Optimal. Leaf size=267 \[ \frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 x^5 (a x+1)^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]
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Rubi [A] time = 0.22, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6160, 6150, 88} \[ \frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 x^5 (a x+1)^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 88
Rule 6150
Rule 6160
Rubi steps
\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {e^{-3 \tanh ^{-1}(a x)} x^5}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {x^5}{(1-a x) (1+a x)^4} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \left (-\frac {1}{a^5}-\frac {1}{16 a^5 (-1+a x)}-\frac {1}{2 a^5 (1+a x)^4}+\frac {9}{4 a^5 (1+a x)^3}-\frac {31}{8 a^5 (1+a x)^2}+\frac {49}{16 a^5 (1+a x)}\right ) \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)^3}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)^2}+\frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (1+a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 110, normalized size = 0.41 \[ \frac {\sqrt {1-a^2 x^2} \left (-48 a^4 x^4-144 a^3 x^3+42 a^2 x^2+270 a x-3 (a x+1)^3 \log (1-a x)+147 (a x+1)^3 \log (a x+1)+140\right )}{48 a^2 c^2 x (a x+1)^3 \sqrt {c-\frac {c}{a^2 x^2}}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} a^{6} x^{6} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{7} c^{3} x^{7} + 3 \, a^{6} c^{3} x^{6} + a^{5} c^{3} x^{5} - 5 \, a^{4} c^{3} x^{4} - 5 \, a^{3} c^{3} x^{3} + a^{2} c^{3} x^{2} + 3 \, a c^{3} x + c^{3}}, x\right ) \]
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 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 176, normalized size = 0.66 \[ -\frac {\left (48 x^{4} a^{4}+3 \ln \left (a x -1\right ) x^{3} a^{3}-147 a^{3} x^{3} \ln \left (a x +1\right )+144 x^{3} a^{3}+9 \ln \left (a x -1\right ) x^{2} a^{2}-441 \ln \left (a x +1\right ) x^{2} a^{2}-42 a^{2} x^{2}+9 \ln \left (a x -1\right ) x a -441 a x \ln \left (a x +1\right )-270 a x +3 \ln \left (a x -1\right )-147 \ln \left (a x +1\right )-140\right ) \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}{48 \left (a x +1\right ) a^{6} x^{5} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}}} \]
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 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (a\,x+1\right )}^3} \,d x \]
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 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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