3.198 \(\int \frac{2-2 x-x^2}{(2+x^2) \sqrt{1+x^3}} \, dx\)

Optimal. Leaf size=16 \[ 2 \tan ^{-1}\left (\frac{x+1}{\sqrt{x^3+1}}\right ) \]

[Out]

2*ArcTan[(1 + x)/Sqrt[1 + x^3]]

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Rubi [A]  time = 0.0767319, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2146, 203} \[ 2 \tan ^{-1}\left (\frac{x+1}{\sqrt{x^3+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 - 2*x - x^2)/((2 + x^2)*Sqrt[1 + x^3]),x]

[Out]

2*ArcTan[(1 + x)/Sqrt[1 + x^3]]

Rule 2146

Int[((f_) + (g_.)*(x_) + (h_.)*(x_)^2)/(((c_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> -Dist[g
/e, Subst[Int[1/(1 + a*x^2), x], x, (1 + (2*h*x)/g)/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, e, f, g, h}, x] &&
EqQ[b*g^3 - 8*a*h^3, 0] && EqQ[g^2 + 2*f*h, 0] && EqQ[b*c*g - 4*a*e*h, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{2-2 x-x^2}{\left (2+x^2\right ) \sqrt{1+x^3}} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{1+x}{\sqrt{1+x^3}}\right )\\ &=2 \tan ^{-1}\left (\frac{1+x}{\sqrt{1+x^3}}\right )\\ \end{align*}

Mathematica [C]  time = 0.828515, size = 296, normalized size = 18.5 \[ \frac{2 \sqrt{\frac{x+1}{1+\sqrt [3]{-1}}} \sqrt{x^2-x+1} \left (\frac{\sqrt{3} \left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right ) F\left (\sin ^{-1}\left (\sqrt{\frac{(-1)^{2/3} x+1}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{(-1)^{2/3} x+1}-\frac{3 i \left (\sqrt{2}-i\right ) \Pi \left (\frac{2 \sqrt{3}}{-i-2 \sqrt{2}+\sqrt{3}};\sin ^{-1}\left (\sqrt{\frac{(-1)^{2/3} x+1}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{(-1)^{5/6}+\sqrt{2}}+\frac{3 \left (5+i \sqrt{2}+i \sqrt{3}+\sqrt{6}\right ) \Pi \left (\frac{2 \sqrt{3}}{-i+2 \sqrt{2}+\sqrt{3}};\sin ^{-1}\left (\sqrt{\frac{(-1)^{2/3} x+1}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{5 i+2 \sqrt{2}+\sqrt{3}+2 i \sqrt{6}}\right )}{3 \sqrt{x^3+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - 2*x - x^2)/((2 + x^2)*Sqrt[1 + x^3]),x]

[Out]

(2*Sqrt[(1 + x)/(1 + (-1)^(1/3))]*Sqrt[1 - x + x^2]*((Sqrt[3]*(1 + (-1)^(1/3))*((-1)^(1/3) - x)*EllipticF[ArcS
in[Sqrt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/(1 + (-1)^(2/3)*x) - ((3*I)*(-I + Sqrt[2])*Ellipti
cPi[(2*Sqrt[3])/(-I - 2*Sqrt[2] + Sqrt[3]), ArcSin[Sqrt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/((
-1)^(5/6) + Sqrt[2]) + (3*(5 + I*Sqrt[2] + I*Sqrt[3] + Sqrt[6])*EllipticPi[(2*Sqrt[3])/(-I + 2*Sqrt[2] + Sqrt[
3]), ArcSin[Sqrt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/(5*I + 2*Sqrt[2] + Sqrt[3] + (2*I)*Sqrt[6
])))/(3*Sqrt[1 + x^3])

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Maple [C]  time = 0.036, size = 1640, normalized size = 102.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2-2*x+2)/(x^2+2)/(x^3+1)^(1/2),x)

[Out]

-2*(3/2-1/2*I*3^(1/2))*((1+x)/(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(((1+x)/(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*I*2^(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-I*2^(1/2))*E
llipticPi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(-1-I*2^(1/2)),((-3/2+1/2*I*3^(1/2))/(-3/2-1/
2*I*3^(1/2)))^(1/2))-2^(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-I*2^(1/2))*EllipticPi(((1+x)/(3/2-1/2*I*3^(
1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(-1-I*2^(1/2)),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))*3^(1/2)-3*
(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-I*2^(1/2))*EllipticPi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1
/2))/(-1-I*2^(1/2)),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))+I*(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
-I*2^(1/2))*EllipticPi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(-1-I*2^(1/2)),((-3/2+1/2*I*3^(1
/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))*3^(1/2)+3*I*2^(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)/(I*2^(1/2)-1)*Elliptic
Pi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(I*2^(1/2)-1),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1
/2)))^(1/2))+2^(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)/(I*2^(1/2)-1)*EllipticPi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1
/2),(-3/2+1/2*I*3^(1/2))/(I*2^(1/2)-1),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))*3^(1/2)-3*(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)/(I*2^(1/2)-1)*EllipticPi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(I*2^(
1/2)-1),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))+I*(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)/(I*2^(1/2)-1)*E
llipticPi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(I*2^(1/2)-1),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2
*I*3^(1/2)))^(1/2))*3^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x^{2} + 2 \, x - 2}{\sqrt{x^{3} + 1}{\left (x^{2} + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-2*x+2)/(x^2+2)/(x^3+1)^(1/2),x, algorithm="maxima")

[Out]

-integrate((x^2 + 2*x - 2)/(sqrt(x^3 + 1)*(x^2 + 2)), x)

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Fricas [A]  time = 1.48543, size = 54, normalized size = 3.38 \begin{align*} -\arctan \left (\frac{x^{2} - 2 \, x}{2 \, \sqrt{x^{3} + 1}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-2*x+2)/(x^2+2)/(x^3+1)^(1/2),x, algorithm="fricas")

[Out]

-arctan(1/2*(x^2 - 2*x)/sqrt(x^3 + 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{2 x}{x^{2} \sqrt{x^{3} + 1} + 2 \sqrt{x^{3} + 1}}\, dx - \int \frac{x^{2}}{x^{2} \sqrt{x^{3} + 1} + 2 \sqrt{x^{3} + 1}}\, dx - \int - \frac{2}{x^{2} \sqrt{x^{3} + 1} + 2 \sqrt{x^{3} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2-2*x+2)/(x**2+2)/(x**3+1)**(1/2),x)

[Out]

-Integral(2*x/(x**2*sqrt(x**3 + 1) + 2*sqrt(x**3 + 1)), x) - Integral(x**2/(x**2*sqrt(x**3 + 1) + 2*sqrt(x**3
+ 1)), x) - Integral(-2/(x**2*sqrt(x**3 + 1) + 2*sqrt(x**3 + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{2} + 2 \, x - 2}{\sqrt{x^{3} + 1}{\left (x^{2} + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-2*x+2)/(x^2+2)/(x^3+1)^(1/2),x, algorithm="giac")

[Out]

integrate(-(x^2 + 2*x - 2)/(sqrt(x^3 + 1)*(x^2 + 2)), x)