Integrand size = 20, antiderivative size = 145 \[ \int \frac {x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=-\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {3} \left (1+\sqrt [3]{2} x\right )}{\sqrt {1+x^3}}\right )}{3 \sqrt {3}}+\frac {2 \sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \]
-2/9*2^(2/3)*arctan((1+2^(1/3)*x)*3^(1/2)/(x^3+1)^(1/2))*3^(1/2)+2/9*(1+x) *EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^( 1/2))*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)*3^(3/4)/(x^3+1)^(1/2)/((1+x)/(1+x+ 3^(1/2))^2)^(1/2)
Result contains complex when optimal does not.
Time = 10.32 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.43 \[ \int \frac {x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {2 \sqrt {\frac {1+x}{1+\sqrt [3]{-1}}} \left (-\frac {\left (\sqrt [3]{-1}-x\right ) \sqrt {\frac {\sqrt [3]{-1}-(-1)^{2/3} x}{1+\sqrt [3]{-1}}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right ),\sqrt [3]{-1}\right )}{\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}}}+\frac {i 2^{2/3} \sqrt {1-x+x^2} \operatorname {EllipticPi}\left (\frac {i \sqrt {3}}{\sqrt [3]{-1}+2^{2/3}},\arcsin \left (\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right ),\sqrt [3]{-1}\right )}{\sqrt [3]{-1}+2^{2/3}}\right )}{\sqrt {1+x^3}} \]
(2*Sqrt[(1 + x)/(1 + (-1)^(1/3))]*(-((((-1)^(1/3) - x)*Sqrt[((-1)^(1/3) - (-1)^(2/3)*x)/(1 + (-1)^(1/3))]*EllipticF[ArcSin[Sqrt[(1 + (-1)^(2/3)*x)/( 1 + (-1)^(1/3))]], (-1)^(1/3)])/Sqrt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]) + (I*2^(2/3)*Sqrt[1 - x + x^2]*EllipticPi[(I*Sqrt[3])/((-1)^(1/3) + 2^(2/ 3)), ArcSin[Sqrt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/((-1) ^(1/3) + 2^(2/3))))/Sqrt[1 + x^3]
Time = 0.43 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2564, 759, 2562, 216}
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.
\(\displaystyle \int \frac {x}{\left (x+2^{2/3}\right ) \sqrt {x^3+1}} \, dx\) |
\(\Big \downarrow \) 2564 |
\(\displaystyle \frac {1}{3} \int \frac {1}{\sqrt {x^3+1}}dx-\frac {1}{3} \int \frac {2^{2/3}-2 x}{\left (x+2^{2/3}\right ) \sqrt {x^3+1}}dx\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {2 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {1}{3} \int \frac {2^{2/3}-2 x}{\left (x+2^{2/3}\right ) \sqrt {x^3+1}}dx\) |
\(\Big \downarrow \) 2562 |
\(\displaystyle \frac {2 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {2}{3} 2^{2/3} \int \frac {1}{\frac {3 \left (\sqrt [3]{2} x+1\right )^2}{x^3+1}+1}d\frac {\sqrt [3]{2} x+1}{\sqrt {x^3+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {3} \left (\sqrt [3]{2} x+1\right )}{\sqrt {x^3+1}}\right )}{3 \sqrt {3}}\) |
(-2*2^(2/3)*ArcTan[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[1 + x^3]])/(3*Sqrt[3]) + (2*Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*Elli pticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3*3^( 1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])
3.1.65.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ Symbol] :> Simp[2*(e/d) Subst[Int[1/(1 + 3*a*x^2), x], x, (1 + 2*d*(x/c)) /Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]
Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x _Symbol] :> Simp[(2*d*e + c*f)/(3*c*d) Int[1/Sqrt[a + b*x^3], x], x] + Si mp[(d*e - c*f)/(3*c*d) Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x], x ] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && (EqQ[b*c^3 - 4*a* d^3, 0] || EqQ[b*c^3 + 8*a*d^3, 0]) && NeQ[2*d*e + c*f, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (114 ) = 228\).
Time = 3.69 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.78
method | result | size |
default | \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}-\frac {2 \,2^{\frac {2}{3}} \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \Pi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )}\) | \(258\) |
elliptic | \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}-\frac {2 \,2^{\frac {2}{3}} \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \Pi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )}\) | \(258\) |
2*(3/2-1/2*I*3^(1/2))*((x+1)/(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)*EllipticF(((x+1)/(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))-2*2^(2/3)*(3/2-1/2*I*3^(1/2)) *((x+1)/(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)/(2^(2/3)-1)*EllipticPi(((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^ (1/2))/(2^(2/3)-1),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.57 \[ \int \frac {x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=-\frac {1}{9} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (-\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2 \, x^{5} + 2 \, x^{2} - 2^{\frac {2}{3}} {\left (7 \, x^{4} + 4 \, x\right )} - 2^{\frac {1}{3}} {\left (5 \, x^{3} + 2\right )}\right )} \sqrt {x^{3} + 1}}{12 \, {\left (2 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) + \frac {2}{3} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \]
-1/9*sqrt(3)*2^(2/3)*arctan(-1/12*sqrt(3)*2^(2/3)*(2*x^5 + 2*x^2 - 2^(2/3) *(7*x^4 + 4*x) - 2^(1/3)*(5*x^3 + 2))*sqrt(x^3 + 1)/(2*x^6 + 3*x^3 + 1)) + 2/3*weierstrassPInverse(0, -4, x)
\[ \int \frac {x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\int \frac {x}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 2^{\frac {2}{3}}\right )}\, dx \]
\[ \int \frac {x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\int { \frac {x}{\sqrt {x^{3} + 1} {\left (x + 2^{\frac {2}{3}}\right )}} \,d x } \]
Exception generated. \[ \int \frac {x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[1]%%%} / %%%{%%{[1,0,0]:[1,0,0,-2]%%},[1]%%%} Error: Ba d Argumen
Timed out. \[ \int \frac {x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\int \frac {x}{\sqrt {x^3+1}\,\left (x+2^{2/3}\right )} \,d x \]