3.514 \(\int \frac{1}{x (1+x)^{3/2} (1-x+x^2)^{3/2}} \, dx\)
Optimal. Leaf size=66 \[ \frac{2}{3 \sqrt{x+1} \sqrt{x^2-x+1}}-\frac{2 \sqrt{x^3+1} \tanh ^{-1}\left (\sqrt{x^3+1}\right )}{3 \sqrt{x+1} \sqrt{x^2-x+1}} \]
[Out]
2/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (2*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 = 0.0338986, antiderivative size = 66, normalized size of antiderivative = 1.,
number of steps used = 5, 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} \[ \frac{2}{3 \sqrt{x+1} \sqrt{x^2-x+1}}-\frac{2 \sqrt{x^3+1} \tanh ^{-1}\left (\sqrt{x^3+1}\right )}{3 \sqrt{x+1} \sqrt{x^2-x+1}} \]
Antiderivative was successfully verified.
[In]
Int[1/(x*(1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]
[Out]
2/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (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)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx &=\frac{\sqrt{1+x^3} \int \frac{1}{x \left (1+x^3\right )^{3/2}} \, dx}{\sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\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{\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{\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 \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 = 6.08176, size = 2511, normalized size = 38.05 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(x*(1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]
[Out]
Sqrt[1 + x]*Sqrt[1 - x + x^2]*(2/(9*(1 + x)) - (2*(-2 + x))/(9*(1 - x + x^2))) + 2*(((-I)*(1 + x)*Sqrt[1 - 6/(
(3 - I*Sqrt[3])*(1 + x))]*Sqrt[1 - 6/((3 + I*Sqrt[3])*(1 + x))]*EllipticF[I*ArcSinh[Sqrt[-6/(3 - I*Sqrt[3])]/S
qrt[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]])*(S
qrt[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]] + 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]])), ArcSin[Sqrt[((Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sqrt[3]])*(Sqrt[6*(3 - I*Sqrt[3])] + 6/Sq
rt[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]])*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[Sqr
t[((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]] + 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]])), 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*Sqr
t[3])] - 6/Sqrt[1 + x]))]], (Sqrt[3 - I*Sqrt[3]] + Sqrt[3 + I*Sqrt[3]])^2/(Sqrt[3 - I*Sqrt[3]] - Sqrt[3 + I*Sq
rt[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]))
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Maple [A] time = 0.937, size = 43, normalized size = 0.7 \begin{align*} -{\frac{2}{3\,{x}^{3}+3}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ({\it Artanh} \left ( \sqrt{{x}^{3}+1} \right ) \sqrt{{x}^{3}+1}-1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x/(1+x)^(3/2)/(x^2-x+1)^(3/2),x)
[Out]
-2/3*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(arctanh((x^3+1)^(1/2))*(x^3+1)^(1/2)-1)/(x^3+1)
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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{3}{2}}{\left (x + 1\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)*x), x)
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Fricas [A] time = 1.74596, size = 205, normalized size = 3.11 \begin{align*} -\frac{{\left (x^{3} + 1\right )} \log \left (\sqrt{x^{2} - x + 1} \sqrt{x + 1} + 1\right ) -{\left (x^{3} + 1\right )} \log \left (\sqrt{x^{2} - x + 1} \sqrt{x + 1} - 1\right ) - 2 \, \sqrt{x^{2} - x + 1} \sqrt{x + 1}}{3 \,{\left (x^{3} + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="fricas")
[Out]
-1/3*((x^3 + 1)*log(sqrt(x^2 - x + 1)*sqrt(x + 1) + 1) - (x^3 + 1)*log(sqrt(x^2 - x + 1)*sqrt(x + 1) - 1) - 2*
sqrt(x^2 - x + 1)*sqrt(x + 1))/(x^3 + 1)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (x + 1\right )^{\frac{3}{2}} \left (x^{2} - x + 1\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)**(3/2)/(x**2-x+1)**(3/2),x)
[Out]
Integral(1/(x*(x + 1)**(3/2)*(x**2 - x + 1)**(3/2)), x)
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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{3}{2}}{\left (x + 1\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="giac")
[Out]
integrate(1/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)*x), x)