Optimal. Leaf size=269 \[ -\frac {\sqrt {2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}+\frac {x}{15 b^2 \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{3/2}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{5/2}}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{7/2} \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{15 d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{7/2} \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{15 d x} \]
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Rubi [A] time = 0.05, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4829, 4824} \[ -\frac {\sqrt {2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}+\frac {x}{15 b^2 \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{3/2}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{5/2}}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{7/2} \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{15 d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{7/2} \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{15 d x} \]
Antiderivative was successfully verified.
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Rule 4824
Rule 4829
Rubi steps
\begin {align*} \int \frac {1}{\left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{7/2}} \, dx &=\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{3/2}}-\frac {\int \frac {1}{\left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{3/2}} \, dx}{15 b^2}\\ &=\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{3/2}}-\frac {\sqrt {2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \cos ^{-1}\left (-1+d x^2\right )}}+\frac {2 \left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (-1+d x^2\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{15 d x}+\frac {2 \left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } \cos \left (\frac {1}{2} \cos ^{-1}\left (-1+d x^2\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{15 d x}\\ \end {align*}
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Mathematica [A] time = 0.57, size = 309, normalized size = 1.15 \[ \frac {2 \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \left (-a^2 \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )+\sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{5/2} C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )+\sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{5/2} S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )+a b \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )-2 a b \cos ^{-1}\left (d x^2-1\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )+b^2 \cos ^{-1}\left (d x^2-1\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )+3 b^2 \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )-b^2 \cos ^{-1}\left (d x^2-1\right )^2 \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )\right )}{15 b^3 d x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{5/2}} \]
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 [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arccos \left (d \,x^{2}-1\right )\right )^{\frac {7}{2}}}\, dx \]
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 \[ \int \frac {1}{{\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2-1\right )\right )}^{7/2}} \,d x \]
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 \[ \int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} - 1 \right )}\right )^{\frac {7}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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