Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac {32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}-\frac {128 d^2 \sqrt [4]{c+d x}}{45 (b c-a d)^3 \sqrt [4]{a+b x}} \]
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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} \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=-\frac {128 d^2 \sqrt [4]{c+d x}}{45 \sqrt [4]{a+b x} (b c-a d)^3}+\frac {32 d \sqrt [4]{c+d x}}{45 (a+b x)^{5/4} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}-\frac {(8 d) \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx}{9 (b c-a d)} \\ & = -\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac {32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}+\frac {\left (32 d^2\right ) \int \frac {1}{(a+b x)^{5/4} (c+d x)^{3/4}} \, dx}{45 (b c-a d)^2} \\ & = -\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac {32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}-\frac {128 d^2 \sqrt [4]{c+d x}}{45 (b c-a d)^3 \sqrt [4]{a+b x}} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x} \left (45 a^2 d^2-18 a b d (c-4 d x)+b^2 \left (5 c^2-8 c d x+32 d^2 x^2\right )\right )}{45 (b c-a d)^3 (a+b x)^{9/4}} \]
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Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {4 \left (d x +c \right )^{\frac {1}{4}} \left (32 d^{2} x^{2} b^{2}+72 x a b \,d^{2}-8 x \,b^{2} c d +45 a^{2} d^{2}-18 a b c d +5 b^{2} c^{2}\right )}{45 \left (b x +a \right )^{\frac {9}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
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Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (83) = 166\).
Time = 0.24 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.49 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=-\frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} + 5 \, b^{2} c^{2} - 18 \, a b c d + 45 \, a^{2} d^{2} - 8 \, {\left (b^{2} c d - 9 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{45 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \]
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\[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {13}{4}} \left (c + d x\right )^{\frac {3}{4}}}\, dx \]
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\[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]
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Time = 1.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\frac {{\left (c+d\,x\right )}^{1/4}\,\left (\frac {128\,d^2\,x^2}{45\,{\left (a\,d-b\,c\right )}^3}+\frac {180\,a^2\,d^2-72\,a\,b\,c\,d+20\,b^2\,c^2}{45\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {32\,d\,x\,\left (9\,a\,d-b\,c\right )}{45\,b\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,{\left (a+b\,x\right )}^{1/4}+\frac {a^2\,{\left (a+b\,x\right )}^{1/4}}{b^2}+\frac {2\,a\,x\,{\left (a+b\,x\right )}^{1/4}}{b}} \]
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