Integrand size = 31, antiderivative size = 30 \[ \int \frac {2-2 x-x^2}{\left (2+d+d x+x^2\right ) \sqrt {1+x^3}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {1+d} (1+x)}{\sqrt {1+x^3}}\right )}{\sqrt {1+d}} \] Output:
2*arctan((1+d)^(1/2)*(1+x)/(x^3+1)^(1/2))/(1+d)^(1/2)
Time = 2.63 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {2-2 x-x^2}{\left (2+d+d x+x^2\right ) \sqrt {1+x^3}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {1+d} \sqrt {1+x^3}}{1-x+x^2}\right )}{\sqrt {1+d}} \] Input:
Integrate[(2 - 2*x - x^2)/((2 + d + d*x + x^2)*Sqrt[1 + x^3]),x]
Output:
(2*ArcTan[(Sqrt[1 + d]*Sqrt[1 + x^3])/(1 - x + x^2)])/Sqrt[1 + d]
Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, 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.
\(\displaystyle \int \frac {-x^2-2 x+2}{\sqrt {x^3+1} \left (d x+d+x^2+2\right )} \, dx\) |
\(\Big \downarrow \) 2570 |
\(\displaystyle -4 \int \frac {1}{-\frac {2 (d+1) (x+1)^2}{x^3+1}-2}d\frac {x+1}{\sqrt {x^3+1}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 \arctan \left (\frac {\sqrt {d+1} (x+1)}{\sqrt {x^3+1}}\right )}{\sqrt {d+1}}\) |
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 = 0.76 (sec) , antiderivative size = 4397, normalized size of antiderivative = 146.57
method | result | size |
default | \(\text {Expression too large to display}\) | \(4397\) |
elliptic | \(\text {Expression too large to display}\) | \(4602\) |
Input:
int((-x^2-2*x+2)/(d*x+x^2+d+2)/(x^3+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-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))-3/2/(d^2-4*d-8)^(1/2)*(1/(3/ 2-1/2*I*3^(1/2))*x+1/(3/2-1/2*I*3^(1/2)))^(1/2)*(1/(-3/2-1/2*I*3^(1/2))*x- 1/2/(-3/2-1/2*I*3^(1/2))-1/2*I/(-3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(-3/ 2+1/2*I*3^(1/2))*x-1/2/(-3/2+1/2*I*3^(1/2))+1/2*I/(-3/2+1/2*I*3^(1/2))*3^( 1/2))^(1/2)/(x^3+1)^(1/2)/(-1+1/2*d-1/2*(d^2-4*d-8)^(1/2))*EllipticPi(((x+ 1)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(-1+1/2*d-1/2*(d^2-4*d- 8)^(1/2)),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))*d^2-1/2*I/(d^ 2-4*d-8)^(1/2)*(1/(3/2-1/2*I*3^(1/2))*x+1/(3/2-1/2*I*3^(1/2)))^(1/2)*(1/(- 3/2-1/2*I*3^(1/2))*x-1/2/(-3/2-1/2*I*3^(1/2))-1/2*I/(-3/2-1/2*I*3^(1/2))*3 ^(1/2))^(1/2)*(1/(-3/2+1/2*I*3^(1/2))*x-1/2/(-3/2+1/2*I*3^(1/2))+1/2*I/(-3 /2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3+1)^(1/2)/(-1+1/2*d+1/2*(d^2-4*d-8)^( 1/2))*EllipticPi(((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(- 1+1/2*d+1/2*(d^2-4*d-8)^(1/2)),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2))) ^(1/2))*d^2*3^(1/2)+3/2*(1/(3/2-1/2*I*3^(1/2))*x+1/(3/2-1/2*I*3^(1/2)))^(1 /2)*(1/(-3/2-1/2*I*3^(1/2))*x-1/2/(-3/2-1/2*I*3^(1/2))-1/2*I/(-3/2-1/2*I*3 ^(1/2))*3^(1/2))^(1/2)*(1/(-3/2+1/2*I*3^(1/2))*x-1/2/(-3/2+1/2*I*3^(1/2))+ 1/2*I/(-3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3+1)^(1/2)/(-1+1/2*d-1/2*(...
Time = 0.14 (sec) , antiderivative size = 181, normalized size of antiderivative = 6.03 \[ \int \frac {2-2 x-x^2}{\left (2+d+d x+x^2\right ) \sqrt {1+x^3}} \, dx=\left [-\frac {\sqrt {-d - 1} \log \left (-\frac {2 \, {\left (3 \, d + 4\right )} x^{3} - x^{4} - {\left (d^{2} + 2 \, d + 4\right )} x^{2} - d^{2} + 4 \, \sqrt {x^{3} + 1} {\left ({\left (d + 2\right )} x - x^{2} + d\right )} \sqrt {-d - 1} - 2 \, {\left (d^{2} + 2 \, d\right )} x + 4 \, d + 4}{2 \, d x^{3} + x^{4} + {\left (d^{2} + 2 \, d + 4\right )} x^{2} + d^{2} + 2 \, {\left (d^{2} + 2 \, d\right )} x + 4 \, d + 4}\right )}{2 \, {\left (d + 1\right )}}, -\frac {\arctan \left (-\frac {\sqrt {x^{3} + 1} {\left ({\left (d + 2\right )} x - x^{2} + d\right )} \sqrt {d + 1}}{2 \, {\left ({\left (d + 1\right )} x^{3} + d + 1\right )}}\right )}{\sqrt {d + 1}}\right ] \] Input:
integrate((-x^2-2*x+2)/(d*x+x^2+d+2)/(x^3+1)^(1/2),x, algorithm="fricas")
Output:
[-1/2*sqrt(-d - 1)*log(-(2*(3*d + 4)*x^3 - x^4 - (d^2 + 2*d + 4)*x^2 - d^2 + 4*sqrt(x^3 + 1)*((d + 2)*x - x^2 + d)*sqrt(-d - 1) - 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))/(d + 1), -arctan(-1/2*sqrt(x^3 + 1)*((d + 2)*x - x^2 + d)*sqrt(d + 1)/((d + 1)*x^3 + d + 1))/sqrt(d + 1)]
\[ \int \frac {2-2 x-x^2}{\left (2+d+d x+x^2\right ) \sqrt {1+x^3}} \, dx=- \int \frac {2 x}{d x \sqrt {x^{3} + 1} + d \sqrt {x^{3} + 1} + x^{2} \sqrt {x^{3} + 1} + 2 \sqrt {x^{3} + 1}}\, dx - \int \frac {x^{2}}{d x \sqrt {x^{3} + 1} + d \sqrt {x^{3} + 1} + x^{2} \sqrt {x^{3} + 1} + 2 \sqrt {x^{3} + 1}}\, dx - \int \left (- \frac {2}{d x \sqrt {x^{3} + 1} + d \sqrt {x^{3} + 1} + x^{2} \sqrt {x^{3} + 1} + 2 \sqrt {x^{3} + 1}}\right )\, dx \] 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(x**3 + 1) + d*sqrt(x**3 + 1) + x**2*sqrt(x**3 + 1) + 2*sqrt(x**3 + 1)), x) - Integral(x**2/(d*x*sqrt(x**3 + 1) + d*sqrt(x**3 + 1) + x**2*sqrt(x**3 + 1) + 2*sqrt(x**3 + 1)), x) - Integral(-2/(d*x*sqr t(x**3 + 1) + d*sqrt(x**3 + 1) + x**2*sqrt(x**3 + 1) + 2*sqrt(x**3 + 1)), 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*(d+2)>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 = 22.61 (sec) , antiderivative size = 632, normalized size of antiderivative = 21.07 \[ \int \frac {2-2 x-x^2}{\left (2+d+d x+x^2\right ) \sqrt {1+x^3}} \, dx=\text {Too large to display} \] Input:
int(-(2*x + x^2 - 2)/((x^3 + 1)^(1/2)*(d + d*x + x^2 + 2)),x)
Output:
- (2*((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)*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)))/(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) - (2*((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)*ellip ticPi(((3^(1/2)*1i)/2 + 3/2)/((d^2 - 4*d - 8)^(1/2)/2 - d/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 - (d^2 - 4*d - 8)^(1/2)/2) + 4))/((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)*(d^2 - 4*d - 8)^(1/2)*((d^2 - 4*d - 8)^(1/ 2)/2 - d/2 + 1)) - (2*((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)/(d/2 + (d^2 - 4*d - 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 + (d^2 - 4*d - 8)^(1/2)/2) + 4))/((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)*1...
\[ \int \frac {2-2 x-x^2}{\left (2+d+d x+x^2\right ) \sqrt {1+x^3}} \, dx=2 \left (\int \frac {\sqrt {x^{3}+1}}{d \,x^{4}+x^{5}+d \,x^{3}+2 x^{3}+d x +x^{2}+d +2}d x \right )-\left (\int \frac {\sqrt {x^{3}+1}\, x^{2}}{d \,x^{4}+x^{5}+d \,x^{3}+2 x^{3}+d x +x^{2}+d +2}d x \right )-2 \left (\int \frac {\sqrt {x^{3}+1}\, x}{d \,x^{4}+x^{5}+d \,x^{3}+2 x^{3}+d x +x^{2}+d +2}d x \right ) \] 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)