\(\int \frac {(-1+6 x^4) \sqrt {x+2 x^5}}{(1+2 x^4) (1-x^2+4 x^4+4 x^8)} \, dx\) [602]

Optimal result

Integrand size = 47, antiderivative size = 47 \[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=\arctan \left (\frac {\sqrt {x+2 x^5}}{1+2 x^4}\right )-\text {arctanh}\left (\frac {\sqrt {x+2 x^5}}{1+2 x^4}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=\int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx \]

[In]

Int[((-1 + 6*x^4)*Sqrt[x + 2*x^5])/((1 + 2*x^4)*(1 - x^2 + 4*x^4 + 4*x^8)),x]

[Out]

(-4*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][1/(Sqrt[1 + 2*x^8]*(1 - x^2 + 2*x^8)), x], x, Sqrt[x]])/(Sqrt[x]*S
qrt[1 + 2*x^4]) + (3*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][x^2/(Sqrt[1 + 2*x^8]*(1 - x^2 + 2*x^8)), x], x, S
qrt[x]])/(Sqrt[x]*Sqrt[1 + 2*x^4]) + (4*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][1/(Sqrt[1 + 2*x^8]*(1 + x^2 +
2*x^8)), x], x, Sqrt[x]])/(Sqrt[x]*Sqrt[1 + 2*x^4]) + (3*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][x^2/(Sqrt[1 +
2*x^8]*(1 + x^2 + 2*x^8)), x], x, Sqrt[x]])/(Sqrt[x]*Sqrt[1 + 2*x^4])

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {x+2 x^5} \int \frac {\sqrt {x} \left (-1+6 x^4\right )}{\sqrt {1+2 x^4} \left (1-x^2+4 x^4+4 x^8\right )} \, dx}{\sqrt {x} \sqrt {1+2 x^4}} \\ & = \frac {\left (2 \sqrt {x+2 x^5}\right ) \text {Subst}\left (\int \frac {x^2 \left (-1+6 x^8\right )}{\sqrt {1+2 x^8} \left (1-x^4+4 x^8+4 x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}} \\ & = \frac {\left (2 \sqrt {x+2 x^5}\right ) \text {Subst}\left (\int \left (\frac {-4+3 x^2}{2 \sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}+\frac {4+3 x^2}{2 \sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}} \\ & = \frac {\sqrt {x+2 x^5} \text {Subst}\left (\int \frac {-4+3 x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\sqrt {x+2 x^5} \text {Subst}\left (\int \frac {4+3 x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}} \\ & = \frac {\sqrt {x+2 x^5} \text {Subst}\left (\int \left (-\frac {4}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}+\frac {3 x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\sqrt {x+2 x^5} \text {Subst}\left (\int \left (\frac {4}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}+\frac {3 x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}} \\ & = \frac {\left (3 \sqrt {x+2 x^5}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\left (3 \sqrt {x+2 x^5}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}-\frac {\left (4 \sqrt {x+2 x^5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\left (4 \sqrt {x+2 x^5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.67 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.43 \[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=\frac {\sqrt {x+2 x^5} \left (\arctan \left (\frac {\sqrt {x}}{\sqrt {1+2 x^4}}\right )-\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {1+2 x^4}}\right )\right )}{\sqrt {x} \sqrt {1+2 x^4}} \]

[In]

Integrate[((-1 + 6*x^4)*Sqrt[x + 2*x^5])/((1 + 2*x^4)*(1 - x^2 + 4*x^4 + 4*x^8)),x]

[Out]

(Sqrt[x + 2*x^5]*(ArcTan[Sqrt[x]/Sqrt[1 + 2*x^4]] - ArcTanh[Sqrt[x]/Sqrt[1 + 2*x^4]]))/(Sqrt[x]*Sqrt[1 + 2*x^4
])

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(-\arctan \left (\frac {\sqrt {2 x^{5}+x}}{x}\right )-\operatorname {arctanh}\left (\frac {\sqrt {2 x^{5}+x}}{x}\right )\)	\(34\)
trager	\(\frac {\ln \left (-\frac {-2 x^{4}+2 \sqrt {2 x^{5}+x}-x -1}{2 x^{4}-x +1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \sqrt {2 x^{5}+x}}{2 x^{4}+x +1}\right )}{2}\)	\(98\)

[In]

int((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x,method=_RETURNVERBOSE)

[Out]

-arctan((2*x^5+x)^(1/2)/x)-arctanh((2*x^5+x)^(1/2)/x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.28 \[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=-\frac {1}{2} \, \arctan \left (\frac {2 \, x^{4} - x + 1}{2 \, \sqrt {2 \, x^{5} + x}}\right ) + \frac {1}{2} \, \log \left (\frac {2 \, x^{4} + x - 2 \, \sqrt {2 \, x^{5} + x} + 1}{2 \, x^{4} - x + 1}\right ) \]

[In]

integrate((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=\text {Timed out} \]

[In]

integrate((6*x**4-1)*(2*x**5+x)**(1/2)/(2*x**4+1)/(4*x**8+4*x**4-x**2+1),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=\int { \frac {\sqrt {2 \, x^{5} + x} {\left (6 \, x^{4} - 1\right )}}{{\left (4 \, x^{8} + 4 \, x^{4} - x^{2} + 1\right )} {\left (2 \, x^{4} + 1\right )}} \,d x } \]

[In]

integrate((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x, algorithm="maxima")

[Out]

integrate(sqrt(2*x^5 + x)*(6*x^4 - 1)/((4*x^8 + 4*x^4 - x^2 + 1)*(2*x^4 + 1)), x)

Giac [F]

\[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=\int { \frac {\sqrt {2 \, x^{5} + x} {\left (6 \, x^{4} - 1\right )}}{{\left (4 \, x^{8} + 4 \, x^{4} - x^{2} + 1\right )} {\left (2 \, x^{4} + 1\right )}} \,d x } \]

[In]

integrate((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x, algorithm="giac")

[Out]

integrate(sqrt(2*x^5 + x)*(6*x^4 - 1)/((4*x^8 + 4*x^4 - x^2 + 1)*(2*x^4 + 1)), x)

Mupad [B] (verification not implemented)

Time = 6.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.53 \[ \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx=\frac {\ln \left (\frac {x}{2}-\sqrt {2\,x^5+x}+x^4+\frac {1}{2}\right )}{2}-\frac {\ln \left (2\,x^4-x+1\right )}{2}+\frac {\ln \left (x^4-\frac {x}{2}+\frac {1}{2}-\sqrt {2\,x^5+x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {\ln \left (2\,x^4+x+1\right )\,1{}\mathrm {i}}{2} \]

[In]

int(((x + 2*x^5)^(1/2)*(6*x^4 - 1))/((2*x^4 + 1)*(4*x^4 - x^2 + 4*x^8 + 1)),x)

[Out]

log(x/2 - (x + 2*x^5)^(1/2) + x^4 + 1/2)/2 + (log(x^4 - (x + 2*x^5)^(1/2)*1i - x/2 + 1/2)*1i)/2 - (log(x + 2*x
^4 + 1)*1i)/2 - log(2*x^4 - x + 1)/2