Optimal. Leaf size=79 \[ -\frac {\left (2-n^2\right ) e^{n \tanh ^{-1}(a x)}}{a^3 c^2 n \left (4-n^2\right )}-\frac {(n-2 a x) e^{n \tanh ^{-1}(a x)}}{a^3 c^2 \left (4-n^2\right ) \left (1-a^2 x^2\right )} \]
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Rubi [A] time = 0.13, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6147, 6137} \[ -\frac {(n-2 a x) e^{n \tanh ^{-1}(a x)}}{a^3 c^2 \left (4-n^2\right ) \left (1-a^2 x^2\right )}-\frac {\left (2-n^2\right ) e^{n \tanh ^{-1}(a x)}}{a^3 c^2 n \left (4-n^2\right )} \]
Antiderivative was successfully verified.
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Rule 6137
Rule 6147
Rubi steps
\begin {align*} \int \frac {e^{n \tanh ^{-1}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx &=-\frac {e^{n \tanh ^{-1}(a x)} (n-2 a x)}{a^3 c^2 \left (4-n^2\right ) \left (1-a^2 x^2\right )}-\frac {\left (2-n^2\right ) \int \frac {e^{n \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{a^2 c \left (4-n^2\right )}\\ &=-\frac {e^{n \tanh ^{-1}(a x)} \left (2-n^2\right )}{a^3 c^2 n \left (4-n^2\right )}-\frac {e^{n \tanh ^{-1}(a x)} (n-2 a x)}{a^3 c^2 \left (4-n^2\right ) \left (1-a^2 x^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 65, normalized size = 0.82 \[ -\frac {(1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n}{2}-1} \left (-a^2 \left (n^2-2\right ) x^2+2 a n x-2\right )}{a^3 c^2 n \left (n^2-4\right )} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.47, size = 91, normalized size = 1.15 \[ -\frac {{\left (2 \, a n x - {\left (a^{2} n^{2} - 2 \, a^{2}\right )} x^{2} - 2\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{3} c^{2} n^{3} - 4 \, a^{3} c^{2} n - {\left (a^{5} c^{2} n^{3} - 4 \, a^{5} c^{2} n\right )} x^{2}} \]
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 \[ \int \frac {x^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a^{2} c x^{2} - c\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 62, normalized size = 0.78 \[ -\frac {{\mathrm e}^{n \arctanh \left (a x \right )} \left (a^{2} n^{2} x^{2}-2 a^{2} x^{2}-2 n a x +2\right )}{\left (a^{2} x^{2}-1\right ) c^{2} a^{3} n \left (n^{2}-4\right )} \]
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 \[ \int \frac {x^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a^{2} c x^{2} - c\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.17, size = 93, normalized size = 1.18 \[ \frac {{\left (a\,x+1\right )}^{n/2}\,\left (\frac {2}{a^5\,c^2\,n\,\left (n^2-4\right )}-\frac {2\,x}{a^4\,c^2\,\left (n^2-4\right )}+\frac {x^2\,\left (n^2-2\right )}{a^3\,c^2\,n\,\left (n^2-4\right )}\right )}{\left (\frac {1}{a^2}-x^2\right )\,{\left (1-a\,x\right )}^{n/2}} \]
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 {cases} \tilde {\infty } x^{3} e^{- \infty n} & \text {for}\: a = - \frac {1}{x} \\\tilde {\infty } x^{3} e^{\infty n} & \text {for}\: a = \frac {1}{x} \\\tilde {\infty } \int x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}\, dx & \text {for}\: c = 0 \\- \frac {a^{2} x^{2} \operatorname {atanh}{\left (a x \right )}}{4 a^{5} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a^{3} c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} - \frac {2 a x \operatorname {atanh}{\left (a x \right )}}{4 a^{5} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a^{3} c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} - \frac {3 a x}{4 a^{5} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a^{3} c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} - \frac {\operatorname {atanh}{\left (a x \right )}}{4 a^{5} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a^{3} c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} - \frac {2}{4 a^{5} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a^{3} c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} & \text {for}\: n = -2 \\\frac {a^{2} x^{2} \log {\left (x - \frac {1}{a} \right )}}{4 a^{5} c^{2} x^{2} - 4 a^{3} c^{2}} - \frac {a^{2} x^{2} \log {\left (x + \frac {1}{a} \right )}}{4 a^{5} c^{2} x^{2} - 4 a^{3} c^{2}} - \frac {2 a x}{4 a^{5} c^{2} x^{2} - 4 a^{3} c^{2}} - \frac {\log {\left (x - \frac {1}{a} \right )}}{4 a^{5} c^{2} x^{2} - 4 a^{3} c^{2}} + \frac {\log {\left (x + \frac {1}{a} \right )}}{4 a^{5} c^{2} x^{2} - 4 a^{3} c^{2}} & \text {for}\: n = 0 \\\frac {\int \frac {x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}}{a^{4} x^{4} - 2 a^{2} x^{2} + 1}\, dx}{c^{2}} & \text {for}\: n = 2 \\- \frac {a^{2} n^{2} x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{5} c^{2} n^{3} x^{2} - 4 a^{5} c^{2} n x^{2} - a^{3} c^{2} n^{3} + 4 a^{3} c^{2} n} + \frac {2 a^{2} x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{5} c^{2} n^{3} x^{2} - 4 a^{5} c^{2} n x^{2} - a^{3} c^{2} n^{3} + 4 a^{3} c^{2} n} + \frac {2 a n x e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{5} c^{2} n^{3} x^{2} - 4 a^{5} c^{2} n x^{2} - a^{3} c^{2} n^{3} + 4 a^{3} c^{2} n} - \frac {2 e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{5} c^{2} n^{3} x^{2} - 4 a^{5} c^{2} n x^{2} - a^{3} c^{2} n^{3} + 4 a^{3} c^{2} n} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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