3.28.25 \(\int \frac {(-2+x^5) \sqrt {-1+x^5}}{x^6 (1-x^5+a x^{10})} \, dx\)

Optimal. Leaf size=250 \[ \frac {\left (-4 \sqrt {2} a^{3/2}+\sqrt {2} \sqrt {1-4 a} \sqrt {a}+\sqrt {2} \sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5-1}}{\sqrt {2 a-\sqrt {1-4 a}-1}}\right )}{5 \sqrt {1-4 a} \sqrt {2 a-\sqrt {1-4 a}-1}}+\frac {\left (4 \sqrt {2} a^{3/2}+\sqrt {2} \sqrt {1-4 a} \sqrt {a}-\sqrt {2} \sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5-1}}{\sqrt {2 a+\sqrt {1-4 a}-1}}\right )}{5 \sqrt {1-4 a} \sqrt {2 a+\sqrt {1-4 a}-1}}+\frac {2 \sqrt {x^5-1}}{5 x^5} \]

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Rubi [A]  time = 0.73, antiderivative size = 97, normalized size of antiderivative = 0.39, number of steps used = 16, number of rules used = 11, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {6728, 266, 47, 63, 203, 50, 6715, 824, 826, 1164, 628} \begin {gather*} \frac {1}{5} \sqrt {a} \log \left (-\sqrt {a} \left (1-x^5\right )+\sqrt {a}-\sqrt {x^5-1}\right )-\frac {1}{5} \sqrt {a} \log \left (-\sqrt {a} \left (1-x^5\right )+\sqrt {a}+\sqrt {x^5-1}\right )+\frac {2 \sqrt {x^5-1}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-2 + x^5)*Sqrt[-1 + x^5])/(x^6*(1 - x^5 + a*x^10)),x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), 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] && FractionQ[m] && GtQ[m, 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]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (-2+x^5\right ) \sqrt {-1+x^5}}{x^6 \left (1-x^5+a x^{10}\right )} \, dx &=\int \left (-\frac {2 \sqrt {-1+x^5}}{x^6}-\frac {\sqrt {-1+x^5}}{x}+\frac {x^4 \sqrt {-1+x^5} \left (-1+2 a+a x^5\right )}{1-x^5+a x^{10}}\right ) \, dx\\ &=-\left (2 \int \frac {\sqrt {-1+x^5}}{x^6} \, dx\right )-\int \frac {\sqrt {-1+x^5}}{x} \, dx+\int \frac {x^4 \sqrt {-1+x^5} \left (-1+2 a+a x^5\right )}{1-x^5+a x^{10}} \, dx\\ &=-\left (\frac {1}{5} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{x} \, dx,x,x^5\right )\right )+\frac {1}{5} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x} (-1+2 a+a x)}{1-x+a x^2} \, dx,x,x^5\right )-\frac {2}{5} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{x^2} \, dx,x,x^5\right )\\ &=\frac {2 \sqrt {-1+x^5}}{5 x^5}+\frac {\operatorname {Subst}\left (\int \frac {-2 a^2+a^2 x}{\sqrt {-1+x} \left (1-x+a x^2\right )} \, dx,x,x^5\right )}{5 a}\\ &=\frac {2 \sqrt {-1+x^5}}{5 x^5}+\frac {2 \operatorname {Subst}\left (\int \frac {-a^2+a^2 x^2}{a+(-1+2 a) x^2+a x^4} \, dx,x,\sqrt {-1+x^5}\right )}{5 a}\\ &=\frac {2 \sqrt {-1+x^5}}{5 x^5}+\frac {1}{5} \sqrt {a} \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 )+\frac {1}{5} \sqrt {a} \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 )\\ &=\frac {2 \sqrt {-1+x^5}}{5 x^5}+\frac {1}{5} \sqrt {a} \log \left (\sqrt {a}-\sqrt {a} \left (1-x^5\right )-\sqrt {-1+x^5}\right )-\frac {1}{5} \sqrt {a} \log \left (\sqrt {a}-\sqrt {a} \left (1-x^5\right )+\sqrt {-1+x^5}\right )\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 238, normalized size = 0.95 \begin {gather*} \frac {2}{5} \left (-\frac {\sqrt {a} \left (4 a+\sqrt {1-4 a}-1\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5-1}}{\sqrt {-2 a-\sqrt {1-4 a}+1}}\right )}{\sqrt {2-8 a} \sqrt {-2 a-\sqrt {1-4 a}+1}}-\frac {\left (-4 a+\sqrt {1-4 a}+1\right ) \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5-1}}{\sqrt {-2 a+\sqrt {1-4 a}+1}}\right )}{\sqrt {2-8 a} \sqrt {-2 a+\sqrt {1-4 a}+1}}+\tan ^{-1}\left (\sqrt {x^5-1}\right )+\frac {x^5+\sqrt {1-x^5} x^5 \tanh ^{-1}\left (\sqrt {1-x^5}\right )-1}{x^5 \sqrt {x^5-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2 + x^5)*Sqrt[-1 + x^5])/(x^6*(1 - x^5 + a*x^10)),x]

[Out]

(2*(ArcTan[Sqrt[-1 + x^5]] + (-1 + x^5 + x^5*Sqrt[1 - x^5]*ArcTanh[Sqrt[1 - x^5]])/(x^5*Sqrt[-1 + x^5]) - (Sqr
t[a]*(-1 + Sqrt[1 - 4*a] + 4*a)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[-1 + x^5])/Sqrt[1 - Sqrt[1 - 4*a] - 2*a]])/(Sqrt
[2 - 8*a]*Sqrt[1 - Sqrt[1 - 4*a] - 2*a]) - ((1 + Sqrt[1 - 4*a] - 4*a)*Sqrt[a]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[-1
+ x^5])/Sqrt[1 + Sqrt[1 - 4*a] - 2*a]])/(Sqrt[2 - 8*a]*Sqrt[1 + Sqrt[1 - 4*a] - 2*a])))/5

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IntegrateAlgebraic [A]  time = 0.35, size = 250, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {-1+x^5}}{5 x^5}+\frac {\left (\sqrt {2} \sqrt {a}+\sqrt {2} \sqrt {1-4 a} \sqrt {a}-4 \sqrt {2} a^{3/2}\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 {-1-\sqrt {1-4 a}+2 a}}+\frac {\left (-\sqrt {2} \sqrt {a}+\sqrt {2} \sqrt {1-4 a} \sqrt {a}+4 \sqrt {2} a^{3/2}\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 {-1+\sqrt {1-4 a}+2 a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-2 + x^5)*Sqrt[-1 + x^5])/(x^6*(1 - x^5 + a*x^10)),x]

[Out]

(2*Sqrt[-1 + x^5])/(5*x^5) + ((Sqrt[2]*Sqrt[a] + Sqrt[2]*Sqrt[1 - 4*a]*Sqrt[a] - 4*Sqrt[2]*a^(3/2))*ArcTan[(Sq
rt[2]*Sqrt[a]*Sqrt[-1 + x^5])/Sqrt[-1 - Sqrt[1 - 4*a] + 2*a]])/(5*Sqrt[1 - 4*a]*Sqrt[-1 - Sqrt[1 - 4*a] + 2*a]
) + ((-(Sqrt[2]*Sqrt[a]) + Sqrt[2]*Sqrt[1 - 4*a]*Sqrt[a] + 4*Sqrt[2]*a^(3/2))*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[-1
+ x^5])/Sqrt[-1 + Sqrt[1 - 4*a] + 2*a]])/(5*Sqrt[1 - 4*a]*Sqrt[-1 + Sqrt[1 - 4*a] + 2*a])

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fricas [A]  time = 0.50, size = 103, normalized size = 0.41 \begin {gather*} \left [\frac {\sqrt {a} x^{5} \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 ) + 2 \, \sqrt {x^{5} - 1}}{5 \, x^{5}}, \frac {2 \, {\left (\sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {-a} x^{5}}{\sqrt {x^{5} - 1}}\right ) + \sqrt {x^{5} - 1}\right )}}{5 \, x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-2)*(x^5-1)^(1/2)/x^6/(a*x^10-x^5+1),x, algorithm="fricas")

[Out]

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

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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^5-2)*(x^5-1)^(1/2)/x^6/(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]=[39]Precision problem choosing root in common_EXT, current
precision 14Done

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{5}-2\right ) \sqrt {x^{5}-1}}{x^{6} \left (a \,x^{10}-x^{5}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5-2)*(x^5-1)^(1/2)/x^6/(a*x^10-x^5+1),x)

[Out]

int((x^5-2)*(x^5-1)^(1/2)/x^6/(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 {\sqrt {x^{5} - 1} {\left (x^{5} - 2\right )}}{{\left (a x^{10} - x^{5} + 1\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-2)*(x^5-1)^(1/2)/x^6/(a*x^10-x^5+1),x, algorithm="maxima")

[Out]

integrate(sqrt(x^5 - 1)*(x^5 - 2)/((a*x^10 - x^5 + 1)*x^6), x)

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mupad [B]  time = 2.54, size = 60, normalized size = 0.24 \begin {gather*} \frac {2\,\sqrt {x^5-1}}{5\,x^5}+\frac {\sqrt {a}\,\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5 - 1)^(1/2)*(x^5 - 2))/(x^6*(a*x^10 - x^5 + 1)),x)

[Out]

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

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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**5-2)*(x**5-1)**(1/2)/x**6/(a*x**10-x**5+1),x)

[Out]

Timed out

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