Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=\frac {6 (a+b x)^{5/6}}{17 (b c-a d) (c+d x)^{17/6}}+\frac {72 b (a+b x)^{5/6}}{187 (b c-a d)^2 (c+d x)^{11/6}}+\frac {432 b^2 (a+b x)^{5/6}}{935 (b c-a d)^3 (c+d x)^{5/6}} \]
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Time = 0.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=\frac {432 b^2 (a+b x)^{5/6}}{935 (c+d x)^{5/6} (b c-a d)^3}+\frac {72 b (a+b x)^{5/6}}{187 (c+d x)^{11/6} (b c-a d)^2}+\frac {6 (a+b x)^{5/6}}{17 (c+d x)^{17/6} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = \frac {6 (a+b x)^{5/6}}{17 (b c-a d) (c+d x)^{17/6}}+\frac {(12 b) \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{17/6}} \, dx}{17 (b c-a d)} \\ & = \frac {6 (a+b x)^{5/6}}{17 (b c-a d) (c+d x)^{17/6}}+\frac {72 b (a+b x)^{5/6}}{187 (b c-a d)^2 (c+d x)^{11/6}}+\frac {\left (72 b^2\right ) \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{11/6}} \, dx}{187 (b c-a d)^2} \\ & = \frac {6 (a+b x)^{5/6}}{17 (b c-a d) (c+d x)^{17/6}}+\frac {72 b (a+b x)^{5/6}}{187 (b c-a d)^2 (c+d x)^{11/6}}+\frac {432 b^2 (a+b x)^{5/6}}{935 (b c-a d)^3 (c+d x)^{5/6}} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=\frac {6 (a+b x)^{5/6} \left (55 a^2 d^2-10 a b d (17 c+6 d x)+b^2 \left (187 c^2+204 c d x+72 d^2 x^2\right )\right )}{935 (b c-a d)^3 (c+d x)^{17/6}} \]
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Time = 0.92 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(-\frac {6 \left (b x +a \right )^{\frac {5}{6}} \left (72 d^{2} x^{2} b^{2}-60 x a b \,d^{2}+204 x \,b^{2} c d +55 a^{2} d^{2}-170 a b c d +187 b^{2} c^{2}\right )}{935 \left (d x +c \right )^{\frac {17}{6}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
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Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (83) = 166\).
Time = 0.24 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=\frac {6 \, {\left (72 \, b^{2} d^{2} x^{2} + 187 \, b^{2} c^{2} - 170 \, a b c d + 55 \, a^{2} d^{2} + 12 \, {\left (17 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {1}{6}}}{935 \, {\left (b^{3} c^{6} - 3 \, a b^{2} c^{5} d + 3 \, a^{2} b c^{4} d^{2} - a^{3} c^{3} d^{3} + {\left (b^{3} c^{3} d^{3} - 3 \, a b^{2} c^{2} d^{4} + 3 \, a^{2} b c d^{5} - a^{3} d^{6}\right )} x^{3} + 3 \, {\left (b^{3} c^{4} d^{2} - 3 \, a b^{2} c^{3} d^{3} + 3 \, a^{2} b c^{2} d^{4} - a^{3} c d^{5}\right )} x^{2} + 3 \, {\left (b^{3} c^{5} d - 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{3} d^{3} - a^{3} c^{2} d^{4}\right )} x\right )}} \]
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Timed out. \[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {1}{6}} {\left (d x + c\right )}^{\frac {23}{6}}} \,d x } \]
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\[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {1}{6}} {\left (d x + c\right )}^{\frac {23}{6}}} \,d x } \]
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Time = 0.98 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.01 \[ \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{23/6}} \, dx=-\frac {{\left (c+d\,x\right )}^{1/6}\,\left (\frac {330\,a^3\,d^2-1020\,a^2\,b\,c\,d+1122\,a\,b^2\,c^2}{935\,d^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x\,\left (-30\,a^2\,b\,d^2+204\,a\,b^2\,c\,d+1122\,b^3\,c^2\right )}{935\,d^3\,{\left (a\,d-b\,c\right )}^3}+\frac {432\,b^3\,x^3}{935\,d\,{\left (a\,d-b\,c\right )}^3}+\frac {72\,b^2\,x^2\,\left (a\,d+17\,b\,c\right )}{935\,d^2\,{\left (a\,d-b\,c\right )}^3}\right )}{x^3\,{\left (a+b\,x\right )}^{1/6}+\frac {c^3\,{\left (a+b\,x\right )}^{1/6}}{d^3}+\frac {3\,c\,x^2\,{\left (a+b\,x\right )}^{1/6}}{d}+\frac {3\,c^2\,x\,{\left (a+b\,x\right )}^{1/6}}{d^2}} \]
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