Optimal. Leaf size=105 \[ \frac {x \left (a+b x^2\right )}{5 a \left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}}+\frac {4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {8 x \left (a+b x^2\right )}{15 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}} \]
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Rubi [A]
time = 0.02, antiderivative size = 107, normalized size of antiderivative = 1.02, number of steps
used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1103, 198, 197}
\begin {gather*} \frac {x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {8 x \left (a+b x^2\right )}{15 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}} \end {gather*}
Antiderivative was successfully verified.
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Rule 197
Rule 198
Rule 1103
Rubi steps
\begin {align*} \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx &=\frac {\left (1+\frac {b x^2}{a}\right )^{3/2} \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{7/2}} \, dx}{a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ &=\frac {x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {\left (4 \left (1+\frac {b x^2}{a}\right )^{3/2}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/2}} \, dx}{5 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ &=\frac {4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {\left (8 \left (1+\frac {b x^2}{a}\right )^{3/2}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/2}} \, dx}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ &=\frac {4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {8 x \left (a+b x^2\right )}{15 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 49, normalized size = 0.47 \begin {gather*} \frac {\left (a+b x^2\right ) \left (15 a^2 x+20 a b x^3+8 b^2 x^5\right )}{15 a^3 \left (\left (a+b x^2\right )^2\right )^{7/4}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.02, size = 55, normalized size = 0.52
method | result | size |
gosper | \(\frac {\left (b \,x^{2}+a \right ) x \left (8 b^{2} x^{4}+20 a b \,x^{2}+15 a^{2}\right )}{15 a^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {7}{4}}}\) | \(55\) |
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 [A]
time = 0.37, size = 80, normalized size = 0.76 \begin {gather*} \frac {{\left (8 \, b^{2} x^{5} + 20 \, a b x^{3} + 15 \, a^{2} x\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}}}{15 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} \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^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {7}{4}}}\, 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 [B]
time = 4.21, size = 56, normalized size = 0.53 \begin {gather*} \frac {x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/4}\,\left (15\,a^2+20\,a\,b\,x^2+8\,b^2\,x^4\right )}{15\,a^3\,{\left (b\,x^2+a\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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