Integrand size = 28, antiderivative size = 237 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2 \, dx=\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )}{8 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3}{6 e}-\frac {f^2 \left (2 d e-b f^2\right )^2 \left (4 a e^2-b^2 f^2\right )}{16 e^4 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \log \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}{8 e^4} \]
1/8*f^2*(-b*f^2+2*d*e)*(-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^4+1/8*f^2*(-b^2*f^2+4*a*e^2)*(e*x+f*(a+b*x+e^2*x^2/f ^2)^(1/2))/e^3+1/6*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^3/e-1/16*f^2*(-b*f^ 2+2*d*e)^2*(-b^2*f^2+4*a*e^2)/e^4/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2 )^(1/2)))
Time = 0.92 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.74 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2 \, dx=\frac {1}{12} \left (2 x \left (6 d^2+6 a f^2+6 d e x+x \left (3 b f^2+4 e^2 x\right )\right )+\frac {\sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )} \left (-3 b^2 f^5+2 b e f^3 (3 d+e x)+4 e^2 f \left (2 a f^2+e x (3 d+2 e x)\right )\right )}{e^3}+\frac {3 f^2 \left (-2 d e+b f^2\right ) \left (-4 a e^2+b^2 f^2\right ) \text {arctanh}\left (\frac {e x}{f \left (-\sqrt {a}+\sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )}\right )}{e^4}\right ) \]
(2*x*(6*d^2 + 6*a*f^2 + 6*d*e*x + x*(3*b*f^2 + 4*e^2*x)) + (Sqrt[a + x*(b + (e^2*x)/f^2)]*(-3*b^2*f^5 + 2*b*e*f^3*(3*d + e*x) + 4*e^2*f*(2*a*f^2 + e *x*(3*d + 2*e*x))))/e^3 + (3*f^2*(-2*d*e + b*f^2)*(-4*a*e^2 + b^2*f^2)*Arc Tanh[(e*x)/(f*(-Sqrt[a] + Sqrt[a + x*(b + (e^2*x)/f^2)]))])/e^4)/12
Time = 0.44 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.04, 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.
| \(\displaystyle \int \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^2 \, dx\) |
| \(\Big \downarrow \) 2541 |
| \(\displaystyle 2 \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 \left (e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}{\left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle 2 \int \left (-\frac {\left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) f^2}{8 e^3 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}+\frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2}{4 e}+\frac {4 a e^2 f^2-b^2 f^4}{16 e^3}+\frac {\left (4 a e^2-b^2 f^2\right ) \left (2 d e f-b f^3\right )^2}{16 e^3 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2}{32 e^4 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \log \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}{16 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )}{16 e^3}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^3}{12 e}\right )\) |
2*((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^3) + (d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^3/(12*e) + (f^2*(2*d *e - b*f^2)^2*(4*a*e^2 - b^2*f^2))/(32*e^4*(2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]))) + (f^2*(2*d*e - b*f^2)*(4*a*e^2 - b^2* f^2)*Log[2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])]) /(16*e^4))
3.5.74.3.1 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]
Time = 1.15 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.27
| method | result | size |
| default | \(\frac {b \,x^{2} f^{2}}{2}+\frac {e^{2} x^{3}}{3}+a \,f^{2} x +2 f \left (d \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )+e \left (\frac {\left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{3 e^{2}}-\frac {b \,f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{2 e^{2}}\right )\right )+\frac {\left (e x +d \right )^{3}}{3 e}\) | \(300\) |
1/2*b*x^2*f^2+1/3*e^2*x^3+a*f^2*x+2*f*(d*(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*(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))))+1/3*(e*x+d)^3/ e
Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.92 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2 \, dx=\frac {16 \, e^{6} x^{3} + 12 \, {\left (b e^{4} f^{2} + 2 \, d e^{5}\right )} x^{2} + 24 \, {\left (a e^{4} f^{2} + d^{2} e^{4}\right )} x - 3 \, {\left (b^{3} f^{6} + 8 \, a d e^{3} f^{2} - 2 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} f^{4}\right )} \log \left (-b f^{2} - 2 \, e^{2} x + 2 \, e f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) - 2 \, {\left (3 \, b^{2} e f^{5} - 8 \, e^{5} f x^{2} - 2 \, {\left (3 \, b d e^{2} + 4 \, a e^{3}\right )} f^{3} - 2 \, {\left (b e^{3} f^{3} + 6 \, d e^{4} f\right )} x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}{24 \, e^{4}} \]
1/24*(16*e^6*x^3 + 12*(b*e^4*f^2 + 2*d*e^5)*x^2 + 24*(a*e^4*f^2 + d^2*e^4) *x - 3*(b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*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^2*e*f^5 - 8*e^5*f*x^2 - 2*(3*b*d*e^2 + 4*a*e^3)*f^3 - 2*(b*e^3*f^3 + 6*d*e^4*f)*x) *sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))/e^4
Time = 0.71 (sec) , antiderivative size = 490, normalized size of antiderivative = 2.07 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2 \, dx=a f^{2} x + \frac {b f^{2} x^{2}}{2} + d^{2} x + d e x^{2} + 2 d f \left (\begin {cases} \left (\frac {a}{2} - \frac {b^{2} f^{2}}{8 e^{2}}\right ) \left (\begin {cases} \frac {\log {\left (b + \frac {2 e^{2} x}{f^{2}} + 2 \sqrt {\frac {e^{2}}{f^{2}}} \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}} \right )}}{\sqrt {\frac {e^{2}}{f^{2}}}} & \text {for}\: a - \frac {b^{2} f^{2}}{4 e^{2}} \neq 0 \\\frac {\left (\frac {b f^{2}}{2 e^{2}} + x\right ) \log {\left (\frac {b f^{2}}{2 e^{2}} + x \right )}}{\sqrt {\frac {e^{2} \left (\frac {b f^{2}}{2 e^{2}} + x\right )^{2}}{f^{2}}}} & \text {otherwise} \end {cases}\right ) + \left (\frac {b f^{2}}{4 e^{2}} + \frac {x}{2}\right ) \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}} & \text {for}\: \frac {e^{2}}{f^{2}} \neq 0 \\\frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + \frac {2 e^{2} x^{3}}{3} + 2 e f \left (\begin {cases} \left (- \frac {a b f^{2}}{12 e^{2}} - \frac {b f^{2} \left (\frac {a}{3} - \frac {b^{2} f^{2}}{8 e^{2}}\right )}{2 e^{2}}\right ) \left (\begin {cases} \frac {\log {\left (b + \frac {2 e^{2} x}{f^{2}} + 2 \sqrt {\frac {e^{2}}{f^{2}}} \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}} \right )}}{\sqrt {\frac {e^{2}}{f^{2}}}} & \text {for}\: a - \frac {b^{2} f^{2}}{4 e^{2}} \neq 0 \\\frac {\left (\frac {b f^{2}}{2 e^{2}} + x\right ) \log {\left (\frac {b f^{2}}{2 e^{2}} + x \right )}}{\sqrt {\frac {e^{2} \left (\frac {b f^{2}}{2 e^{2}} + x\right )^{2}}{f^{2}}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}} \left (\frac {b f^{2} x}{12 e^{2}} + \frac {x^{2}}{3} + \frac {f^{2} \left (\frac {a}{3} - \frac {b^{2} f^{2}}{8 e^{2}}\right )}{e^{2}}\right ) & \text {for}\: \frac {e^{2}}{f^{2}} \neq 0 \\\frac {2 \left (- \frac {a \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a + b x\right )^{\frac {5}{2}}}{5}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]
a*f**2*x + b*f**2*x**2/2 + d**2*x + d*e*x**2 + 2*d*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*e**2*x**3/3 + 2*e*f*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))
Exception generated. \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2 \, dx=\text {Exception raised: ValueError} \]
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 = 236, normalized size of antiderivative = 1.00 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2 \, dx=\frac {1}{2} \, b f^{2} x^{2} + \frac {2}{3} \, e^{2} x^{3} + a f^{2} x + d e x^{2} + d^{2} x + \frac {1}{12} \, \sqrt {b f^{2} x + e^{2} x^{2} + a f^{2}} {\left (2 \, {\left (\frac {4 \, e x {\left | f \right |}}{f} + \frac {b e^{3} f^{3} {\left | f \right |} + 6 \, d e^{4} f {\left | f \right |}}{e^{4} f^{2}}\right )} x - \frac {3 \, b^{2} e f^{5} {\left | f \right |} - 6 \, b d e^{2} f^{3} {\left | f \right |} - 8 \, a e^{3} f^{3} {\left | f \right |}}{e^{4} f^{2}}\right )} - \frac {{\left (b^{3} f^{5} {\left | f \right |} - 2 \, b^{2} d e f^{3} {\left | f \right |} - 4 \, a b e^{2} f^{3} {\left | f \right |} + 8 \, a d e^{3} f {\left | f \right |}\right )} \log \left ({\left | -b f^{2} - 2 \, {\left (x {\left | e \right |} - \sqrt {b f^{2} x + e^{2} x^{2} + a f^{2}}\right )} {\left | e \right |} \right |}\right )}{8 \, e^{3} {\left | e \right |}} \]
1/2*b*f^2*x^2 + 2/3*e^2*x^3 + a*f^2*x + d*e*x^2 + d^2*x + 1/12*sqrt(b*f^2* x + e^2*x^2 + a*f^2)*(2*(4*e*x*abs(f)/f + (b*e^3*f^3*abs(f) + 6*d*e^4*f*ab s(f))/(e^4*f^2))*x - (3*b^2*e*f^5*abs(f) - 6*b*d*e^2*f^3*abs(f) - 8*a*e^3* f^3*abs(f))/(e^4*f^2)) - 1/8*(b^3*f^5*abs(f) - 2*b^2*d*e*f^3*abs(f) - 4*a* b*e^2*f^3*abs(f) + 8*a*d*e^3*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^3*abs(e))
Timed out. \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2 \, dx=\int {\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^2 \,d x \]