3.41 \(\int x \sqrt{x+x^2} \, dx\)

Optimal. Leaf size=48 \[ \frac{1}{3} \left (x^2+x\right )^{3/2}-\frac{1}{8} (2 x+1) \sqrt{x^2+x}+\frac{1}{8} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2+x}}\right ) \]

[Out]

-((1 + 2*x)*Sqrt[x + x^2])/8 + (x + x^2)^(3/2)/3 + ArcTanh[x/Sqrt[x + x^2]]/8

________________________________________________________________________________________

Rubi [A]  time = 0.0099011, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {640, 612, 620, 206} \[ \frac{1}{3} \left (x^2+x\right )^{3/2}-\frac{1}{8} (2 x+1) \sqrt{x^2+x}+\frac{1}{8} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2+x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[x + x^2],x]

[Out]

-((1 + 2*x)*Sqrt[x + x^2])/8 + (x + x^2)^(3/2)/3 + ArcTanh[x/Sqrt[x + x^2]]/8

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

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

Rubi steps

\begin{align*} \int x \sqrt{x+x^2} \, dx &=\frac{1}{3} \left (x+x^2\right )^{3/2}-\frac{1}{2} \int \sqrt{x+x^2} \, dx\\ &=-\frac{1}{8} (1+2 x) \sqrt{x+x^2}+\frac{1}{3} \left (x+x^2\right )^{3/2}+\frac{1}{16} \int \frac{1}{\sqrt{x+x^2}} \, dx\\ &=-\frac{1}{8} (1+2 x) \sqrt{x+x^2}+\frac{1}{3} \left (x+x^2\right )^{3/2}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{x+x^2}}\right )\\ &=-\frac{1}{8} (1+2 x) \sqrt{x+x^2}+\frac{1}{3} \left (x+x^2\right )^{3/2}+\frac{1}{8} \tanh ^{-1}\left (\frac{x}{\sqrt{x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0306674, size = 43, normalized size = 0.9 \[ \frac{1}{24} \sqrt{x (x+1)} \left (8 x^2+2 x+\frac{3 \sinh ^{-1}\left (\sqrt{x}\right )}{\sqrt{x+1} \sqrt{x}}-3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[x + x^2],x]

[Out]

(Sqrt[x*(1 + x)]*(-3 + 2*x + 8*x^2 + (3*ArcSinh[Sqrt[x]])/(Sqrt[x]*Sqrt[1 + x])))/24

________________________________________________________________________________________

Maple [A]  time = 0.045, size = 38, normalized size = 0.8 \begin{align*}{\frac{1}{3} \left ({x}^{2}+x \right ) ^{{\frac{3}{2}}}}-{\frac{1+2\,x}{8}\sqrt{{x}^{2}+x}}+{\frac{1}{16}\ln \left ( x+{\frac{1}{2}}+\sqrt{{x}^{2}+x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^2+x)^(3/2)-1/8*(1+2*x)*(x^2+x)^(1/2)+1/16*ln(x+1/2+(x^2+x)^(1/2))

________________________________________________________________________________________

Maxima [A]  time = 1.14893, size = 62, normalized size = 1.29 \begin{align*} \frac{1}{3} \,{\left (x^{2} + x\right )}^{\frac{3}{2}} - \frac{1}{4} \, \sqrt{x^{2} + x} x - \frac{1}{8} \, \sqrt{x^{2} + x} + \frac{1}{16} \, \log \left (2 \, x + 2 \, \sqrt{x^{2} + x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^2 + x)^(3/2) - 1/4*sqrt(x^2 + x)*x - 1/8*sqrt(x^2 + x) + 1/16*log(2*x + 2*sqrt(x^2 + x) + 1)

________________________________________________________________________________________

Fricas [A]  time = 1.94107, size = 104, normalized size = 2.17 \begin{align*} \frac{1}{24} \,{\left (8 \, x^{2} + 2 \, x - 3\right )} \sqrt{x^{2} + x} - \frac{1}{16} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} + x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*x^2 + 2*x - 3)*sqrt(x^2 + x) - 1/16*log(-2*x + 2*sqrt(x^2 + x) - 1)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{x \left (x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(x*(x + 1)), x)

________________________________________________________________________________________

Giac [A]  time = 1.1555, size = 51, normalized size = 1.06 \begin{align*} \frac{1}{24} \,{\left (2 \,{\left (4 \, x + 1\right )} x - 3\right )} \sqrt{x^{2} + x} - \frac{1}{16} \, \log \left ({\left | -2 \, x + 2 \, \sqrt{x^{2} + x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(2*(4*x + 1)*x - 3)*sqrt(x^2 + x) - 1/16*log(abs(-2*x + 2*sqrt(x^2 + x) - 1))