Integrand size = 58, antiderivative size = 171 \[ \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=-\frac {\left (d^2-a f^2\right )^2 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-2+n}}{4 e f (2-n)}-\frac {\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n}{2 e f n}+\frac {\left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{2+n}}{4 e f (2+n)} \]
-1/4*(-a*f^2+d^2)^2*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(-2+n)/e/f /(2-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))^n/e/f/ n+1/4*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(2+n)/e/f/(2+n)
Time = 2.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^n \left (\frac {2 \left (-d^2+a f^2\right )}{n}+\frac {\left (d^2-a f^2\right )^2}{(-2+n) \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}+\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}{2+n}\right )}{4 e f} \]
Integrate[Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2]*(d + e*x + f*Sqrt[a + (2 *d*e*x)/f^2 + (e^2*x^2)/f^2])^n,x]
((d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^n*((2*(-d^2 + a*f^2))/n + ( d^2 - a*f^2)^2/((-2 + n)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2) + (d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2/(2 + n)))/(4*e*f)
Time = 0.41 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2546, 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.
\(\displaystyle \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n \, dx\) |
\(\Big \downarrow \) 2546 |
\(\displaystyle \frac {2 \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-3} \left (d^2-a f^2-\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^2\right )^2}{8 e}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-3} \left (d^2-a f^2-\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^2\right )^2d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{4 e f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\left (d^2-a f^2\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-3}-2 \left (d^2-a f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-1}+\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n+1}\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{4 e f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\left (d^2-a f^2\right )^2 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-2}}{2-n}-\frac {2 \left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{n}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+2}}{n+2}}{4 e f}\) |
Int[Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2]*(d + e*x + f*Sqrt[a + (2*d*e*x )/f^2 + (e^2*x^2)/f^2])^n,x]
(-(((d^2 - a*f^2)^2*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^ (-2 + n))/(2 - n)) - (2*(d^2 - a*f^2)*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n)/n + (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f ^2])^(2 + n)/(2 + n))/(4*e*f)
3.6.15.3.1 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_.)*(x_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*S qrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Simp[(2/f^(2*m) )*(i/c)^m Subst[Int[x^n*((d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e *x^2)^(2*m + 1)/(-2*d*e + b*f^2 + 2*e*x)^(2*(m + 1))), x], x, d + e*x + f*S qrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n}, x] && Eq Q[e^2 - c*f^2, 0] && EqQ[c*g - a*i, 0] && EqQ[c*h - b*i, 0] && IntegerQ[2*m ] && (IntegerQ[m] || GtQ[i/c, 0])
\[\int \sqrt {a +\frac {2 d e x}{f^{2}}+\frac {e^{2} x^{2}}{f^{2}}}\, {\left (d +e x +f \sqrt {a +\frac {2 d e x}{f^{2}}+\frac {e^{2} x^{2}}{f^{2}}}\right )}^{n}d x\]
Time = 0.33 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.71 \[ \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {{\left (e^{2} n^{2} x^{2} + a f^{2} n^{2} + 2 \, d e n^{2} x - 2 \, a f^{2} + 2 \, d^{2} - 2 \, {\left (e f n x + d f n\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}}\right )} {\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n}}{e f n^{3} - 4 \, e f n} \]
integrate((a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^ 2/f^2)^(1/2))^n,x, algorithm="fricas")
(e^2*n^2*x^2 + a*f^2*n^2 + 2*d*e*n^2*x - 2*a*f^2 + 2*d^2 - 2*(e*f*n*x + d* f*n)*sqrt((e^2*x^2 + a*f^2 + 2*d*e*x)/f^2))*(e*x + f*sqrt((e^2*x^2 + a*f^2 + 2*d*e*x)/f^2) + d)^n/(e*f*n^3 - 4*e*f*n)
Exception generated. \[ \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\text {Exception raised: HeuristicGCDFailed} \]
integrate((a+2*d*e*x/f**2+e**2*x**2/f**2)**(1/2)*(d+e*x+f*(a+2*d*e*x/f**2+ e**2*x**2/f**2)**(1/2))**n,x)
\[ \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int { \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n} \,d x } \]
integrate((a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^ 2/f^2)^(1/2))^n,x, algorithm="maxima")
integrate(sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*(e*x + sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*f + d)^n, x)
\[ \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int { \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n} \,d x } \]
integrate((a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^ 2/f^2)^(1/2))^n,x, algorithm="giac")
integrate(sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*(e*x + sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*f + d)^n, x)
Timed out. \[ \int \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int {\left (d+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}}+e\,x\right )}^n\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}} \,d x \]
int((d + f*(a + (e^2*x^2)/f^2 + (2*d*e*x)/f^2)^(1/2) + e*x)^n*(a + (e^2*x^ 2)/f^2 + (2*d*e*x)/f^2)^(1/2),x)