Integrand size = 58, antiderivative size = 297 \[ \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=-\frac {\left (d^2-a f^2\right )^4 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-4+n}}{16 e f^3 (4-n)}+\frac {\left (d^2-a f^2\right )^3 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-2+n}}{4 e f^3 (2-n)}+\frac {3 \left (d^2-a f^2\right )^2 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n}{8 e f^3 n}-\frac {\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{2+n}}{4 e f^3 (2+n)}+\frac {\left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{4+n}}{16 e f^3 (4+n)} \]
-1/16*(-a*f^2+d^2)^4*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(-4+n)/e/ f^3/(4-n)+1/4*(-a*f^2+d^2)^3*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^( -2+n)/e/f^3/(2-n)+3/8*(-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))^n/e/f^3/n-1/4*(-a*f^2+d^2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/ 2))^(2+n)/e/f^3/(2+n)+1/16*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(4+ n)/e/f^3/(4+n)
Time = 6.54 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.77 \[ \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^n \left (\frac {6 \left (d^2-a f^2\right )^2}{n}+\frac {\left (d^2-a f^2\right )^4}{(-4+n) \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^4}-\frac {4 \left (d^2-a f^2\right )^3}{(-2+n) \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}-\frac {4 \left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}{2+n}+\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^4}{4+n}\right )}{16 e f^3} \]
Integrate[(a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2)^(3/2)*(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*((6*(d^2 - a*f^2)^2)/n + (d^2 - a*f^2)^4/((-4 + n)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^4) - (4*(d^2 - a*f^2)^3)/((-2 + n)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f ^2])^2) - (4*(d^2 - a*f^2)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2 )/(2 + n) + (d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^4/(4 + n)))/(16* e*f^3)
Time = 0.47 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.90, 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.
| \(\displaystyle \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n \, dx\) |
| \(\Big \downarrow \) 2546 |
| \(\displaystyle \frac {2 \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-5} \left (d^2-a f^2-\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^2\right )^4}{32 e}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{f^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-5} \left (d^2-a f^2-\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^2\right )^4d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{16 e f^3}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {\int \left (\left (d^2-a f^2\right )^4 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-5}-4 \left (d^2-a f^2\right )^3 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-3}+6 \left (d^2-a f^2\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-1}-4 \left (d^2-a f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n+1}+\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n+3}\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{16 e f^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\left (d^2-a f^2\right )^4 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-4}}{4-n}+\frac {4 \left (d^2-a f^2\right )^3 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-2}}{2-n}+\frac {6 \left (d^2-a f^2\right )^2 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{n}-\frac {4 \left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+2}}{n+2}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+4}}{n+4}}{16 e f^3}\) |
Int[(a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2)^(3/2)*(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)^4*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^ (-4 + n))/(4 - n)) + (4*(d^2 - a*f^2)^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^ 2 + (e^2*x^2)/f^2])^(-2 + n))/(2 - n) + (6*(d^2 - a*f^2)^2*(d + e*x + f*Sq rt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n)/n - (4*(d^2 - a*f^2)*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(2 + n))/(2 + n) + (d + e*x + f *Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(4 + n)/(4 + n))/(16*e*f^3)
3.6.14.3.1 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])
\[\int \left (a +\frac {2 d e x}{f^{2}}+\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} {\left (d +e x +f \sqrt {a +\frac {2 d e x}{f^{2}}+\frac {e^{2} x^{2}}{f^{2}}}\right )}^{n}d x\]
Time = 0.30 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.27 \[ \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\frac {{\left (a^{2} f^{4} n^{4} + 24 \, a^{2} f^{4} - 48 \, a d^{2} f^{2} + {\left (e^{4} n^{4} - 4 \, e^{4} n^{2}\right )} x^{4} + 24 \, d^{4} + 4 \, {\left (d e^{3} n^{4} - 4 \, d e^{3} n^{2}\right )} x^{3} - 4 \, {\left (4 \, a^{2} f^{4} - 3 \, a d^{2} f^{2}\right )} n^{2} + 2 \, {\left ({\left (a e^{2} f^{2} + 2 \, d^{2} e^{2}\right )} n^{4} - 2 \, {\left (5 \, a e^{2} f^{2} + d^{2} e^{2}\right )} n^{2}\right )} x^{2} + 4 \, {\left (a d e f^{2} n^{4} - 2 \, {\left (5 \, a d e f^{2} - 3 \, d^{3} e\right )} n^{2}\right )} x - 4 \, {\left (a d f^{3} n^{3} + {\left (e^{3} f n^{3} - 4 \, e^{3} f n\right )} x^{3} + 3 \, {\left (d e^{2} f n^{3} - 4 \, d e^{2} f n\right )} x^{2} - 2 \, {\left (5 \, a d f^{3} - 3 \, d^{3} f\right )} n + {\left ({\left (a e f^{3} + 2 \, d^{2} e f\right )} n^{3} - 2 \, {\left (5 \, a e f^{3} + d^{2} e f\right )} n\right )} x\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}}\right )} {\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n}}{e f^{3} n^{5} - 20 \, e f^{3} n^{3} + 64 \, e f^{3} n} \]
integrate((a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^ 2/f^2)^(1/2))^n,x, algorithm="fricas")
(a^2*f^4*n^4 + 24*a^2*f^4 - 48*a*d^2*f^2 + (e^4*n^4 - 4*e^4*n^2)*x^4 + 24* d^4 + 4*(d*e^3*n^4 - 4*d*e^3*n^2)*x^3 - 4*(4*a^2*f^4 - 3*a*d^2*f^2)*n^2 + 2*((a*e^2*f^2 + 2*d^2*e^2)*n^4 - 2*(5*a*e^2*f^2 + d^2*e^2)*n^2)*x^2 + 4*(a *d*e*f^2*n^4 - 2*(5*a*d*e*f^2 - 3*d^3*e)*n^2)*x - 4*(a*d*f^3*n^3 + (e^3*f* n^3 - 4*e^3*f*n)*x^3 + 3*(d*e^2*f*n^3 - 4*d*e^2*f*n)*x^2 - 2*(5*a*d*f^3 - 3*d^3*f)*n + ((a*e*f^3 + 2*d^2*e*f)*n^3 - 2*(5*a*e*f^3 + d^2*e*f)*n)*x)*sq rt((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^3*n^5 - 20*e*f^3*n^3 + 64*e*f^3*n)
Exception generated. \[ \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\text {Exception raised: HeuristicGCDFailed} \]
integrate((a+2*d*e*x/f**2+e**2*x**2/f**2)**(3/2)*(d+e*x+f*(a+2*d*e*x/f**2+ e**2*x**2/f**2)**(1/2))**n,x)
\[ \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int { {\left (\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}\right )}^{\frac {3}{2}} {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n} \,d x } \]
integrate((a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^ 2/f^2)^(1/2))^n,x, algorithm="maxima")
integrate((e^2*x^2/f^2 + a + 2*d*e*x/f^2)^(3/2)*(e*x + sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*f + d)^n, x)
\[ \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int { {\left (\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}\right )}^{\frac {3}{2}} {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a + \frac {2 \, d e x}{f^{2}}} f + d\right )}^{n} \,d x } \]
integrate((a+2*d*e*x/f^2+e^2*x^2/f^2)^(3/2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^ 2/f^2)^(1/2))^n,x, algorithm="giac")
integrate((e^2*x^2/f^2 + a + 2*d*e*x/f^2)^(3/2)*(e*x + sqrt(e^2*x^2/f^2 + a + 2*d*e*x/f^2)*f + d)^n, x)
Timed out. \[ \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx=\int {\left (d+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}}+e\,x\right )}^n\,{\left (a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}\right )}^{3/2} \,d x \]
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)^(3/2),x)