\(\int x (a^2+2 a b x^2+b^2 x^4)^2 \, dx\) [426]

Optimal result

Integrand size = 22, antiderivative size = 16 \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {\left (a+b x^2\right )^5}{10 b} \]

[Out]

1/10*(b*x^2+a)^5/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {28, 267} \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {\left (a+b x^2\right )^5}{10 b} \]

[In]

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

[Out]

(a + b*x^2)^5/(10*b)

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 267

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int x \left (a b+b^2 x^2\right )^4 \, dx}{b^4} \\ & = \frac {\left (a+b x^2\right )^5}{10 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {\left (a+b x^2\right )^5}{10 b} \]

[In]

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

[Out]

(a + b*x^2)^5/(10*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(44\) vs. \(2(14)=28\).

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.81

method	result	size
default	\(\frac {1}{10} b^{4} x^{10}+\frac {1}{2} a \,b^{3} x^{8}+a^{2} b^{2} x^{6}+a^{3} b \,x^{4}+\frac {1}{2} a^{4} x^{2}\)	\(45\)
norman	\(\frac {1}{10} b^{4} x^{10}+\frac {1}{2} a \,b^{3} x^{8}+a^{2} b^{2} x^{6}+a^{3} b \,x^{4}+\frac {1}{2} a^{4} x^{2}\)	\(45\)
risch	\(\frac {1}{10} b^{4} x^{10}+\frac {1}{2} a \,b^{3} x^{8}+a^{2} b^{2} x^{6}+a^{3} b \,x^{4}+\frac {1}{2} a^{4} x^{2}\)	\(45\)
parallelrisch	\(\frac {1}{10} b^{4} x^{10}+\frac {1}{2} a \,b^{3} x^{8}+a^{2} b^{2} x^{6}+a^{3} b \,x^{4}+\frac {1}{2} a^{4} x^{2}\)	\(45\)
gosper	\(\frac {x^{2} \left (b^{4} x^{8}+5 a \,b^{3} x^{6}+10 a^{2} b^{2} x^{4}+10 a^{3} b \,x^{2}+5 a^{4}\right )}{10}\)	\(48\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (14) = 28\).

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {1}{10} \, b^{4} x^{10} + \frac {1}{2} \, a b^{3} x^{8} + a^{2} b^{2} x^{6} + a^{3} b x^{4} + \frac {1}{2} \, a^{4} x^{2} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (10) = 20\).

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {a^{4} x^{2}}{2} + a^{3} b x^{4} + a^{2} b^{2} x^{6} + \frac {a b^{3} x^{8}}{2} + \frac {b^{4} x^{10}}{10} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (14) = 28\).

Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {1}{10} \, b^{4} x^{10} + \frac {1}{2} \, a b^{3} x^{8} + a^{2} b^{2} x^{6} + a^{3} b x^{4} + \frac {1}{2} \, a^{4} x^{2} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {1}{10} \, b^{4} x^{10} + \frac {1}{2} \, a b^{3} x^{8} + a^{2} b^{2} x^{6} + a^{3} b x^{4} + \frac {1}{2} \, a^{4} x^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int x \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {a^4\,x^2}{2}+a^3\,b\,x^4+a^2\,b^2\,x^6+\frac {a\,b^3\,x^8}{2}+\frac {b^4\,x^{10}}{10} \]

[In]

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

[Out]

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