3.493 \(\int \frac{\sqrt{1+x} \sqrt{1-x+x^2}}{x} \, dx\)

Optimal. Leaf size=66 \[ \frac{2}{3} \sqrt{x+1} \sqrt{x^2-x+1}-\frac{2 \sqrt{x+1} \sqrt{x^2-x+1} \tanh ^{-1}\left (\sqrt{x^3+1}\right )}{3 \sqrt{x^3+1}} \]

[Out]

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

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Rubi [A]  time = 0.032602, 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, 50, 63, 207} \[ \frac{2}{3} \sqrt{x+1} \sqrt{x^2-x+1}-\frac{2 \sqrt{x+1} \sqrt{x^2-x+1} \tanh ^{-1}\left (\sqrt{x^3+1}\right )}{3 \sqrt{x^3+1}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 + x]*Sqrt[1 - x + x^2])/x,x]

[Out]

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

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 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 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{\sqrt{1+x} \sqrt{1-x+x^2}}{x} \, dx &=\frac{\left (\sqrt{1+x} \sqrt{1-x+x^2}\right ) \int \frac{\sqrt{1+x^3}}{x} \, dx}{\sqrt{1+x^3}}\\ &=\frac{\left (\sqrt{1+x} \sqrt{1-x+x^2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{x} \, dx,x,x^3\right )}{3 \sqrt{1+x^3}}\\ &=\frac{2}{3} \sqrt{1+x} \sqrt{1-x+x^2}+\frac{\left (\sqrt{1+x} \sqrt{1-x+x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+x}} \, dx,x,x^3\right )}{3 \sqrt{1+x^3}}\\ &=\frac{2}{3} \sqrt{1+x} \sqrt{1-x+x^2}+\frac{\left (2 \sqrt{1+x} \sqrt{1-x+x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sqrt{1+x^3}\right )}{3 \sqrt{1+x^3}}\\ &=\frac{2}{3} \sqrt{1+x} \sqrt{1-x+x^2}-\frac{2 \sqrt{1+x} \sqrt{1-x+x^2} \tanh ^{-1}\left (\sqrt{1+x^3}\right )}{3 \sqrt{1+x^3}}\\ \end{align*}

Mathematica [C]  time = 0.411064, size = 197, normalized size = 2.98 \[ \frac{\sqrt{x+1} \left (2 \left (x^2-x+1\right )+\frac{3 i \sqrt{2} \sqrt{\frac{-2 i x+\sqrt{3}+i}{\sqrt{3}+3 i}} \sqrt{\frac{2 i x+\sqrt{3}-i}{\sqrt{3}-3 i}} \Pi \left (\frac{3}{2}-\frac{i \sqrt{3}}{2};i \sinh ^{-1}\left (\sqrt{2} \sqrt{-\frac{i (x+1)}{3 i+\sqrt{3}}}\right )|\frac{3 i+\sqrt{3}}{3 i-\sqrt{3}}\right )}{\sqrt{-\frac{i (x+1)}{\sqrt{3}+3 i}}}\right )}{3 \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + x]*(2*(1 - x + x^2) + ((3*I)*Sqrt[2]*Sqrt[(I + Sqrt[3] - (2*I)*x)/(3*I + Sqrt[3])]*Sqrt[(-I + Sqrt[3
] + (2*I)*x)/(-3*I + Sqrt[3])]*EllipticPi[3/2 - (I/2)*Sqrt[3], I*ArcSinh[Sqrt[2]*Sqrt[((-I)*(1 + x))/(3*I + Sq
rt[3])]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[((-I)*(1 + x))/(3*I + Sqrt[3])]))/(3*Sqrt[1 - x + x^2])

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Maple [A]  time = 0.091, size = 43, normalized size = 0.7 \begin{align*} -{\frac{2}{3}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ( -\sqrt{{x}^{3}+1}+{\it Artanh} \left ( \sqrt{{x}^{3}+1} \right ) \right ){\frac{1}{\sqrt{{x}^{3}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - x + 1} \sqrt{x + 1}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^2 - x + 1)*sqrt(x + 1)/x, x)

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Fricas [A]  time = 1.70636, size = 169, normalized size = 2.56 \begin{align*} \frac{2}{3} \, \sqrt{x^{2} - x + 1} \sqrt{x + 1} - \frac{1}{3} \, \log \left (\sqrt{x^{2} - x + 1} \sqrt{x + 1} + 1\right ) + \frac{1}{3} \, \log \left (\sqrt{x^{2} - x + 1} \sqrt{x + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(x^2 - x + 1)*sqrt(x + 1) - 1/3*log(sqrt(x^2 - x + 1)*sqrt(x + 1) + 1) + 1/3*log(sqrt(x^2 - x + 1)*sqr
t(x + 1) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x + 1} \sqrt{x^{2} - x + 1}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**(1/2)*(x**2-x+1)**(1/2)/x,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - x + 1} \sqrt{x + 1}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/2)*(x^2-x+1)^(1/2)/x,x, algorithm="giac")

[Out]

integrate(sqrt(x^2 - x + 1)*sqrt(x + 1)/x, x)