Integrand size = 28, antiderivative size = 303 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx=\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \left (e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )}{8 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2}{16 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^4}{8 e}-\frac {f^2 \left (2 d e-b f^2\right )^3 \left (4 a e^2-b^2 f^2\right )}{32 e^5 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {3 f^2 \left (2 d e-b f^2\right )^2 \left (4 a e^2-b^2 f^2\right ) \log \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}{32 e^5} \]
3/32*f^2*(-b*f^2+2*d*e)^2*(-b^2*f^2+4*a*e^2)*ln(b*f^2+2*e*(e*x+f*(a+x*(b*f ^2+e^2*x)/f^2)^(1/2)))/e^5+1/8*f^2*(-b*f^2+2*d*e)*(-b^2*f^2+4*a*e^2)*(e*x+ f*(a+b*x+e^2*x^2/f^2)^(1/2))/e^4+1/16*f^2*(-b^2*f^2+4*a*e^2)*(d+e*x+f*(a+b *x+e^2*x^2/f^2)^(1/2))^2/e^3+1/8*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^4/e-1 /32*f^2*(-b*f^2+2*d*e)^3*(-b^2*f^2+4*a*e^2)/e^5/(b*f^2+2*e*(e*x+f*(a+x*(b* f^2+e^2*x)/f^2)^(1/2)))
Time = 1.27 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.86 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx=\frac {1}{16} \left (8 x \left (2 d^3+3 d^2 e x+e x \left (3 a f^2+2 x \left (b f^2+e^2 x\right )\right )+d \left (6 a f^2+x \left (3 b f^2+4 e^2 x\right )\right )\right )+\frac {\sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )} \left (3 b^3 f^7-2 b^2 e f^5 (6 d+e x)+4 b e^2 f^3 \left (3 d^2-2 a f^2+2 d e x+2 e^2 x^2\right )+8 e^3 f \left (2 a f^2 (2 d+e x)+e x \left (3 d^2+4 d e x+2 e^2 x^2\right )\right )\right )}{e^4}+\frac {3 \left (4 a e^2-b^2 f^2\right ) \left (-2 d e f+b f^3\right )^2 \text {arctanh}\left (\frac {e x}{f \left (-\sqrt {a}+\sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )}\right )}{e^5}\right ) \]
(8*x*(2*d^3 + 3*d^2*e*x + e*x*(3*a*f^2 + 2*x*(b*f^2 + e^2*x)) + d*(6*a*f^2 + x*(3*b*f^2 + 4*e^2*x))) + (Sqrt[a + x*(b + (e^2*x)/f^2)]*(3*b^3*f^7 - 2 *b^2*e*f^5*(6*d + e*x) + 4*b*e^2*f^3*(3*d^2 - 2*a*f^2 + 2*d*e*x + 2*e^2*x^ 2) + 8*e^3*f*(2*a*f^2*(2*d + e*x) + e*x*(3*d^2 + 4*d*e*x + 2*e^2*x^2))))/e ^4 + (3*(4*a*e^2 - b^2*f^2)*(-2*d*e*f + b*f^3)^2*ArcTanh[(e*x)/(f*(-Sqrt[a ] + Sqrt[a + x*(b + (e^2*x)/f^2)]))])/e^5)/16
Time = 0.52 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2541, 1195, 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.
| \(\displaystyle \int \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^3 \, dx\) |
| \(\Big \downarrow \) 2541 |
| \(\displaystyle 2 \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^3 \left (e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}{\left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle 2 \int \left (\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^3}{32 e^4 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )}{16 e^4}+\frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^3}{4 e}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}{16 e^3}-\frac {3 \left (4 a e^2-b^2 f^2\right ) \left (2 d e f-b f^3\right )^2}{32 e^4 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^3}{64 e^5 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}+\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2 \log \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}{64 e^5}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )}{16 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^2}{32 e^3}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^4}{16 e}\right )\) |
2*((f^2*(2*d*e - b*f^2)*(4*a*e^2 - b^2*f^2)*(d + e*x + f*Sqrt[a + b*x + (e ^2*x^2)/f^2]))/(16*e^4) + (f^2*(4*a*e^2 - b^2*f^2)*(d + e*x + f*Sqrt[a + b *x + (e^2*x^2)/f^2])^2)/(32*e^3) + (d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f ^2])^4/(16*e) + (f^2*(2*d*e - b*f^2)^3*(4*a*e^2 - b^2*f^2))/(64*e^5*(2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]))) + (3*f^2*(2*d *e - b*f^2)^2*(4*a*e^2 - b^2*f^2)*Log[2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqr t[a + b*x + (e^2*x^2)/f^2])])/(64*e^5))
3.5.73.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(808\) vs. \(2(283)=566\).
Time = 1.26 (sec) , antiderivative size = 809, normalized size of antiderivative = 2.67
| method | result | size |
| default | \(f^{3} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 \left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{16 e^{2}}\right )+3 f^{2} \left (\frac {e^{3} x^{4}}{4 f^{2}}+\frac {\left (\frac {d \,e^{2}}{f^{2}}+b e \right ) x^{3}}{3}+\frac {\left (a e +b d \right ) x^{2}}{2}+a d x \right )+3 f \left (d^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )+e^{2} \left (\frac {x \left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{4 e^{2}}-\frac {5 b \,f^{2} \left (\frac {\left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{3 e^{2}}-\frac {b \,f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{2 e^{2}}\right )}{8 e^{2}}-\frac {a \,f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{4 e^{2}}\right )+2 e d \left (\frac {\left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{3 e^{2}}-\frac {b \,f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{2 e^{2}}\right )\right )+\frac {\left (e x +d \right )^{4}}{4 e}\) | \(809\) |
f^3*(1/8*(b+2*e^2*x/f^2)/e^2*f^2*(a+b*x+e^2*x^2/f^2)^(3/2)+3/16*(4*e^2/f^2 *a-b^2)/e^2*f^2*(1/4*(b+2*e^2*x/f^2)/e^2*f^2*(a+b*x+e^2*x^2/f^2)^(1/2)+1/8 *(4*e^2/f^2*a-b^2)/e^2*f^2*ln((1/2*b+e^2*x/f^2)/(e^2/f^2)^(1/2)+(a+b*x+e^2 *x^2/f^2)^(1/2))/(e^2/f^2)^(1/2)))+3*f^2*(1/4*e^3/f^2*x^4+1/3*(d*e^2/f^2+b *e)*x^3+1/2*(a*e+b*d)*x^2+a*d*x)+3*f*(d^2*(1/4*(b+2*e^2*x/f^2)/e^2*f^2*(a+ b*x+e^2*x^2/f^2)^(1/2)+1/8*(4*e^2/f^2*a-b^2)/e^2*f^2*ln((1/2*b+e^2*x/f^2)/ (e^2/f^2)^(1/2)+(a+b*x+e^2*x^2/f^2)^(1/2))/(e^2/f^2)^(1/2))+e^2*(1/4*x*(a+ b*x+e^2*x^2/f^2)^(3/2)/e^2*f^2-5/8*b/e^2*f^2*(1/3*(a+b*x+e^2*x^2/f^2)^(3/2 )/e^2*f^2-1/2*b/e^2*f^2*(1/4*(b+2*e^2*x/f^2)/e^2*f^2*(a+b*x+e^2*x^2/f^2)^( 1/2)+1/8*(4*e^2/f^2*a-b^2)/e^2*f^2*ln((1/2*b+e^2*x/f^2)/(e^2/f^2)^(1/2)+(a +b*x+e^2*x^2/f^2)^(1/2))/(e^2/f^2)^(1/2)))-1/4*a/e^2*f^2*(1/4*(b+2*e^2*x/f ^2)/e^2*f^2*(a+b*x+e^2*x^2/f^2)^(1/2)+1/8*(4*e^2/f^2*a-b^2)/e^2*f^2*ln((1/ 2*b+e^2*x/f^2)/(e^2/f^2)^(1/2)+(a+b*x+e^2*x^2/f^2)^(1/2))/(e^2/f^2)^(1/2)) )+2*e*d*(1/3*(a+b*x+e^2*x^2/f^2)^(3/2)/e^2*f^2-1/2*b/e^2*f^2*(1/4*(b+2*e^2 *x/f^2)/e^2*f^2*(a+b*x+e^2*x^2/f^2)^(1/2)+1/8*(4*e^2/f^2*a-b^2)/e^2*f^2*ln ((1/2*b+e^2*x/f^2)/(e^2/f^2)^(1/2)+(a+b*x+e^2*x^2/f^2)^(1/2))/(e^2/f^2)^(1 /2))))+1/4*(e*x+d)^4/e
Time = 0.33 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.14 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx=\frac {32 \, e^{8} x^{4} + 32 \, {\left (b e^{6} f^{2} + 2 \, d e^{7}\right )} x^{3} + 48 \, {\left (d^{2} e^{6} + {\left (b d e^{5} + a e^{6}\right )} f^{2}\right )} x^{2} + 32 \, {\left (3 \, a d e^{5} f^{2} + d^{3} e^{5}\right )} x + 3 \, {\left (b^{4} f^{8} - 16 \, a d^{2} e^{4} f^{2} - 4 \, {\left (b^{3} d e + a b^{2} e^{2}\right )} f^{6} + 4 \, {\left (b^{2} d^{2} e^{2} + 4 \, a b d e^{3}\right )} f^{4}\right )} \log \left (-b f^{2} - 2 \, e^{2} x + 2 \, e f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \, {\left (3 \, b^{3} e f^{7} + 16 \, e^{7} f x^{3} - 4 \, {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} f^{5} + 4 \, {\left (3 \, b d^{2} e^{3} + 8 \, a d e^{4}\right )} f^{3} + 8 \, {\left (b e^{5} f^{3} + 4 \, d e^{6} f\right )} x^{2} - 2 \, {\left (b^{2} e^{3} f^{5} - 12 \, d^{2} e^{5} f - 4 \, {\left (b d e^{4} + 2 \, a e^{5}\right )} f^{3}\right )} x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}{32 \, e^{5}} \]
1/32*(32*e^8*x^4 + 32*(b*e^6*f^2 + 2*d*e^7)*x^3 + 48*(d^2*e^6 + (b*d*e^5 + a*e^6)*f^2)*x^2 + 32*(3*a*d*e^5*f^2 + d^3*e^5)*x + 3*(b^4*f^8 - 16*a*d^2* e^4*f^2 - 4*(b^3*d*e + a*b^2*e^2)*f^6 + 4*(b^2*d^2*e^2 + 4*a*b*d*e^3)*f^4) *log(-b*f^2 - 2*e^2*x + 2*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) + 2*( 3*b^3*e*f^7 + 16*e^7*f*x^3 - 4*(3*b^2*d*e^2 + 2*a*b*e^3)*f^5 + 4*(3*b*d^2* e^3 + 8*a*d*e^4)*f^3 + 8*(b*e^5*f^3 + 4*d*e^6*f)*x^2 - 2*(b^2*e^3*f^5 - 12 *d^2*e^5*f - 4*(b*d*e^4 + 2*a*e^5)*f^3)*x)*sqrt((b*f^2*x + e^2*x^2 + a*f^2 )/f^2))/e^5
Time = 1.14 (sec) , antiderivative size = 1363, normalized size of antiderivative = 4.50 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx=\text {Too large to display} \]
3*a*d*f**2*x + 3*a*e*f**2*x**2/2 + a*f**3*Piecewise(((a/2 - b**2*f**2/(8*e **2))*Piecewise((log(b + 2*e**2*x/f**2 + 2*sqrt(e**2/f**2)*sqrt(a + b*x + e**2*x**2/f**2))/sqrt(e**2/f**2), Ne(a - b**2*f**2/(4*e**2), 0)), ((b*f**2 /(2*e**2) + x)*log(b*f**2/(2*e**2) + x)/sqrt(e**2*(b*f**2/(2*e**2) + x)**2 /f**2), True)) + (b*f**2/(4*e**2) + x/2)*sqrt(a + b*x + e**2*x**2/f**2), N e(e**2/f**2, 0)), (2*(a + b*x)**(3/2)/(3*b), Ne(b, 0)), (sqrt(a)*x, True)) + 3*b*d*f**2*x**2/2 + b*e*f**2*x**3 + b*f**3*Piecewise(((-a*b*f**2/(12*e* *2) - b*f**2*(a/3 - b**2*f**2/(8*e**2))/(2*e**2))*Piecewise((log(b + 2*e** 2*x/f**2 + 2*sqrt(e**2/f**2)*sqrt(a + b*x + e**2*x**2/f**2))/sqrt(e**2/f** 2), Ne(a - b**2*f**2/(4*e**2), 0)), ((b*f**2/(2*e**2) + x)*log(b*f**2/(2*e **2) + x)/sqrt(e**2*(b*f**2/(2*e**2) + x)**2/f**2), True)) + sqrt(a + b*x + e**2*x**2/f**2)*(b*f**2*x/(12*e**2) + x**2/3 + f**2*(a/3 - b**2*f**2/(8* e**2))/e**2), Ne(e**2/f**2, 0)), (2*(-a*(a + b*x)**(3/2)/3 + (a + b*x)**(5 /2)/5)/b**2, Ne(b, 0)), (sqrt(a)*x**2/2, True)) + d**3*x + 3*d**2*e*x**2/2 + 3*d**2*f*Piecewise(((a/2 - b**2*f**2/(8*e**2))*Piecewise((log(b + 2*e** 2*x/f**2 + 2*sqrt(e**2/f**2)*sqrt(a + b*x + e**2*x**2/f**2))/sqrt(e**2/f** 2), Ne(a - b**2*f**2/(4*e**2), 0)), ((b*f**2/(2*e**2) + x)*log(b*f**2/(2*e **2) + x)/sqrt(e**2*(b*f**2/(2*e**2) + x)**2/f**2), True)) + (b*f**2/(4*e* *2) + x/2)*sqrt(a + b*x + e**2*x**2/f**2), Ne(e**2/f**2, 0)), (2*(a + b*x) **(3/2)/(3*b), Ne(b, 0)), (sqrt(a)*x, True)) + 2*d*e**2*x**3 + 6*d*e*f*...
Exception generated. \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b^2*f^2-4*a*e^2>0)', see `assume ?` for mor
Time = 0.35 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.31 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx=b e f^{2} x^{3} + e^{3} x^{4} + \frac {3}{2} \, b d f^{2} x^{2} + \frac {3}{2} \, a e f^{2} x^{2} + 2 \, d e^{2} x^{3} + 3 \, a d f^{2} x + \frac {3}{2} \, d^{2} e x^{2} + d^{3} x + \frac {1}{16} \, \sqrt {b f^{2} x + e^{2} x^{2} + a f^{2}} {\left (2 \, {\left (4 \, {\left (\frac {2 \, e^{2} x {\left | f \right |}}{f} + \frac {b e^{6} f^{4} {\left | f \right |} + 4 \, d e^{7} f^{2} {\left | f \right |}}{e^{6} f^{3}}\right )} x - \frac {b^{2} e^{4} f^{6} {\left | f \right |} - 4 \, b d e^{5} f^{4} {\left | f \right |} - 8 \, a e^{6} f^{4} {\left | f \right |} - 12 \, d^{2} e^{6} f^{2} {\left | f \right |}}{e^{6} f^{3}}\right )} x + \frac {3 \, b^{3} e^{2} f^{8} {\left | f \right |} - 12 \, b^{2} d e^{3} f^{6} {\left | f \right |} - 8 \, a b e^{4} f^{6} {\left | f \right |} + 12 \, b d^{2} e^{4} f^{4} {\left | f \right |} + 32 \, a d e^{5} f^{4} {\left | f \right |}}{e^{6} f^{3}}\right )} + \frac {3 \, {\left (b^{4} f^{7} {\left | f \right |} - 4 \, b^{3} d e f^{5} {\left | f \right |} - 4 \, a b^{2} e^{2} f^{5} {\left | f \right |} + 4 \, b^{2} d^{2} e^{2} f^{3} {\left | f \right |} + 16 \, a b d e^{3} f^{3} {\left | f \right |} - 16 \, a d^{2} e^{4} f {\left | f \right |}\right )} \log \left ({\left | -b f^{2} - 2 \, {\left (x {\left | e \right |} - \sqrt {b f^{2} x + e^{2} x^{2} + a f^{2}}\right )} {\left | e \right |} \right |}\right )}{32 \, e^{4} {\left | e \right |}} \]
b*e*f^2*x^3 + e^3*x^4 + 3/2*b*d*f^2*x^2 + 3/2*a*e*f^2*x^2 + 2*d*e^2*x^3 + 3*a*d*f^2*x + 3/2*d^2*e*x^2 + d^3*x + 1/16*sqrt(b*f^2*x + e^2*x^2 + a*f^2) *(2*(4*(2*e^2*x*abs(f)/f + (b*e^6*f^4*abs(f) + 4*d*e^7*f^2*abs(f))/(e^6*f^ 3))*x - (b^2*e^4*f^6*abs(f) - 4*b*d*e^5*f^4*abs(f) - 8*a*e^6*f^4*abs(f) - 12*d^2*e^6*f^2*abs(f))/(e^6*f^3))*x + (3*b^3*e^2*f^8*abs(f) - 12*b^2*d*e^3 *f^6*abs(f) - 8*a*b*e^4*f^6*abs(f) + 12*b*d^2*e^4*f^4*abs(f) + 32*a*d*e^5* f^4*abs(f))/(e^6*f^3)) + 3/32*(b^4*f^7*abs(f) - 4*b^3*d*e*f^5*abs(f) - 4*a *b^2*e^2*f^5*abs(f) + 4*b^2*d^2*e^2*f^3*abs(f) + 16*a*b*d*e^3*f^3*abs(f) - 16*a*d^2*e^4*f*abs(f))*log(abs(-b*f^2 - 2*(x*abs(e) - sqrt(b*f^2*x + e^2* x^2 + a*f^2))*abs(e)))/(e^4*abs(e))
Timed out. \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx=\int {\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^3 \,d x \]