Integrand size = 20, antiderivative size = 146 \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {3+2 \sqrt {3}} (1+x)}{\sqrt {1+x^3}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}+\frac {\sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \]
1/3*(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^(1/4)/(x^3+1)^(1/2)/((1 +x)/(1+x+3^(1/2))^2)^(1/2)+arctan((1+x)*(3+2*3^(1/2))^(1/2)/(x^3+1)^(1/2)) /(9+6*3^(1/2))^(1/2)
Result contains complex when optimal does not.
Time = 20.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=-\frac {4 \sqrt {2} \sqrt {\frac {i (1+x)}{3 i+\sqrt {3}}} \sqrt {1-x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{3 i+(1+2 i) \sqrt {3}},\arcsin \left (\frac {\sqrt {i+\sqrt {3}-2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{\left (3 i+(1+2 i) \sqrt {3}\right ) \sqrt {1+x^3}} \]
(-4*Sqrt[2]*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*Sqrt[1 - x + x^2]*EllipticPi [(2*Sqrt[3])/(3*I + (1 + 2*I)*Sqrt[3]), ArcSin[Sqrt[I + Sqrt[3] - (2*I)*x] /(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])])/((3*I + (1 + 2*I)*Sqrt[ 3])*Sqrt[1 + x^3])
Time = 0.41 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2560, 27, 759, 2565, 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.
\(\displaystyle \int \frac {1}{\left (x+\sqrt {3}+1\right ) \sqrt {x^3+1}} \, dx\) |
\(\Big \downarrow \) 2560 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {x^3+1}}dx}{2 \sqrt {3}}-\frac {\int \frac {6 \left (x-\sqrt {3}+1\right )}{\left (x+\sqrt {3}+1\right ) \sqrt {x^3+1}}dx}{12 \sqrt {3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {x^3+1}}dx}{2 \sqrt {3}}-\frac {\int \frac {x-\sqrt {3}+1}{\left (x+\sqrt {3}+1\right ) \sqrt {x^3+1}}dx}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {\sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {\int \frac {x-\sqrt {3}+1}{\left (x+\sqrt {3}+1\right ) \sqrt {x^3+1}}dx}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 2565 |
\(\displaystyle \frac {\int \frac {1}{\frac {\left (3+2 \sqrt {3}\right ) (x+1)^2}{x^3+1}+1}d\frac {x+1}{\sqrt {x^3+1}}}{\sqrt {3}}+\frac {\sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {\arctan \left (\frac {\sqrt {3+2 \sqrt {3}} (x+1)}{\sqrt {x^3+1}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}\) |
ArcTan[(Sqrt[3 + 2*Sqrt[3]]*(1 + x))/Sqrt[1 + x^3]]/Sqrt[3*(3 + 2*Sqrt[3]) ] + (Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*Ell ipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3^(3 /4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])
3.1.10.3.1 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[((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[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Simp[-6 *a*(d^3/(c*(b*c^3 - 28*a*d^3))) Int[1/Sqrt[a + b*x^3], x], x] + Simp[1/(c *(b*c^3 - 28*a*d^3)) Int[Simp[c*(b*c^3 - 22*a*d^3) + 6*a*d^4*x, x]/((c + d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b^2*c^6 - 20 *a*b*c^3*d^3 - 8*a^2*d^6, 0]
Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ Symbol] :> With[{k = Simplify[(d*e + 2*c*f)/(c*f)]}, Simp[(1 + k)*(e/d) S ubst[Int[1/(1 + (3 + 2*k)*a*x^2), x], x, (1 + (1 + k)*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^2*c ^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6, 0] && EqQ[6*a*d^4*e - c*f*(b*c^3 - 22*a*d^ 3), 0]
Time = 2.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \Pi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\) | \(132\) |
elliptic | \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \Pi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\) | \(132\) |
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)*3^(1/2)*EllipticPi(((x+1)/(3/2-1/2*I*3^(1/2)))^( 1/2),1/3*(-3/2+1/2*I*3^(1/2))*3^(1/2),((-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 = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=-\frac {1}{6} \, \sqrt {2 \, \sqrt {3} - 3} \arctan \left (\frac {{\left (\sqrt {3} {\left (x^{2} - 4 \, x - 2\right )} - 6 \, x - 6\right )} \sqrt {2 \, \sqrt {3} - 3}}{6 \, \sqrt {x^{3} + 1}}\right ) + \frac {1}{3} \, \sqrt {3} {\rm weierstrassPInverse}\left (0, -4, x\right ) \]
-1/6*sqrt(2*sqrt(3) - 3)*arctan(1/6*(sqrt(3)*(x^2 - 4*x - 2) - 6*x - 6)*sq rt(2*sqrt(3) - 3)/sqrt(x^3 + 1)) + 1/3*sqrt(3)*weierstrassPInverse(0, -4, x)
\[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=\int \frac {1}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 1 + \sqrt {3}\right )}\, dx \]
\[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=\int { \frac {1}{\sqrt {x^{3} + 1} {\left (x + \sqrt {3} + 1\right )}} \,d x } \]
Exception generated. \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{1,[2 ]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%} Error: Bad Argument Value
Timed out. \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=\text {Hanged} \]