3.3 \(\int x^2 \sqrt{1+x} \, dx\)

Optimal. Leaf size=34 \[ \frac{2}{7} (x+1)^{7/2}-\frac{4}{5} (x+1)^{5/2}+\frac{2}{3} (x+1)^{3/2} \]

[Out]

(2*(1 + x)^(3/2))/3 - (4*(1 + x)^(5/2))/5 + (2*(1 + x)^(7/2))/7

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Rubi [A]  time = 0.0051121, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {43} \[ \frac{2}{7} (x+1)^{7/2}-\frac{4}{5} (x+1)^{5/2}+\frac{2}{3} (x+1)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[1 + x],x]

[Out]

(2*(1 + x)^(3/2))/3 - (4*(1 + x)^(5/2))/5 + (2*(1 + x)^(7/2))/7

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^2 \sqrt{1+x} \, dx &=\int \left (\sqrt{1+x}-2 (1+x)^{3/2}+(1+x)^{5/2}\right ) \, dx\\ &=\frac{2}{3} (1+x)^{3/2}-\frac{4}{5} (1+x)^{5/2}+\frac{2}{7} (1+x)^{7/2}\\ \end{align*}

Mathematica [A]  time = 0.0056155, size = 21, normalized size = 0.62 \[ \frac{2}{105} (x+1)^{3/2} \left (15 x^2-12 x+8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[1 + x],x]

[Out]

(2*(1 + x)^(3/2)*(8 - 12*x + 15*x^2))/105

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Maple [A]  time = 0.003, size = 18, normalized size = 0.5 \begin{align*}{\frac{30\,{x}^{2}-24\,x+16}{105} \left ( 1+x \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(1+x)^(3/2)*(15*x^2-12*x+8)

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Maxima [A]  time = 0.93806, size = 30, normalized size = 0.88 \begin{align*} \frac{2}{7} \,{\left (x + 1\right )}^{\frac{7}{2}} - \frac{4}{5} \,{\left (x + 1\right )}^{\frac{5}{2}} + \frac{2}{3} \,{\left (x + 1\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(x + 1)^(7/2) - 4/5*(x + 1)^(5/2) + 2/3*(x + 1)^(3/2)

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Fricas [A]  time = 0.417573, size = 62, normalized size = 1.82 \begin{align*} \frac{2}{105} \,{\left (15 \, x^{3} + 3 \, x^{2} - 4 \, x + 8\right )} \sqrt{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*x^3 + 3*x^2 - 4*x + 8)*sqrt(x + 1)

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Sympy [A]  time = 1.19745, size = 48, normalized size = 1.41 \begin{align*} \frac{2 x^{3} \sqrt{x + 1}}{7} + \frac{2 x^{2} \sqrt{x + 1}}{35} - \frac{8 x \sqrt{x + 1}}{105} + \frac{16 \sqrt{x + 1}}{105} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**3*sqrt(x + 1)/7 + 2*x**2*sqrt(x + 1)/35 - 8*x*sqrt(x + 1)/105 + 16*sqrt(x + 1)/105

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Giac [A]  time = 1.06609, size = 30, normalized size = 0.88 \begin{align*} \frac{2}{7} \,{\left (x + 1\right )}^{\frac{7}{2}} - \frac{4}{5} \,{\left (x + 1\right )}^{\frac{5}{2}} + \frac{2}{3} \,{\left (x + 1\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(x + 1)^(7/2) - 4/5*(x + 1)^(5/2) + 2/3*(x + 1)^(3/2)