3.521 \(\int \frac{1}{x (1+x)^{5/2} (1-x+x^2)^{5/2}} \, dx\)
Optimal. Leaf size=96 \[ \frac{2}{3 \sqrt{x+1} \sqrt{x^2-x+1}}+\frac{2}{9 \sqrt{x+1} \sqrt{x^2-x+1} \left (x^3+1\right )}-\frac{2 \sqrt{x^3+1} \tanh ^{-1}\left (\sqrt{x^3+1}\right )}{3 \sqrt{x+1} \sqrt{x^2-x+1}} \]
[Out]
2/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2]) + 2/(9*Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3)) - (2*Sqrt[1 + x^3]*ArcTanh
[Sqrt[1 + x^3]])/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2])
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Rubi [A] time = 0.0368954, antiderivative size = 96, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used =
{915, 266, 51, 63, 207} \[ \frac{2}{3 \sqrt{x+1} \sqrt{x^2-x+1}}+\frac{2}{9 \sqrt{x+1} \sqrt{x^2-x+1} \left (x^3+1\right )}-\frac{2 \sqrt{x^3+1} \tanh ^{-1}\left (\sqrt{x^3+1}\right )}{3 \sqrt{x+1} \sqrt{x^2-x+1}} \]
Antiderivative was successfully verified.
[In]
Int[1/(x*(1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]
[Out]
2/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2]) + 2/(9*Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3)) - (2*Sqrt[1 + x^3]*ArcTanh
[Sqrt[1 + x^3]])/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2])
Rule 915
Int[((g_.)*(x_))^(n_)*((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[((d
+ e*x)^FracPart[p]*(a + b*x + c*x^2)^FracPart[p])/(a*d + c*e*x^3)^FracPart[p], Int[(g*x)^n*(a*d + c*e*x^3)^p,
x], x] /; FreeQ[{a, b, c, d, e, g, m, n, p}, x] && EqQ[m - p, 0] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 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 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx &=\frac{\sqrt{1+x^3} \int \frac{1}{x \left (1+x^3\right )^{5/2}} \, dx}{\sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{\sqrt{1+x^3} \operatorname{Subst}\left (\int \frac{1}{x (1+x)^{5/2}} \, dx,x,x^3\right )}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2}{9 \sqrt{1+x} \sqrt{1-x+x^2} \left (1+x^3\right )}+\frac{\sqrt{1+x^3} \operatorname{Subst}\left (\int \frac{1}{x (1+x)^{3/2}} \, dx,x,x^3\right )}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2}{3 \sqrt{1+x} \sqrt{1-x+x^2}}+\frac{2}{9 \sqrt{1+x} \sqrt{1-x+x^2} \left (1+x^3\right )}+\frac{\sqrt{1+x^3} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+x}} \, dx,x,x^3\right )}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2}{3 \sqrt{1+x} \sqrt{1-x+x^2}}+\frac{2}{9 \sqrt{1+x} \sqrt{1-x+x^2} \left (1+x^3\right )}+\frac{\left (2 \sqrt{1+x^3}\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sqrt{1+x^3}\right )}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2}{3 \sqrt{1+x} \sqrt{1-x+x^2}}+\frac{2}{9 \sqrt{1+x} \sqrt{1-x+x^2} \left (1+x^3\right )}-\frac{2 \sqrt{1+x^3} \tanh ^{-1}\left (\sqrt{1+x^3}\right )}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ \end{align*}
Mathematica [C] time = 6.08822, size = 2539, normalized size = 26.45 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(x*(1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]
[Out]
Sqrt[1 + x]*Sqrt[1 - x + x^2]*(2/(81*(1 + x)^2) + 22/(81*(1 + x)) - (2*(-1 + x))/(27*(1 - x + x^2)^2) - (2*(-2
1 + 11*x))/(81*(1 - x + x^2))) + 2*(((-I)*(1 + x)*Sqrt[1 - 6/((3 - I*Sqrt[3])*(1 + x))]*Sqrt[1 - 6/((3 + I*Sqr
t[3])*(1 + x))]*EllipticF[I*ArcSinh[Sqrt[-6/(3 - I*Sqrt[3])]/Sqrt[1 + x]], (3 - I*Sqrt[3])/(3 + I*Sqrt[3])])/(
Sqrt[6]*Sqrt[-(3 - I*Sqrt[3])^(-1)]*Sqrt[3 - 3*(1 + x) + (1 + x)^2]) + (Sqrt[3/2]*(Sqrt[1/2 - (I/2)/Sqrt[3]] +
Sqrt[1/2 + (I/2)/Sqrt[3]])*(1 + x)*(-Sqrt[1/2 - (I/2)/Sqrt[3]] + 1/Sqrt[1 + x])^2*Sqrt[(Sqrt[(2*(3 - I*Sqrt[3
]))/3]*(-Sqrt[1/2 + (I/2)/Sqrt[3]] + 1/Sqrt[1 + x]))/((Sqrt[1/2 - (I/2)/Sqrt[3]] + Sqrt[1/2 + (I/2)/Sqrt[3]])*
(-Sqrt[1/2 - (I/2)/Sqrt[3]] + 1/Sqrt[1 + x]))]*Sqrt[(Sqrt[(2*(3 - I*Sqrt[3]))/3]*(Sqrt[1/2 + (I/2)/Sqrt[3]] +
1/Sqrt[1 + x]))/((Sqrt[1/2 - (I/2)/Sqrt[3]] - Sqrt[1/2 + (I/2)/Sqrt[3]])*(-Sqrt[1/2 - (I/2)/Sqrt[3]] + 1/Sqrt[
1 + x]))]*Sqrt[((Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] + 6/Sqrt[1 + x]))/((Sqrt[
3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] - 6/Sqrt[1 + x]))]*((1 + Sqrt[1/2 - (I/2)/Sqrt[
3]])*EllipticF[ArcSin[Sqrt[((Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] + 6/Sqrt[1 +
x]))/((Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] - 6/Sqrt[1 + x]))]], (Sqrt[3 - I*Sq
rt[3]] + Sqrt[3 + I*Sqrt[3]])^2/(Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])^2] - Sqrt[(2*(3 - I*Sqrt[3]))/3]*E
llipticPi[((-1 + Sqrt[1/2 - (I/2)/Sqrt[3]])*(Sqrt[1/2 - (I/2)/Sqrt[3]] + Sqrt[1/2 + (I/2)/Sqrt[3]]))/((-1 - Sq
rt[1/2 - (I/2)/Sqrt[3]])*(-Sqrt[1/2 - (I/2)/Sqrt[3]] + Sqrt[1/2 + (I/2)/Sqrt[3]])), ArcSin[Sqrt[((Sqrt[3 - I*S
qrt[3]] - Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] + 6/Sqrt[1 + x]))/((Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*S
qrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] - 6/Sqrt[1 + x]))]], (Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])^2/(Sqrt[3 -
I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])^2]))/(Sqrt[3 - I*Sqrt[3]]*(-1 - Sqrt[1/2 - (I/2)/Sqrt[3]])*(1 - Sqrt[1/2 -
(I/2)/Sqrt[3]])*(Sqrt[1/2 - (I/2)/Sqrt[3]] - Sqrt[1/2 + (I/2)/Sqrt[3]])*Sqrt[3 - 3*(1 + x) + (1 + x)^2]) - (Sq
rt[3/2]*(Sqrt[1/2 - (I/2)/Sqrt[3]] + Sqrt[1/2 + (I/2)/Sqrt[3]])*(1 + x)*(-Sqrt[1/2 - (I/2)/Sqrt[3]] + 1/Sqrt[1
+ x])^2*Sqrt[(Sqrt[(2*(3 - I*Sqrt[3]))/3]*(-Sqrt[1/2 + (I/2)/Sqrt[3]] + 1/Sqrt[1 + x]))/((Sqrt[1/2 - (I/2)/Sq
rt[3]] + Sqrt[1/2 + (I/2)/Sqrt[3]])*(-Sqrt[1/2 - (I/2)/Sqrt[3]] + 1/Sqrt[1 + x]))]*Sqrt[(Sqrt[(2*(3 - I*Sqrt[3
]))/3]*(Sqrt[1/2 + (I/2)/Sqrt[3]] + 1/Sqrt[1 + x]))/((Sqrt[1/2 - (I/2)/Sqrt[3]] - Sqrt[1/2 + (I/2)/Sqrt[3]])*(
-Sqrt[1/2 - (I/2)/Sqrt[3]] + 1/Sqrt[1 + x]))]*Sqrt[((Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I
*Sqrt[3])] + 6/Sqrt[1 + x]))/((Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] - 6/Sqrt[1
+ x]))]*((-1 + Sqrt[1/2 - (I/2)/Sqrt[3]])*EllipticF[ArcSin[Sqrt[((Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])*(
Sqrt[6*(3 - I*Sqrt[3])] + 6/Sqrt[1 + x]))/((Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])
] - 6/Sqrt[1 + x]))]], (Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])^2/(Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]
])^2] - Sqrt[(2*(3 - I*Sqrt[3]))/3]*EllipticPi[((1 + Sqrt[1/2 - (I/2)/Sqrt[3]])*(Sqrt[1/2 - (I/2)/Sqrt[3]] + S
qrt[1/2 + (I/2)/Sqrt[3]]))/((1 - Sqrt[1/2 - (I/2)/Sqrt[3]])*(-Sqrt[1/2 - (I/2)/Sqrt[3]] + Sqrt[1/2 + (I/2)/Sqr
t[3]])), ArcSin[Sqrt[((Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] + 6/Sqrt[1 + x]))/(
(Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] - 6/Sqrt[1 + x]))]], (Sqrt[3 - I*Sqrt[3]]
+ Sqrt[3 + I*Sqrt[3]])^2/(Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])^2]))/(Sqrt[3 - I*Sqrt[3]]*(-1 - Sqrt[1/2
- (I/2)/Sqrt[3]])*(1 - Sqrt[1/2 - (I/2)/Sqrt[3]])*(Sqrt[1/2 - (I/2)/Sqrt[3]] - Sqrt[1/2 + (I/2)/Sqrt[3]])*Sqr
t[3 - 3*(1 + x) + (1 + x)^2]))
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Maple [A] time = 1.008, size = 69, normalized size = 0.7 \begin{align*} -{\frac{2}{9\,{x}^{3}+9} \left ( 3\,{\it Artanh} \left ( \sqrt{{x}^{3}+1} \right ) \sqrt{{x}^{3}+1}{x}^{3}-3\,{x}^{3}+3\,{\it Artanh} \left ( \sqrt{{x}^{3}+1} \right ) \sqrt{{x}^{3}+1}-4 \right ){\frac{1}{\sqrt{1+x}}}{\frac{1}{\sqrt{{x}^{2}-x+1}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x)
[Out]
-2/9*(3*arctanh((x^3+1)^(1/2))*(x^3+1)^(1/2)*x^3-3*x^3+3*arctanh((x^3+1)^(1/2))*(x^3+1)^(1/2)-4)/(x^3+1)/(x^2-
x+1)^(1/2)/(1+x)^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{2} - x + 1\right )}^{\frac{5}{2}}{\left (x + 1\right )}^{\frac{5}{2}} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="maxima")
[Out]
integrate(1/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)*x), x)
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Fricas [A] time = 1.73839, size = 258, normalized size = 2.69 \begin{align*} \frac{2 \,{\left (3 \, x^{3} + 4\right )} \sqrt{x^{2} - x + 1} \sqrt{x + 1} - 3 \,{\left (x^{6} + 2 \, x^{3} + 1\right )} \log \left (\sqrt{x^{2} - x + 1} \sqrt{x + 1} + 1\right ) + 3 \,{\left (x^{6} + 2 \, x^{3} + 1\right )} \log \left (\sqrt{x^{2} - x + 1} \sqrt{x + 1} - 1\right )}{9 \,{\left (x^{6} + 2 \, x^{3} + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="fricas")
[Out]
1/9*(2*(3*x^3 + 4)*sqrt(x^2 - x + 1)*sqrt(x + 1) - 3*(x^6 + 2*x^3 + 1)*log(sqrt(x^2 - x + 1)*sqrt(x + 1) + 1)
+ 3*(x^6 + 2*x^3 + 1)*log(sqrt(x^2 - x + 1)*sqrt(x + 1) - 1))/(x^6 + 2*x^3 + 1)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)**(5/2)/(x**2-x+1)**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{2} - x + 1\right )}^{\frac{5}{2}}{\left (x + 1\right )}^{\frac{5}{2}} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="giac")
[Out]
integrate(1/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)*x), x)