Optimal. Leaf size=284 \[ \frac{2 b d^{3/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right ),-\frac{1-c}{c+1}\right )}{3 \sqrt{1-c} \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac{2 b d \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x}-\frac{2 b d^{3/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{c+1}\right )}{3 \sqrt{1-c} \sqrt{-c^2-2 c d x^2-d^2 x^4+1}} \]
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Rubi [A] time = 0.253468, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {4842, 12, 1123, 1140, 493, 424, 419} \[ -\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac{2 b d \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x}+\frac{2 b d^{3/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{c+1}\right )}{3 \sqrt{1-c} \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{2 b d^{3/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{c+1}\right )}{3 \sqrt{1-c} \sqrt{-c^2-2 c d x^2-d^2 x^4+1}} \]
Antiderivative was successfully verified.
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Rule 4842
Rule 12
Rule 1123
Rule 1140
Rule 493
Rule 424
Rule 419
Rubi steps
\begin{align*} \int \frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x^4} \, dx &=-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}+\frac{1}{3} b \int \frac{2 d}{x^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}+\frac{1}{3} (2 b d) \int \frac{1}{x^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{3 \left (1-c^2\right ) x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac{(2 b d) \int \frac{d^2 x^2}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx}{3 \left (1-c^2\right )}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{3 \left (1-c^2\right ) x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac{\left (2 b d^3\right ) \int \frac{x^2}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx}{3 \left (1-c^2\right )}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{3 \left (1-c^2\right ) x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac{\left (2 b d^3 \sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}} \sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}\right ) \int \frac{x^2}{\sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}} \sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}} \, dx}{3 \left (1-c^2\right ) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{3 \left (1-c^2\right ) x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}+\frac{\left (2 b (1+c) d^2 \sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}} \sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}\right ) \int \frac{1}{\sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}} \sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}} \, dx}{3 \left (1-c^2\right ) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}-\frac{\left (2 b (1+c) d^2 \sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}} \sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}\right ) \int \frac{\sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}}}{\sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}} \, dx}{3 \left (1-c^2\right ) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{3 \left (1-c^2\right ) x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac{2 b d^{3/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{1+\frac{d x^2}{1+c}} E\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{1+c}\right )}{3 \sqrt{1-c} \sqrt{1-c^2-2 c d x^2-d^2 x^4}}+\frac{2 b d^{3/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{1+\frac{d x^2}{1+c}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{1+c}\right )}{3 \sqrt{1-c} \sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ \end{align*}
Mathematica [C] time = 0.396864, size = 243, normalized size = 0.86 \[ \frac{2 i b (1-c) d^2 \sqrt{1-\frac{d x^2}{-c-1}} \sqrt{1-\frac{d x^2}{1-c}} \left (E\left (i \sinh ^{-1}\left (\sqrt{-\frac{d}{-c-1}} x\right )|\frac{-c-1}{1-c}\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{-\frac{d}{-c-1}}\right ),\frac{-c-1}{1-c}\right )\right )}{3 (c-1) (c+1) \sqrt{-\frac{d}{-c-1}} \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{a}{3 x^3}+\frac{2 b d \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (c^2-1\right ) x}-\frac{b \sin ^{-1}\left (c+d x^2\right )}{3 x^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.013, size = 207, normalized size = 0.7 \begin{align*} -{\frac{a}{3\,{x}^{3}}}+b \left ( -{\frac{\arcsin \left ( d{x}^{2}+c \right ) }{3\,{x}^{3}}}+{\frac{2\,d}{3} \left ({\frac{1}{ \left ({c}^{2}-1 \right ) x}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}-2\,{\frac{{d}^{2} \left ( -{c}^{2}+1 \right ) }{ \left ({c}^{2}-1 \right ) \sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1} \left ( -2\,dc+2\,d \right ) }\sqrt{1+{\frac{d{x}^{2}}{-1+c}}}\sqrt{1+{\frac{d{x}^{2}}{1+c}}} \left ({\it EllipticF} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ) -{\it EllipticE} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ) \right ){\frac{1}{\sqrt{-{\frac{d}{-1+c}}}}}} \right ) } \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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{b \arcsin \left (d x^{2} + c\right ) + a}{x^{4}}, 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{a + b \operatorname{asin}{\left (c + d x^{2} \right )}}{x^{4}}\, 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{b \arcsin \left (d x^{2} + c\right ) + a}{x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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