\(\int \frac {e+f x}{x \sqrt {1+x^3}} \, dx\) [214]

Optimal result

Integrand size = 18, antiderivative size = 120 \[ \int \frac {e+f x}{x \sqrt {1+x^3}} \, dx=-\frac {2}{3} e \text {arctanh}\left (\sqrt {1+x^3}\right )+\frac {2 \sqrt {2+\sqrt {3}} f (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \] Output:

-2/3*e*arctanh((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+x+3^(1/2))^2)^(1/2)*EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2 
)+2*I)*3^(3/4)/((1+x)/(1+x+3^(1/2))^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 = 10.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.28 \[ \int \frac {e+f x}{x \sqrt {1+x^3}} \, dx=-\frac {2}{3} e \text {arctanh}\left (\sqrt {1+x^3}\right )+f x \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-x^3\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 120, 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, 759, 798, 73, 220}

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

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

(-2*e*ArcTanh[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 + Sqr 
t[3] + x)], -7 - 4*Sqrt[3]])/(3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sq 
rt[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 220

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

rule 759

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] & 
& PosQ[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] (verified)

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

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.44

method	result	size
meijerg	\(f x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )+\frac {e \left (\left (-2 \ln \left (2\right )+3 \ln \left (x \right )\right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right )\right )}{3 \sqrt {\pi }}\)	\(53\)
default	\(\frac {2 f \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}-\frac {2 e \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\)	\(129\)
elliptic	\(\frac {2 f \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}-\frac {2 e \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\)	\(129\)

Input:

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

Output:

f*x*hypergeom([1/3,1/2],[4/3],-x^3)+1/3*e/Pi^(1/2)*((-2*ln(2)+3*ln(x))*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 = 30, normalized size of antiderivative = 0.25 \[ \int \frac {e+f x}{x \sqrt {1+x^3}} \, dx=\frac {1}{3} \, e \log \left (\frac {x^{3} - 2 \, \sqrt {x^{3} + 1} + 2}{x^{3}}\right ) + 2 \, f {\rm weierstrassPInverse}\left (0, -4, x\right ) \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35 \[ \int \frac {e+f x}{x \sqrt {1+x^3}} \, dx=- \frac {2 e \operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} + \frac {f x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \] Input:

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

Output:

-2*e*asinh(x**(-3/2))/3 + f*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**3*ex 
p_polar(I*pi))/(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 = 0.06 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.72 \[ \int \frac {e+f x}{x \sqrt {1+x^3}} \, dx=\frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (f\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-e\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:

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

Output:

((3^(1/2)*1i + 3)*((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, as 
in(((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)*1i)/2 
 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2))/(x^3 - x*(((3^(1/2)*1i)/2 - 1/2 
)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1 
/2))^(1/2)
 

Reduce [F]

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

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

Output:

(3*int(sqrt(x**3 + 1)/(x**3 + 1),x)*f + log(sqrt(x**3 + 1) - 1)*e - log(sq 
rt(x**3 + 1) + 1)*e)/3