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

Optimal. Leaf size=170 \[ \frac {2 \left (e-2^{2/3} f\right ) \tanh ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{2} x+1\right )}{\sqrt {-x^3-1}}\right )}{3 \sqrt {3}}+\frac {2 \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x-\sqrt {3}+1\right )^2}} \left (\sqrt [3]{2} e+f\right ) F\left (\sin ^{-1}\left (\frac {x+\sqrt {3}+1}{x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {-\frac {x+1}{\left (x-\sqrt {3}+1\right )^2}} \sqrt {-x^3-1}} \]

[Out]

2/9*(e-2^(2/3)*f)*arctanh((1+2^(1/3)*x)*3^(1/2)/(-x^3-1)^(1/2))*3^(1/2)+2/9*(2^(1/3)*e+f)*(1+x)*EllipticF((1+x
+3^(1/2))/(1+x-3^(1/2)),2*I-I*3^(1/2))*(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 = 0.22, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2139, 219, 2137, 206} \[ \frac {2 \left (e-2^{2/3} f\right ) \tanh ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{2} x+1\right )}{\sqrt {-x^3-1}}\right )}{3 \sqrt {3}}+\frac {2 \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x-\sqrt {3}+1\right )^2}} \left (\sqrt [3]{2} e+f\right ) F\left (\sin ^{-1}\left (\frac {x+\sqrt {3}+1}{x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {-\frac {x+1}{\left (x-\sqrt {3}+1\right )^2}} \sqrt {-x^3-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e - 2^(2/3)*f)*ArcTanh[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[-1 - x^3]])/(3*Sqrt[3]) + (2*Sqrt[2 - Sqrt[3]]*(2^(1
/3)*e + f)*(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 206

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

Rule 219

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 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[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 {e+f x}{\left (2^{2/3}+x\right ) \sqrt {-1-x^3}} \, dx &=\frac {1}{6} \left (\sqrt [3]{2} e-2 f\right ) \int \frac {2^{2/3}-2 x}{\left (2^{2/3}+x\right ) \sqrt {-1-x^3}} \, dx+\frac {1}{3} \left (\sqrt [3]{2} e+f\right ) \int \frac {1}{\sqrt {-1-x^3}} \, dx\\ &=\frac {2 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{2} e+f\right ) (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 (e-2^{2/3} f\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 (e-2^{2/3} f\right ) \tanh ^{-1}\left (\frac {\sqrt {3} \left (1+\sqrt [3]{2} x\right )}{\sqrt {-1-x^3}}\right )}{3 \sqrt {3}}+\frac {2 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{2} e+f\right ) (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 = 0.46, size = 342, normalized size = 2.01 \[ \frac {2 \sqrt [6]{2} \sqrt {\frac {i (x+1)}{\sqrt {3}+3 i}} \left (f \sqrt {2 i x+\sqrt {3}-i} \left (\left (3 \sqrt [3]{2}+4 i \sqrt {3}+i \sqrt [3]{2} \sqrt {3}\right ) x+i \sqrt [3]{2} \sqrt {3}-2 i \sqrt {3}-3 \sqrt [3]{2}-6\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )-2 \sqrt {3} \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^2-x+1} \left (\sqrt [3]{2} e-2 f\right ) \Pi \left (\frac {2 \sqrt {3}}{i+2 i 2^{2/3}+\sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\sqrt {3} \left (i+2 i 2^{2/3}+\sqrt {3}\right ) \sqrt {-2 i x+\sqrt {3}+i} \sqrt {-x^3-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*2^(1/6)*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*(f*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])] - 2*Sqrt[3]*(2^(1/3)*e - 2*f)*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]*(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 = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (f x^{3} + e x^{2} - 2^{\frac {2}{3}} {\left (f x^{2} + e x\right )} + 2 \cdot 2^{\frac {1}{3}} {\left (f x + e\right )}\right )} \sqrt {-x^{3} - 1}}{x^{6} + 5 \, x^{3} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(f*x^3 + e*x^2 - 2^(2/3)*(f*x^2 + e*x) + 2*2^(1/3)*(f*x + e))*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((f*x+e)/(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 [A]  time = 0.04, size = 255, normalized size = 1.50 \[ -\frac {2 i \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x +1}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, f \EllipticF \left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}-1}}-\frac {2 i \left (e -2^{\frac {2}{3}} f \right ) \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x +1}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{2^{\frac {2}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}-1}\, \left (2^{\frac {2}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*I*f*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)*EllipticF(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))-2/3*I*(e-2^(2/3)*f)*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)/(2^(2/3)+1/2+1/2*I*3^(1/2))*Ellipt
icPi(1/3*3^(1/2)*(I*(x-1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),I*3^(1/2)/(2^(2/3)+1/2+1/2*I*3^(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 \[ \int \frac {f x + e}{\sqrt {-x^{3} - 1} {\left (x + 2^{\frac {2}{3}}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e + f*x)/((- 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 \[ \int \frac {e + f x}{\sqrt {- \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 2^{\frac {2}{3}}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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