3.9.39 \(\int \frac {\sqrt {1+x^3} (2+2 x^3+x^6)}{x^7 (-1+x^6)} \, dx\) [839]

3.9.39.1 Optimal result
3.9.39.2 Mathematica [A] (verified)
3.9.39.3 Rubi [C] (warning: unable to verify)
3.9.39.4 Maple [A] (verified)
3.9.39.5 Fricas [A] (verification not implemented)
3.9.39.6 Sympy [B] (verification not implemented)
3.9.39.7 Maxima [F]
3.9.39.8 Giac [A] (verification not implemented)
3.9.39.9 Mupad [B] (verification not implemented)

3.9.39.1 Optimal result

Integrand size = 30, antiderivative size = 63 \[ \int \frac {\sqrt {1+x^3} \left (2+2 x^3+x^6\right )}{x^7 \left (-1+x^6\right )} \, dx=\frac {\sqrt {1+x^3} \left (2+5 x^3\right )}{6 x^6}+\frac {5}{2} \text {arctanh}\left (\sqrt {1+x^3}\right )-\frac {5}{3} \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+x^3}}{\sqrt {2}}\right ) \]

output
1/6*(x^3+1)^(1/2)*(5*x^3+2)/x^6+5/2*arctanh((x^3+1)^(1/2))-5/3*2^(1/2)*arc 
tanh(1/2*(x^3+1)^(1/2)*2^(1/2))
 
3.9.39.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {1+x^3} \left (2+2 x^3+x^6\right )}{x^7 \left (-1+x^6\right )} \, dx=\frac {1}{6} \left (\frac {\sqrt {1+x^3} \left (2+5 x^3\right )}{x^6}+15 \text {arctanh}\left (\sqrt {1+x^3}\right )-10 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+x^3}}{\sqrt {2}}\right )\right ) \]

input
Integrate[(Sqrt[1 + x^3]*(2 + 2*x^3 + x^6))/(x^7*(-1 + x^6)),x]
 
output
((Sqrt[1 + x^3]*(2 + 5*x^3))/x^6 + 15*ArcTanh[Sqrt[1 + x^3]] - 10*Sqrt[2]* 
ArcTanh[Sqrt[1 + x^3]/Sqrt[2]])/6
 
3.9.39.3 Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 3.54 (sec) , antiderivative size = 1234, normalized size of antiderivative = 19.59, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1388, 7276, 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.

\(\displaystyle \int \frac {\sqrt {x^3+1} \left (x^6+2 x^3+2\right )}{x^7 \left (x^6-1\right )} \, dx\)

\(\Big \downarrow \) 1388

\(\displaystyle \int \frac {x^6+2 x^3+2}{x^7 \left (x^3-1\right ) \sqrt {x^3+1}}dx\)

\(\Big \downarrow \) 7276

\(\displaystyle \int \left (\frac {5}{3 (x-1) \sqrt {x^3+1}}-\frac {5}{x \sqrt {x^3+1}}-\frac {2}{x^7 \sqrt {x^3+1}}-\frac {4}{x^4 \sqrt {x^3+1}}+\frac {5 (2 x+1)}{3 \left (x^2+x+1\right ) \sqrt {x^3+1}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5 i (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \arctan \left (\frac {\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}}}{\sqrt {\frac {(3-6 i)-(2-3 i) \sqrt {3}}{(4+6 i)-(2+4 i) \sqrt {3}}} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}}}\right )}{3 \sqrt {2} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {10 \sqrt {\frac {(6-3 i)-(3-2 i) \sqrt {3}}{(-6-4 i)+(4+2 i) \sqrt {3}}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \arctan \left (\frac {\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}}}{\sqrt {\frac {(6-3 i)-(3-2 i) \sqrt {3}}{(-6-4 i)+(4+2 i) \sqrt {3}}} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}}}\right )}{3^{3/4} \left (3 i-\sqrt {3}\right ) \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {5 (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \text {arctanh}\left (\frac {\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}}}{\sqrt {2} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}}}\right )}{3 \sqrt {2} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {5}{2} \text {arctanh}\left (\sqrt {x^3+1}\right )-\frac {20 \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} \left (1+(2+i) \sqrt {3}\right ) \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {20 \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} \left (1+(2-i) \sqrt {3}\right ) \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {10 (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 {2+\sqrt {3}} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {40 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (i+(1+2 i) \sqrt {3}\right )^2}{\left (1-(2+i) \sqrt {3}\right )^2},\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \left (7+i \sqrt {3}\right ) \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {40 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (1+(2+i) \sqrt {3}\right )^2}{\left (i-(1+2 i) \sqrt {3}\right )^2},\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \left (7-i \sqrt {3}\right ) \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {20 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticPi}\left (97+56 \sqrt {3},\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 {5 \sqrt {x^3+1}}{6 x^3}+\frac {\sqrt {x^3+1}}{3 x^6}\)

input
Int[(Sqrt[1 + x^3]*(2 + 2*x^3 + x^6))/(x^7*(-1 + x^6)),x]
 
output
Sqrt[1 + x^3]/(3*x^6) + (5*Sqrt[1 + x^3])/(6*x^3) + (((5*I)/3)*(1 + x)*Sqr 
t[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*ArcTan[(3^(1/4)*Sqrt[(1 + x)/(1 + Sqr 
t[3] + x)^2])/(Sqrt[((3 - 6*I) - (2 - 3*I)*Sqrt[3])/((4 + 6*I) - (2 + 4*I) 
*Sqrt[3])]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2])])/(Sqrt[2]*Sqrt[(1 + x 
)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) + (10*Sqrt[((6 - 3*I) - (3 - 2*I)*Sq 
rt[3])/((-6 - 4*I) + (4 + 2*I)*Sqrt[3])]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + S 
qrt[3] + x)^2]*ArcTan[(3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2])/(Sqrt[(( 
6 - 3*I) - (3 - 2*I)*Sqrt[3])/((-6 - 4*I) + (4 + 2*I)*Sqrt[3])]*Sqrt[(1 - 
x + x^2)/(1 + Sqrt[3] + x)^2])])/(3^(3/4)*(3*I - Sqrt[3])*Sqrt[(1 + x)/(1 
+ Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (5*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt 
[3] + x)^2]*ArcTanh[Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]/(Sqrt[2]*Sqrt[(1 - x 
 + x^2)/(1 + Sqrt[3] + x)^2])])/(3*Sqrt[2]*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^ 
2]*Sqrt[1 + x^3]) + (5*ArcTanh[Sqrt[1 + x^3]])/2 - (10*(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*3^(1/4)*Sqrt[2 + Sqrt[3]]*Sqrt[(1 + x)/(1 + 
Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (20*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*3^(1/4)*(1 + (2 - I)*Sqrt[3])*Sqrt[(1 + x)/( 
1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (20*Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 
 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 ...
 

3.9.39.3.1 Defintions of rubi rules used

rule 1388
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7276
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 
3.9.39.4 Maple [A] (verified)

Time = 3.69 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08

method result size
risch \(\frac {5 x^{6}+7 x^{3}+2}{6 x^{6} \sqrt {x^{3}+1}}-\frac {5 \ln \left (\sqrt {x^{3}+1}-1\right )}{4}-\frac {5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{3}+1}\, \sqrt {2}}{2}\right )}{3}+\frac {5 \ln \left (\sqrt {x^{3}+1}+1\right )}{4}\) \(68\)
trager \(\frac {\sqrt {x^{3}+1}\, \left (5 x^{3}+2\right )}{6 x^{6}}+\frac {5 \ln \left (-\frac {x^{3}+2 \sqrt {x^{3}+1}+2}{x^{3}}\right )}{4}+\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{3}+4 \sqrt {x^{3}+1}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\left (x -1\right ) \left (x^{2}+x +1\right )}\right )}{6}\) \(96\)
default \(\frac {-20 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{3}+1}\, \sqrt {2}}{2}\right ) \sqrt {2}\, x^{6}-15 \ln \left (\sqrt {x^{3}+1}-1\right ) x^{6}+15 \ln \left (\sqrt {x^{3}+1}+1\right ) x^{6}+10 \sqrt {x^{3}+1}\, x^{3}+4 \sqrt {x^{3}+1}}{12 \left (\sqrt {x^{3}+1}-1\right )^{2} \left (\sqrt {x^{3}+1}+1\right )^{2}}\) \(98\)
pseudoelliptic \(\frac {-20 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{3}+1}\, \sqrt {2}}{2}\right ) \sqrt {2}\, x^{6}-15 \ln \left (\sqrt {x^{3}+1}-1\right ) x^{6}+15 \ln \left (\sqrt {x^{3}+1}+1\right ) x^{6}+10 \sqrt {x^{3}+1}\, x^{3}+4 \sqrt {x^{3}+1}}{12 \left (\sqrt {x^{3}+1}-1\right )^{2} \left (\sqrt {x^{3}+1}+1\right )^{2}}\) \(98\)
elliptic \(\text {Expression too large to display}\) \(920\)

input
int((x^3+1)^(1/2)*(x^6+2*x^3+2)/x^7/(x^6-1),x,method=_RETURNVERBOSE)
 
output
1/6*(5*x^6+7*x^3+2)/x^6/(x^3+1)^(1/2)-5/4*ln((x^3+1)^(1/2)-1)-5/3*2^(1/2)* 
arctanh(1/2*(x^3+1)^(1/2)*2^(1/2))+5/4*ln((x^3+1)^(1/2)+1)
 
3.9.39.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {1+x^3} \left (2+2 x^3+x^6\right )}{x^7 \left (-1+x^6\right )} \, dx=\frac {10 \, \sqrt {2} x^{6} \log \left (\frac {x^{3} - 2 \, \sqrt {2} \sqrt {x^{3} + 1} + 3}{x^{3} - 1}\right ) + 15 \, x^{6} \log \left (\sqrt {x^{3} + 1} + 1\right ) - 15 \, x^{6} \log \left (\sqrt {x^{3} + 1} - 1\right ) + 2 \, {\left (5 \, x^{3} + 2\right )} \sqrt {x^{3} + 1}}{12 \, x^{6}} \]

input
integrate((x^3+1)^(1/2)*(x^6+2*x^3+2)/x^7/(x^6-1),x, algorithm="fricas")
 
output
1/12*(10*sqrt(2)*x^6*log((x^3 - 2*sqrt(2)*sqrt(x^3 + 1) + 3)/(x^3 - 1)) + 
15*x^6*log(sqrt(x^3 + 1) + 1) - 15*x^6*log(sqrt(x^3 + 1) - 1) + 2*(5*x^3 + 
 2)*sqrt(x^3 + 1))/x^6
 
3.9.39.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (58) = 116\).

Time = 20.18 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.00 \[ \int \frac {\sqrt {1+x^3} \left (2+2 x^3+x^6\right )}{x^7 \left (-1+x^6\right )} \, dx=\frac {5 \sqrt {2} \left (\log {\left (\sqrt {x^{3} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {x^{3} + 1} + \sqrt {2} \right )}\right )}{6} - \frac {5 \log {\left (\sqrt {x^{3} + 1} - 1 \right )}}{4} + \frac {5 \log {\left (\sqrt {x^{3} + 1} + 1 \right )}}{4} + \frac {5}{12 \left (\sqrt {x^{3} + 1} + 1\right )} - \frac {1}{12 \left (\sqrt {x^{3} + 1} + 1\right )^{2}} + \frac {5}{12 \left (\sqrt {x^{3} + 1} - 1\right )} + \frac {1}{12 \left (\sqrt {x^{3} + 1} - 1\right )^{2}} \]

input
integrate((x**3+1)**(1/2)*(x**6+2*x**3+2)/x**7/(x**6-1),x)
 
output
5*sqrt(2)*(log(sqrt(x**3 + 1) - sqrt(2)) - log(sqrt(x**3 + 1) + sqrt(2)))/ 
6 - 5*log(sqrt(x**3 + 1) - 1)/4 + 5*log(sqrt(x**3 + 1) + 1)/4 + 5/(12*(sqr 
t(x**3 + 1) + 1)) - 1/(12*(sqrt(x**3 + 1) + 1)**2) + 5/(12*(sqrt(x**3 + 1) 
 - 1)) + 1/(12*(sqrt(x**3 + 1) - 1)**2)
 
3.9.39.7 Maxima [F]

\[ \int \frac {\sqrt {1+x^3} \left (2+2 x^3+x^6\right )}{x^7 \left (-1+x^6\right )} \, dx=\int { \frac {{\left (x^{6} + 2 \, x^{3} + 2\right )} \sqrt {x^{3} + 1}}{{\left (x^{6} - 1\right )} x^{7}} \,d x } \]

input
integrate((x^3+1)^(1/2)*(x^6+2*x^3+2)/x^7/(x^6-1),x, algorithm="maxima")
 
output
integrate((x^6 + 2*x^3 + 2)*sqrt(x^3 + 1)/((x^6 - 1)*x^7), x)
 
3.9.39.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {1+x^3} \left (2+2 x^3+x^6\right )}{x^7 \left (-1+x^6\right )} \, dx=\frac {5}{6} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, \sqrt {x^{3} + 1} \right |}}{2 \, {\left (\sqrt {2} + \sqrt {x^{3} + 1}\right )}}\right ) + \frac {5 \, {\left (x^{3} + 1\right )}^{\frac {3}{2}} - 3 \, \sqrt {x^{3} + 1}}{6 \, x^{6}} + \frac {5}{4} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) - \frac {5}{4} \, \log \left ({\left | \sqrt {x^{3} + 1} - 1 \right |}\right ) \]

input
integrate((x^3+1)^(1/2)*(x^6+2*x^3+2)/x^7/(x^6-1),x, algorithm="giac")
 
output
5/6*sqrt(2)*log(1/2*abs(-2*sqrt(2) + 2*sqrt(x^3 + 1))/(sqrt(2) + sqrt(x^3 
+ 1))) + 1/6*(5*(x^3 + 1)^(3/2) - 3*sqrt(x^3 + 1))/x^6 + 5/4*log(sqrt(x^3 
+ 1) + 1) - 5/4*log(abs(sqrt(x^3 + 1) - 1))
 
3.9.39.9 Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 724, normalized size of antiderivative = 11.49 \[ \int \frac {\sqrt {1+x^3} \left (2+2 x^3+x^6\right )}{x^7 \left (-1+x^6\right )} \, dx=\text {Too large to display} \]

input
int(((x^3 + 1)^(1/2)*(2*x^3 + x^6 + 2))/(x^7*(x^6 - 1)),x)
 
output
(5*(x^3 + 1)^(1/2))/(6*x^3) + (x^3 + 1)^(1/2)/(3*x^6) + (15*((3^(1/2)*1i)/ 
2 + 3/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1 
)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/ 
2 + 3/2))^(1/2)*ellipticPi((3^(1/2)*1i)/2 + 3/2, asin(((x + 1)/((3^(1/2)*1 
i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(2*( 
x^3 - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i 
)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)) - (5*((3^(1/2)*1i)/2 + 3/2)*((x 
+ (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1)/((3^(1/2)*1 
i)/2 + 3/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/ 
2)*ellipticPi((3^(1/2)*1i)/4 + 3/4, asin(((x + 1)/((3^(1/2)*1i)/2 + 3/2))^ 
(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(3*(x^3 - x*(((3^ 
(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*(( 
3^(1/2)*1i)/2 + 1/2))^(1/2)) - (10*((3^(1/2)*1i)/2 + 3/2)*((x + (3^(1/2)*1 
i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1)/((3^(1/2)*1i)/2 + 3/2)) 
^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticP 
i(((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 + 1/2), asin(((x + 1)/((3^(1/2)*1 
i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(3*( 
(3^(1/2)*1i)/2 + 1/2)*(x^3 - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1 
/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)) + (10*((3 
^(1/2)*1i)/2 + 3/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))...