Integrand size = 17, antiderivative size = 61 \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\frac {2 \left (c d^2+a e^2\right ) \sqrt {d+e x}}{e^3}-\frac {4 c d (d+e x)^{3/2}}{3 e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3} \]
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Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {711} \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (a e^2+c d^2\right )}{e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3}-\frac {4 c d (d+e x)^{3/2}}{3 e^3} \]
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Rule 711
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c d^2+a e^2}{e^2 \sqrt {d+e x}}-\frac {2 c d \sqrt {d+e x}}{e^2}+\frac {c (d+e x)^{3/2}}{e^2}\right ) \, dx \\ & = \frac {2 \left (c d^2+a e^2\right ) \sqrt {d+e x}}{e^3}-\frac {4 c d (d+e x)^{3/2}}{3 e^3}+\frac {2 c (d+e x)^{5/2}}{5 e^3} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (15 a e^2+c \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3} \]
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Time = 1.90 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {2 \left (3 \left (c \,x^{2}+5 a \right ) e^{2}-4 x c d e +8 c \,d^{2}\right ) \sqrt {e x +d}}{15 e^{3}}\) | \(40\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (3 c \,x^{2} e^{2}-4 x c d e +15 e^{2} a +8 c \,d^{2}\right )}{15 e^{3}}\) | \(41\) |
trager | \(\frac {2 \sqrt {e x +d}\, \left (3 c \,x^{2} e^{2}-4 x c d e +15 e^{2} a +8 c \,d^{2}\right )}{15 e^{3}}\) | \(41\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (3 c \,x^{2} e^{2}-4 x c d e +15 e^{2} a +8 c \,d^{2}\right )}{15 e^{3}}\) | \(41\) |
derivativedivides | \(\frac {\frac {2 c \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {4 c d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a \,e^{2} \sqrt {e x +d}+2 c \,d^{2} \sqrt {e x +d}}{e^{3}}\) | \(52\) |
default | \(\frac {\frac {2 c \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {4 c d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a \,e^{2} \sqrt {e x +d}+2 c \,d^{2} \sqrt {e x +d}}{e^{3}}\) | \(52\) |
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none
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.66 \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, c e^{2} x^{2} - 4 \, c d e x + 8 \, c d^{2} + 15 \, a e^{2}\right )} \sqrt {e x + d}}{15 \, e^{3}} \]
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Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 a \sqrt {d + e x} + \frac {2 c \left (d^{2} \sqrt {d + e x} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{2}}}{e} & \text {for}\: e \neq 0 \\\frac {a x + \frac {c x^{3}}{3}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
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none
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} a + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c}{e^{2}}\right )}}{15 \, e} \]
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none
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} a + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c}{e^{2}}\right )}}{15 \, e} \]
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Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{\sqrt {d+e x}} \, dx=\frac {2\,\sqrt {d+e\,x}\,\left (3\,c\,{\left (d+e\,x\right )}^2+15\,a\,e^2+15\,c\,d^2-10\,c\,d\,\left (d+e\,x\right )\right )}{15\,e^3} \]
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