Integrand size = 35, antiderivative size = 38 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {1-d} (1-x)}{\sqrt {1-x^3}}\right )}{\sqrt {1-d}} \] Output:
-2*arctan((1-d)^(1/2)*(1-x)/(-x^3+1)^(1/2))/(1-d)^(1/2)
Time = 3.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {-1+d} \sqrt {1-x^3}}{1+x+x^2}\right )}{\sqrt {-1+d}} \] Input:
Integrate[(2 + 2*x - x^2)/((2 - d + d*x + x^2)*Sqrt[1 - x^3]),x]
Output:
(-2*ArcTanh[(Sqrt[-1 + d]*Sqrt[1 - x^3])/(1 + x + x^2)])/Sqrt[-1 + d]
Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2570, 217}
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.
\(\displaystyle \int \frac {-x^2+2 x+2}{\sqrt {1-x^3} \left (d x-d+x^2+2\right )} \, dx\) |
\(\Big \downarrow \) 2570 |
\(\displaystyle 4 \int \frac {1}{-\frac {2 (1-d) (1-x)^2}{1-x^3}-2}d\frac {1-x}{\sqrt {1-x^3}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {2 \arctan \left (\frac {\sqrt {1-d} (1-x)}{\sqrt {1-x^3}}\right )}{\sqrt {1-d}}\) |
Input:
Int[(2 + 2*x - x^2)/((2 - d + d*x + x^2)*Sqrt[1 - x^3]),x]
Output:
(-2*ArcTan[(Sqrt[1 - d]*(1 - x))/Sqrt[1 - x^3]])/Sqrt[1 - d]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((f_) + (g_.)*(x_) + (h_.)*(x_)^2)/(((c_) + (d_.)*(x_) + (e_.)*(x_)^2)* Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Simp[-2*g*h Subst[Int[1/(2*e*h - (b*d*f - 2*a*e*h)*x^2), x], x, (1 + 2*h*(x/g))/Sqrt[a + b*x^3]], x] /; Fre eQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b*d*f - 2*a*e*h, 0] && EqQ[b*g^3 - 8 *a*h^3, 0] && EqQ[g^2 + 2*f*h, 0] && EqQ[b*d*f + b*c*g - 4*a*e*h, 0]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.73 (sec) , antiderivative size = 1908, normalized size of antiderivative = 50.21
method | result | size |
default | \(\text {Expression too large to display}\) | \(1908\) |
elliptic | \(\text {Expression too large to display}\) | \(1919\) |
Input:
int((-x^2+2*x+2)/(d*x+x^2-d+2)/(-x^3+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3*I*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((x-1)/(-3/2+1/2*I*3 ^(1/2)))^(1/2)*(-I*(x+1/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(-x^3+1)^(1/2)*Ell ipticF(1/3*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),(I*3^(1/2)/(-3/ 2+1/2*I*3^(1/2)))^(1/2))+1/3*I/(d^2+4*d-8)^(1/2)*3^(1/2)*(I*x*3^(1/2)+1/2* I*3^(1/2)+3/2)^(1/2)*(1/(-3/2+1/2*I*3^(1/2))*x-1/(-3/2+1/2*I*3^(1/2)))^(1/ 2)*(-I*x*3^(1/2)-1/2*I*3^(1/2)+3/2)^(1/2)/(-x^3+1)^(1/2)/(-1/2+1/2*I*3^(1/ 2)+1/2*d-1/2*(d^2+4*d-8)^(1/2))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2-1/2*I*3^( 1/2))*3^(1/2))^(1/2),I*3^(1/2)/(-1/2+1/2*I*3^(1/2)+1/2*d-1/2*(d^2+4*d-8)^( 1/2)),(I*3^(1/2)/(-3/2+1/2*I*3^(1/2)))^(1/2))*d^2-1/3*I*3^(1/2)*(I*x*3^(1/ 2)+1/2*I*3^(1/2)+3/2)^(1/2)*(1/(-3/2+1/2*I*3^(1/2))*x-1/(-3/2+1/2*I*3^(1/2 )))^(1/2)*(-I*x*3^(1/2)-1/2*I*3^(1/2)+3/2)^(1/2)/(-x^3+1)^(1/2)/(-1/2+1/2* I*3^(1/2)+1/2*d-1/2*(d^2+4*d-8)^(1/2))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2-1/ 2*I*3^(1/2))*3^(1/2))^(1/2),I*3^(1/2)/(-1/2+1/2*I*3^(1/2)+1/2*d-1/2*(d^2+4 *d-8)^(1/2)),(I*3^(1/2)/(-3/2+1/2*I*3^(1/2)))^(1/2))*d+4/3*I/(d^2+4*d-8)^( 1/2)*3^(1/2)*(I*x*3^(1/2)+1/2*I*3^(1/2)+3/2)^(1/2)*(1/(-3/2+1/2*I*3^(1/2)) *x-1/(-3/2+1/2*I*3^(1/2)))^(1/2)*(-I*x*3^(1/2)-1/2*I*3^(1/2)+3/2)^(1/2)/(- x^3+1)^(1/2)/(-1/2+1/2*I*3^(1/2)+1/2*d-1/2*(d^2+4*d-8)^(1/2))*EllipticPi(1 /3*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),I*3^(1/2)/(-1/2+1/2*I*3 ^(1/2)+1/2*d-1/2*(d^2+4*d-8)^(1/2)),(I*3^(1/2)/(-3/2+1/2*I*3^(1/2)))^(1/2) )*d-2/3*I*3^(1/2)*(I*x*3^(1/2)+1/2*I*3^(1/2)+3/2)^(1/2)*(1/(-3/2+1/2*I*...
Time = 0.14 (sec) , antiderivative size = 191, normalized size of antiderivative = 5.03 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=\left [\frac {\log \left (-\frac {2 \, {\left (3 \, d - 4\right )} x^{3} - x^{4} - {\left (d^{2} - 2 \, d + 4\right )} x^{2} - 4 \, \sqrt {-x^{3} + 1} {\left ({\left (d - 2\right )} x - x^{2} - d\right )} \sqrt {d - 1} - d^{2} + 2 \, {\left (d^{2} - 2 \, d\right )} x - 4 \, d + 4}{2 \, d x^{3} + x^{4} + {\left (d^{2} - 2 \, d + 4\right )} x^{2} + d^{2} - 2 \, {\left (d^{2} - 2 \, d\right )} x - 4 \, d + 4}\right )}{2 \, \sqrt {d - 1}}, -\frac {\sqrt {-d + 1} \arctan \left (-\frac {\sqrt {-x^{3} + 1} {\left ({\left (d - 2\right )} x - x^{2} - d\right )} \sqrt {-d + 1}}{2 \, {\left ({\left (d - 1\right )} x^{3} - d + 1\right )}}\right )}{d - 1}\right ] \] Input:
integrate((-x^2+2*x+2)/(d*x+x^2-d+2)/(-x^3+1)^(1/2),x, algorithm="fricas")
Output:
[1/2*log(-(2*(3*d - 4)*x^3 - x^4 - (d^2 - 2*d + 4)*x^2 - 4*sqrt(-x^3 + 1)* ((d - 2)*x - x^2 - d)*sqrt(d - 1) - d^2 + 2*(d^2 - 2*d)*x - 4*d + 4)/(2*d* x^3 + x^4 + (d^2 - 2*d + 4)*x^2 + d^2 - 2*(d^2 - 2*d)*x - 4*d + 4))/sqrt(d - 1), -sqrt(-d + 1)*arctan(-1/2*sqrt(-x^3 + 1)*((d - 2)*x - x^2 - d)*sqrt (-d + 1)/((d - 1)*x^3 - d + 1))/(d - 1)]
\[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=- \int \left (- \frac {2 x}{d x \sqrt {1 - x^{3}} - d \sqrt {1 - x^{3}} + x^{2} \sqrt {1 - x^{3}} + 2 \sqrt {1 - x^{3}}}\right )\, dx - \int \frac {x^{2}}{d x \sqrt {1 - x^{3}} - d \sqrt {1 - x^{3}} + x^{2} \sqrt {1 - x^{3}} + 2 \sqrt {1 - x^{3}}}\, dx - \int \left (- \frac {2}{d x \sqrt {1 - x^{3}} - d \sqrt {1 - x^{3}} + x^{2} \sqrt {1 - x^{3}} + 2 \sqrt {1 - x^{3}}}\right )\, dx \] Input:
integrate((-x**2+2*x+2)/(d*x+x**2-d+2)/(-x**3+1)**(1/2),x)
Output:
-Integral(-2*x/(d*x*sqrt(1 - x**3) - d*sqrt(1 - x**3) + x**2*sqrt(1 - x**3 ) + 2*sqrt(1 - x**3)), x) - Integral(x**2/(d*x*sqrt(1 - x**3) - d*sqrt(1 - x**3) + x**2*sqrt(1 - x**3) + 2*sqrt(1 - x**3)), x) - Integral(-2/(d*x*sq rt(1 - x**3) - d*sqrt(1 - x**3) + x**2*sqrt(1 - x**3) + 2*sqrt(1 - x**3)), x)
Exception generated. \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((-x^2+2*x+2)/(d*x+x^2-d+2)/(-x^3+1)^(1/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(d^2-4*(2-d)>0)', see `assume?` f or more de
\[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=\int { -\frac {x^{2} - 2 \, x - 2}{\sqrt {-x^{3} + 1} {\left (d x + x^{2} - d + 2\right )}} \,d x } \] Input:
integrate((-x^2+2*x+2)/(d*x+x^2-d+2)/(-x^3+1)^(1/2),x, algorithm="giac")
Output:
integrate(-(x^2 - 2*x - 2)/(sqrt(-x^3 + 1)*(d*x + x^2 - d + 2)), x)
Time = 22.92 (sec) , antiderivative size = 677, normalized size of antiderivative = 17.82 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=\text {Too large to display} \] Input:
int((2*x - x^2 + 2)/((1 - x^3)^(1/2)*(d*x - d + x^2 + 2)),x)
Output:
(2*((3^(1/2)*1i)/2 + 3/2)*(x^3 - 1)^(1/2)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3 ^(1/2)*1i)/2 - 3/2))^(1/2)*((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)*ellipticF(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)))/((1 - x^3)^(1/2)*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) + x^3)^(1/2)) + (2* ((3^(1/2)*1i)/2 + 3/2)*(x^3 - 1)^(1/2)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1 /2)*1i)/2 - 3/2))^(1/2)*((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)*ellipticPi(((3^(1/2)*1i)/2 + 3/2)/(d/2 - (4*d + d^2 - 8)^(1/2)/2 + 1), 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))*(d + (d + 2)*(d/2 - (4*d + d^2 - 8)^(1/2)/2) - 4))/((1 - x^3)^(1/2)*(((3^(1/2)*1i) /2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i) /2 + 1/2) + 1) + x^3)^(1/2)*(d/2 - (4*d + d^2 - 8)^(1/2)/2 + 1)*(4*d + d^2 - 8)^(1/2)) - (2*((3^(1/2)*1i)/2 + 3/2)*(x^3 - 1)^(1/2)*(-(x - (3^(1/2)*1 i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((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)*elliptic Pi(((3^(1/2)*1i)/2 + 3/2)/(d/2 + (4*d + d^2 - 8)^(1/2)/2 + 1), 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))*(d + (d + 2)*(d/2 + (4*d + d^2 - 8)^(1/2)/2) - 4))/((1 - x^3)...
\[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {1-x^3}} \, dx=-2 \left (\int \frac {\sqrt {-x^{3}+1}}{d \,x^{4}+x^{5}-d \,x^{3}+2 x^{3}-d x -x^{2}+d -2}d x \right )+\int \frac {\sqrt {-x^{3}+1}\, x^{2}}{d \,x^{4}+x^{5}-d \,x^{3}+2 x^{3}-d x -x^{2}+d -2}d x -2 \left (\int \frac {\sqrt {-x^{3}+1}\, x}{d \,x^{4}+x^{5}-d \,x^{3}+2 x^{3}-d x -x^{2}+d -2}d x \right ) \] Input:
int((-x^2+2*x+2)/(d*x+x^2-d+2)/(-x^3+1)^(1/2),x)
Output:
- 2*int(sqrt( - x**3 + 1)/(d*x**4 - d*x**3 - d*x + d + x**5 + 2*x**3 - x* *2 - 2),x) + int((sqrt( - x**3 + 1)*x**2)/(d*x**4 - d*x**3 - d*x + d + x** 5 + 2*x**3 - x**2 - 2),x) - 2*int((sqrt( - x**3 + 1)*x)/(d*x**4 - d*x**3 - d*x + d + x**5 + 2*x**3 - x**2 - 2),x)