3.1469 \(\int (a^2+2 a b x+b^2 x^2)^2 \, dx\)

Optimal. Leaf size=14 \[ \frac{(a+b x)^5}{5 b} \]

[Out]

(a + b*x)^5/(5*b)

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Rubi [A]  time = 0.0023447, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {27, 32} \[ \frac{(a+b x)^5}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(a + b*x)^5/(5*b)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.001292, size = 14, normalized size = 1. \[ \frac{(a+b x)^5}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*x)^5/(5*b)

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Maple [B]  time = 0.039, size = 43, normalized size = 3.1 \begin{align*}{\frac{{b}^{4}{x}^{5}}{5}}+a{b}^{3}{x}^{4}+2\,{b}^{2}{a}^{2}{x}^{3}+2\,{a}^{3}b{x}^{2}+{a}^{4}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^4*x^5+a*b^3*x^4+2*b^2*a^2*x^3+2*a^3*b*x^2+a^4*x

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Maxima [B]  time = 1.16626, size = 72, normalized size = 5.14 \begin{align*} \frac{1}{5} \, b^{4} x^{5} + a b^{3} x^{4} + \frac{4}{3} \, a^{2} b^{2} x^{3} + a^{4} x + \frac{2}{3} \,{\left (b^{2} x^{3} + 3 \, a b x^{2}\right )} a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^4*x^5 + a*b^3*x^4 + 4/3*a^2*b^2*x^3 + a^4*x + 2/3*(b^2*x^3 + 3*a*b*x^2)*a^2

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Fricas [B]  time = 1.39525, size = 85, normalized size = 6.07 \begin{align*} \frac{1}{5} x^{5} b^{4} + x^{4} b^{3} a + 2 x^{3} b^{2} a^{2} + 2 x^{2} b a^{3} + x a^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*b^4 + x^4*b^3*a + 2*x^3*b^2*a^2 + 2*x^2*b*a^3 + x*a^4

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Sympy [B]  time = 0.065736, size = 42, normalized size = 3. \begin{align*} a^{4} x + 2 a^{3} b x^{2} + 2 a^{2} b^{2} x^{3} + a b^{3} x^{4} + \frac{b^{4} x^{5}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*x + 2*a**3*b*x**2 + 2*a**2*b**2*x**3 + a*b**3*x**4 + b**4*x**5/5

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Giac [B]  time = 1.14409, size = 57, normalized size = 4.07 \begin{align*} \frac{1}{5} \, b^{4} x^{5} + a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{3} + 2 \, a^{3} b x^{2} + a^{4} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^4*x^5 + a*b^3*x^4 + 2*a^2*b^2*x^3 + 2*a^3*b*x^2 + a^4*x