Integrand size = 37, antiderivative size = 152 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}-\frac {15 e^2 \sqrt {c d^2-a e^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}} \]
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Time = 0.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {640, 43, 52, 65, 214} \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {15 e^2 \sqrt {c d^2-a e^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3} \]
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Rule 43
Rule 52
Rule 65
Rule 214
Rule 640
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^{5/2}}{(a e+c d x)^3} \, dx \\ & = -\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {(5 e) \int \frac {(d+e x)^{3/2}}{(a e+c d x)^2} \, dx}{4 c d} \\ & = -\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {\left (15 e^2\right ) \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{8 c^2 d^2} \\ & = \frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {\left (15 e^2 \left (c d^2-a e^2\right )\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 c^3 d^3} \\ & = \frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {\left (15 e \left (c d^2-a e^2\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {c d^2}{e}+a e+\frac {c d x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 c^3 d^3} \\ & = \frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}-\frac {15 e^2 \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}} \\ \end{align*}
Time = 0.49 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x} \left (-15 a^2 e^4+5 a c d e^2 (d-5 e x)+c^2 d^2 \left (2 d^2+9 d e x-8 e^2 x^2\right )\right )}{4 c^3 d^3 (a e+c d x)^2}-\frac {15 e^2 \sqrt {-c d^2+a e^2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 c^{7/2} d^{7/2}} \]
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Time = 11.84 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.08
method | result | size |
pseudoelliptic | \(-\frac {15 \left (e^{2} \left (e^{2} a -c \,d^{2}\right ) \left (c d x +a e \right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )-\left (-\frac {2 d^{2} \left (-4 x^{2} e^{2}+\frac {9}{2} d e x +d^{2}\right ) c^{2}}{15}-\frac {a d \,e^{2} \left (-5 e x +d \right ) c}{3}+a^{2} e^{4}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, \sqrt {e x +d}\right )}{4 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, c^{3} d^{3} \left (c d x +a e \right )^{2}}\) | \(164\) |
derivativedivides | \(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} d \,e^{2} a c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {15 \left (e^{2} a -c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{c^{3} d^{3}}\right )\) | \(173\) |
default | \(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} d \,e^{2} a c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {15 \left (e^{2} a -c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{c^{3} d^{3}}\right )\) | \(173\) |
risch | \(\frac {2 e^{2} \textit {\_O1} \sqrt {e x +d}}{d^{3}}-\frac {\left (e^{2} a -c \,d^{2}\right ) e^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{2} d^{3} \textit {\_Z}^{6}+\left (3 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) \textit {\_Z}^{4}+\left (3 a^{2} d \,e^{4}-6 a \,d^{3} e^{2} c +3 d^{5} c^{2}\right ) \textit {\_Z}^{2}+a^{3} c^{2} e^{6} \textit {\_O1} -3 a^{2} d^{2} e^{4}+3 a \,d^{4} e^{2} c -d^{6} c^{2}\right )}{\sum }\frac {\left (d^{2} \left (3 \textit {\_R}^{4}-3 \textit {\_R}^{2} d +d^{2}\right ) c^{2}-d \,e^{2} a c \left (-3 \textit {\_R}^{2}+2 d \right )+a^{2} e^{4}\right ) \ln \left (\sqrt {e x +d}-\textit {\_R} \right )}{\textit {\_R} \left (c d \,\textit {\_R}^{2}+e^{2} a -c \,d^{2}\right )^{2}}\right )}{3 c^{4} d^{4}}\) | \(234\) |
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Time = 0.33 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.89 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\left [\frac {15 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - {\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}, -\frac {15 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - {\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}\right ] \]
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Timed out. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.34 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2 \, \sqrt {e x + d} e^{2}}{c^{3} d^{3}} + \frac {15 \, {\left (c d^{2} e^{2} - a e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, \sqrt {-c^{2} d^{3} + a c d e^{2}} c^{3} d^{3}} - \frac {9 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{3} e^{2} - 7 \, \sqrt {e x + d} c^{2} d^{4} e^{2} - 9 \, {\left (e x + d\right )}^{\frac {3}{2}} a c d e^{4} + 14 \, \sqrt {e x + d} a c d^{2} e^{4} - 7 \, \sqrt {e x + d} a^{2} e^{6}}{4 \, {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2} c^{3} d^{3}} \]
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Time = 0.17 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.58 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2\,e^2\,\sqrt {d+e\,x}}{c^3\,d^3}-\frac {\left (\frac {9\,c^2\,d^3\,e^2}{4}-\frac {9\,a\,c\,d\,e^4}{4}\right )\,{\left (d+e\,x\right )}^{3/2}-\sqrt {d+e\,x}\,\left (\frac {7\,a^2\,e^6}{4}-\frac {7\,a\,c\,d^2\,e^4}{2}+\frac {7\,c^2\,d^4\,e^2}{4}\right )}{c^5\,d^7-\left (2\,c^5\,d^6-2\,a\,c^4\,d^4\,e^2\right )\,\left (d+e\,x\right )+c^5\,d^5\,{\left (d+e\,x\right )}^2-2\,a\,c^4\,d^5\,e^2+a^2\,c^3\,d^3\,e^4}-\frac {15\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e^2\,\sqrt {a\,e^2-c\,d^2}\,\sqrt {d+e\,x}}{a\,e^4-c\,d^2\,e^2}\right )\,\sqrt {a\,e^2-c\,d^2}}{4\,c^{7/2}\,d^{7/2}} \]
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