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

Optimal. Leaf size=158 \[ \frac {2 \left (2-3\ 2^{2/3}\right ) \tan ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{2} x+1\right )}{\sqrt {x^3+1}}\right )}{3 \sqrt {3}}+\frac {2 \left (3+2 \sqrt [3]{2}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}} \]

[Out]

2/9*(2-3*2^(2/3))*arctan((1+2^(1/3)*x)*3^(1/2)/(x^3+1)^(1/2))*3^(1/2)+2/9*(3+2*2^(1/3))*(1+x)*EllipticF((1+x-3
^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)*3^(3/4)/(x^3+
1)^(1/2)/((1+x)/(1+x+3^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2139, 218, 2137, 203} \[ \frac {2 \left (2-3\ 2^{2/3}\right ) \tan ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{2} x+1\right )}{\sqrt {x^3+1}}\right )}{3 \sqrt {3}}+\frac {2 \left (3+2 \sqrt [3]{2}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2 - 3*2^(2/3))*ArcTan[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[1 + x^3]])/(3*Sqrt[3]) + (2*(3 + 2*2^(1/3))*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)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^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 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

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]

Rule 2139

Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(2*d*e + c*f)/(3*c
*d), Int[1/Sqrt[a + b*x^3], x], x] + Dist[(d*e - c*f)/(3*c*d), Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x]
, 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[b*c^3 + 8*a*d^3,
0]) && NeQ[2*d*e + c*f, 0]

Rubi steps

\begin {align*} \int \frac {2+3 x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx &=\frac {1}{3} \left (-3+\sqrt [3]{2}\right ) \int \frac {2^{2/3}-2 x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx+\frac {1}{3} \left (3+2 \sqrt [3]{2}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx\\ &=\frac {2 \left (3+2 \sqrt [3]{2}\right ) \sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}+\frac {1}{3} \left (2 \left (2-3\ 2^{2/3}\right )\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 \left (2-3\ 2^{2/3}\right ) \tan ^{-1}\left (\frac {\sqrt {3} \left (1+\sqrt [3]{2} x\right )}{\sqrt {1+x^3}}\right )}{3 \sqrt {3}}+\frac {2 \left (3+2 \sqrt [3]{2}\right ) \sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.52, size = 336, normalized size = 2.13 \[ \frac {2 \sqrt [6]{2} \sqrt {\frac {i (x+1)}{\sqrt {3}+3 i}} \left (3 \sqrt {2 i x+\sqrt {3}-i} \left (\left (3 \sqrt [3]{2}+4 i \sqrt {3}+i \sqrt [3]{2} \sqrt {3}\right ) x+i \sqrt [3]{2} \sqrt {3}-2 i \sqrt {3}-3 \sqrt [3]{2}-6\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )-4 \sqrt {3} \left (\sqrt [3]{2}-3\right ) \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^2-x+1} \Pi \left (\frac {2 \sqrt {3}}{i+2 i 2^{2/3}+\sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\sqrt {3} \left (i+2 i 2^{2/3}+\sqrt {3}\right ) \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^3+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*2^(1/6)*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*(3*Sqrt[-I + Sqrt[3] + (2*I)*x]*(-6 - 3*2^(1/3) - (2*I)*Sqrt[3] +
I*2^(1/3)*Sqrt[3] + (3*2^(1/3) + (4*I)*Sqrt[3] + I*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])] - 4*Sqrt[3]*(-3 + 2^(1/3))*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]/(S
qrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])]))/(Sqrt[3]*(I + (2*I)*2^(2/3) + Sqrt[3])*Sqrt[I + Sqrt[3] - (2*
I)*x]*Sqrt[1 + x^3])

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{3} + 2 \, x^{2} - 2^{\frac {2}{3}} {\left (3 \, x^{2} + 2 \, x\right )} + 2 \cdot 2^{\frac {1}{3}} {\left (3 \, x + 2\right )}\right )} \sqrt {x^{3} + 1}}{x^{6} + 5 \, x^{3} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^3 + 2*x^2 - 2^(2/3)*(3*x^2 + 2*x) + 2*2^(1/3)*(3*x + 2))*sqrt(x^3 + 1)/(x^6 + 5*x^3 + 4), x)

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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+3*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 [B]  time = 0.04, size = 262, normalized size = 1.66 \[ \frac {6 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {2 \left (2-3 \,2^{\frac {2}{3}}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticPi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {3\,x+2}{\sqrt {x^3+1}\,\left (x+2^{2/3}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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