Optimal. Leaf size=149 \[ \frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \text {ArcCos}\left (-1+d x^2\right )\right )}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcCos}\left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \text {ArcCos}\left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}} \]
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Rubi [A]
time = 0.01, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps
used = 1, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {4911}
\begin {gather*} -\frac {x \cos \left (\frac {a}{2 b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcCos}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \text {ArcCos}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \text {ArcCos}\left (d x^2-1\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 4911
Rubi steps
\begin {align*} \int \frac {1}{\left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^2} \, dx &=\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}\\ \end {align*}
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Mathematica [A]
time = 0.26, size = 131, normalized size = 0.88 \begin {gather*} \frac {\sqrt {-d x^2 \left (-2+d x^2\right )} \left (\frac {b}{a+b \text {ArcCos}\left (-1+d x^2\right )}+\frac {\sin \left (\frac {1}{2} \text {ArcCos}\left (-1+d x^2\right )\right ) \left (\cos \left (\frac {a}{2 b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcCos}\left (-1+d x^2\right )}{2 b}\right )+\sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \text {ArcCos}\left (-1+d x^2\right )}{2 b}\right )\right )}{-2+d x^2}\right )}{2 b^2 d x} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arccos \left (d \,x^{2}-1\right )\right )^{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} - 1 \right )}\right )^{2}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \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 \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2-1\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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