3.6.21 \(\int \frac {1}{x (1+x)^{5/2} (1-x+x^2)^{5/2}} \, dx\) [521]

Optimal. Leaf size=96 \[ \frac {2}{3 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {2 \sqrt {1+x^3} \tanh ^{-1}\left (\sqrt {1+x^3}\right )}{3 \sqrt {1+x} \sqrt {1-x+x^2}} \]

[Out]

2/3/(1+x)^(1/2)/(x^2-x+1)^(1/2)+2/9/(x^3+1)/(1+x)^(1/2)/(x^2-x+1)^(1/2)-2/3*arctanh((x^3+1)^(1/2))*(x^3+1)^(1/
2)/(1+x)^(1/2)/(x^2-x+1)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 = {929, 272, 53, 65, 213} \begin {gather*} \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}} \end {gather*}

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 53

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 65

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 213

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

Rule 272

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 929

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]

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} \text {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} \text {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} \text {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 ) \text {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*}

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Mathematica [A]
time = 10.09, size = 95, normalized size = 0.99 \begin {gather*} \frac {\frac {2 \left (4+3 x^3\right )}{3 (1+x)^{3/2} \left (1-x+x^2\right )}-2 (1+x) \sqrt {\frac {1-x+x^2}{(1+x)^2}} \tanh ^{-1}\left (\frac {1}{(1+x)^{3/2} \sqrt {\frac {1-x+x^2}{(1+x)^2}}}\right )}{3 \sqrt {1-x+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]

[Out]

((2*(4 + 3*x^3))/(3*(1 + x)^(3/2)*(1 - x + x^2)) - 2*(1 + x)*Sqrt[(1 - x + x^2)/(1 + x)^2]*ArcTanh[1/((1 + x)^
(3/2)*Sqrt[(1 - x + x^2)/(1 + x)^2])])/(3*Sqrt[1 - x + x^2])

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Maple [A]
time = 0.11, size = 69, normalized size = 0.72

method	result	size
elliptic	\(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {2}{9 \left (x^{3}+1\right )^{\frac {3}{2}}}+\frac {2}{3 \sqrt {x^{3}+1}}-\frac {2 \arctanh \left (\sqrt {x^{3}+1}\right )}{3}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(60\)
default	\(-\frac {2 \left (3 \sqrt {x^{3}+1}\, \arctanh \left (\sqrt {x^{3}+1}\right ) x^{3}-3 x^{3}+3 \arctanh \left (\sqrt {x^{3}+1}\right ) \sqrt {x^{3}+1}-4\right )}{9 \left (x^{3}+1\right ) \sqrt {x^{2}-x +1}\, \sqrt {1+x}}\)	\(69\)

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,method=_RETURNVERBOSE)

[Out]

-2/9*(3*(x^3+1)^(1/2)*arctanh((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 = 0.00, 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(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 = 1.41, size = 101, normalized size = 1.05 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (x + 1\right )^{\frac {5}{2}} \left (x^{2} - x + 1\right )^{\frac {5}{2}}}\, dx \end {gather*}

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]

Integral(1/(x*(x + 1)**(5/2)*(x**2 - x + 1)**(5/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (x+1\right )}^{5/2}\,{\left (x^2-x+1\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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