\(\int \frac {1+3 x^4}{(-1-a x+x^4) \sqrt {-x+x^5}} \, dx\) [283]

Optimal result

Integrand size = 30, antiderivative size = 26 \[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {-x+x^5}}\right )}{\sqrt {a}} \]

[Out]

-2*arctanh(a^(1/2)*x/(x^5-x)^(1/2))/a^(1/2)

Rubi [F]

\[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=\int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx \]

[In]

Int[(1 + 3*x^4)/((-1 - a*x + x^4)*Sqrt[-x + x^5]),x]

[Out]

((3 + 3*I)*x^2*Sqrt[-(((-1)^(3/4)*(1 + (-1)^(1/4)*x)^2)/x)]*Sqrt[(I*(1 - x^4))/x^2]*EllipticF[ArcSin[Sqrt[((-1
)^(3/4)*(Sqrt[2] - 2*(-1)^(1/4)*x + I*Sqrt[2]*x^2))/x]/2], -2*(1 - Sqrt[2])])/(Sqrt[2*(2 + Sqrt[2])]*(1 + (-1)
^(1/4)*x)*Sqrt[-x + x^5]) - ((3 + 3*I)*x^2*Sqrt[((-1)^(3/4)*(1 - (-1)^(1/4)*x)^2)/x]*Sqrt[(I*(1 - x^4))/x^2]*E
llipticF[ArcSin[Sqrt[-(((-1)^(3/4)*(Sqrt[2] + 2*(-1)^(1/4)*x + I*Sqrt[2]*x^2))/x)]/2], -2*(1 - Sqrt[2])])/(Sqr
t[2*(2 + Sqrt[2])]*(1 - (-1)^(1/4)*x)*Sqrt[-x + x^5]) - (8*Sqrt[x]*Sqrt[-1 + x^4]*Defer[Subst][Defer[Int][1/((
1 + a*x^2 - x^8)*Sqrt[-1 + x^8]), x], x, Sqrt[x]])/Sqrt[-x + x^5] - (6*a*Sqrt[x]*Sqrt[-1 + x^4]*Defer[Subst][D
efer[Int][x^2/((1 + a*x^2 - x^8)*Sqrt[-1 + x^8]), x], x, Sqrt[x]])/Sqrt[-x + x^5]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt {-1+x^4}\right ) \int \frac {1+3 x^4}{\sqrt {x} \sqrt {-1+x^4} \left (-1-a x+x^4\right )} \, dx}{\sqrt {-x+x^5}} \\ & = \frac {\left (2 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \frac {1+3 x^8}{\sqrt {-1+x^8} \left (-1-a x^2+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}} \\ & = \frac {\left (2 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \left (\frac {3}{\sqrt {-1+x^8}}+\frac {4+3 a x^2}{\sqrt {-1+x^8} \left (-1-a x^2+x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}} \\ & = \frac {\left (2 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \frac {4+3 a x^2}{\sqrt {-1+x^8} \left (-1-a x^2+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}}+\frac {\left (6 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}} \\ & = \frac {\left (2 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \left (-\frac {4}{\left (1+a x^2-x^8\right ) \sqrt {-1+x^8}}-\frac {3 a x^2}{\left (1+a x^2-x^8\right ) \sqrt {-1+x^8}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}}+\frac {\left (3 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \frac {1-\sqrt [4]{-1} x^2}{\sqrt {-1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}}+\frac {\left (3 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \frac {1+\sqrt [4]{-1} x^2}{\sqrt {-1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}} \\ & = \frac {(3+3 i) x^2 \sqrt {-\frac {(-1)^{3/4} \left (1+\sqrt [4]{-1} x\right )^2}{x}} \sqrt {\frac {i \left (1-x^4\right )}{x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {(-1)^{3/4} \left (\sqrt {2}-2 \sqrt [4]{-1} x+i \sqrt {2} x^2\right )}{x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{\sqrt {2 \left (2+\sqrt {2}\right )} \left (1+\sqrt [4]{-1} x\right ) \sqrt {-x+x^5}}-\frac {(3+3 i) x^2 \sqrt {\frac {(-1)^{3/4} \left (1-\sqrt [4]{-1} x\right )^2}{x}} \sqrt {\frac {i \left (1-x^4\right )}{x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {(-1)^{3/4} \left (\sqrt {2}+2 \sqrt [4]{-1} x+i \sqrt {2} x^2\right )}{x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{\sqrt {2 \left (2+\sqrt {2}\right )} \left (1-\sqrt [4]{-1} x\right ) \sqrt {-x+x^5}}-\frac {\left (8 \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1+a x^2-x^8\right ) \sqrt {-1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}}-\frac {\left (6 a \sqrt {x} \sqrt {-1+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+a x^2-x^8\right ) \sqrt {-1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x+x^5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {-x+x^5}}\right )}{\sqrt {a}} \]

[In]

Integrate[(1 + 3*x^4)/((-1 - a*x + x^4)*Sqrt[-x + x^5]),x]

[Out]

(-2*ArcTanh[(Sqrt[a]*x)/Sqrt[-x + x^5]])/Sqrt[a]

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{5}-x}}{x \sqrt {a}}\right )}{\sqrt {a}}\)	\(23\)

[In]

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

[Out]

-2/a^(1/2)*arctanh((x^5-x)^(1/2)/x/a^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (20) = 40\).

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 5.08 \[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=\left [\frac {\log \left (\frac {x^{8} + 6 \, a x^{5} + a^{2} x^{2} - 2 \, x^{4} - 4 \, \sqrt {x^{5} - x} {\left (x^{4} + a x - 1\right )} \sqrt {a} - 6 \, a x + 1}{x^{8} - 2 \, a x^{5} + a^{2} x^{2} - 2 \, x^{4} + 2 \, a x + 1}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {x^{5} - x} {\left (x^{4} + a x - 1\right )} \sqrt {-a}}{2 \, {\left (a x^{5} - a x\right )}}\right )}{a}\right ] \]

[In]

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

[Out]

[1/2*log((x^8 + 6*a*x^5 + a^2*x^2 - 2*x^4 - 4*sqrt(x^5 - x)*(x^4 + a*x - 1)*sqrt(a) - 6*a*x + 1)/(x^8 - 2*a*x^
5 + a^2*x^2 - 2*x^4 + 2*a*x + 1))/sqrt(a), sqrt(-a)*arctan(1/2*sqrt(x^5 - x)*(x^4 + a*x - 1)*sqrt(-a)/(a*x^5 -
a*x))/a]

Sympy [F(-1)]

Timed out. \[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=\int { \frac {3 \, x^{4} + 1}{\sqrt {x^{5} - x} {\left (x^{4} - a x - 1\right )}} \,d x } \]

[In]

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

[Out]

integrate((3*x^4 + 1)/(sqrt(x^5 - x)*(x^4 - a*x - 1)), x)

Giac [F]

\[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=\int { \frac {3 \, x^{4} + 1}{\sqrt {x^{5} - x} {\left (x^{4} - a x - 1\right )}} \,d x } \]

[In]

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

[Out]

integrate((3*x^4 + 1)/(sqrt(x^5 - x)*(x^4 - a*x - 1)), x)

Mupad [B] (verification not implemented)

Time = 5.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {1+3 x^4}{\left (-1-a x+x^4\right ) \sqrt {-x+x^5}} \, dx=\frac {\ln \left (\frac {a\,x-2\,\sqrt {a}\,\sqrt {x^5-x}+x^4-1}{-x^4+a\,x+1}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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