Optimal. Leaf size=89 \[ -\frac {1}{2 a x^2}-\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c}}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^2+c x^4\right )}{4 a^2} \]
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Rubi [A]
time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1599, 1128,
723, 814, 648, 632, 212, 642} \begin {gather*} -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c}}+\frac {b \log \left (a+b x^2+c x^4\right )}{4 a^2}-\frac {b \log (x)}{a^2}-\frac {1}{2 a x^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 632
Rule 642
Rule 648
Rule 723
Rule 814
Rule 1128
Rule 1599
Rubi steps
\begin {align*} \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx &=\int \frac {1}{x^3 \left (a+b x^2+c x^4\right )} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )\\ &=-\frac {1}{2 a x^2}+\frac {\text {Subst}\left (\int \frac {-b-c x}{x \left (a+b x+c x^2\right )} \, dx,x,x^2\right )}{2 a}\\ &=-\frac {1}{2 a x^2}+\frac {\text {Subst}\left (\int \left (-\frac {b}{a x}+\frac {b^2-a c+b c x}{a \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )}{2 a}\\ &=-\frac {1}{2 a x^2}-\frac {b \log (x)}{a^2}+\frac {\text {Subst}\left (\int \frac {b^2-a c+b c x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 a^2}\\ &=-\frac {1}{2 a x^2}-\frac {b \log (x)}{a^2}+\frac {b \text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^2}+\frac {\left (b^2-2 a c\right ) \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^2}\\ &=-\frac {1}{2 a x^2}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^2+c x^4\right )}{4 a^2}-\frac {\left (b^2-2 a c\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 a^2}\\ &=-\frac {1}{2 a x^2}-\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c}}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^2+c x^4\right )}{4 a^2}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 135, normalized size = 1.52 \begin {gather*} \frac {-\frac {2 a}{x^2}-4 b \log (x)+\frac {\left (b^2-2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}+\frac {\left (-b^2+2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}}{4 a^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.03, size = 85, normalized size = 0.96
method | result | size |
default | \(-\frac {-\frac {b \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2}+\frac {2 \left (a c -\frac {b^{2}}{2}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 a^{2}}-\frac {1}{2 a \,x^{2}}-\frac {b \ln \left (x \right )}{a^{2}}\) | \(85\) |
risch | \(-\frac {1}{2 a \,x^{2}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\left (4 a^{3} c -a^{2} b^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c +b^{3}\right ) \textit {\_Z} +c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (10 a^{3} c -3 a^{2} b^{2}\right ) \textit {\_R}^{2}-4 \textit {\_R} a b c +2 c^{2}\right ) x^{2}-a^{3} b \,\textit {\_R}^{2}+\left (a^{2} c -2 a \,b^{2}\right ) \textit {\_R} +2 b c \right )\right )}{2}\) | \(124\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 293, normalized size = 3.29 \begin {gather*} \left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} x^{2} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} x^{2} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 3.52, size = 94, normalized size = 1.06 \begin {gather*} \frac {b \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{2}} - \frac {b \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{2}} + \frac {b x^{2} - a}{2 \, a^{2} x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.91, size = 2033, normalized size = 22.84 \begin {gather*} \frac {\mathrm {atan}\left (\frac {16\,a^6\,x^2\,\left (\frac {\left (a^2\,c^2-9\,a\,b^2\,c+3\,b^4\right )\,\left (\frac {c^5}{a^3}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {6\,b\,c^4}{a^2}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {20\,a^3\,c^4+2\,a^2\,b^2\,c^3}{a^3}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )}{2\,a^3\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}-\frac {\left (\frac {\left (2\,a\,c-b^2\right )\,\left (\frac {20\,a^3\,c^4+2\,a^2\,b^2\,c^3}{a^3}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )}{2\,a^3\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )\,\left (2\,a\,c-b^2\right )}{8\,a^5\,\sqrt {4\,a\,c-b^2}\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )\,\left (2\,a\,c-b^2\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}-\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )\,{\left (2\,a\,c-b^2\right )}^2}{32\,a^7\,\left (4\,a\,c-b^2\right )\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{8\,a^3\,c^2\,\left (a^2\,c^2+24\,a\,b^2\,c-6\,b^4\right )}+\frac {\left (\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {\left (2\,a\,c-b^2\right )\,\left (\frac {20\,a^3\,c^4+2\,a^2\,b^2\,c^3}{a^3}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )}{2\,a^3\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )\,\left (2\,a\,c-b^2\right )}{8\,a^5\,\sqrt {4\,a\,c-b^2}\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}-\frac {\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )\,{\left (2\,a\,c-b^2\right )}^3}{64\,a^9\,{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {\left (\frac {6\,b\,c^4}{a^2}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {20\,a^3\,c^4+2\,a^2\,b^2\,c^3}{a^3}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (40\,a^4\,b\,c^3-12\,a^3\,b^3\,c^2\right )}{2\,a^3\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )\,\left (2\,a\,c-b^2\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}\right )\,\left (13\,a^2\,b\,c^2-15\,a\,b^3\,c+3\,b^5\right )}{8\,a^3\,c^2\,\sqrt {4\,a\,c-b^2}\,\left (a^2\,c^2+24\,a\,b^2\,c-6\,b^4\right )}\right )\,{\left (4\,a\,c-b^2\right )}^{3/2}}{4\,a^2\,c^4-4\,a\,b^2\,c^3+b^4\,c^2}-\frac {2\,a^3\,\left (4\,a\,c-b^2\right )\,\left (13\,a^2\,b\,c^2-15\,a\,b^3\,c+3\,b^5\right )\,\left (\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {\left (\frac {4\,a^3\,b\,c^3-4\,a^2\,b^3\,c^2}{a^3}+\frac {2\,a\,b^2\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a^3\,c-4\,a^2\,b^2}\right )\,\left (2\,a\,c-b^2\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}+\frac {b^2\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )\,\left (2\,a\,c-b^2\right )}{2\,a\,\sqrt {4\,a\,c-b^2}\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}+\frac {\left (2\,a\,c-b^2\right )\,\left (\frac {a^2\,c^4-4\,a\,b^2\,c^3}{a^3}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {4\,a^3\,b\,c^3-4\,a^2\,b^3\,c^2}{a^3}+\frac {2\,a\,b^2\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a^3\,c-4\,a^2\,b^2}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}-\frac {b^2\,c^2\,{\left (2\,a\,c-b^2\right )}^3}{16\,a^5\,{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{c^2\,\left (a^2\,c^2+24\,a\,b^2\,c-6\,b^4\right )\,\left (4\,a^2\,c^4-4\,a\,b^2\,c^3+b^4\,c^2\right )}+\frac {2\,a^3\,{\left (4\,a\,c-b^2\right )}^{3/2}\,\left (a^2\,c^2-9\,a\,b^2\,c+3\,b^4\right )\,\left (\frac {b\,c^4}{a^3}-\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {a^2\,c^4-4\,a\,b^2\,c^3}{a^3}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {4\,a^3\,b\,c^3-4\,a^2\,b^3\,c^2}{a^3}+\frac {2\,a\,b^2\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a^3\,c-4\,a^2\,b^2}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}+\frac {\left (2\,a\,c-b^2\right )\,\left (\frac {\left (\frac {4\,a^3\,b\,c^3-4\,a^2\,b^3\,c^2}{a^3}+\frac {2\,a\,b^2\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a^3\,c-4\,a^2\,b^2}\right )\,\left (2\,a\,c-b^2\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}+\frac {b^2\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )\,\left (2\,a\,c-b^2\right )}{2\,a\,\sqrt {4\,a\,c-b^2}\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{4\,a^2\,\sqrt {4\,a\,c-b^2}}+\frac {b^2\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )\,{\left (2\,a\,c-b^2\right )}^2}{8\,a^3\,\left (4\,a\,c-b^2\right )\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}\right )}{c^2\,\left (a^2\,c^2+24\,a\,b^2\,c-6\,b^4\right )\,\left (4\,a^2\,c^4-4\,a\,b^2\,c^3+b^4\,c^2\right )}\right )\,\left (2\,a\,c-b^2\right )}{2\,a^2\,\sqrt {4\,a\,c-b^2}}-\frac {b\,\ln \left (x\right )}{a^2}-\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (2\,b^3-8\,a\,b\,c\right )}{2\,\left (16\,a^3\,c-4\,a^2\,b^2\right )}-\frac {1}{2\,a\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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