Integrand size = 19, antiderivative size = 136 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=-\frac {4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}+\frac {16 d (c+d x)^{9/4}}{119 (b c-a d)^2 (a+b x)^{17/4}}-\frac {128 d^2 (c+d x)^{9/4}}{1547 (b c-a d)^3 (a+b x)^{13/4}}+\frac {512 d^3 (c+d x)^{9/4}}{13923 (b c-a d)^4 (a+b x)^{9/4}} \]
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Time = 0.03 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=\frac {512 d^3 (c+d x)^{9/4}}{13923 (a+b x)^{9/4} (b c-a d)^4}-\frac {128 d^2 (c+d x)^{9/4}}{1547 (a+b x)^{13/4} (b c-a d)^3}+\frac {16 d (c+d x)^{9/4}}{119 (a+b x)^{17/4} (b c-a d)^2}-\frac {4 (c+d x)^{9/4}}{21 (a+b x)^{21/4} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}-\frac {(4 d) \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx}{7 (b c-a d)} \\ & = -\frac {4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}+\frac {16 d (c+d x)^{9/4}}{119 (b c-a d)^2 (a+b x)^{17/4}}+\frac {\left (32 d^2\right ) \int \frac {(c+d x)^{5/4}}{(a+b x)^{17/4}} \, dx}{119 (b c-a d)^2} \\ & = -\frac {4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}+\frac {16 d (c+d x)^{9/4}}{119 (b c-a d)^2 (a+b x)^{17/4}}-\frac {128 d^2 (c+d x)^{9/4}}{1547 (b c-a d)^3 (a+b x)^{13/4}}-\frac {\left (128 d^3\right ) \int \frac {(c+d x)^{5/4}}{(a+b x)^{13/4}} \, dx}{1547 (b c-a d)^3} \\ & = -\frac {4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}+\frac {16 d (c+d x)^{9/4}}{119 (b c-a d)^2 (a+b x)^{17/4}}-\frac {128 d^2 (c+d x)^{9/4}}{1547 (b c-a d)^3 (a+b x)^{13/4}}+\frac {512 d^3 (c+d x)^{9/4}}{13923 (b c-a d)^4 (a+b x)^{9/4}} \\ \end{align*}
Time = 1.70 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=\frac {4 (c+d x)^{9/4} \left (1547 a^3 d^3+357 a^2 b d^2 (-9 c+4 d x)+21 a b^2 d \left (117 c^2-72 c d x+32 d^2 x^2\right )+b^3 \left (-663 c^3+468 c^2 d x-288 c d^2 x^2+128 d^3 x^3\right )\right )}{13923 (b c-a d)^4 (a+b x)^{21/4}} \]
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Time = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.26
method | result | size |
gosper | \(\frac {4 \left (d x +c \right )^{\frac {9}{4}} \left (128 d^{3} x^{3} b^{3}+672 x^{2} a \,b^{2} d^{3}-288 x^{2} b^{3} c \,d^{2}+1428 x \,a^{2} b \,d^{3}-1512 x a \,b^{2} c \,d^{2}+468 x \,b^{3} c^{2} d +1547 a^{3} d^{3}-3213 a^{2} b c \,d^{2}+2457 a \,b^{2} c^{2} d -663 b^{3} c^{3}\right )}{13923 \left (b x +a \right )^{\frac {21}{4}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}\) | \(171\) |
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Leaf count of result is larger than twice the leaf count of optimal. 649 vs. \(2 (112) = 224\).
Time = 0.37 (sec) , antiderivative size = 649, normalized size of antiderivative = 4.77 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=\frac {4 \, {\left (128 \, b^{3} d^{5} x^{5} - 663 \, b^{3} c^{5} + 2457 \, a b^{2} c^{4} d - 3213 \, a^{2} b c^{3} d^{2} + 1547 \, a^{3} c^{2} d^{3} - 32 \, {\left (b^{3} c d^{4} - 21 \, a b^{2} d^{5}\right )} x^{4} + 4 \, {\left (5 \, b^{3} c^{2} d^{3} - 42 \, a b^{2} c d^{4} + 357 \, a^{2} b d^{5}\right )} x^{3} - {\left (15 \, b^{3} c^{3} d^{2} - 105 \, a b^{2} c^{2} d^{3} + 357 \, a^{2} b c d^{4} - 1547 \, a^{3} d^{5}\right )} x^{2} - 2 \, {\left (429 \, b^{3} c^{4} d - 1701 \, a b^{2} c^{3} d^{2} + 2499 \, a^{2} b c^{2} d^{3} - 1547 \, a^{3} c d^{4}\right )} x\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{13923 \, {\left (a^{6} b^{4} c^{4} - 4 \, a^{7} b^{3} c^{3} d + 6 \, a^{8} b^{2} c^{2} d^{2} - 4 \, a^{9} b c d^{3} + a^{10} d^{4} + {\left (b^{10} c^{4} - 4 \, a b^{9} c^{3} d + 6 \, a^{2} b^{8} c^{2} d^{2} - 4 \, a^{3} b^{7} c d^{3} + a^{4} b^{6} d^{4}\right )} x^{6} + 6 \, {\left (a b^{9} c^{4} - 4 \, a^{2} b^{8} c^{3} d + 6 \, a^{3} b^{7} c^{2} d^{2} - 4 \, a^{4} b^{6} c d^{3} + a^{5} b^{5} d^{4}\right )} x^{5} + 15 \, {\left (a^{2} b^{8} c^{4} - 4 \, a^{3} b^{7} c^{3} d + 6 \, a^{4} b^{6} c^{2} d^{2} - 4 \, a^{5} b^{5} c d^{3} + a^{6} b^{4} d^{4}\right )} x^{4} + 20 \, {\left (a^{3} b^{7} c^{4} - 4 \, a^{4} b^{6} c^{3} d + 6 \, a^{5} b^{5} c^{2} d^{2} - 4 \, a^{6} b^{4} c d^{3} + a^{7} b^{3} d^{4}\right )} x^{3} + 15 \, {\left (a^{4} b^{6} c^{4} - 4 \, a^{5} b^{5} c^{3} d + 6 \, a^{6} b^{4} c^{2} d^{2} - 4 \, a^{7} b^{3} c d^{3} + a^{8} b^{2} d^{4}\right )} x^{2} + 6 \, {\left (a^{5} b^{5} c^{4} - 4 \, a^{6} b^{4} c^{3} d + 6 \, a^{7} b^{3} c^{2} d^{2} - 4 \, a^{8} b^{2} c d^{3} + a^{9} b d^{4}\right )} x\right )}} \]
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Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=\text {Timed out} \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {25}{4}}} \,d x } \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {25}{4}}} \,d x } \]
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Time = 1.50 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.76 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx=\frac {{\left (c+d\,x\right )}^{1/4}\,\left (\frac {x^2\,\left (6188\,a^3\,d^5-1428\,a^2\,b\,c\,d^4+420\,a\,b^2\,c^2\,d^3-60\,b^3\,c^3\,d^2\right )}{13923\,b^5\,{\left (a\,d-b\,c\right )}^4}-\frac {-6188\,a^3\,c^2\,d^3+12852\,a^2\,b\,c^3\,d^2-9828\,a\,b^2\,c^4\,d+2652\,b^3\,c^5}{13923\,b^5\,{\left (a\,d-b\,c\right )}^4}+\frac {x\,\left (12376\,a^3\,c\,d^4-19992\,a^2\,b\,c^2\,d^3+13608\,a\,b^2\,c^3\,d^2-3432\,b^3\,c^4\,d\right )}{13923\,b^5\,{\left (a\,d-b\,c\right )}^4}+\frac {512\,d^5\,x^5}{13923\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {128\,d^4\,x^4\,\left (21\,a\,d-b\,c\right )}{13923\,b^3\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,d^3\,x^3\,\left (357\,a^2\,d^2-42\,a\,b\,c\,d+5\,b^2\,c^2\right )}{13923\,b^4\,{\left (a\,d-b\,c\right )}^4}\right )}{x^5\,{\left (a+b\,x\right )}^{1/4}+\frac {a^5\,{\left (a+b\,x\right )}^{1/4}}{b^5}+\frac {10\,a^2\,x^3\,{\left (a+b\,x\right )}^{1/4}}{b^2}+\frac {10\,a^3\,x^2\,{\left (a+b\,x\right )}^{1/4}}{b^3}+\frac {5\,a\,x^4\,{\left (a+b\,x\right )}^{1/4}}{b}+\frac {5\,a^4\,x\,{\left (a+b\,x\right )}^{1/4}}{b^4}} \]
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