Optimal. Leaf size=227 \[ \frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x}{8 b^2 \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}-\frac{\sqrt{-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^2} \]
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Rubi [A] time = 0.0483044, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4828, 4816} \[ \frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x}{8 b^2 \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}-\frac{\sqrt{-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^2} \]
Antiderivative was successfully verified.
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Rule 4828
Rule 4816
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^3} \, dx &=-\frac{\sqrt{-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^2}+\frac{x}{8 b^2 \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )}-\frac{\int \frac{1}{a+b \sin ^{-1}\left (1+d x^2\right )} \, dx}{8 b^2}\\ &=-\frac{\sqrt{-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^2}+\frac{x}{8 b^2 \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 )}{16 b^3 \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 )}{16 b^3 \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 = 0.536223, size = 187, normalized size = 0.82 \[ \frac{x \left (\left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )+\left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x}{8 b^2 \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}-\frac{\sqrt{-d x^2 \left (d x^2+2\right )}}{4 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.057, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}+1 \right ) \right ) ^{-3}\, 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{b d x \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a d x - 2 \, \sqrt{-d x^{2} - 2} b \sqrt{d} -{\left (b^{4} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right )^{2} + 2 \, a b^{3} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a^{2} b^{2} d\right )} \int \frac{1}{b^{3} \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a b^{2}}\,{d x}}{8 \,{\left (b^{4} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right )^{2} + 2 \, a b^{3} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a^{2} b^{2} 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^{3} \arcsin \left (d x^{2} + 1\right )^{3} + 3 \, a b^{2} \arcsin \left (d x^{2} + 1\right )^{2} + 3 \, a^{2} b \arcsin \left (d x^{2} + 1\right ) + a^{3}}, 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 )^{3}}\, 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 )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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