3.5.25 \(\int x^2 (a^2+2 a b x^2+b^2 x^4)^2 \, dx\) [425]

Optimal. Leaf size=56 \[ \frac {a^4 x^3}{3}+\frac {4}{5} a^3 b x^5+\frac {6}{7} a^2 b^2 x^7+\frac {4}{9} a b^3 x^9+\frac {b^4 x^{11}}{11} \]

[Out]

1/3*a^4*x^3+4/5*a^3*b*x^5+6/7*a^2*b^2*x^7+4/9*a*b^3*x^9+1/11*b^4*x^11

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Rubi [A]
time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {28, 276} \begin {gather*} \frac {a^4 x^3}{3}+\frac {4}{5} a^3 b x^5+\frac {6}{7} a^2 b^2 x^7+\frac {4}{9} a b^3 x^9+\frac {b^4 x^{11}}{11} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*x^3)/3 + (4*a^3*b*x^5)/5 + (6*a^2*b^2*x^7)/7 + (4*a*b^3*x^9)/9 + (b^4*x^11)/11

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 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]

Rubi steps

\begin {align*} \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx &=\frac {\int x^2 \left (a b+b^2 x^2\right )^4 \, dx}{b^4}\\ &=\frac {\int \left (a^4 b^4 x^2+4 a^3 b^5 x^4+6 a^2 b^6 x^6+4 a b^7 x^8+b^8 x^{10}\right ) \, dx}{b^4}\\ &=\frac {a^4 x^3}{3}+\frac {4}{5} a^3 b x^5+\frac {6}{7} a^2 b^2 x^7+\frac {4}{9} a b^3 x^9+\frac {b^4 x^{11}}{11}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 56, normalized size = 1.00 \begin {gather*} \frac {a^4 x^3}{3}+\frac {4}{5} a^3 b x^5+\frac {6}{7} a^2 b^2 x^7+\frac {4}{9} a b^3 x^9+\frac {b^4 x^{11}}{11} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*x^3)/3 + (4*a^3*b*x^5)/5 + (6*a^2*b^2*x^7)/7 + (4*a*b^3*x^9)/9 + (b^4*x^11)/11

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Maple [A]
time = 0.05, size = 47, normalized size = 0.84

method	result	size
default	\(\frac {1}{3} a^{4} x^{3}+\frac {4}{5} a^{3} b \,x^{5}+\frac {6}{7} a^{2} b^{2} x^{7}+\frac {4}{9} a \,b^{3} x^{9}+\frac {1}{11} b^{4} x^{11}\)	\(47\)
norman	\(\frac {1}{3} a^{4} x^{3}+\frac {4}{5} a^{3} b \,x^{5}+\frac {6}{7} a^{2} b^{2} x^{7}+\frac {4}{9} a \,b^{3} x^{9}+\frac {1}{11} b^{4} x^{11}\)	\(47\)
risch	\(\frac {1}{3} a^{4} x^{3}+\frac {4}{5} a^{3} b \,x^{5}+\frac {6}{7} a^{2} b^{2} x^{7}+\frac {4}{9} a \,b^{3} x^{9}+\frac {1}{11} b^{4} x^{11}\)	\(47\)
gosper	\(\frac {x^{3} \left (315 b^{4} x^{8}+1540 a \,b^{3} x^{6}+2970 a^{2} b^{2} x^{4}+2772 a^{3} b \,x^{2}+1155 a^{4}\right )}{3465}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/3*a^4*x^3+4/5*a^3*b*x^5+6/7*a^2*b^2*x^7+4/9*a*b^3*x^9+1/11*b^4*x^11

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Maxima [A]
time = 0.28, size = 46, normalized size = 0.82 \begin {gather*} \frac {1}{11} \, b^{4} x^{11} + \frac {4}{9} \, a b^{3} x^{9} + \frac {6}{7} \, a^{2} b^{2} x^{7} + \frac {4}{5} \, a^{3} b x^{5} + \frac {1}{3} \, a^{4} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*b^4*x^11 + 4/9*a*b^3*x^9 + 6/7*a^2*b^2*x^7 + 4/5*a^3*b*x^5 + 1/3*a^4*x^3

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Fricas [A]
time = 0.34, size = 46, normalized size = 0.82 \begin {gather*} \frac {1}{11} \, b^{4} x^{11} + \frac {4}{9} \, a b^{3} x^{9} + \frac {6}{7} \, a^{2} b^{2} x^{7} + \frac {4}{5} \, a^{3} b x^{5} + \frac {1}{3} \, a^{4} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*b^4*x^11 + 4/9*a*b^3*x^9 + 6/7*a^2*b^2*x^7 + 4/5*a^3*b*x^5 + 1/3*a^4*x^3

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Sympy [A]
time = 0.01, size = 53, normalized size = 0.95 \begin {gather*} \frac {a^{4} x^{3}}{3} + \frac {4 a^{3} b x^{5}}{5} + \frac {6 a^{2} b^{2} x^{7}}{7} + \frac {4 a b^{3} x^{9}}{9} + \frac {b^{4} x^{11}}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*x**3/3 + 4*a**3*b*x**5/5 + 6*a**2*b**2*x**7/7 + 4*a*b**3*x**9/9 + b**4*x**11/11

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Giac [A]
time = 4.28, size = 46, normalized size = 0.82 \begin {gather*} \frac {1}{11} \, b^{4} x^{11} + \frac {4}{9} \, a b^{3} x^{9} + \frac {6}{7} \, a^{2} b^{2} x^{7} + \frac {4}{5} \, a^{3} b x^{5} + \frac {1}{3} \, a^{4} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*b^4*x^11 + 4/9*a*b^3*x^9 + 6/7*a^2*b^2*x^7 + 4/5*a^3*b*x^5 + 1/3*a^4*x^3

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Mupad [B]
time = 0.02, size = 46, normalized size = 0.82 \begin {gather*} \frac {a^4\,x^3}{3}+\frac {4\,a^3\,b\,x^5}{5}+\frac {6\,a^2\,b^2\,x^7}{7}+\frac {4\,a\,b^3\,x^9}{9}+\frac {b^4\,x^{11}}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^4*x^3)/3 + (b^4*x^11)/11 + (4*a^3*b*x^5)/5 + (4*a*b^3*x^9)/9 + (6*a^2*b^2*x^7)/7

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