3.1.6 \(\int x^2 (a x+b x^3)^2 \, dx\) [6]

Optimal. Leaf size=30 \[ \frac {a^2 x^5}{5}+\frac {2}{7} a b x^7+\frac {b^2 x^9}{9} \]

[Out]

1/5*a^2*x^5+2/7*a*b*x^7+1/9*b^2*x^9

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1598, 276} \begin {gather*} \frac {a^2 x^5}{5}+\frac {2}{7} a b x^7+\frac {b^2 x^9}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a*x + b*x^3)^2,x]

[Out]

(a^2*x^5)/5 + (2*a*b*x^7)/7 + (b^2*x^9)/9

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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+b x^3\right )^2 \, dx &=\int x^4 \left (a+b x^2\right )^2 \, dx\\ &=\int \left (a^2 x^4+2 a b x^6+b^2 x^8\right ) \, dx\\ &=\frac {a^2 x^5}{5}+\frac {2}{7} a b x^7+\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^5}{5}+\frac {2}{7} a b x^7+\frac {b^2 x^9}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a*x + b*x^3)^2,x]

[Out]

(a^2*x^5)/5 + (2*a*b*x^7)/7 + (b^2*x^9)/9

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Maple [A]
time = 0.34, size = 25, normalized size = 0.83

method	result	size
default	\(\frac {1}{5} a^{2} x^{5}+\frac {2}{7} a b \,x^{7}+\frac {1}{9} b^{2} x^{9}\)	\(25\)
norman	\(\frac {1}{5} a^{2} x^{5}+\frac {2}{7} a b \,x^{7}+\frac {1}{9} b^{2} x^{9}\)	\(25\)
risch	\(\frac {1}{5} a^{2} x^{5}+\frac {2}{7} a b \,x^{7}+\frac {1}{9} b^{2} x^{9}\)	\(25\)
gosper	\(\frac {x^{5} \left (35 b^{2} x^{4}+90 a b \,x^{2}+63 a^{2}\right )}{315}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*x^5+2/7*a*b*x^7+1/9*b^2*x^9

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Maxima [A]
time = 0.29, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {2}{7} \, a b x^{7} + \frac {1}{5} \, a^{2} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^2*x^9 + 2/7*a*b*x^7 + 1/5*a^2*x^5

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Fricas [A]
time = 1.42, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {2}{7} \, a b x^{7} + \frac {1}{5} \, a^{2} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^2*x^9 + 2/7*a*b*x^7 + 1/5*a^2*x^5

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Sympy [A]
time = 0.01, size = 26, normalized size = 0.87 \begin {gather*} \frac {a^{2} x^{5}}{5} + \frac {2 a b x^{7}}{7} + \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,x)

[Out]

a**2*x**5/5 + 2*a*b*x**7/7 + b**2*x**9/9

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Giac [A]
time = 1.16, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {2}{7} \, a b x^{7} + \frac {1}{5} \, a^{2} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^2*x^9 + 2/7*a*b*x^7 + 1/5*a^2*x^5

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Mupad [B]
time = 0.04, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^2\,x^5}{5}+\frac {2\,a\,b\,x^7}{7}+\frac {b^2\,x^9}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a*x + b*x^3)^2,x)

[Out]

(a^2*x^5)/5 + (b^2*x^9)/9 + (2*a*b*x^7)/7

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