a x \right )\right ) \sqrt {\pi }\, \sqrt {\arccos \left (a x \right )}+15 \pi \FresnelC \left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )}{128 a^{2} \sqrt {\pi }} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
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 \[ \int x\,{\mathrm {acos}\left (a\,x\right )}^{5/2} \,d x \]
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 \[ \int x \operatorname {acos}^{\frac {5}{2}}{\left (a x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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