3.3.50 \(\int x^5 (a^2+2 a b x^2+b^2 x^4)^2 \, dx\)

Optimal. Leaf size=53 \[ \frac {a^2 \left (a+b x^2\right )^5}{10 b^3}+\frac {\left (a+b x^2\right )^7}{14 b^3}-\frac {a \left (a+b x^2\right )^6}{6 b^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 266, 43} \begin {gather*} \frac {a^2 \left (a+b x^2\right )^5}{10 b^3}+\frac {\left (a+b x^2\right )^7}{14 b^3}-\frac {a \left (a+b x^2\right )^6}{6 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(a^2*(a + b*x^2)^5)/(10*b^3) - (a*(a + b*x^2)^6)/(6*b^3) + (a + b*x^2)^7/(14*b^3)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]]

Rubi steps

\begin {align*} \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx &=\frac {\int x^5 \left (a b+b^2 x^2\right )^4 \, dx}{b^4}\\ &=\frac {\operatorname {Subst}\left (\int x^2 \left (a b+b^2 x\right )^4 \, dx,x,x^2\right )}{2 b^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2 \left (a b+b^2 x\right )^4}{b^2}-\frac {2 a \left (a b+b^2 x\right )^5}{b^3}+\frac {\left (a b+b^2 x\right )^6}{b^4}\right ) \, dx,x,x^2\right )}{2 b^4}\\ &=\frac {a^2 \left (a+b x^2\right )^5}{10 b^3}-\frac {a \left (a+b x^2\right )^6}{6 b^3}+\frac {\left (a+b x^2\right )^7}{14 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 56, normalized size = 1.06 \begin {gather*} \frac {a^4 x^6}{6}+\frac {1}{2} a^3 b x^8+\frac {3}{5} a^2 b^2 x^{10}+\frac {1}{3} a b^3 x^{12}+\frac {b^4 x^{14}}{14} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(a^4*x^6)/6 + (a^3*b*x^8)/2 + (3*a^2*b^2*x^10)/5 + (a*b^3*x^12)/3 + (b^4*x^14)/14

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

IntegrateAlgebraic[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^2, x]

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fricas [A]  time = 0.87, size = 46, normalized size = 0.87 \begin {gather*} \frac {1}{14} x^{14} b^{4} + \frac {1}{3} x^{12} b^{3} a + \frac {3}{5} x^{10} b^{2} a^{2} + \frac {1}{2} x^{8} b a^{3} + \frac {1}{6} x^{6} a^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

1/14*x^14*b^4 + 1/3*x^12*b^3*a + 3/5*x^10*b^2*a^2 + 1/2*x^8*b*a^3 + 1/6*x^6*a^4

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giac [A]  time = 0.15, size = 46, normalized size = 0.87 \begin {gather*} \frac {1}{14} \, b^{4} x^{14} + \frac {1}{3} \, a b^{3} x^{12} + \frac {3}{5} \, a^{2} b^{2} x^{10} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{6} \, a^{4} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

1/14*b^4*x^14 + 1/3*a*b^3*x^12 + 3/5*a^2*b^2*x^10 + 1/2*a^3*b*x^8 + 1/6*a^4*x^6

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maple [A]  time = 0.00, size = 47, normalized size = 0.89 \begin {gather*} \frac {1}{14} b^{4} x^{14}+\frac {1}{3} a \,b^{3} x^{12}+\frac {3}{5} a^{2} b^{2} x^{10}+\frac {1}{2} a^{3} b \,x^{8}+\frac {1}{6} a^{4} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

1/14*b^4*x^14+1/3*a*b^3*x^12+3/5*a^2*b^2*x^10+1/2*a^3*b*x^8+1/6*a^4*x^6

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maxima [A]  time = 1.34, size = 46, normalized size = 0.87 \begin {gather*} \frac {1}{14} \, b^{4} x^{14} + \frac {1}{3} \, a b^{3} x^{12} + \frac {3}{5} \, a^{2} b^{2} x^{10} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{6} \, a^{4} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

1/14*b^4*x^14 + 1/3*a*b^3*x^12 + 3/5*a^2*b^2*x^10 + 1/2*a^3*b*x^8 + 1/6*a^4*x^6

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mupad [B]  time = 0.02, size = 46, normalized size = 0.87 \begin {gather*} \frac {a^4\,x^6}{6}+\frac {a^3\,b\,x^8}{2}+\frac {3\,a^2\,b^2\,x^{10}}{5}+\frac {a\,b^3\,x^{12}}{3}+\frac {b^4\,x^{14}}{14} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

(a^4*x^6)/6 + (b^4*x^14)/14 + (a^3*b*x^8)/2 + (a*b^3*x^12)/3 + (3*a^2*b^2*x^10)/5

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sympy [A]  time = 0.08, size = 49, normalized size = 0.92 \begin {gather*} \frac {a^{4} x^{6}}{6} + \frac {a^{3} b x^{8}}{2} + \frac {3 a^{2} b^{2} x^{10}}{5} + \frac {a b^{3} x^{12}}{3} + \frac {b^{4} x^{14}}{14} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

a**4*x**6/6 + a**3*b*x**8/2 + 3*a**2*b**2*x**10/5 + a*b**3*x**12/3 + b**4*x**14/14

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