Optimal. Leaf size=246 \[ -\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}-\frac {4 b^{2/3} (11 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{14/3}}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}} \]
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Rubi [A]
time = 0.10, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {468, 296, 331,
206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {4 b^{2/3} (11 A b-5 a B) \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{14/3}}-\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 206
Rule 210
Rule 296
Rule 331
Rule 468
Rule 631
Rule 642
Rule 648
Rubi steps
\begin {align*} \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx &=\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {(11 A b-5 a B) \int \frac {1}{x^6 \left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {(4 (11 A b-5 a B)) \int \frac {1}{x^6 \left (a+b x^3\right )} \, dx}{9 a^2 b}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}-\frac {(4 (11 A b-5 a B)) \int \frac {1}{x^3 \left (a+b x^3\right )} \, dx}{9 a^3}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {(4 b (11 A b-5 a B)) \int \frac {1}{a+b x^3} \, dx}{9 a^4}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {(4 b (11 A b-5 a B)) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 a^{14/3}}+\frac {(4 b (11 A b-5 a B)) \int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{27 a^{14/3}}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {\left (2 b^{2/3} (11 A b-5 a B)\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{27 a^{14/3}}+\frac {(2 b (11 A b-5 a B)) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{9 a^{13/3}}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}}+\frac {\left (4 b^{2/3} (11 A b-5 a B)\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{9 a^{14/3}}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}-\frac {4 b^{2/3} (11 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{14/3}}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}}\\ \end {align*}
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Mathematica [A]
time = 0.11, size = 210, normalized size = 0.85 \begin {gather*} \frac {-\frac {54 a^{5/3} A}{x^5}-\frac {135 a^{2/3} (-3 A b+a B)}{x^2}-\frac {45 a^{5/3} b (-A b+a B) x}{\left (a+b x^3\right )^2}-\frac {15 a^{2/3} b (-17 A b+11 a B) x}{a+b x^3}-40 \sqrt {3} b^{2/3} (11 A b-5 a B) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+40 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+20 b^{2/3} (-11 A b+5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{270 a^{14/3}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.31, size = 174, normalized size = 0.71
method | result | size |
default | \(\frac {b \left (\frac {\left (\frac {17}{18} b^{2} A -\frac {11}{18} a b B \right ) x^{4}+\frac {a \left (10 A b -7 B a \right ) x}{9}}{\left (b \,x^{3}+a \right )^{2}}+\frac {4 \left (11 A b -5 B a \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{9}\right )}{a^{4}}-\frac {A}{5 a^{3} x^{5}}-\frac {-3 A b +B a}{2 a^{4} x^{2}}\) | \(174\) |
risch | \(\frac {\frac {2 b^{2} \left (11 A b -5 B a \right ) x^{9}}{9 a^{4}}+\frac {16 b \left (11 A b -5 B a \right ) x^{6}}{45 a^{3}}+\frac {\left (11 A b -5 B a \right ) x^{3}}{10 a^{2}}-\frac {A}{5 a}}{x^{5} \left (b \,x^{3}+a \right )^{2}}+\frac {4 \left (\munderset {\textit {\_R} =\RootOf \left (a^{14} \textit {\_Z}^{3}-1331 A^{3} b^{5}+1815 A^{2} B a \,b^{4}-825 A \,B^{2} a^{2} b^{3}+125 B^{3} a^{3} b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{14}+3993 A^{3} b^{5}-5445 A^{2} B a \,b^{4}+2475 A \,B^{2} a^{2} b^{3}-375 B^{3} a^{3} b^{2}\right ) x +\left (-121 A^{2} a^{5} b^{3}+110 A B \,a^{6} b^{2}-25 B^{2} a^{7} b \right ) \textit {\_R} \right )\right )}{27}\) | \(221\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 221, normalized size = 0.90 \begin {gather*} -\frac {20 \, {\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{9} + 32 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{6} + 18 \, A a^{3} + 9 \, {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{3}}{90 \, {\left (a^{4} b^{2} x^{11} + 2 \, a^{5} b x^{8} + a^{6} x^{5}\right )}} - \frac {4 \, \sqrt {3} {\left (5 \, B a - 11 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, {\left (5 \, B a - 11 \, A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {4 \, {\left (5 \, B a - 11 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.18, size = 384, normalized size = 1.56 \begin {gather*} -\frac {60 \, {\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{9} + 96 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{6} + 54 \, A a^{3} + 27 \, {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{3} + 40 \, \sqrt {3} {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 20 \, {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} - a b x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 40 \, {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right )}{270 \, {\left (a^{4} b^{2} x^{11} + 2 \, a^{5} b x^{8} + a^{6} x^{5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.57, size = 173, normalized size = 0.70 \begin {gather*} \operatorname {RootSum} {\left (19683 t^{3} a^{14} - 85184 A^{3} b^{5} + 116160 A^{2} B a b^{4} - 52800 A B^{2} a^{2} b^{3} + 8000 B^{3} a^{3} b^{2}, \left ( t \mapsto t \log {\left (- \frac {27 t a^{5}}{- 44 A b^{2} + 20 B a b} + x \right )} \right )\right )} + \frac {- 18 A a^{3} + x^{9} \cdot \left (220 A b^{3} - 100 B a b^{2}\right ) + x^{6} \cdot \left (352 A a b^{2} - 160 B a^{2} b\right ) + x^{3} \cdot \left (99 A a^{2} b - 45 B a^{3}\right )}{90 a^{6} x^{5} + 180 a^{5} b x^{8} + 90 a^{4} b^{2} x^{11}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.69, size = 229, normalized size = 0.93 \begin {gather*} -\frac {4 \, \sqrt {3} {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 11 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{5}} + \frac {4 \, {\left (5 \, B a b - 11 \, A b^{2}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a^{5}} - \frac {2 \, {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 11 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \, a^{5}} - \frac {11 \, B a b^{2} x^{4} - 17 \, A b^{3} x^{4} + 14 \, B a^{2} b x - 20 \, A a b^{2} x}{18 \, {\left (b x^{3} + a\right )}^{2} a^{4}} - \frac {5 \, B a x^{3} - 15 \, A b x^{3} + 2 \, A a}{10 \, a^{4} x^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.58, size = 207, normalized size = 0.84 \begin {gather*} \frac {\frac {x^3\,\left (11\,A\,b-5\,B\,a\right )}{10\,a^2}-\frac {A}{5\,a}+\frac {2\,b^2\,x^9\,\left (11\,A\,b-5\,B\,a\right )}{9\,a^4}+\frac {16\,b\,x^6\,\left (11\,A\,b-5\,B\,a\right )}{45\,a^3}}{a^2\,x^5+2\,a\,b\,x^8+b^2\,x^{11}}+\frac {4\,b^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}}-\frac {4\,b^{2/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}}+\frac {4\,b^{2/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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