Integrand size = 26, antiderivative size = 178 \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}+\frac {64 (16 b c-11 a d) \left (a+b x^2\right )^{3/4}}{55 a^4 e^3 (e x)^{7/2}}-\frac {256 (16 b c-11 a d) \left (a+b x^2\right )^{7/4}}{385 a^5 e^3 (e x)^{7/2}} \]
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Time = 0.06 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {464, 279, 270} \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {256 \left (a+b x^2\right )^{7/4} (16 b c-11 a d)}{385 a^5 e^3 (e x)^{7/2}}+\frac {64 \left (a+b x^2\right )^{3/4} (16 b c-11 a d)}{55 a^4 e^3 (e x)^{7/2}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}} \]
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Rule 270
Rule 279
Rule 464
Rubi steps \begin{align*} \text {integral}& = -\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {(16 b c-11 a d) \int \frac {1}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx}{11 a e^2} \\ & = -\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {(12 (16 b c-11 a d)) \int \frac {1}{(e x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx}{55 a^2 e^2} \\ & = -\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {(96 (16 b c-11 a d)) \int \frac {1}{(e x)^{9/2} \sqrt [4]{a+b x^2}} \, dx}{55 a^3 e^2} \\ & = -\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}+\frac {64 (16 b c-11 a d) \left (a+b x^2\right )^{3/4}}{55 a^4 e^3 (e x)^{7/2}}+\frac {(128 (16 b c-11 a d)) \int \frac {\left (a+b x^2\right )^{3/4}}{(e x)^{9/2}} \, dx}{55 a^4 e^2} \\ & = -\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}+\frac {64 (16 b c-11 a d) \left (a+b x^2\right )^{3/4}}{55 a^4 e^3 (e x)^{7/2}}-\frac {256 (16 b c-11 a d) \left (a+b x^2\right )^{7/4}}{385 a^5 e^3 (e x)^{7/2}} \\ \end{align*}
Time = 1.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.65 \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 x \left (35 a^4 c-80 a^3 b c x^2+55 a^4 d x^2+320 a^2 b^2 c x^4-220 a^3 b d x^4+2560 a b^3 c x^6-1760 a^2 b^2 d x^6+2048 b^4 c x^8-1408 a b^3 d x^8\right )}{385 a^5 (e x)^{13/2} \left (a+b x^2\right )^{5/4}} \]
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Time = 3.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(-\frac {2 x \left (-1408 a \,b^{3} d \,x^{8}+2048 b^{4} c \,x^{8}-1760 a^{2} b^{2} d \,x^{6}+2560 a \,b^{3} c \,x^{6}-220 a^{3} b d \,x^{4}+320 a^{2} b^{2} c \,x^{4}+55 a^{4} d \,x^{2}-80 a^{3} b c \,x^{2}+35 a^{4} c \right )}{385 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{5} \left (e x \right )^{\frac {13}{2}}}\) | \(110\) |
risch | \(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (-66 a b d \,x^{4}+117 b^{2} c \,x^{4}+11 a^{2} d \,x^{2}-30 a b c \,x^{2}+7 a^{2} c \right )}{77 a^{5} x^{5} e^{6} \sqrt {e x}}+\frac {2 b^{2} \left (14 x^{2} a b d -19 b^{2} c \,x^{2}+15 a^{2} d -20 a b c \right ) x}{5 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{5} e^{6} \sqrt {e x}}\) | \(123\) |
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Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.80 \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 \, {\left (128 \, {\left (16 \, b^{4} c - 11 \, a b^{3} d\right )} x^{8} + 160 \, {\left (16 \, a b^{3} c - 11 \, a^{2} b^{2} d\right )} x^{6} + 35 \, a^{4} c + 20 \, {\left (16 \, a^{2} b^{2} c - 11 \, a^{3} b d\right )} x^{4} - 5 \, {\left (16 \, a^{3} b c - 11 \, a^{4} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{385 \, {\left (a^{5} b^{2} e^{7} x^{10} + 2 \, a^{6} b e^{7} x^{8} + a^{7} e^{7} x^{6}\right )}} \]
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Timed out. \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=\text {Timed out} \]
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\[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {13}{2}}} \,d x } \]
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\[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {13}{2}}} \,d x } \]
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Time = 6.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.88 \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {64\,x^6\,\left (11\,a\,d-16\,b\,c\right )}{77\,a^4\,e^6}-\frac {2\,c}{11\,a\,b^2\,e^6}+\frac {8\,x^4\,\left (11\,a\,d-16\,b\,c\right )}{77\,a^3\,b\,e^6}-\frac {x^2\,\left (110\,a^4\,d-160\,a^3\,b\,c\right )}{385\,a^5\,b^2\,e^6}+\frac {256\,b\,x^8\,\left (11\,a\,d-16\,b\,c\right )}{385\,a^5\,e^6}\right )}{x^9\,\sqrt {e\,x}+\frac {a^2\,x^5\,\sqrt {e\,x}}{b^2}+\frac {2\,a\,x^7\,\sqrt {e\,x}}{b}} \]
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