3.8.90 \(\int \frac {-x+3 x^5}{\sqrt {x+x^5} (a-x^2+2 a x^4+a x^8)} \, dx\) [790]

3.8.90.1 Optimal result

Integrand size = 39, antiderivative size = 60 \[ \int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {x+x^5}}{\sqrt [4]{a} \left (1+x^4\right )}\right )}{\sqrt [4]{a}}-\frac {\text {arctanh}\left (\frac {\sqrt {x+x^5}}{\sqrt [4]{a} \left (1+x^4\right )}\right )}{\sqrt [4]{a}} \]

output

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

3.8.90.2 Mathematica [F]

\[ \int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx=\int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx \]

input

Integrate[(-x + 3*x^5)/(Sqrt[x + x^5]*(a - x^2 + 2*a*x^4 + a*x^8)),x]
 

output

Integrate[(-x + 3*x^5)/(Sqrt[x + x^5]*(a - x^2 + 2*a*x^4 + a*x^8)), x]
 

3.8.90.3 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.

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

\(\Big \downarrow \) 2027

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

Int[(-x + 3*x^5)/(Sqrt[x + x^5]*(a - x^2 + 2*a*x^4 + a*x^8)),x]
 

output

$Aborted
 

3.8.90.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 2027

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7293

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

3.8.90.4 Maple [A] (verified)

Time = 2.81 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(-\frac {\left (\frac {1}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\left (\frac {1}{a}\right )^{\frac {1}{4}} x +\sqrt {x \left (x^{4}+1\right )}}{-\left (\frac {1}{a}\right )^{\frac {1}{4}} x +\sqrt {x \left (x^{4}+1\right )}}\right )+2 \arctan \left (\frac {\sqrt {x \left (x^{4}+1\right )}}{x \left (\frac {1}{a}\right )^{\frac {1}{4}}}\right )\right )}{2}\)	\(69\)

input

int((3*x^5-x)/(x^5+x)^(1/2)/(a*x^8+2*a*x^4-x^2+a),x,method=_RETURNVERBOSE)
 

output

-1/2*(1/a)^(1/4)*(ln(((1/a)^(1/4)*x+(x*(x^4+1))^(1/2))/(-(1/a)^(1/4)*x+(x* 
(x^4+1))^(1/2)))+2*arctan((x*(x^4+1))^(1/2)/x/(1/a)^(1/4)))
 

3.8.90.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 341, normalized size of antiderivative = 5.68 \[ \int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx=-\frac {\log \left (\frac {a x^{8} + 2 \, a x^{4} + x^{2} + 2 \, \sqrt {x^{5} + x} {\left (a^{\frac {1}{4}} x + \frac {a x^{4} + a}{a^{\frac {1}{4}}}\right )} + a + \frac {2 \, {\left (a x^{5} + a x\right )}}{\sqrt {a}}}{a x^{8} + 2 \, a x^{4} - x^{2} + a}\right )}{4 \, a^{\frac {1}{4}}} + \frac {\log \left (\frac {a x^{8} + 2 \, a x^{4} + x^{2} - 2 \, \sqrt {x^{5} + x} {\left (a^{\frac {1}{4}} x + \frac {a x^{4} + a}{a^{\frac {1}{4}}}\right )} + a + \frac {2 \, {\left (a x^{5} + a x\right )}}{\sqrt {a}}}{a x^{8} + 2 \, a x^{4} - x^{2} + a}\right )}{4 \, a^{\frac {1}{4}}} - \frac {i \, \log \left (\frac {a x^{8} + 2 \, a x^{4} + x^{2} - 2 \, \sqrt {x^{5} + x} {\left (i \, a^{\frac {1}{4}} x + \frac {-i \, a x^{4} - i \, a}{a^{\frac {1}{4}}}\right )} + a - \frac {2 \, {\left (a x^{5} + a x\right )}}{\sqrt {a}}}{a x^{8} + 2 \, a x^{4} - x^{2} + a}\right )}{4 \, a^{\frac {1}{4}}} + \frac {i \, \log \left (\frac {a x^{8} + 2 \, a x^{4} + x^{2} - 2 \, \sqrt {x^{5} + x} {\left (-i \, a^{\frac {1}{4}} x + \frac {i \, a x^{4} + i \, a}{a^{\frac {1}{4}}}\right )} + a - \frac {2 \, {\left (a x^{5} + a x\right )}}{\sqrt {a}}}{a x^{8} + 2 \, a x^{4} - x^{2} + a}\right )}{4 \, a^{\frac {1}{4}}} \]

input

integrate((3*x^5-x)/(x^5+x)^(1/2)/(a*x^8+2*a*x^4-x^2+a),x, algorithm="fric 
as")
 

output

-1/4*log((a*x^8 + 2*a*x^4 + x^2 + 2*sqrt(x^5 + x)*(a^(1/4)*x + (a*x^4 + a) 
/a^(1/4)) + a + 2*(a*x^5 + a*x)/sqrt(a))/(a*x^8 + 2*a*x^4 - x^2 + a))/a^(1 
/4) + 1/4*log((a*x^8 + 2*a*x^4 + x^2 - 2*sqrt(x^5 + x)*(a^(1/4)*x + (a*x^4 
 + a)/a^(1/4)) + a + 2*(a*x^5 + a*x)/sqrt(a))/(a*x^8 + 2*a*x^4 - x^2 + a)) 
/a^(1/4) - 1/4*I*log((a*x^8 + 2*a*x^4 + x^2 - 2*sqrt(x^5 + x)*(I*a^(1/4)*x 
 + (-I*a*x^4 - I*a)/a^(1/4)) + a - 2*(a*x^5 + a*x)/sqrt(a))/(a*x^8 + 2*a*x 
^4 - x^2 + a))/a^(1/4) + 1/4*I*log((a*x^8 + 2*a*x^4 + x^2 - 2*sqrt(x^5 + x 
)*(-I*a^(1/4)*x + (I*a*x^4 + I*a)/a^(1/4)) + a - 2*(a*x^5 + a*x)/sqrt(a))/ 
(a*x^8 + 2*a*x^4 - x^2 + a))/a^(1/4)
 

3.8.90.6 Sympy [F(-1)]

Timed out. \[ \int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx=\text {Timed out} \]

input

integrate((3*x**5-x)/(x**5+x)**(1/2)/(a*x**8+2*a*x**4-x**2+a),x)
 

output

Timed out
 

3.8.90.7 Maxima [F]

\[ \int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx=\int { \frac {3 \, x^{5} - x}{{\left (a x^{8} + 2 \, a x^{4} - x^{2} + a\right )} \sqrt {x^{5} + x}} \,d x } \]

input

integrate((3*x^5-x)/(x^5+x)^(1/2)/(a*x^8+2*a*x^4-x^2+a),x, algorithm="maxi 
ma")
 

output

integrate((3*x^5 - x)/((a*x^8 + 2*a*x^4 - x^2 + a)*sqrt(x^5 + x)), x)
 

3.8.90.8 Giac [F]

\[ \int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx=\int { \frac {3 \, x^{5} - x}{{\left (a x^{8} + 2 \, a x^{4} - x^{2} + a\right )} \sqrt {x^{5} + x}} \,d x } \]

input

integrate((3*x^5-x)/(x^5+x)^(1/2)/(a*x^8+2*a*x^4-x^2+a),x, algorithm="giac 
")
 

output

sage0*x
 

3.8.90.9 Mupad [B] (verification not implemented)

Time = 8.91 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.63 \[ \int \frac {-x+3 x^5}{\sqrt {x+x^5} \left (a-x^2+2 a x^4+a x^8\right )} \, dx=\frac {\ln \left (\frac {x-2\,a^{1/4}\,\sqrt {x^5+x}+\sqrt {a}+\sqrt {a}\,x^4}{\sqrt {a}-x+\sqrt {a}\,x^4}\right )}{2\,a^{1/4}}+\frac {\ln \left (\frac {x-\sqrt {a}-\sqrt {a}\,x^4+a^{1/4}\,\sqrt {x^5+x}\,2{}\mathrm {i}}{x+\sqrt {a}+\sqrt {a}\,x^4}\right )\,1{}\mathrm {i}}{2\,a^{1/4}} \]

input

int(-(x - 3*x^5)/((x + x^5)^(1/2)*(a + 2*a*x^4 + a*x^8 - x^2)),x)
 

output

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