\(\int (a^2+2 a b x+b^2 x^2)^p \, dx\) [1747]

Optimal result

Integrand size = 18, antiderivative size = 34 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b (1+2 p)} \]

[Out]

(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^p/b/(1+2*p)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {623} \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b (2 p+1)} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^p,x]

[Out]

((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^p)/(b*(1 + 2*p))

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b (1+2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(a+b x) \left ((a+b x)^2\right )^p}{b (1+2 p)} \]

[In]

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

[Out]

((a + b*x)*((a + b*x)^2)^p)/(b*(1 + 2*p))

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {\left (b x +a \right ) \left (\left (b x +a \right )^{2}\right )^{p}}{b \left (1+2 p \right )}\)	\(26\)
gosper	\(\frac {\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{b \left (1+2 p \right )}\)	\(35\)
parallelrisch	\(\frac {x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a b +\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2}}{\left (1+2 p \right ) a b}\)	\(60\)
norman	\(\frac {x \,{\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{1+2 p}+\frac {a \,{\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{b \left (1+2 p \right )}\)	\(63\)

[In]

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

[Out]

(b*x+a)/b/(1+2*p)*((b*x+a)^2)^p

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \, b p + b} \]

[In]

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

[Out]

(b*x + a)*(b^2*x^2 + 2*a*b*x + a^2)^p/(2*b*p + b)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (29) = 58\).

Time = 0.48 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.94 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\begin {cases} \frac {x}{\sqrt {a^{2}}} & \text {for}\: b = 0 \wedge p = - \frac {1}{2} \\x \left (a^{2}\right )^{p} & \text {for}\: b = 0 \\\frac {\left (\frac {a}{b} + x\right ) \log {\left (\frac {a}{b} + x \right )}}{\sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} & \text {for}\: p = - \frac {1}{2} \\\frac {a \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{2 b p + b} + \frac {b x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{2 b p + b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x/sqrt(a**2), Eq(b, 0) & Eq(p, -1/2)), (x*(a**2)**p, Eq(b, 0)), ((a/b + x)*log(a/b + x)/sqrt(b**2*(
a/b + x)**2), Eq(p, -1/2)), (a*(a**2 + 2*a*b*x + b**2*x**2)**p/(2*b*p + b) + b*x*(a**2 + 2*a*b*x + b**2*x**2)*
*p/(2*b*p + b), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b x + a\right )}^{2 \, p}}{b {\left (2 \, p + 1\right )}} \]

[In]

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

[Out]

(b*x + a)*(b*x + a)^(2*p)/(b*(2*p + 1))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.50 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a}{2 \, b p + b} \]

[In]

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

[Out]

((b^2*x^2 + 2*a*b*x + a^2)^p*b*x + (b^2*x^2 + 2*a*b*x + a^2)^p*a)/(2*b*p + b)

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\left (\frac {x}{2\,p+1}+\frac {a}{b\,\left (2\,p+1\right )}\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^p,x)

[Out]

(x/(2*p + 1) + a/(b*(2*p + 1)))*(a^2 + b^2*x^2 + 2*a*b*x)^p