Optimal. Leaf size=213 \[ \frac{35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 \left (a+b x^2\right )^5 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{9/2} \sqrt{b} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \]
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Rubi [A] time = 0.07216, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1088, 199, 205} \[ \frac{35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 \left (a+b x^2\right )^5 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{9/2} \sqrt{b} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 1088
Rule 199
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac{\left (2 a b+2 b^2 x^2\right )^5 \int \frac{1}{\left (2 a b+2 b^2 x^2\right )^5} \, dx}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac{x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{\left (7 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac{1}{\left (2 a b+2 b^2 x^2\right )^4} \, dx}{16 a b \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac{x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{\left (35 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac{1}{\left (2 a b+2 b^2 x^2\right )^3} \, dx}{192 a^2 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac{x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{\left (35 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac{1}{\left (2 a b+2 b^2 x^2\right )^2} \, dx}{512 a^3 b^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac{x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{\left (35 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac{1}{2 a b+2 b^2 x^2} \, dx}{2048 a^4 b^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac{x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac{35 \left (a+b x^2\right )^5 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{9/2} \sqrt{b} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.0382371, size = 105, normalized size = 0.49 \[ \frac{\sqrt{a} \sqrt{b} x \left (511 a^2 b x^2+279 a^3+385 a b^2 x^4+105 b^3 x^6\right )+105 \left (a+b x^2\right )^4 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{384 a^{9/2} \sqrt{b} \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.22, size = 169, normalized size = 0.8 \begin{align*}{\frac{b{x}^{2}+a}{384\,{a}^{4}} \left ( 105\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{8}{b}^{4}+105\,\sqrt{ab}{x}^{7}{b}^{3}+420\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{6}a{b}^{3}+385\,\sqrt{ab}{x}^{5}a{b}^{2}+630\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{4}{a}^{2}{b}^{2}+511\,\sqrt{ab}{x}^{3}{a}^{2}b+420\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{2}{a}^{3}b+279\,\sqrt{ab}x{a}^{3}+105\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){a}^{4} \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \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 [A] time = 1.5553, size = 691, normalized size = 3.24 \begin{align*} \left [\frac{210 \, a b^{4} x^{7} + 770 \, a^{2} b^{3} x^{5} + 1022 \, a^{3} b^{2} x^{3} + 558 \, a^{4} b x - 105 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{768 \,{\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}, \frac{105 \, a b^{4} x^{7} + 385 \, a^{2} b^{3} x^{5} + 511 \, a^{3} b^{2} x^{3} + 279 \, a^{4} b x + 105 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{384 \,{\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}\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^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac{5}{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*} \mathit{sage}_{0} x \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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