\(\int (a^2+2 a b x+b^2 x^2)^2 \, dx\) [1469]

Optimal result

Integrand size = 18, antiderivative size = 14 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {(a+b x)^5}{5 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {27, 32} \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {(a+b x)^5}{5 b} \]

[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*} \text {integral}& = \int (a+b x)^4 \, dx \\ & = \frac {(a+b x)^5}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {(a+b x)^5}{5 b} \]

[In]

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

[Out]

(a + b*x)^5/(5*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(42\) vs. \(2(12)=24\).

Time = 2.33 (sec) , antiderivative size = 43, normalized size of antiderivative = 3.07

method	result	size
default	\(\frac {1}{5} b^{4} x^{5}+a \,b^{3} x^{4}+2 a^{2} b^{2} x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(43\)
norman	\(\frac {1}{5} b^{4} x^{5}+a \,b^{3} x^{4}+2 a^{2} b^{2} x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(43\)
risch	\(\frac {1}{5} b^{4} x^{5}+a \,b^{3} x^{4}+2 a^{2} b^{2} x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(43\)
parallelrisch	\(\frac {1}{5} b^{4} x^{5}+a \,b^{3} x^{4}+2 a^{2} b^{2} x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(43\)
gosper	\(\frac {x \left (b^{4} x^{4}+5 a \,b^{3} x^{3}+10 a^{2} b^{2} x^{2}+10 a^{3} b x +5 a^{4}\right )}{5}\)	\(44\)

[In]

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

[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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (12) = 24\).

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 3.00 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{5} \, b^{4} x^{5} + a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{3} + 2 \, a^{3} b x^{2} + a^{4} x \]

[In]

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

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (8) = 16\).

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 3.00 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=a^{4} x + 2 a^{3} b x^{2} + 2 a^{2} b^{2} x^{3} + a b^{3} x^{4} + \frac {b^{4} x^{5}}{5} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (12) = 24\).

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 3.79 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{5} \, b^{4} x^{5} + a b^{3} x^{4} + \frac {4}{3} \, a^{2} b^{2} x^{3} + a^{4} x + \frac {2}{3} \, {\left (b^{2} x^{3} + 3 \, a b x^{2}\right )} a^{2} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (12) = 24\).

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 3.00 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{5} \, b^{4} x^{5} + a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{3} + 2 \, a^{3} b x^{2} + a^{4} x \]

[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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 3.00 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=a^4\,x+2\,a^3\,b\,x^2+2\,a^2\,b^2\,x^3+a\,b^3\,x^4+\frac {b^4\,x^5}{5} \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

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