Integrand size = 21, antiderivative size = 253 \[ \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2 \left (1+x^3\right )}{\sqrt {1+x} \left (1+\sqrt {3}+x\right ) \sqrt {1-x+x^2}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}+\frac {2 \sqrt {2} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \]
2*(x^3+1)/(1+x+3^(1/2))/(1+x)^(1/2)/(x^2-x+1)^(1/2)+2/3*EllipticF((1+x-3^( 1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*2^(1/2)*(1+x)^(1/2)*((x^2-x+1)/(1+x+3^( 1/2))^2)^(1/2)*3^(3/4)/(x^2-x+1)^(1/2)/((1+x)/(1+x+3^(1/2))^2)^(1/2)-3^(1/ 4)*EllipticE((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*(1+x)^(1/2)*(1/2*6 ^(1/2)-1/2*2^(1/2))*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)/(x^2-x+1)^(1/2)/((1+ x)/(1+x+3^(1/2))^2)^(1/2)
Result contains complex when optimal does not.
Time = 21.74 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.48 \[ \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {(1+x)^{3/2} \left (\frac {12 \sqrt {-\frac {i}{3 i+\sqrt {3}}} \left (1-x+x^2\right )}{(1+x)^2}+\frac {3 \sqrt {2} \left (1-i \sqrt {3}\right ) \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} E\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {1+x}}+\frac {i \sqrt {2} \left (3 i+\sqrt {3}\right ) \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {1+x}}\right )}{6 \sqrt {-\frac {i}{3 i+\sqrt {3}}} \sqrt {1-x+x^2}} \]
((1 + x)^(3/2)*((12*Sqrt[(-I)/(3*I + Sqrt[3])]*(1 - x + x^2))/(1 + x)^2 + (3*Sqrt[2]*(1 - I*Sqrt[3])*Sqrt[(3*I + Sqrt[3] - (6*I)/(1 + x))/(3*I + Sqr t[3])]*Sqrt[(-3*I + Sqrt[3] + (6*I)/(1 + x))/(-3*I + Sqrt[3])]*EllipticE[I *ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[1 + x] + (I*Sqrt[2]*(3*I + Sqrt[3])*Sqrt[(3*I + Sqrt[3] - (6*I)/(1 + x))/(3*I + Sqrt[3])]*Sqrt[(-3*I + Sqrt[3] + (6*I)/(1 + x))/(-3 *I + Sqrt[3])]*EllipticF[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x ]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[1 + x]))/(6*Sqrt[(-I)/(3*I + Sq rt[3])]*Sqrt[1 - x + x^2])
Time = 0.34 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1210, 832, 759, 2416}
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.
| \(\displaystyle \int \frac {x}{\sqrt {x+1} \sqrt {x^2-x+1}} \, dx\) |
| \(\Big \downarrow \) 1210 |
| \(\displaystyle \frac {\sqrt {x^3+1} \int \frac {x}{\sqrt {x^3+1}}dx}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 832 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx-\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {x^3+1}}dx\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 2416 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {2 \sqrt {x^3+1}}{x+\sqrt {3}+1}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
(Sqrt[1 + x^3]*((2*Sqrt[1 + x^3])/(1 + Sqrt[3] + x) - (3^(1/4)*Sqrt[2 - Sq rt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(Sqrt[(1 + x)/(1 + Sq rt[3] + x)^2]*Sqrt[1 + x^3]) - (2*(1 - Sqrt[3])*Sqrt[2 + Sqrt[3]]*(1 + x)* Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x) /(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])))/(Sqrt[1 + x]*Sqrt[1 - x + x^2])
3.6.5.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] ], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r) Int[1/Sqrt[a + b*x ^3], x], x] + Simp[1/r Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x ]] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p]) Int[(f + g*x)^n*(a*d + c* e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) ]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( (1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt [3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) *s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
Time = 0.68 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.80
| method | result | size |
| elliptic | \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) E\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right ) \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{\sqrt {x^{3}+1}\, \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\) | \(202\) |
| default | \(\frac {\sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \left (-3+i \sqrt {3}\right ) \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \left (i E\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) \sqrt {3}-i F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) \sqrt {3}+3 E\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )\right )}{2 x^{3}+2}\) | \(275\) |
2*(3/2-1/2*I*3^(1/2))*((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2-1/2*I*3^(1 /2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2 )))^(1/2)/(x^3+1)^(1/2)*((-3/2-1/2*I*3^(1/2))*EllipticE(((1+x)/(3/2-1/2*I* 3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))+(1/2+1/ 2*I*3^(1/2))*EllipticF(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1 /2))/(-3/2-1/2*I*3^(1/2)))^(1/2)))*((1+x)*(x^2-x+1))^(1/2)/(1+x)^(1/2)/(x^ 2-x+1)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.04 \[ \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=-2 \, {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\right ) \]
\[ \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {x}{\sqrt {x + 1} \sqrt {x^{2} - x + 1}}\, dx \]
\[ \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int { \frac {x}{\sqrt {x^{2} - x + 1} \sqrt {x + 1}} \,d x } \]
\[ \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int { \frac {x}{\sqrt {x^{2} - x + 1} \sqrt {x + 1}} \,d x } \]
Timed out. \[ \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {x}{\sqrt {x+1}\,\sqrt {x^2-x+1}} \,d x \]