3.7 \(\int x (a x+b x^3)^2 \, dx\)

Optimal. Leaf size=30 \[ \frac {a^2 x^4}{4}+\frac {1}{3} a b x^6+\frac {b^2 x^8}{8} \]

[Out]

1/4*a^2*x^4+1/3*a*b*x^6+1/8*b^2*x^8

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1584, 266, 43} \[ \frac {a^2 x^4}{4}+\frac {1}{3} a b x^6+\frac {b^2 x^8}{8} \]

Antiderivative was successfully verified.

[In]

Int[x*(a*x + b*x^3)^2,x]

[Out]

(a^2*x^4)/4 + (a*b*x^6)/3 + (b^2*x^8)/8

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])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1584

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

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Mathematica [A]  time = 0.00, size = 30, normalized size = 1.00 \[ \frac {a^2 x^4}{4}+\frac {1}{3} a b x^6+\frac {b^2 x^8}{8} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a*x + b*x^3)^2,x]

[Out]

(a^2*x^4)/4 + (a*b*x^6)/3 + (b^2*x^8)/8

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fricas [A]  time = 0.80, size = 24, normalized size = 0.80 \[ \frac {1}{8} x^{8} b^{2} + \frac {1}{3} x^{6} b a + \frac {1}{4} x^{4} a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^8*b^2 + 1/3*x^6*b*a + 1/4*x^4*a^2

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giac [A]  time = 0.16, size = 24, normalized size = 0.80 \[ \frac {1}{8} \, b^{2} x^{8} + \frac {1}{3} \, a b x^{6} + \frac {1}{4} \, a^{2} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*x^8 + 1/3*a*b*x^6 + 1/4*a^2*x^4

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maple [A]  time = 0.04, size = 25, normalized size = 0.83 \[ \frac {1}{8} b^{2} x^{8}+\frac {1}{3} a b \,x^{6}+\frac {1}{4} a^{2} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4+1/3*a*b*x^6+1/8*b^2*x^8

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maxima [A]  time = 1.34, size = 24, normalized size = 0.80 \[ \frac {1}{8} \, b^{2} x^{8} + \frac {1}{3} \, a b x^{6} + \frac {1}{4} \, a^{2} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*x^8 + 1/3*a*b*x^6 + 1/4*a^2*x^4

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mupad [B]  time = 0.03, size = 24, normalized size = 0.80 \[ \frac {a^2\,x^4}{4}+\frac {a\,b\,x^6}{3}+\frac {b^2\,x^8}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a*x + b*x^3)^2,x)

[Out]

(a^2*x^4)/4 + (b^2*x^8)/8 + (a*b*x^6)/3

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sympy [A]  time = 0.07, size = 24, normalized size = 0.80 \[ \frac {a^{2} x^{4}}{4} + \frac {a b x^{6}}{3} + \frac {b^{2} x^{8}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**4/4 + a*b*x**6/3 + b**2*x**8/8

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