Optimal. Leaf size=78 \[ -\frac {15}{16 a^3 x^2}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac {15 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{7/2}} \]
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Rubi [A]
time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 296, 331,
211} \begin {gather*} -\frac {15 \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{7/2}}-\frac {15}{16 a^3 x^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 281
Rule 296
Rule 331
Rubi steps
\begin {align*} \int \frac {1}{x^3 \left (a+c x^4\right )^3} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (a+c x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5 \text {Subst}\left (\int \frac {1}{x^2 \left (a+c x^2\right )^2} \, dx,x,x^2\right )}{8 a}\\ &=\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}+\frac {15 \text {Subst}\left (\int \frac {1}{x^2 \left (a+c x^2\right )} \, dx,x,x^2\right )}{16 a^2}\\ &=-\frac {15}{16 a^3 x^2}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac {(15 c) \text {Subst}\left (\int \frac {1}{a+c x^2} \, dx,x,x^2\right )}{16 a^3}\\ &=-\frac {15}{16 a^3 x^2}+\frac {1}{8 a x^2 \left (a+c x^4\right )^2}+\frac {5}{16 a^2 x^2 \left (a+c x^4\right )}-\frac {15 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{7/2}}\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 105, normalized size = 1.35 \begin {gather*} \frac {-\frac {\sqrt {a} \left (8 a^2+25 a c x^4+15 c^2 x^8\right )}{x^2 \left (a+c x^4\right )^2}+15 \sqrt {c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+15 \sqrt {c} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{16 a^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.16, size = 58, normalized size = 0.74
method | result | size |
default | \(-\frac {c \left (\frac {\frac {7}{8} c \,x^{6}+\frac {9}{8} a \,x^{2}}{\left (x^{4} c +a \right )^{2}}+\frac {15 \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{8 \sqrt {a c}}\right )}{2 a^{3}}-\frac {1}{2 a^{3} x^{2}}\) | \(58\) |
risch | \(\frac {-\frac {15 c^{2} x^{8}}{16 a^{3}}-\frac {25 c \,x^{4}}{16 a^{2}}-\frac {1}{2 a}}{x^{2} \left (x^{4} c +a \right )^{2}}+\frac {15 \left (\munderset {\textit {\_R} =\RootOf \left (a^{7} \textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (-5 \textit {\_R}^{2} a^{7}-4 c \right ) x^{2}-a^{4} \textit {\_R} \right )\right )}{32}\) | \(82\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 75, normalized size = 0.96 \begin {gather*} -\frac {15 \, c^{2} x^{8} + 25 \, a c x^{4} + 8 \, a^{2}}{16 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}} - \frac {15 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 218, normalized size = 2.79 \begin {gather*} \left [-\frac {30 \, c^{2} x^{8} + 50 \, a c x^{4} - 15 \, {\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} x^{2}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} - 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) + 16 \, a^{2}}{32 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}}, -\frac {15 \, c^{2} x^{8} + 25 \, a c x^{4} - 15 \, {\left (c^{2} x^{10} + 2 \, a c x^{6} + a^{2} x^{2}\right )} \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) + 8 \, a^{2}}{16 \, {\left (a^{3} c^{2} x^{10} + 2 \, a^{4} c x^{6} + a^{5} x^{2}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.29, size = 121, normalized size = 1.55 \begin {gather*} \frac {15 \sqrt {- \frac {c}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {c}{a^{7}}}}{c} + x^{2} \right )}}{32} - \frac {15 \sqrt {- \frac {c}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {c}{a^{7}}}}{c} + x^{2} \right )}}{32} + \frac {- 8 a^{2} - 25 a c x^{4} - 15 c^{2} x^{8}}{16 a^{5} x^{2} + 32 a^{4} c x^{6} + 16 a^{3} c^{2} x^{10}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.52, size = 61, normalized size = 0.78 \begin {gather*} -\frac {15 \, c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} - \frac {7 \, c^{2} x^{6} + 9 \, a c x^{2}}{16 \, {\left (c x^{4} + a\right )}^{2} a^{3}} - \frac {1}{2 \, a^{3} x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.05, size = 72, normalized size = 0.92 \begin {gather*} -\frac {\frac {1}{2\,a}+\frac {25\,c\,x^4}{16\,a^2}+\frac {15\,c^2\,x^8}{16\,a^3}}{a^2\,x^2+2\,a\,c\,x^6+c^2\,x^{10}}-\frac {15\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{16\,a^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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