Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=-\frac {4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac {32 d}{21 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}+\frac {128 d^2 \sqrt [4]{a+b x}}{21 (b c-a d)^3 \sqrt [4]{c+d x}} \]
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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} \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\frac {128 d^2 \sqrt [4]{a+b x}}{21 \sqrt [4]{c+d x} (b c-a d)^3}+\frac {32 d}{21 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac {(8 d) \int \frac {1}{(a+b x)^{7/4} (c+d x)^{5/4}} \, dx}{7 (b c-a d)} \\ & = -\frac {4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac {32 d}{21 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}+\frac {\left (32 d^2\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{5/4}} \, dx}{21 (b c-a d)^2} \\ & = -\frac {4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac {32 d}{21 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}+\frac {128 d^2 \sqrt [4]{a+b x}}{21 (b c-a d)^3 \sqrt [4]{c+d x}} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\frac {84 a^2 d^2+56 a b d (c+4 d x)+4 b^2 \left (-3 c^2+8 c d x+32 d^2 x^2\right )}{21 (b c-a d)^3 (a+b x)^{7/4} \sqrt [4]{c+d x}} \]
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Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(-\frac {4 \left (32 d^{2} x^{2} b^{2}+56 x a b \,d^{2}+8 x \,b^{2} c d +21 a^{2} d^{2}+14 a b c d -3 b^{2} c^{2}\right )}{21 \left (b x +a \right )^{\frac {7}{4}} \left (d x +c \right )^{\frac {1}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
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Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (83) = 166\).
Time = 0.32 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.70 \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 14 \, a b c d + 21 \, a^{2} d^{2} + 8 \, {\left (b^{2} c d + 7 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{21 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \]
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\[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {11}{4}} \left (c + d x\right )^{\frac {5}{4}}}\, dx \]
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\[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \]
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\[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{11/4}\,{\left (c+d\,x\right )}^{5/4}} \,d x \]
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