3.2.27 \(\int \frac {e+f x}{(1+\sqrt {3}+x) \sqrt {1+x^3}} \, dx\) [127]

Optimal. Leaf size=173 \[ \frac {\left (e-f-\sqrt {3} f\right ) \tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1+x)}{\sqrt {1+x^3}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}+\frac {\sqrt {2+\sqrt {3}} \left (e-\left (1-\sqrt {3}\right ) f\right ) (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2166, 224, 2165, 209} \begin {gather*} \frac {\sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (e-\left (1-\sqrt {3}\right ) f\right ) F\left (\text {ArcSin}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {\text {ArcTan}\left (\frac {\sqrt {3+2 \sqrt {3}} (x+1)}{\sqrt {x^3+1}}\right ) \left (e-\sqrt {3} f-f\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e - f - Sqrt[3]*f)*ArcTan[(Sqrt[3 + 2*Sqrt[3]]*(1 + x))/Sqrt[1 + x^3]])/Sqrt[3*(3 + 2*Sqrt[3])] + (Sqrt[2 +
Sqrt[3]]*(e - (1 - Sqrt[3])*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^(3/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])

Rule 209

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

Rule 224

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

Rule 2165

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> With[{k = Simplify[(d*e
+ 2*c*f)/(c*f)]}, Dist[(1 + k)*(e/d), Subst[Int[1/(1 + (3 + 2*k)*a*x^2), x], x, (1 + (1 + k)*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^2*c^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6
, 0] && EqQ[6*a*d^4*e - c*f*(b*c^3 - 22*a*d^3), 0]

Rule 2166

Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[-(6*a*d^4*e - c*f*
(b*c^3 - 22*a*d^3))/(c*d*(b*c^3 - 28*a*d^3)), Int[1/Sqrt[a + b*x^3], x], x] + Dist[(d*e - c*f)/(c*d*(b*c^3 - 2
8*a*d^3)), Int[(c*(b*c^3 - 22*a*d^3) + 6*a*d^4*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^2*c^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6, 0] && NeQ[6*a*d^4*e - c*f*(b*c^3 - 2
2*a*d^3), 0]

Rubi steps

\begin {align*} \int \frac {e+f x}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx &=\frac {\left (e-\left (1-\sqrt {3}\right ) f\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{2 \sqrt {3}}+\frac {\left (e-\left (1+\sqrt {3}\right ) f\right ) \int \frac {\left (1+\sqrt {3}\right ) \left (-22+\left (1+\sqrt {3}\right )^3\right )+6 x}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx}{\left (1+\sqrt {3}\right ) \left (-28+\left (1+\sqrt {3}\right )^3\right )}\\ &=\frac {\sqrt {2+\sqrt {3}} \left (e-\left (1-\sqrt {3}\right ) 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^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}-\frac {\left (12 \left (e-\left (1+\sqrt {3}\right ) f\right )\right ) \text {Subst}\left (\int \frac {1}{1+\left (3+2 \sqrt {3}\right ) x^2} \, dx,x,\frac {1+x}{\sqrt {1+x^3}}\right )}{\left (1+\sqrt {3}\right ) \left (-28+\left (1+\sqrt {3}\right )^3\right )}\\ &=\frac {\left (e-f-\sqrt {3} f\right ) \tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1+x)}{\sqrt {1+x^3}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}+\frac {\sqrt {2+\sqrt {3}} \left (e-\left (1-\sqrt {3}\right ) 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^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 20.36, size = 291, normalized size = 1.68 \begin {gather*} \frac {2 \sqrt {\frac {2}{3}} \sqrt {\frac {i (1+x)}{3 i+\sqrt {3}}} \left (3 f \sqrt {-i+\sqrt {3}+2 i x} \left ((-2-i)-\sqrt {3}+\left ((1+2 i)+i \sqrt {3}\right ) x\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {i+\sqrt {3}-2 i x}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )+2 \left (-\sqrt {3} e+\left (3+\sqrt {3}\right ) f\right ) \sqrt {i+\sqrt {3}-2 i x} \sqrt {1-x+x^2} \Pi \left (\frac {2 \sqrt {3}}{3 i+(1+2 i) \sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {i+\sqrt {3}-2 i x}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\left (3 i+(1+2 i) \sqrt {3}\right ) \sqrt {i+\sqrt {3}-2 i x} \sqrt {1+x^3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.29, size = 260, normalized size = 1.50

method	result	size
default	\(\frac {2 f \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {2 \left (e -f -f \sqrt {3}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \EllipticPi \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\)	\(260\)
elliptic	\(\frac {2 f \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {2 \left (e -f -f \sqrt {3}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \EllipticPi \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\)	\(260\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)/(1+x+3^(1/2))/(x^3+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*f*(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))+2/3*(e-f-f*3^(1/2))*(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)*3^(1/2)*EllipticPi(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),1/3*(-3/2+1/2*I*3^(1/2))*3^(1/2),((-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.17, size = 746, normalized size = 4.31 \begin {gather*} \left [-\frac {1}{3} \, {\left (\sqrt {3} {\left (f - e\right )} - 3 \, f\right )} {\rm weierstrassPInverse}\left (0, -4, x\right ) + \frac {1}{12} \, \sqrt {6 \, f e - 2 \, \sqrt {3} {\left (f^{2} + f e + e^{2}\right )} + 3 \, e^{2}} \log \left (-\frac {2 \, f^{2} x^{8} - 32 \, f^{2} x^{7} + 224 \, f^{2} x^{6} - 32 \, f^{2} x^{5} + 224 \, f^{2} x^{4} + 448 \, f^{2} x^{3} + 128 \, f^{2} x^{2} + 256 \, f^{2} x + 4 \, {\left (f x^{6} - 18 \, f x^{5} + 12 \, f x^{4} - 40 \, f x^{3} - 36 \, f x^{2} - 24 \, f x + 2 \, {\left (x^{6} - 9 \, x^{5} + 21 \, x^{4} - 4 \, x^{3} + 12 \, x + 4\right )} e + \sqrt {3} {\left (f x^{6} - 6 \, f x^{5} + 24 \, f x^{4} + 8 \, f x^{3} + 12 \, f x^{2} + 24 \, f x + {\left (x^{6} - 12 \, x^{5} + 18 \, x^{4} - 16 \, x^{3} - 12 \, x^{2} - 8\right )} e + 16 \, f\right )} - 32 \, f\right )} \sqrt {x^{3} + 1} \sqrt {6 \, f e - 2 \, \sqrt {3} {\left (f^{2} + f e + e^{2}\right )} + 3 \, e^{2}} + 224 \, f^{2} - {\left (x^{8} - 16 \, x^{7} + 112 \, x^{6} - 16 \, x^{5} + 112 \, x^{4} + 224 \, x^{3} + 64 \, x^{2} + 128 \, x + 112\right )} e^{2} + 2 \, {\left (f x^{8} - 16 \, f x^{7} + 112 \, f x^{6} - 16 \, f x^{5} + 112 \, f x^{4} + 224 \, f x^{3} + 64 \, f x^{2} + 128 \, f x + 112 \, f\right )} e - 16 \, \sqrt {3} {\left (2 \, f^{2} x^{7} - 4 \, f^{2} x^{6} + 12 \, f^{2} x^{5} + 10 \, f^{2} x^{4} + 4 \, f^{2} x^{3} + 12 \, f^{2} x^{2} + 8 \, f^{2} x + 8 \, f^{2} - {\left (x^{7} - 2 \, x^{6} + 6 \, x^{5} + 5 \, x^{4} + 2 \, x^{3} + 6 \, x^{2} + 4 \, x + 4\right )} e^{2} + 2 \, {\left (f x^{7} - 2 \, f x^{6} + 6 \, f x^{5} + 5 \, f x^{4} + 2 \, f x^{3} + 6 \, f x^{2} + 4 \, f x + 4 \, f\right )} e\right )}}{x^{8} + 8 \, x^{7} + 16 \, x^{6} - 16 \, x^{5} - 56 \, x^{4} + 32 \, x^{3} + 64 \, x^{2} - 64 \, x + 16}\right ), -\frac {1}{3} \, {\left (\sqrt {3} {\left (f - e\right )} - 3 \, f\right )} {\rm weierstrassPInverse}\left (0, -4, x\right ) - \frac {1}{6} \, \sqrt {-6 \, f e + 2 \, \sqrt {3} {\left (f^{2} + f e + e^{2}\right )} - 3 \, e^{2}} \arctan \left (-\frac {{\left (3 \, f x^{2} - 6 \, f x - 6 \, {\left (x + 1\right )} e - \sqrt {3} {\left (f x^{2} + 2 \, f x - {\left (x^{2} - 4 \, x - 2\right )} e + 4 \, f\right )}\right )} \sqrt {x^{3} + 1} \sqrt {-6 \, f e + 2 \, \sqrt {3} {\left (f^{2} + f e + e^{2}\right )} - 3 \, e^{2}}}{6 \, {\left (2 \, f^{2} x^{3} + 2 \, f^{2} - {\left (x^{3} + 1\right )} e^{2} + 2 \, {\left (f x^{3} + f\right )} e\right )}}\right )\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(sqrt(3)*(f - e) - 3*f)*weierstrassPInverse(0, -4, x) + 1/12*sqrt(6*f*e - 2*sqrt(3)*(f^2 + f*e + e^2) +
3*e^2)*log(-(2*f^2*x^8 - 32*f^2*x^7 + 224*f^2*x^6 - 32*f^2*x^5 + 224*f^2*x^4 + 448*f^2*x^3 + 128*f^2*x^2 + 256
*f^2*x + 4*(f*x^6 - 18*f*x^5 + 12*f*x^4 - 40*f*x^3 - 36*f*x^2 - 24*f*x + 2*(x^6 - 9*x^5 + 21*x^4 - 4*x^3 + 12*
x + 4)*e + sqrt(3)*(f*x^6 - 6*f*x^5 + 24*f*x^4 + 8*f*x^3 + 12*f*x^2 + 24*f*x + (x^6 - 12*x^5 + 18*x^4 - 16*x^3
- 12*x^2 - 8)*e + 16*f) - 32*f)*sqrt(x^3 + 1)*sqrt(6*f*e - 2*sqrt(3)*(f^2 + f*e + e^2) + 3*e^2) + 224*f^2 - (
x^8 - 16*x^7 + 112*x^6 - 16*x^5 + 112*x^4 + 224*x^3 + 64*x^2 + 128*x + 112)*e^2 + 2*(f*x^8 - 16*f*x^7 + 112*f*
x^6 - 16*f*x^5 + 112*f*x^4 + 224*f*x^3 + 64*f*x^2 + 128*f*x + 112*f)*e - 16*sqrt(3)*(2*f^2*x^7 - 4*f^2*x^6 + 1
2*f^2*x^5 + 10*f^2*x^4 + 4*f^2*x^3 + 12*f^2*x^2 + 8*f^2*x + 8*f^2 - (x^7 - 2*x^6 + 6*x^5 + 5*x^4 + 2*x^3 + 6*x
^2 + 4*x + 4)*e^2 + 2*(f*x^7 - 2*f*x^6 + 6*f*x^5 + 5*f*x^4 + 2*f*x^3 + 6*f*x^2 + 4*f*x + 4*f)*e))/(x^8 + 8*x^7
+ 16*x^6 - 16*x^5 - 56*x^4 + 32*x^3 + 64*x^2 - 64*x + 16)), -1/3*(sqrt(3)*(f - e) - 3*f)*weierstrassPInverse(
0, -4, x) - 1/6*sqrt(-6*f*e + 2*sqrt(3)*(f^2 + f*e + e^2) - 3*e^2)*arctan(-1/6*(3*f*x^2 - 6*f*x - 6*(x + 1)*e
- sqrt(3)*(f*x^2 + 2*f*x - (x^2 - 4*x - 2)*e + 4*f))*sqrt(x^3 + 1)*sqrt(-6*f*e + 2*sqrt(3)*(f^2 + f*e + e^2) -
3*e^2)/(2*f^2*x^3 + 2*f^2 - (x^3 + 1)*e^2 + 2*(f*x^3 + f)*e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e + f x}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 1 + \sqrt {3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{1,[2]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%} Error: Bad
Argument Va

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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