Integrand size = 17, antiderivative size = 61 \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\frac {2 \left (c f^2+a g^2\right ) \sqrt {f+g x}}{g^3}-\frac {4 c f (f+g x)^{3/2}}{3 g^3}+\frac {2 c (f+g x)^{5/2}}{5 g^3} \]
[Out]
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} \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\frac {2 \sqrt {f+g x} \left (a g^2+c f^2\right )}{g^3}+\frac {2 c (f+g x)^{5/2}}{5 g^3}-\frac {4 c f (f+g x)^{3/2}}{3 g^3} \]
[In]
[Out]
Rule 711
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c f^2+a g^2}{g^2 \sqrt {f+g x}}-\frac {2 c f \sqrt {f+g x}}{g^2}+\frac {c (f+g x)^{3/2}}{g^2}\right ) \, dx \\ & = \frac {2 \left (c f^2+a g^2\right ) \sqrt {f+g x}}{g^3}-\frac {4 c f (f+g x)^{3/2}}{3 g^3}+\frac {2 c (f+g x)^{5/2}}{5 g^3} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\frac {2 \sqrt {f+g x} \left (15 a g^2+c \left (8 f^2-4 f g x+3 g^2 x^2\right )\right )}{15 g^3} \]
[In]
[Out]
Time = 0.40 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {2 \left (3 \left (c \,x^{2}+5 a \right ) g^{2}-4 c f x g +8 c \,f^{2}\right ) \sqrt {g x +f}}{15 g^{3}}\) | \(40\) |
gosper | \(\frac {2 \sqrt {g x +f}\, \left (3 c \,x^{2} g^{2}-4 c f x g +15 a \,g^{2}+8 c \,f^{2}\right )}{15 g^{3}}\) | \(41\) |
trager | \(\frac {2 \sqrt {g x +f}\, \left (3 c \,x^{2} g^{2}-4 c f x g +15 a \,g^{2}+8 c \,f^{2}\right )}{15 g^{3}}\) | \(41\) |
risch | \(\frac {2 \sqrt {g x +f}\, \left (3 c \,x^{2} g^{2}-4 c f x g +15 a \,g^{2}+8 c \,f^{2}\right )}{15 g^{3}}\) | \(41\) |
derivativedivides | \(\frac {\frac {2 c \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {4 c f \left (g x +f \right )^{\frac {3}{2}}}{3}+2 a \,g^{2} \sqrt {g x +f}+2 c \,f^{2} \sqrt {g x +f}}{g^{3}}\) | \(52\) |
default | \(\frac {\frac {2 c \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {4 c f \left (g x +f \right )^{\frac {3}{2}}}{3}+2 a \,g^{2} \sqrt {g x +f}+2 c \,f^{2} \sqrt {g x +f}}{g^{3}}\) | \(52\) |
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.66 \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (3 \, c g^{2} x^{2} - 4 \, c f g x + 8 \, c f^{2} + 15 \, a g^{2}\right )} \sqrt {g x + f}}{15 \, g^{3}} \]
[In]
[Out]
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\begin {cases} \frac {2 a \sqrt {f + g x} + \frac {2 c \left (f^{2} \sqrt {f + g x} - \frac {2 f \left (f + g x\right )^{\frac {3}{2}}}{3} + \frac {\left (f + g x\right )^{\frac {5}{2}}}{5}\right )}{g^{2}}}{g} & \text {for}\: g \neq 0 \\\frac {a x + \frac {c x^{3}}{3}}{\sqrt {f}} & \text {otherwise} \end {cases} \]
[In]
[Out]
none
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {g x + f} a + \frac {{\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} c}{g^{2}}\right )}}{15 \, g} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {g x + f} a + \frac {{\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} c}{g^{2}}\right )}}{15 \, g} \]
[In]
[Out]
Time = 11.83 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{\sqrt {f+g x}} \, dx=\frac {2\,\sqrt {f+g\,x}\,\left (3\,c\,{\left (f+g\,x\right )}^2+15\,a\,g^2+15\,c\,f^2-10\,c\,f\,\left (f+g\,x\right )\right )}{15\,g^3} \]
[In]
[Out]