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]

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

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.39, size = 22, normalized size = 0.65 \[ \frac {2}{105} \, {\left (15 \, x^{3} + 3 \, x^{2} - 4 \, x + 8\right )} \sqrt {x + 1} \]

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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giac [A]  time = 0.01, size = 22, normalized size = 0.65 \[ \frac {2}{7} \, {\left (x + 1\right )}^{\frac {7}{2}} - \frac {4}{5} \, {\left (x + 1\right )}^{\frac {5}{2}} + \frac {2}{3} \, {\left (x + 1\right )}^{\frac {3}{2}} \]

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)

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maple [A]  time = 0.00, size = 18, normalized size = 0.53 \[ \frac {2 \left (x +1\right )^{\frac {3}{2}} \left (15 x^{2}-12 x +8\right )}{105} \]

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 = 0.55, size = 22, normalized size = 0.65 \[ \frac {2}{7} \, {\left (x + 1\right )}^{\frac {7}{2}} - \frac {4}{5} \, {\left (x + 1\right )}^{\frac {5}{2}} + \frac {2}{3} \, {\left (x + 1\right )}^{\frac {3}{2}} \]

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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mupad [B]  time = 0.07, size = 19, normalized size = 0.56 \[ -\frac {2\,{\left (x+1\right )}^{3/2}\,\left (42\,x-15\,{\left (x+1\right )}^2+7\right )}{105} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x + 1)^(1/2),x)

[Out]

-(2*(x + 1)^(3/2)*(42*x - 15*(x + 1)^2 + 7))/105

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sympy [A]  time = 1.39, size = 48, normalized size = 1.41 \[ \frac {2 x^{3} \sqrt {x + 1}}{7} + \frac {2 x^{2} \sqrt {x + 1}}{35} - \frac {8 x \sqrt {x + 1}}{105} + \frac {16 \sqrt {x + 1}}{105} \]

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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