Integrand size = 33, antiderivative size = 107 \[ \int \left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n \, dx=\frac {\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-1+n}}{2 e (1-n)}+\frac {\left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{1+n}}{2 e (1+n)} \]
1/2*(-a*f^2+d^2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(-1+n)/e/(1-n )+1/2*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(1+n)/e/(1+n)
Time = 0.01 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.83 \[ \int \left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n \, dx=\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^{-1+n} \left (\frac {-d^2+a f^2}{-1+n}+\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}{1+n}\right )}{2 e} \]
((d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^(-1 + n)*((-d^2 + a*f^2)/(- 1 + n) + (d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2/(1 + n)))/(2*e)
Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2543, 2541, 27, 244, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}+d+e x\right )^n \, dx\) |
\(\Big \downarrow \) 2543 |
\(\displaystyle \int \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^ndx\) |
\(\Big \downarrow \) 2541 |
\(\displaystyle 2 \int -\frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-2} \left (d^2-a f^2-\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^2\right )}{4 e}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-2} \left (d^2-a f^2-\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^2\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{2 e}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-2}-\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^n\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {\left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-1}}{1-n}-\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{n+1}}{2 e}\) |
-1/2*(-(((d^2 - a*f^2)*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2 ])^(-1 + n))/(1 - n)) - (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^ 2])^(1 + n)/(1 + n))/e
3.6.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c _.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2 Subst[Int[(g + h*x^n)^p*((d ^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x )^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e , f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
Int[((g_.) + (h_.)*((u_) + (f_.)*Sqrt[v_])^(n_))^(p_.), x_Symbol] :> Int[(g + h*(ExpandToSum[u, x] + f*Sqrt[ExpandToSum[v, x]])^n)^p, x] /; FreeQ[{f, g, h, n}, x] && LinearQ[u, x] && QuadraticQ[v, x] && !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x]) && EqQ[Coefficient[u, x, 1]^2 - Coefficient[v, x, 2]*f^2, 0] && IntegerQ[p]
\[\int {\left (d +e x +f \sqrt {\frac {a \,f^{2}+e x \left (e x +2 d \right )}{f^{2}}}\right )}^{n}d x\]
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.75 \[ \int \left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n \, dx=\frac {{\left (f n \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} - e x - d\right )} {\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n}}{e n^{2} - e} \]
(f*n*sqrt((e^2*x^2 + a*f^2 + 2*d*e*x)/f^2) - e*x - d)*(e*x + f*sqrt((e^2*x ^2 + a*f^2 + 2*d*e*x)/f^2) + d)^n/(e*n^2 - e)
Timed out. \[ \int \left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n \, dx=\text {Timed out} \]
\[ \int \left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n \, dx=\int { {\left (e x + f \left (\frac {\sqrt {a f^{2} + {\left (e x + 2 \, d\right )} e x}}{f}\right ) + d\right )}^{n} \,d x } \]
\[ \int \left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n \, dx=\int { {\left (e x + f \sqrt {\frac {a f^{2} + {\left (e x + 2 \, d\right )} e x}{f^{2}}} + d\right )}^{n} \,d x } \]
Timed out. \[ \int \left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n \, dx=\int {\left (d+e\,x+f\,\sqrt {\frac {a\,f^2+e\,x\,\left (2\,d+e\,x\right )}{f^2}}\right )}^n \,d x \]