Optimal. Leaf size=97 \[ -\frac {3 (5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}-\frac {x (7 A b-3 a B)}{8 a^3 \left (a+b x^2\right )}-\frac {A}{a^3 x}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2} \]
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Rubi [A] time = 0.10, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {456, 453, 205} \begin {gather*} -\frac {x (7 A b-3 a B)}{8 a^3 \left (a+b x^2\right )}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}-\frac {3 (5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}-\frac {A}{a^3 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 453
Rule 456
Rubi steps
\begin {align*} \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx &=-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {1}{4} \int \frac {-\frac {4 A}{a}+\frac {3 (A b-a B) x^2}{a^2}}{x^2 \left (a+b x^2\right )^2} \, dx\\ &=-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}+\frac {1}{8} \int \frac {\frac {8 A}{a^2}-\frac {(7 A b-3 a B) x^2}{a^3}}{x^2 \left (a+b x^2\right )} \, dx\\ &=-\frac {A}{a^3 x}-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac {(3 (5 A b-a B)) \int \frac {1}{a+b x^2} \, dx}{8 a^3}\\ &=-\frac {A}{a^3 x}-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac {3 (5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 96, normalized size = 0.99 \begin {gather*} \frac {3 (a B-5 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}+\frac {x (3 a B-7 A b)}{8 a^3 \left (a+b x^2\right )}-\frac {A}{a^3 x}+\frac {x (a B-A b)}{4 a^2 \left (a+b x^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.46, size = 324, normalized size = 3.34 \begin {gather*} \left [-\frac {16 \, A a^{3} b - 6 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 10 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac {8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 82, normalized size = 0.85 \begin {gather*} \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {A}{a^{3} x} + \frac {3 \, B a b x^{3} - 7 \, A b^{2} x^{3} + 5 \, B a^{2} x - 9 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 125, normalized size = 1.29 \begin {gather*} -\frac {7 A \,b^{2} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{3}}+\frac {3 B b \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {9 A b x}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {5 B x}{8 \left (b \,x^{2}+a \right )^{2} a}-\frac {15 A b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{3}}+\frac {3 B \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{2}}-\frac {A}{a^{3} x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.43, size = 96, normalized size = 0.99 \begin {gather*} \frac {3 \, {\left (B a b - 5 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} + 5 \, {\left (B a^{2} - 5 \, A a b\right )} x^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} + \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.19, size = 113, normalized size = 1.16 \begin {gather*} -\frac {\frac {A}{a}+\frac {5\,x^2\,\left (5\,A\,b-B\,a\right )}{8\,a^2}+\frac {3\,b\,x^4\,\left (5\,A\,b-B\,a\right )}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,x\,\left (5\,A\,b-B\,a\right )}{\sqrt {a}\,\left (15\,A\,b-3\,B\,a\right )}\right )\,\left (5\,A\,b-B\,a\right )}{8\,a^{7/2}\,\sqrt {b}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.67, size = 194, normalized size = 2.00 \begin {gather*} - \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (- \frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (\frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {- 8 A a^{2} + x^{4} \left (- 15 A b^{2} + 3 B a b\right ) + x^{2} \left (- 25 A a b + 5 B a^{2}\right )}{8 a^{5} x + 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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