Integrand size = 15, antiderivative size = 97 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=\frac {3 a^4}{b^5 \left (b+a \sqrt [3]{x}\right )}-\frac {3}{4 b^2 x^{4/3}}+\frac {2 a}{b^3 x}-\frac {9 a^2}{2 b^4 x^{2/3}}+\frac {12 a^3}{b^5 \sqrt [3]{x}}-\frac {15 a^4 \log \left (b+a \sqrt [3]{x}\right )}{b^6}+\frac {5 a^4 \log (x)}{b^6} \]
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Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {269, 272, 46} \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=-\frac {15 a^4 \log \left (a \sqrt [3]{x}+b\right )}{b^6}+\frac {5 a^4 \log (x)}{b^6}+\frac {3 a^4}{b^5 \left (a \sqrt [3]{x}+b\right )}+\frac {12 a^3}{b^5 \sqrt [3]{x}}-\frac {9 a^2}{2 b^4 x^{2/3}}+\frac {2 a}{b^3 x}-\frac {3}{4 b^2 x^{4/3}} \]
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Rule 46
Rule 269
Rule 272
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\left (b+a \sqrt [3]{x}\right )^2 x^{7/3}} \, dx \\ & = 3 \text {Subst}\left (\int \frac {1}{x^5 (b+a x)^2} \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (\frac {1}{b^2 x^5}-\frac {2 a}{b^3 x^4}+\frac {3 a^2}{b^4 x^3}-\frac {4 a^3}{b^5 x^2}+\frac {5 a^4}{b^6 x}-\frac {a^5}{b^5 (b+a x)^2}-\frac {5 a^5}{b^6 (b+a x)}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {3 a^4}{b^5 \left (b+a \sqrt [3]{x}\right )}-\frac {3}{4 b^2 x^{4/3}}+\frac {2 a}{b^3 x}-\frac {9 a^2}{2 b^4 x^{2/3}}+\frac {12 a^3}{b^5 \sqrt [3]{x}}-\frac {15 a^4 \log \left (b+a \sqrt [3]{x}\right )}{b^6}+\frac {5 a^4 \log (x)}{b^6} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=\frac {\frac {b \left (-3 b^4+5 a b^3 \sqrt [3]{x}-10 a^2 b^2 x^{2/3}+30 a^3 b x+60 a^4 x^{4/3}\right )}{\left (b+a \sqrt [3]{x}\right ) x^{4/3}}-60 a^4 \log \left (b+a \sqrt [3]{x}\right )+20 a^4 \log (x)}{4 b^6} \]
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Time = 13.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(\frac {3 a^{4}}{b^{5} \left (b +a \,x^{\frac {1}{3}}\right )}-\frac {3}{4 b^{2} x^{\frac {4}{3}}}+\frac {2 a}{b^{3} x}-\frac {9 a^{2}}{2 b^{4} x^{\frac {2}{3}}}+\frac {12 a^{3}}{b^{5} x^{\frac {1}{3}}}-\frac {15 a^{4} \ln \left (b +a \,x^{\frac {1}{3}}\right )}{b^{6}}+\frac {5 a^{4} \ln \left (x \right )}{b^{6}}\) | \(84\) |
default | \(\frac {3 a^{4}}{b^{5} \left (b +a \,x^{\frac {1}{3}}\right )}-\frac {3}{4 b^{2} x^{\frac {4}{3}}}+\frac {2 a}{b^{3} x}-\frac {9 a^{2}}{2 b^{4} x^{\frac {2}{3}}}+\frac {12 a^{3}}{b^{5} x^{\frac {1}{3}}}-\frac {15 a^{4} \ln \left (b +a \,x^{\frac {1}{3}}\right )}{b^{6}}+\frac {5 a^{4} \ln \left (x \right )}{b^{6}}\) | \(84\) |
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Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.53 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=\frac {20 \, a^{4} b^{3} x^{2} + 8 \, a b^{6} x - 60 \, {\left (a^{7} x^{3} + a^{4} b^{3} x^{2}\right )} \log \left (a x^{\frac {1}{3}} + b\right ) + 60 \, {\left (a^{7} x^{3} + a^{4} b^{3} x^{2}\right )} \log \left (x^{\frac {1}{3}}\right ) + 3 \, {\left (20 \, a^{6} b x^{2} + 15 \, a^{3} b^{4} x - b^{7}\right )} x^{\frac {2}{3}} - 6 \, {\left (5 \, a^{5} b^{2} x^{2} + 3 \, a^{2} b^{5} x\right )} x^{\frac {1}{3}}}{4 \, {\left (a^{3} b^{6} x^{3} + b^{9} x^{2}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (95) = 190\).
Time = 1.98 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.51 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {4}{3}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {3}{4 b^{2} x^{\frac {4}{3}}} & \text {for}\: a = 0 \\- \frac {1}{2 a^{2} x^{2}} & \text {for}\: b = 0 \\\frac {20 a^{5} x^{3} \log {\left (x \right )}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} - \frac {60 a^{5} x^{3} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} + \frac {20 a^{4} b x^{\frac {8}{3}} \log {\left (x \right )}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} - \frac {60 a^{4} b x^{\frac {8}{3}} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} + \frac {60 a^{4} b x^{\frac {8}{3}}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} + \frac {30 a^{3} b^{2} x^{\frac {7}{3}}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} - \frac {10 a^{2} b^{3} x^{2}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} + \frac {5 a b^{4} x^{\frac {5}{3}}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} - \frac {3 b^{5} x^{\frac {4}{3}}}{4 a b^{6} x^{3} + 4 b^{7} x^{\frac {8}{3}}} & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=-\frac {15 \, a^{4} \log \left (a + \frac {b}{x^{\frac {1}{3}}}\right )}{b^{6}} - \frac {3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{4}}{4 \, b^{6}} + \frac {5 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{3} a}{b^{6}} - \frac {15 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{2} a^{2}}{b^{6}} + \frac {30 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} a^{3}}{b^{6}} - \frac {3 \, a^{5}}{{\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} b^{6}} \]
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Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=-\frac {15 \, a^{4} \log \left ({\left | a x^{\frac {1}{3}} + b \right |}\right )}{b^{6}} + \frac {5 \, a^{4} \log \left ({\left | x \right |}\right )}{b^{6}} + \frac {60 \, a^{4} b x^{\frac {4}{3}} + 30 \, a^{3} b^{2} x - 10 \, a^{2} b^{3} x^{\frac {2}{3}} + 5 \, a b^{4} x^{\frac {1}{3}} - 3 \, b^{5}}{4 \, {\left (a x^{\frac {1}{3}} + b\right )} b^{6} x^{\frac {4}{3}}} \]
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Time = 5.96 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^3} \, dx=\frac {\frac {5\,a\,x^{1/3}}{4\,b^2}-\frac {3}{4\,b}+\frac {15\,a^3\,x}{2\,b^4}-\frac {5\,a^2\,x^{2/3}}{2\,b^3}+\frac {15\,a^4\,x^{4/3}}{b^5}}{a\,x^{5/3}+b\,x^{4/3}}-\frac {30\,a^4\,\mathrm {atanh}\left (\frac {2\,a\,x^{1/3}}{b}+1\right )}{b^6} \]
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