3.5.55 \(\int \frac {1}{x^5 \sqrt {1+x^3}} \, dx\) [455]

3.5.55.1 Optimal result
3.5.55.2 Mathematica [C] (verified)
3.5.55.3 Rubi [A] (verified)
3.5.55.4 Maple [C] (verified)
3.5.55.5 Fricas [C] (verification not implemented)
3.5.55.6 Sympy [A] (verification not implemented)
3.5.55.7 Maxima [F]
3.5.55.8 Giac [F]
3.5.55.9 Mupad [B] (verification not implemented)

3.5.55.1 Optimal result

Integrand size = 13, antiderivative size = 262 \[ \int \frac {1}{x^5 \sqrt {1+x^3}} \, dx=-\frac {\sqrt {1+x^3}}{4 x^4}+\frac {5 \sqrt {1+x^3}}{8 x}-\frac {5 \sqrt {1+x^3}}{8 \left (1+\sqrt {3}+x\right )}+\frac {5 \sqrt [4]{3} \sqrt {2-\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{16 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}-\frac {5 (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{4 \sqrt {2} \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \]

output
-1/4*(x^3+1)^(1/2)/x^4+5/8*(x^3+1)^(1/2)/x-5/8*(x^3+1)^(1/2)/(1+x+3^(1/2)) 
-5/24*(1+x)*EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*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)+5/16*3^(1/4)*(1+x)*EllipticE((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)/(x^3+ 
1)^(1/2)/((1+x)/(1+x+3^(1/2))^2)^(1/2)
 
3.5.55.2 Mathematica [C] (verified)

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

Time = 10.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.08 \[ \int \frac {1}{x^5 \sqrt {1+x^3}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},-\frac {1}{3},-x^3\right )}{4 x^4} \]

input
Integrate[1/(x^5*Sqrt[1 + x^3]),x]
 
output
-1/4*Hypergeometric2F1[-4/3, 1/2, -1/3, -x^3]/x^4
 
3.5.55.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {847, 847, 832, 759, 2416}

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.

\(\displaystyle \int \frac {1}{x^5 \sqrt {x^3+1}} \, dx\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {5}{8} \int \frac {1}{x^2 \sqrt {x^3+1}}dx-\frac {\sqrt {x^3+1}}{4 x^4}\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {5}{8} \left (\frac {1}{2} \int \frac {x}{\sqrt {x^3+1}}dx-\frac {\sqrt {x^3+1}}{x}\right )-\frac {\sqrt {x^3+1}}{4 x^4}\)

\(\Big \downarrow \) 832

\(\displaystyle -\frac {5}{8} \left (\frac {1}{2} \left (\int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx-\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {x^3+1}}dx\right )-\frac {\sqrt {x^3+1}}{x}\right )-\frac {\sqrt {x^3+1}}{4 x^4}\)

\(\Big \downarrow \) 759

\(\displaystyle -\frac {5}{8} \left (\frac {1}{2} \left (\int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\right )-\frac {\sqrt {x^3+1}}{x}\right )-\frac {\sqrt {x^3+1}}{4 x^4}\)

\(\Big \downarrow \) 2416

\(\displaystyle -\frac {5}{8} \left (\frac {1}{2} \left (-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {2 \sqrt {x^3+1}}{x+\sqrt {3}+1}\right )-\frac {\sqrt {x^3+1}}{x}\right )-\frac {\sqrt {x^3+1}}{4 x^4}\)

input
Int[1/(x^5*Sqrt[1 + x^3]),x]
 
output
-1/4*Sqrt[1 + x^3]/x^4 - (5*(-(Sqrt[1 + x^3]/x) + ((2*Sqrt[1 + x^3])/(1 + 
Sqrt[3] + x) - (3^(1/4)*Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + 
Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 
- 4*Sqrt[3]])/(Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (2*(1 - 
Sqrt[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^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]))/2))/8
 

3.5.55.3.1 Defintions of rubi rules used

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 832
Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && PosQ[a]
 

rule 847
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 2416
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S 
imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt 
[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) 
*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 
3.5.55.4 Maple [C] (verified)

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

Time = 3.95 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.06

method result size
meijerg \(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {4}{3},\frac {1}{2};-\frac {1}{3};-x^{3}\right )}{4 x^{4}}\) \(17\)
default \(-\frac {\sqrt {x^{3}+1}}{4 x^{4}}+\frac {5 \sqrt {x^{3}+1}}{8 x}-\frac {5 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) E\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{8 \sqrt {x^{3}+1}}\) \(198\)
risch \(\frac {5 x^{6}+3 x^{3}-2}{8 x^{4} \sqrt {x^{3}+1}}-\frac {5 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) E\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{8 \sqrt {x^{3}+1}}\) \(198\)
elliptic \(-\frac {\sqrt {x^{3}+1}}{4 x^{4}}+\frac {5 \sqrt {x^{3}+1}}{8 x}-\frac {5 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) E\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{8 \sqrt {x^{3}+1}}\) \(198\)

input
int(1/x^5/(x^3+1)^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/4/x^4*hypergeom([-4/3,1/2],[-1/3],-x^3)
 
3.5.55.5 Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.13 \[ \int \frac {1}{x^5 \sqrt {1+x^3}} \, dx=\frac {5 \, x^{4} {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\right ) + {\left (5 \, x^{3} - 2\right )} \sqrt {x^{3} + 1}}{8 \, x^{4}} \]

input
integrate(1/x^5/(x^3+1)^(1/2),x, algorithm="fricas")
 
output
1/8*(5*x^4*weierstrassZeta(0, -4, weierstrassPInverse(0, -4, x)) + (5*x^3 
- 2)*sqrt(x^3 + 1))/x^4
 
3.5.55.6 Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.14 \[ \int \frac {1}{x^5 \sqrt {1+x^3}} \, dx=\frac {\Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{2} \\ - \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{4} \Gamma \left (- \frac {1}{3}\right )} \]

input
integrate(1/x**5/(x**3+1)**(1/2),x)
 
output
gamma(-4/3)*hyper((-4/3, 1/2), (-1/3,), x**3*exp_polar(I*pi))/(3*x**4*gamm 
a(-1/3))
 
3.5.55.7 Maxima [F]

\[ \int \frac {1}{x^5 \sqrt {1+x^3}} \, dx=\int { \frac {1}{\sqrt {x^{3} + 1} x^{5}} \,d x } \]

input
integrate(1/x^5/(x^3+1)^(1/2),x, algorithm="maxima")
 
output
integrate(1/(sqrt(x^3 + 1)*x^5), x)
 
3.5.55.8 Giac [F]

\[ \int \frac {1}{x^5 \sqrt {1+x^3}} \, dx=\int { \frac {1}{\sqrt {x^{3} + 1} x^{5}} \,d x } \]

input
integrate(1/x^5/(x^3+1)^(1/2),x, algorithm="giac")
 
output
integrate(1/(sqrt(x^3 + 1)*x^5), x)
 
3.5.55.9 Mupad [B] (verification not implemented)

Time = 5.47 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^5 \sqrt {1+x^3}} \, dx=\frac {5\,\sqrt {x^3+1}}{8\,x}-\frac {\sqrt {x^3+1}}{4\,x^4}+\frac {5\,\left (\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\left (-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {E}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{8\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

input
int(1/(x^5*(x^3 + 1)^(1/2)),x)
 
output
(5*(x^3 + 1)^(1/2))/(8*x) - (x^3 + 1)^(1/2)/(4*x^4) + (5*(((3^(1/2)*1i)/2 
- 1/2)*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)) - ((3^(1/2)*1i)/2 - 3/2)*ellipticE(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)))*((3^(1/2)*1i)/2 + 3/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^ 
(1/2)*1i)/2 - 3/2))^(1/2)*((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))/(8*(x^3 - x*(((3^(1/2)*1 
i)/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))