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

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

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Rubi [A]  time = 0.38, 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 = {6715, 826, 1164, 628} \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 628

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 826

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 1164

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 6715

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} \operatorname {Subst}\left (\int \frac {-2+x}{\sqrt {-1+x} \left (1-x+a x^2\right )} \, dx,x,x^5\right )\\ &=\frac {2}{5} \operatorname {Subst}\left (\int \frac {-1+x^2}{a+(-1+2 a) x^2+a x^4} \, dx,x,\sqrt {-1+x^5}\right )\\ &=\frac {\operatorname {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 {\operatorname {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 = 0.19, size = 158, normalized size = 0.82 \begin {gather*} \frac {\left (\sqrt {1-4 a}-1\right ) \sqrt {-2 a+\sqrt {1-4 a}+1} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5-1}}{\sqrt {-2 a+\sqrt {1-4 a}+1}}\right )-\left (\sqrt {1-4 a}+1\right ) \sqrt {-2 a-\sqrt {1-4 a}+1} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5-1}}{\sqrt {-2 a-\sqrt {1-4 a}+1}}\right )}{5 \sqrt {2} a^{3/2}} \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])*Sqrt[1 - Sqrt[1 - 4*a] - 2*a]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[-1 + x^5])/Sqrt[1 - Sqrt[1
- 4*a] - 2*a]]) + (-1 + Sqrt[1 - 4*a])*Sqrt[1 + Sqrt[1 - 4*a] - 2*a]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[-1 + x^5])/
Sqrt[1 + Sqrt[1 - 4*a] - 2*a]])/(5*Sqrt[2]*a^(3/2))

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IntegrateAlgebraic [A]  time = 0.31, size = 193, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2} \left (1+\sqrt {1-4 a}-4 a\right ) \tan ^{-1}\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 ) \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, 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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:a: r
ecursive definition (in sto)  Error: Bad Argument ValueWarning, need to choose a branch for the root of a poly
nomial with parameters. This might be wrong.The choice was done assuming [a]=[6]a: recursive definition (in st
o)  Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial with parameters. Th
is might be wrong.The choice was done assuming [a]=[-31]Done

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maple [F]  time = 0.05, 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*} \int \frac {{\left (x^{5} - 2\right )} x^{4}}{{\left (a x^{10} - x^{5} + 1\right )} \sqrt {x^{5} - 1}}\,{d x} \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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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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sympy [F(-1)]  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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