Integrand size = 33, antiderivative size = 107 \[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-1+n}}{2 e (1-n)}+\frac {\left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{1+n}}{2 e (1+n)} \]
[Out]
Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2141, 12, 14} \[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {\left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-1}}{2 e (1-n)}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{2 e (n+1)} \]
[In]
[Out]
Rule 12
Rule 14
Rule 2141
Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^{-2+n} \left (d^2 e-\left (-a e+\frac {2 d^2 e}{f^2}\right ) f^2+e x^2\right )}{4 e^2} \, dx,x,d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right ) \\ & = \frac {\text {Subst}\left (\int x^{-2+n} \left (d^2 e-\left (-a e+\frac {2 d^2 e}{f^2}\right ) f^2+e x^2\right ) \, dx,x,d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )}{2 e^2} \\ & = \frac {\text {Subst}\left (\int \left (-e \left (d^2-a f^2\right ) x^{-2+n}+e x^n\right ) \, dx,x,d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )}{2 e^2} \\ & = \frac {\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-1+n}}{2 e (1-n)}+\frac {\left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{1+n}}{2 e (1+n)} \\ \end{align*}
Time = 1.00 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.83 \[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^{-1+n} \left (\frac {-d^2+a f^2}{-1+n}+\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}{1+n}\right )}{2 e} \]
[In]
[Out]
\[\int {\left (d +e x +f \sqrt {a +\frac {2 d e x}{f^{2}}+\frac {e^{2} x^{2}}{f^{2}}}\right )}^{n}d x\]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.75 \[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {{\left (f n \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} - e x - d\right )} {\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n}}{e n^{2} - e} \]
[In]
[Out]
\[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int \left (d + e x + f \sqrt {a + \frac {2 d e x}{f^{2}} + \frac {e^{2} x^{2}}{f^{2}}}\right )^{n}\, dx \]
[In]
[Out]
\[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int { {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n} \,d x } \]
[In]
[Out]
\[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int { {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n} \,d x } \]
[In]
[Out]
Timed out. \[ \int \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int {\left (d+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}}+e\,x\right )}^n \,d x \]
[In]
[Out]