3.3.8 \(\int x^2 (a x^2+b x^3)^2 \, dx\) [208]

Optimal. Leaf size=30 \[ \frac {a^2 x^7}{7}+\frac {1}{4} a b x^8+\frac {b^2 x^9}{9} \]

[Out]

1/7*a^2*x^7+1/4*a*b*x^8+1/9*b^2*x^9

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Rubi [A]
time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 45} \begin {gather*} \frac {a^2 x^7}{7}+\frac {1}{4} a b x^8+\frac {b^2 x^9}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a*x^2 + b*x^3)^2,x]

[Out]

(a^2*x^7)/7 + (a*b*x^8)/4 + (b^2*x^9)/9

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])

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int x^2 \left (a x^2+b x^3\right )^2 \, dx &=\int x^6 (a+b x)^2 \, dx\\ &=\int \left (a^2 x^6+2 a b x^7+b^2 x^8\right ) \, dx\\ &=\frac {a^2 x^7}{7}+\frac {1}{4} a b x^8+\frac {b^2 x^9}{9}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} \frac {a^2 x^7}{7}+\frac {1}{4} a b x^8+\frac {b^2 x^9}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a*x^2 + b*x^3)^2,x]

[Out]

(a^2*x^7)/7 + (a*b*x^8)/4 + (b^2*x^9)/9

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Maple [A]
time = 0.43, size = 25, normalized size = 0.83

method	result	size
gosper	\(\frac {x^{7} \left (28 b^{2} x^{2}+63 a b x +36 a^{2}\right )}{252}\)	\(25\)
default	\(\frac {1}{7} a^{2} x^{7}+\frac {1}{4} a b \,x^{8}+\frac {1}{9} b^{2} x^{9}\)	\(25\)
norman	\(\frac {1}{7} a^{2} x^{7}+\frac {1}{4} a b \,x^{8}+\frac {1}{9} b^{2} x^{9}\)	\(25\)
risch	\(\frac {1}{7} a^{2} x^{7}+\frac {1}{4} a b \,x^{8}+\frac {1}{9} b^{2} x^{9}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^3+a*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/7*a^2*x^7+1/4*a*b*x^8+1/9*b^2*x^9

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Maxima [A]
time = 0.28, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {1}{4} \, a b x^{8} + \frac {1}{7} \, a^{2} x^{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a*x^2)^2,x, algorithm="maxima")

[Out]

1/9*b^2*x^9 + 1/4*a*b*x^8 + 1/7*a^2*x^7

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Fricas [A]
time = 1.28, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {1}{4} \, a b x^{8} + \frac {1}{7} \, a^{2} x^{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a*x^2)^2,x, algorithm="fricas")

[Out]

1/9*b^2*x^9 + 1/4*a*b*x^8 + 1/7*a^2*x^7

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Sympy [A]
time = 0.01, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^{2} x^{7}}{7} + \frac {a b x^{8}}{4} + \frac {b^{2} x^{9}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**3+a*x**2)**2,x)

[Out]

a**2*x**7/7 + a*b*x**8/4 + b**2*x**9/9

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Giac [A]
time = 1.44, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {1}{4} \, a b x^{8} + \frac {1}{7} \, a^{2} x^{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a*x^2)^2,x, algorithm="giac")

[Out]

1/9*b^2*x^9 + 1/4*a*b*x^8 + 1/7*a^2*x^7

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Mupad [B]
time = 0.04, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^2\,x^7}{7}+\frac {a\,b\,x^8}{4}+\frac {b^2\,x^9}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a*x^2 + b*x^3)^2,x)

[Out]

(a^2*x^7)/7 + (b^2*x^9)/9 + (a*b*x^8)/4

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