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

Optimal. Leaf size=37 \[ \frac {2\ 2^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{2} x+1\right )}{\sqrt {x^3+1}}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2137, 203} \[ \frac {2\ 2^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{2} x+1\right )}{\sqrt {x^3+1}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*2^(2/3)*ArcTan[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[1 + x^3]])/Sqrt[3]

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])

Rule 2137

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(2*e)/d, Subst[Int[
1/(1 + 3*a*x^2), x], x, (1 + (2*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*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rubi steps

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

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Mathematica [C]  time = 0.44, size = 326, normalized size = 8.81 \[ -\frac {4 \sqrt [6]{2} \sqrt {\frac {i (x+1)}{\sqrt {3}+3 i}} \left (\sqrt {2 i x+\sqrt {3}-i} \left (\left (-3 i \sqrt [3]{2}+4 \sqrt {3}+\sqrt [3]{2} \sqrt {3}\right ) x+\sqrt [3]{2} \sqrt {3}-2 \sqrt {3}+3 i \sqrt [3]{2}+6 i\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )-6 i \sqrt {3} \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^2-x+1} \Pi \left (\frac {2 \sqrt {3}}{i+2 i 2^{2/3}+\sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\sqrt {3} \left (1+2\ 2^{2/3}-i \sqrt {3}\right ) \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^3+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*2^(1/6)*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*(Sqrt[-I + Sqrt[3] + (2*I)*x]*(6*I + (3*I)*2^(1/3) - 2*Sqrt[3] +
2^(1/3)*Sqrt[3] + ((-3*I)*2^(1/3) + 4*Sqrt[3] + 2^(1/3)*Sqrt[3])*x)*EllipticF[ArcSin[Sqrt[I + Sqrt[3] - (2*I)
*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])] - (6*I)*Sqrt[3]*Sqrt[I + Sqrt[3] - (2*I)*x]*Sqrt[1 - x +
x^2]*EllipticPi[(2*Sqrt[3])/(I + (2*I)*2^(2/3) + Sqrt[3]), ArcSin[Sqrt[I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4)
)], (2*Sqrt[3])/(3*I + Sqrt[3])]))/(Sqrt[3]*(1 + 2*2^(2/3) - I*Sqrt[3])*Sqrt[I + Sqrt[3] - (2*I)*x]*Sqrt[1 + x
^3])

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fricas [B]  time = 1.57, size = 75, normalized size = 2.03 \[ \frac {1}{3} \, \sqrt {6} 2^{\frac {1}{6}} \arctan \left (-\frac {\sqrt {6} 2^{\frac {1}{6}} {\left (2 \, x^{5} + 2 \, x^{2} - 2^{\frac {2}{3}} {\left (7 \, x^{4} + 4 \, x\right )} - 2^{\frac {1}{3}} {\left (5 \, x^{3} + 2\right )}\right )} \sqrt {x^{3} + 1}}{12 \, {\left (2 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(6)*2^(1/6)*arctan(-1/12*sqrt(6)*2^(1/6)*(2*x^5 + 2*x^2 - 2^(2/3)*(7*x^4 + 4*x) - 2^(1/3)*(5*x^3 + 2))
*sqrt(x^3 + 1)/(2*x^6 + 3*x^3 + 1))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{1,[1]%%%} / %%%{%%{[1,0,0]:[1,0,0,-2]%%},[1]%%%} Error: Bad Arg
ument Value

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maple [C]  time = 0.07, size = 258, normalized size = 6.97 \[ -\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {6 \,2^{\frac {2}{3}} \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticPi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(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))+6*2^(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)/(2^(2/3)-1)*EllipticPi(((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2),(-3/2+1/2*I*3^(1/2))/(2^(2/3)-1),((-3/2+1/2*I
*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {2 \, x - 2^{\frac {2}{3}}}{\sqrt {x^{3} + 1} {\left (x + 2^{\frac {2}{3}}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.59, size = 70, normalized size = 1.89 \[ \frac {2^{2/3}\,\sqrt {3}\,\ln \left (\frac {\left (\sqrt {3}\,1{}\mathrm {i}+\sqrt {x^3+1}+2^{1/3}\,\sqrt {3}\,x\,1{}\mathrm {i}\right )\,{\left (\sqrt {3}\,1{}\mathrm {i}-\sqrt {x^3+1}+2^{1/3}\,\sqrt {3}\,x\,1{}\mathrm {i}\right )}^3}{{\left (x+2^{2/3}\right )}^6}\right )\,1{}\mathrm {i}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(2/3)*3^(1/2)*log(((3^(1/2)*1i + (x^3 + 1)^(1/2) + 2^(1/3)*3^(1/2)*x*1i)*(3^(1/2)*1i - (x^3 + 1)^(1/2) + 2^
(1/3)*3^(1/2)*x*1i)^3)/(x + 2^(2/3))^6)*1i)/3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {2^{\frac {2}{3}}}{x \sqrt {x^{3} + 1} + 2^{\frac {2}{3}} \sqrt {x^{3} + 1}}\right )\, dx - \int \frac {2 x}{x \sqrt {x^{3} + 1} + 2^{\frac {2}{3}} \sqrt {x^{3} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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