\(\int \frac {2-5 x^3}{\sqrt {1-x^3} (1+x^2-x^5)} \, dx\) [278]

Optimal result

Integrand size = 31, antiderivative size = 16 \[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=2 \arctan \left (x \sqrt {1-x^3}\right ) \] Output:

2*arctan(x*(-x^3+1)^(1/2))
 

Mathematica [A] (verified)

Time = 2.42 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=2 \arctan \left (x \sqrt {1-x^3}\right ) \] Input:

Integrate[(2 - 5*x^3)/(Sqrt[1 - x^3]*(1 + x^2 - x^5)),x]
 

Output:

2*ArcTan[x*Sqrt[1 - x^3]]
 

Rubi [F]

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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(2 - 5*x^3)/(Sqrt[1 - x^3]*(1 + x^2 - x^5)),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 3.88

method	result	size
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{5}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 x \sqrt {-x^{3}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{5}-x^{2}-1}\right )\)	\(62\)
default	\(-i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{5}-\textit {\_Z}^{2}-1\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{3}-1\right ) \sqrt {2}\, \sqrt {i \left (1+2 x -i \sqrt {3}\right )}\, \sqrt {\frac {-1+x}{i \sqrt {3}-3}}\, \sqrt {-\frac {i \left (i \sqrt {3}+2 x +1\right )}{2}}\, \left (-2 \underline {\hspace {1.25 ex}}\alpha ^{4}+\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}+i \sqrt {3}\, \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}\right )\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {3 \underline {\hspace {1.25 ex}}\alpha ^{3}}{2}-i \underline {\hspace {1.25 ex}}\alpha ^{4} \sqrt {3}-\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{2}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {3}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{2}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{2 \sqrt {-x^{3}+1}}\right )\)	\(191\)
elliptic	\(-i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{5}-\textit {\_Z}^{2}-1\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{3}-1\right ) \sqrt {2}\, \sqrt {i \left (1+2 x -i \sqrt {3}\right )}\, \sqrt {\frac {-1+x}{i \sqrt {3}-3}}\, \sqrt {-\frac {i \left (i \sqrt {3}+2 x +1\right )}{2}}\, \left (-2 \underline {\hspace {1.25 ex}}\alpha ^{4}+\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}+i \sqrt {3}\, \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}\right )\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {3 \underline {\hspace {1.25 ex}}\alpha ^{3}}{2}-i \underline {\hspace {1.25 ex}}\alpha ^{4} \sqrt {3}-\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{2}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {3}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{2}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{2 \sqrt {-x^{3}+1}}\right )\)	\(191\)

Input:

int((-5*x^3+2)/(-x^3+1)^(1/2)/(-x^5+x^2+1),x,method=_RETURNVERBOSE)
 

Output:

RootOf(_Z^2+1)*ln((RootOf(_Z^2+1)*x^5-RootOf(_Z^2+1)*x^2+2*x*(-x^3+1)^(1/2 
)+RootOf(_Z^2+1))/(x^5-x^2-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.94 \[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=\arctan \left (\frac {{\left (x^{5} - x^{2} + 1\right )} \sqrt {-x^{3} + 1}}{2 \, {\left (x^{4} - x\right )}}\right ) \] Input:

integrate((-5*x^3+2)/(-x^3+1)^(1/2)/(-x^5+x^2+1),x, algorithm="fricas")
 

Output:

arctan(1/2*(x^5 - x^2 + 1)*sqrt(-x^3 + 1)/(x^4 - x))
 

Sympy [F]

\[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=\int \frac {5 x^{3} - 2}{\sqrt {- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{5} - x^{2} - 1\right )}\, dx \] Input:

integrate((-5*x**3+2)/(-x**3+1)**(1/2)/(-x**5+x**2+1),x)
 

Output:

Integral((5*x**3 - 2)/(sqrt(-(x - 1)*(x**2 + x + 1))*(x**5 - x**2 - 1)), x 
)
 

Maxima [F]

\[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=\int { \frac {5 \, x^{3} - 2}{{\left (x^{5} - x^{2} - 1\right )} \sqrt {-x^{3} + 1}} \,d x } \] Input:

integrate((-5*x^3+2)/(-x^3+1)^(1/2)/(-x^5+x^2+1),x, algorithm="maxima")
 

Output:

integrate((5*x^3 - 2)/((x^5 - x^2 - 1)*sqrt(-x^3 + 1)), x)
 

Giac [F]

\[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=\int { \frac {5 \, x^{3} - 2}{{\left (x^{5} - x^{2} - 1\right )} \sqrt {-x^{3} + 1}} \,d x } \] Input:

integrate((-5*x^3+2)/(-x^3+1)^(1/2)/(-x^5+x^2+1),x, algorithm="giac")
 

Output:

integrate((5*x^3 - 2)/((x^5 - x^2 - 1)*sqrt(-x^3 + 1)), x)
 

Mupad [B] (verification not implemented)

Time = 23.07 (sec) , antiderivative size = 170, normalized size of antiderivative = 10.62 \[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=\sum _{k=1}^5\frac {\sqrt {6}\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (x-1\right )}\,\Pi \left (-\frac {3+\sqrt {3}\,1{}\mathrm {i}}{2\,\left (\mathrm {root}\left (z^5-z^2-1,z,k\right )-1\right )};\mathrm {asin}\left (\frac {\sqrt {6}\,\sqrt {\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (x-1\right )}}{6}\right )\middle |\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {3\,x+3-\sqrt {3}\,x\,1{}\mathrm {i}+\sqrt {3}\,1{}\mathrm {i}}\,\sqrt {3\,x+3+\sqrt {3}\,x\,1{}\mathrm {i}-\sqrt {3}\,1{}\mathrm {i}}}{18\,\left (\mathrm {root}\left (z^5-z^2-1,z,k\right )-1\right )\,\sqrt {1-x^3}\,\mathrm {root}\left (z^5-z^2-1,z,k\right )} \] Input:

int(-(5*x^3 - 2)/((1 - x^3)^(1/2)*(x^2 - x^5 + 1)),x)
 

Output:

symsum((6^(1/2)*((3^(1/2)*1i)/2 + 3/2)*((3^(1/2)*1i - 3)*(x - 1))^(1/2)*el 
lipticPi(-(3^(1/2)*1i + 3)/(2*(root(z^5 - z^2 - 1, z, k) - 1)), asin((6^(1 
/2)*((3^(1/2)*1i - 3)*(x - 1))^(1/2))/6), (3^(1/2)*1i)/2 + 1/2)*(3*x - 3^( 
1/2)*x*1i + 3^(1/2)*1i + 3)^(1/2)*(3*x + 3^(1/2)*x*1i - 3^(1/2)*1i + 3)^(1 
/2))/(18*(root(z^5 - z^2 - 1, z, k) - 1)*(1 - x^3)^(1/2)*root(z^5 - z^2 - 
1, z, k)), k, 1, 5)
 

Reduce [F]

\[ \int \frac {2-5 x^3}{\sqrt {1-x^3} \left (1+x^2-x^5\right )} \, dx=2 \left (\int \frac {\sqrt {-x^{3}+1}}{x^{8}-2 x^{5}-x^{3}+x^{2}+1}d x \right )-5 \left (\int \frac {\sqrt {-x^{3}+1}\, x^{3}}{x^{8}-2 x^{5}-x^{3}+x^{2}+1}d x \right ) \] Input:

int((-5*x^3+2)/(-x^3+1)^(1/2)/(-x^5+x^2+1),x)
 

Output:

2*int(sqrt( - x**3 + 1)/(x**8 - 2*x**5 - x**3 + x**2 + 1),x) - 5*int((sqrt 
( - x**3 + 1)*x**3)/(x**8 - 2*x**5 - x**3 + x**2 + 1),x)