3.25.7 \(\int \frac {x^4 (-2+x^5)}{\sqrt {-1+x^5} (1-x^5+a x^{10})} \, dx\) [2407]

Optimal. Leaf size=193 \[ \frac {\sqrt {2} \left (1+\sqrt {1-4 a}-4 a\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {-1+x^5}}{\sqrt {-1-\sqrt {1-4 a}+2 a}}\right )}{5 \sqrt {1-4 a} \sqrt {a} \sqrt {-1-\sqrt {1-4 a}+2 a}}+\frac {\sqrt {2} \left (-1+\sqrt {1-4 a}+4 a\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {-1+x^5}}{\sqrt {-1+\sqrt {1-4 a}+2 a}}\right )}{5 \sqrt {1-4 a} \sqrt {a} \sqrt {-1+\sqrt {1-4 a}+2 a}} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 81, normalized size of antiderivative = 0.42, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6847, 840, 1178, 642} \begin {gather*} \frac {\log \left (-\sqrt {a} \left (1-x^5\right )+\sqrt {a}-\sqrt {x^5-1}\right )}{5 \sqrt {a}}-\frac {\log \left (-\sqrt {a} \left (1-x^5\right )+\sqrt {a}+\sqrt {x^5-1}\right )}{5 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(-2 + x^5))/(Sqrt[-1 + x^5]*(1 - x^5 + a*x^10)),x]

[Out]

Log[Sqrt[a] - Sqrt[a]*(1 - x^5) - Sqrt[-1 + x^5]]/(5*Sqrt[a]) - Log[Sqrt[a] - Sqrt[a]*(1 - x^5) + Sqrt[-1 + x^
5]]/(5*Sqrt[a])

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 840

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1178

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {align*} \int \frac {x^4 \left (-2+x^5\right )}{\sqrt {-1+x^5} \left (1-x^5+a x^{10}\right )} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {-2+x}{\sqrt {-1+x} \left (1-x+a x^2\right )} \, dx,x,x^5\right )\\ &=\frac {2}{5} \text {Subst}\left (\int \frac {-1+x^2}{a+(-1+2 a) x^2+a x^4} \, dx,x,\sqrt {-1+x^5}\right )\\ &=\frac {\text {Subst}\left (\int \frac {\frac {1}{\sqrt {a}}+2 x}{-1-\frac {x}{\sqrt {a}}-x^2} \, dx,x,\sqrt {-1+x^5}\right )}{5 \sqrt {a}}+\frac {\text {Subst}\left (\int \frac {\frac {1}{\sqrt {a}}-2 x}{-1+\frac {x}{\sqrt {a}}-x^2} \, dx,x,\sqrt {-1+x^5}\right )}{5 \sqrt {a}}\\ &=\frac {\log \left (\sqrt {a}-\sqrt {a} \left (1-x^5\right )-\sqrt {-1+x^5}\right )}{5 \sqrt {a}}-\frac {\log \left (\sqrt {a}-\sqrt {a} \left (1-x^5\right )+\sqrt {-1+x^5}\right )}{5 \sqrt {a}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 169, normalized size = 0.88 \begin {gather*} \frac {\frac {\left (1+\sqrt {1-4 a}-4 a\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {-1+x^5}}{\sqrt {-1-\sqrt {1-4 a}+2 a}}\right )}{\sqrt {-1-\sqrt {1-4 a}+2 a}}+\frac {\left (-1+\sqrt {1-4 a}+4 a\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {-1+x^5}}{\sqrt {-1+\sqrt {1-4 a}+2 a}}\right )}{\sqrt {-1+\sqrt {1-4 a}+2 a}}}{5 \sqrt {\frac {1}{2}-2 a} \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(-2 + x^5))/(Sqrt[-1 + x^5]*(1 - x^5 + a*x^10)),x]

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (x^{5}-2\right )}{\sqrt {x^{5}-1}\, \left (a \,x^{10}-x^{5}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(x^5-2)/(x^5-1)^(1/2)/(a*x^10-x^5+1),x)

[Out]

int(x^4*(x^5-2)/(x^5-1)^(1/2)/(a*x^10-x^5+1),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(x^5-2)/(x^5-1)^(1/2)/(a*x^10-x^5+1),x, algorithm="maxima")

[Out]

integrate((x^5 - 2)*x^4/((a*x^10 - x^5 + 1)*sqrt(x^5 - 1)), x)

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Fricas [A]
time = 0.36, size = 74, normalized size = 0.38 \begin {gather*} \left [\frac {\log \left (\frac {a x^{10} - 2 \, \sqrt {x^{5} - 1} \sqrt {a} x^{5} + x^{5} - 1}{a x^{10} - x^{5} + 1}\right )}{5 \, \sqrt {a}}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{5}}{\sqrt {x^{5} - 1}}\right )}{5 \, a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(x^5-2)/(x^5-1)^(1/2)/(a*x^10-x^5+1),x, algorithm="fricas")

[Out]

[1/5*log((a*x^10 - 2*sqrt(x^5 - 1)*sqrt(a)*x^5 + x^5 - 1)/(a*x^10 - x^5 + 1))/sqrt(a), 2/5*sqrt(-a)*arctan(sqr
t(-a)*x^5/sqrt(x^5 - 1))/a]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(x**5-2)/(x**5-1)**(1/2)/(a*x**10-x**5+1),x)

[Out]

Timed out

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Giac [A]
time = 0.46, size = 34, normalized size = 0.18 \begin {gather*} \frac {\pi + 2 \, \arctan \left (\frac {{\left (x^{5} - 1\right )} a + a}{\sqrt {x^{5} - 1} \sqrt {-a}}\right )}{5 \, \sqrt {-a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(x^5-2)/(x^5-1)^(1/2)/(a*x^10-x^5+1),x, algorithm="giac")

[Out]

1/5*(pi + 2*arctan(((x^5 - 1)*a + a)/(sqrt(x^5 - 1)*sqrt(-a))))/sqrt(-a)

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Mupad [B]
time = 2.24, size = 47, normalized size = 0.24 \begin {gather*} \frac {\ln \left (\frac {a\,x^{10}+x^5-2\,\sqrt {a}\,x^5\,\sqrt {x^5-1}-1}{4\,a\,x^{10}-4\,x^5+4}\right )}{5\,\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(x^5 - 2))/((x^5 - 1)^(1/2)*(a*x^10 - x^5 + 1)),x)

[Out]

log((a*x^10 + x^5 - 2*a^(1/2)*x^5*(x^5 - 1)^(1/2) - 1)/(4*a*x^10 - 4*x^5 + 4))/(5*a^(1/2))

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