Optimal. Leaf size=87 \[ \frac {1}{16 a c^3 (1-a x)}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{8 a c^3 (1-a x)^4}+\frac {\tanh ^{-1}(a x)}{16 a c^3} \]
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Rubi [A] time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 = {6140, 44, 207} \[ \frac {1}{16 a c^3 (1-a x)}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{8 a c^3 (1-a x)^4}+\frac {\tanh ^{-1}(a x)}{16 a c^3} \]
Antiderivative was successfully verified.
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Rule 44
Rule 207
Rule 6140
Rubi steps
\begin {align*} \int \frac {e^{4 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac {\int \frac {1}{(1-a x)^5 (1+a x)} \, dx}{c^3}\\ &=\frac {\int \left (-\frac {1}{2 (-1+a x)^5}+\frac {1}{4 (-1+a x)^4}-\frac {1}{8 (-1+a x)^3}+\frac {1}{16 (-1+a x)^2}-\frac {1}{16 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3}\\ &=\frac {1}{8 a c^3 (1-a x)^4}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{16 a c^3 (1-a x)}-\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{16 c^3}\\ &=\frac {1}{8 a c^3 (1-a x)^4}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{16 a c^3 (1-a x)}+\frac {\tanh ^{-1}(a x)}{16 a c^3}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 52, normalized size = 0.60 \[ \frac {-3 a^3 x^3+12 a^2 x^2-19 a x+3 (a x-1)^4 \tanh ^{-1}(a x)+16}{48 a c^3 (a x-1)^4} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.59, size = 147, normalized size = 1.69 \[ -\frac {6 \, a^{3} x^{3} - 24 \, a^{2} x^{2} + 38 \, a x - 3 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) - 32}{96 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 68, normalized size = 0.78 \[ \frac {\log \left ({\left | a x + 1 \right |}\right )}{32 \, a c^{3}} - \frac {\log \left ({\left | a x - 1 \right |}\right )}{32 \, a c^{3}} - \frac {3 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 19 \, a x - 16}{48 \, {\left (a x - 1\right )}^{4} a c^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 90, normalized size = 1.03 \[ \frac {1}{8 c^{3} a \left (a x -1\right )^{4}}-\frac {1}{12 c^{3} a \left (a x -1\right )^{3}}+\frac {1}{16 c^{3} a \left (a x -1\right )^{2}}-\frac {1}{16 c^{3} a \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{32 a \,c^{3}}+\frac {\ln \left (a x +1\right )}{32 a \,c^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 102, normalized size = 1.17 \[ -\frac {3 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 19 \, a x - 16}{48 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac {\log \left (a x + 1\right )}{32 \, a c^{3}} - \frac {\log \left (a x - 1\right )}{32 \, a c^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.10, size = 83, normalized size = 0.95 \[ \frac {\mathrm {atanh}\left (a\,x\right )}{16\,a\,c^3}-\frac {\frac {19\,x}{48}-\frac {a\,x^2}{4}-\frac {1}{3\,a}+\frac {a^2\,x^3}{16}}{a^4\,c^3\,x^4-4\,a^3\,c^3\,x^3+6\,a^2\,c^3\,x^2-4\,a\,c^3\,x+c^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.60, size = 99, normalized size = 1.14 \[ - \frac {3 a^{3} x^{3} - 12 a^{2} x^{2} + 19 a x - 16}{48 a^{5} c^{3} x^{4} - 192 a^{4} c^{3} x^{3} + 288 a^{3} c^{3} x^{2} - 192 a^{2} c^{3} x + 48 a c^{3}} - \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{32} - \frac {\log {\left (x + \frac {1}{a} \right )}}{32}}{a c^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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