Optimal. Leaf size=205 \[ -\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{\sqrt{-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )} \]
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Rubi [A] time = 0.0257812, antiderivative size = 205, normalized size of antiderivative = 1., 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 = {4825} \[ -\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{\sqrt{-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )} \]
Antiderivative was successfully verified.
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Rule 4825
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \sin ^{-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 \sin ^{-1}\left (1+d x^2\right )\right )}-\frac{x \text{Ci}\left (\frac{a+b \sin ^{-1}\left (1+d x^2\right )}{2 b}\right ) \left (\cos \left (\frac{a}{2 b}\right )+\sin \left (\frac{a}{2 b}\right )\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}+\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (1+d x^2\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}\\ \end{align*}
Mathematica [A] time = 1.33235, size = 164, normalized size = 0.8 \[ -\frac{\frac{x^2 \left (\left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )+\left (\sin \left (\frac{a}{2 b}\right )-\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )}+\frac{2 b \sqrt{-d x^2 \left (d x^2+2\right )}}{d \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}}{4 b^2 x} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.058, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}+1 \right ) \right ) ^{-2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b^{2} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a b d\right )} \sqrt{d} \int \frac{\sqrt{-d x^{2} - 2} x}{a b d x^{2} + 2 \, a b +{\left (b^{2} d x^{2} + 2 \, b^{2}\right )} \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right )}\,{d x} - \sqrt{-d x^{2} - 2} \sqrt{d}}{2 \,{\left (b^{2} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a b d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{2} \arcsin \left (d x^{2} + 1\right )^{2} + 2 \, a b \arcsin \left (d x^{2} + 1\right ) + a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{asin}{\left (d x^{2} + 1 \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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