3.1.53 \(\int x^2 (a+b x)^2 \, dx\) [53]

Optimal. Leaf size=30 \[ \frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \]

[Out]

1/3*a^2*x^3+1/2*a*b*x^4+1/5*b^2*x^5

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Rubi [A]
time = 0.01, antiderivative size = 30, 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 = {45} \begin {gather*} \frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)^2,x]

[Out]

(a^2*x^3)/3 + (a*b*x^4)/2 + (b^2*x^5)/5

Rule 45

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 (a+b x)^2 \, dx &=\int \left (a^2 x^2+2 a b x^3+b^2 x^4\right ) \, dx\\ &=\frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} \frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x)^2,x]

[Out]

(a^2*x^3)/3 + (a*b*x^4)/2 + (b^2*x^5)/5

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Mathics [A]
time = 1.66, size = 23, normalized size = 0.77 \begin {gather*} x^3 \left (\frac {a^2}{3}+\frac {a b x}{2}+\frac {b^2 x^2}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^2*(a + b*x)^2,x]')

[Out]

x ^ 3 (a ^ 2 / 3 + a b x / 2 + b ^ 2 x ^ 2 / 5)

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Maple [A]
time = 0.07, size = 25, normalized size = 0.83

method	result	size
gosper	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
default	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
norman	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
risch	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/3*a^2*x^3+1/2*a*b*x^4+1/5*b^2*x^5

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Maxima [A]
time = 0.25, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^2,x, algorithm="maxima")

[Out]

1/5*b^2*x^5 + 1/2*a*b*x^4 + 1/3*a^2*x^3

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Fricas [A]
time = 0.31, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^2,x, algorithm="fricas")

[Out]

1/5*b^2*x^5 + 1/2*a*b*x^4 + 1/3*a^2*x^3

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Sympy [A]
time = 0.03, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^{2} x^{3}}{3} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{5}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)**2,x)

[Out]

a**2*x**3/3 + a*b*x**4/2 + b**2*x**5/5

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Giac [A]
time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} \frac {1}{5} x^{5} b^{2}+\frac {1}{2} x^{4} b a+\frac {1}{3} x^{3} a^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^2,x)

[Out]

1/5*b^2*x^5 + 1/2*a*b*x^4 + 1/3*a^2*x^3

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Mupad [B]
time = 0.03, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^2\,x^3}{3}+\frac {a\,b\,x^4}{2}+\frac {b^2\,x^5}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x)^2,x)

[Out]

(a^2*x^3)/3 + (b^2*x^5)/5 + (a*b*x^4)/2

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