Integrand size = 28, antiderivative size = 292 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {b^4 (4 b d-5 a e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^3}-\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^6 (a+b x) (d+e x)^2}+\frac {10 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}+\frac {10 b^3 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^6 (a+b x)} \]
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Time = 0.10 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 45} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {10 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^6 (a+b x) (d+e x)}-\frac {5 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{2 e^6 (a+b x) (d+e x)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{3 e^6 (a+b x) (d+e x)^3}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)}-\frac {b^4 x \sqrt {a^2+2 a b x+b^2 x^2} (4 b d-5 a e)}{e^5 (a+b x)}+\frac {10 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^6 (a+b x)} \]
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Rule 45
Rule 660
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{(d+e x)^4} \, dx}{b^4 \left (a b+b^2 x\right )} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b^9 (4 b d-5 a e)}{e^5}+\frac {b^{10} x}{e^4}-\frac {b^5 (b d-a e)^5}{e^5 (d+e x)^4}+\frac {5 b^6 (b d-a e)^4}{e^5 (d+e x)^3}-\frac {10 b^7 (b d-a e)^3}{e^5 (d+e x)^2}+\frac {10 b^8 (b d-a e)^2}{e^5 (d+e x)}\right ) \, dx}{b^4 \left (a b+b^2 x\right )} \\ & = -\frac {b^4 (4 b d-5 a e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^3}-\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^6 (a+b x) (d+e x)^2}+\frac {10 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}+\frac {10 b^3 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^6 (a+b x)} \\ \end{align*}
Time = 1.07 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\sqrt {(a+b x)^2} \left (-2 a^5 e^5-5 a^4 b e^4 (d+3 e x)-20 a^3 b^2 e^3 \left (d^2+3 d e x+3 e^2 x^2\right )+10 a^2 b^3 d e^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )+10 a b^4 e \left (-13 d^4-27 d^3 e x-9 d^2 e^2 x^2+9 d e^3 x^3+3 e^4 x^4\right )+b^5 \left (47 d^5+81 d^4 e x-9 d^3 e^2 x^2-63 d^2 e^3 x^3-15 d e^4 x^4+3 e^5 x^5\right )+60 b^3 (b d-a e)^2 (d+e x)^3 \log (d+e x)\right )}{6 e^6 (a+b x) (d+e x)^3} \]
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Time = 2.84 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.01
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{4} \left (\frac {1}{2} b e \,x^{2}+5 a e x -4 b d x \right )}{\left (b x +a \right ) e^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-10 a^{3} b^{2} e^{4}+30 d \,e^{3} a^{2} b^{3}-30 d^{2} e^{2} a \,b^{4}+10 b^{5} d^{3} e \right ) x^{2}-\frac {5 b \left (e^{4} a^{4}+4 b \,e^{3} d \,a^{3}-18 b^{2} e^{2} d^{2} a^{2}+20 a \,b^{3} d^{3} e -7 b^{4} d^{4}\right ) x}{2}-\frac {2 a^{5} e^{5}+5 a^{4} b d \,e^{4}+20 a^{3} b^{2} d^{2} e^{3}-110 a^{2} b^{3} d^{3} e^{2}+130 a \,b^{4} d^{4} e -47 b^{5} d^{5}}{6 e}\right )}{\left (b x +a \right ) e^{5} \left (e x +d \right )^{3}}+\frac {10 \sqrt {\left (b x +a \right )^{2}}\, b^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{6}}\) | \(295\) |
default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (180 \ln \left (e x +d \right ) b^{5} d^{4} e x +47 b^{5} d^{5}-2 a^{5} e^{5}-5 a^{4} b d \,e^{4}-20 a^{3} b^{2} d^{2} e^{3}+110 a^{2} b^{3} d^{3} e^{2}-130 a \,b^{4} d^{4} e -60 a^{3} b^{2} d \,e^{4} x -15 a^{4} b \,e^{5} x -360 \ln \left (e x +d \right ) a \,b^{4} d^{3} e^{2} x +180 \ln \left (e x +d \right ) b^{5} d^{3} e^{2} x^{2}+60 \ln \left (e x +d \right ) a^{2} b^{3} d^{3} e^{2}-120 \ln \left (e x +d \right ) a \,b^{4} d^{4} e +60 \ln \left (e x +d \right ) a^{2} b^{3} e^{5} x^{3}+180 \ln \left (e x +d \right ) a^{2} b^{3} d \,e^{4} x^{2}-360 \ln \left (e x +d \right ) a \,b^{4} d^{2} e^{3} x^{2}+90 x^{3} a \,b^{4} d \,e^{4}+180 x^{2} a^{2} b^{3} d \,e^{4}-90 x^{2} a \,b^{4} d^{2} e^{3}+180 \ln \left (e x +d \right ) a^{2} b^{3} d^{2} e^{3} x +60 \ln \left (e x +d \right ) b^{5} d^{2} e^{3} x^{3}-120 \ln \left (e x +d \right ) a \,b^{4} d \,e^{4} x^{3}-15 x^{4} b^{5} d \,e^{4}-63 x^{3} b^{5} d^{2} e^{3}+60 \ln \left (e x +d \right ) b^{5} d^{5}+81 b^{5} d^{4} e x +3 x^{5} e^{5} b^{5}+30 x^{4} a \,b^{4} e^{5}-60 x^{2} a^{3} b^{2} e^{5}-9 x^{2} b^{5} d^{3} e^{2}+270 x \,a^{2} b^{3} d^{2} e^{3}-270 x a \,b^{4} d^{3} e^{2}\right )}{6 \left (b x +a \right )^{5} e^{6} \left (e x +d \right )^{3}}\) | \(502\) |
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Time = 0.28 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {3 \, b^{5} e^{5} x^{5} + 47 \, b^{5} d^{5} - 130 \, a b^{4} d^{4} e + 110 \, a^{2} b^{3} d^{3} e^{2} - 20 \, a^{3} b^{2} d^{2} e^{3} - 5 \, a^{4} b d e^{4} - 2 \, a^{5} e^{5} - 15 \, {\left (b^{5} d e^{4} - 2 \, a b^{4} e^{5}\right )} x^{4} - 9 \, {\left (7 \, b^{5} d^{2} e^{3} - 10 \, a b^{4} d e^{4}\right )} x^{3} - 3 \, {\left (3 \, b^{5} d^{3} e^{2} + 30 \, a b^{4} d^{2} e^{3} - 60 \, a^{2} b^{3} d e^{4} + 20 \, a^{3} b^{2} e^{5}\right )} x^{2} + 3 \, {\left (27 \, b^{5} d^{4} e - 90 \, a b^{4} d^{3} e^{2} + 90 \, a^{2} b^{3} d^{2} e^{3} - 20 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x + 60 \, {\left (b^{5} d^{5} - 2 \, a b^{4} d^{4} e + a^{2} b^{3} d^{3} e^{2} + {\left (b^{5} d^{2} e^{3} - 2 \, a b^{4} d e^{4} + a^{2} b^{3} e^{5}\right )} x^{3} + 3 \, {\left (b^{5} d^{3} e^{2} - 2 \, a b^{4} d^{2} e^{3} + a^{2} b^{3} d e^{4}\right )} x^{2} + 3 \, {\left (b^{5} d^{4} e - 2 \, a b^{4} d^{3} e^{2} + a^{2} b^{3} d^{2} e^{3}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \]
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\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]
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Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.28 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {10 \, {\left (b^{5} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{4} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{3} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {b^{5} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) - 8 \, b^{5} d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 10 \, a b^{4} e^{4} x \mathrm {sgn}\left (b x + a\right )}{2 \, e^{8}} + \frac {47 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 130 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 110 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 20 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - 2 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right ) + 60 \, {\left (b^{5} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{4} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{3} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{3} b^{2} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 15 \, {\left (7 \, b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 20 \, a b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{4} b e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x}{6 \, {\left (e x + d\right )}^{3} e^{6}} \]
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Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]
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