\(\int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx\) [216]

Optimal result

Integrand size = 18, antiderivative size = 137 \[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=\frac {2}{3} e \arctan \left (\sqrt {-1+x^3}\right )-\frac {2 \sqrt {2-\sqrt {3}} f (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.38 \[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=\frac {2}{3} e \arctan \left (\sqrt {-1+x^3}\right )+\frac {f x \sqrt {1-x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},x^3\right )}{\sqrt {-1+x^3}} \] Input:

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

Output:

(2*e*ArcTan[Sqrt[-1 + x^3]])/3 + (f*x*Sqrt[1 - x^3]*Hypergeometric2F1[1/3, 
 1/2, 4/3, x^3])/Sqrt[-1 + x^3]
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2371, 27, 760, 798, 73, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2371

\(\displaystyle e \int \frac {1}{x \sqrt {x^3-1}}dx+\int \frac {f}{\sqrt {x^3-1}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle e \int \frac {1}{x \sqrt {x^3-1}}dx+f \int \frac {1}{\sqrt {x^3-1}}dx\)

\(\Big \downarrow \) 760

\(\displaystyle e \int \frac {1}{x \sqrt {x^3-1}}dx-\frac {2 \sqrt {2-\sqrt {3}} f (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 798

\(\displaystyle \frac {1}{3} e \int \frac {1}{x^3 \sqrt {x^3-1}}dx^3-\frac {2 \sqrt {2-\sqrt {3}} f (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2}{3} e \int \frac {1}{x^6+1}d\sqrt {x^3-1}-\frac {2 \sqrt {2-\sqrt {3}} f (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2}{3} e \arctan \left (\sqrt {x^3-1}\right )-\frac {2 \sqrt {2-\sqrt {3}} f (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

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

rule 760

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]*Sqrt[(- 
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 
] && NegQ[a]
 

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2371

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Simp[Coeff[Pq, 
x, 0]   Int[1/(x*Sqrt[a + b*x^n]), x], x] + Int[ExpandToSum[(Pq - Coeff[Pq, 
 x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IG 
tQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.68

method	result	size
meijerg	\(\frac {f \sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\sqrt {\operatorname {signum}\left (x^{3}-1\right )}}+\frac {e \sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, \left (\left (-2 \ln \left (2\right )+3 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )\right )}{3 \sqrt {\pi }\, \sqrt {\operatorname {signum}\left (x^{3}-1\right )}}\)	\(93\)
default	\(\frac {2 f \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}+\frac {2 e \arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(129\)
elliptic	\(\frac {2 f \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}+\frac {2 e \arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(129\)

Input:

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

Output:

f/signum(x^3-1)^(1/2)*(-signum(x^3-1))^(1/2)*x*hypergeom([1/3,1/2],[4/3],x 
^3)+1/3*e/Pi^(1/2)/signum(x^3-1)^(1/2)*(-signum(x^3-1))^(1/2)*((-2*ln(2)+3 
*ln(x)+I*Pi)*Pi^(1/2)-2*Pi^(1/2)*ln(1/2+1/2*(-x^3+1)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.19 \[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=\frac {1}{3} \, e \arctan \left (\frac {x^{3} - 2}{2 \, \sqrt {x^{3} - 1}}\right ) + 2 \, f {\rm weierstrassPInverse}\left (0, 4, x\right ) \] Input:

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

Output:

1/3*e*arctan(1/2*(x^3 - 2)/sqrt(x^3 - 1)) + 2*f*weierstrassPInverse(0, 4, 
x)
 

Sympy [A] (verification not implemented)

Time = 1.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.44 \[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=e \left (\begin {cases} \frac {2 i \operatorname {acosh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {2 \operatorname {asin}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {otherwise} \end {cases}\right ) - \frac {i f x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \] Input:

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

Output:

e*Piecewise((2*I*acosh(x**(-3/2))/3, 1/Abs(x**3) > 1), (-2*asin(x**(-3/2)) 
/3, True)) - I*f*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**3)/(3*gamma(4/3 
))
 

Maxima [F]

\[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=\int { \frac {f x + e}{\sqrt {x^{3} - 1} x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=\int { \frac {f x + e}{\sqrt {x^{3} - 1} x} \,d x } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.51 \[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=-\frac {\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (f\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )+e\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (\sqrt {3}-3{}\mathrm {i}\right )\,1{}\mathrm {i}}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:

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

Output:

-((-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + (3^(1/2 
)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(f*ellipticF(asin((-(x - 1)/( 
(3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3 
/2)) + e*ellipticPi((3^(1/2)*1i)/2 + 3/2, asin((-(x - 1)/((3^(1/2)*1i)/2 + 
 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))*(-(x - 1)/ 
((3^(1/2)*1i)/2 + 3/2))^(1/2)*(3^(1/2) - 3i)*1i)/(((3^(1/2)*1i)/2 - 1/2)*( 
(3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 
 1) + x^3)^(1/2)
 

Reduce [F]

\[ \int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx=\frac {\mathit {atan} \left (\frac {\sqrt {x^{3}-1}\, x^{3}-2 \sqrt {x^{3}-1}}{2 x^{3}-2}\right ) e}{3}+\left (\int \frac {\sqrt {x^{3}-1}}{x^{3}-1}d x \right ) f \] Input:

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

Output:

(atan((sqrt(x**3 - 1)*x**3 - 2*sqrt(x**3 - 1))/(2*x**3 - 2))*e + 3*int(sqr 
t(x**3 - 1)/(x**3 - 1),x)*f)/3