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

Optimal. Leaf size=34 \[ \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b (1+2 p)} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {623} \begin {gather*} \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b (2 p+1)} \end {gather*}

Antiderivative was successfully verified.

[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*} \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)}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 23, normalized size = 0.68 \begin {gather*} \frac {(a+b x) \left ((a+b x)^2\right )^p}{b+2 b p} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.65, size = 26, normalized size = 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	\(\frac {\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{b \left (1+2 p \right )}\)	\(35\)
norman	\(\frac {x \,{\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{1+2 p}+\frac {a \,{\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{b \left (1+2 p \right )}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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Maxima [A]
time = 0.30, size = 25, normalized size = 0.74 \begin {gather*} \frac {{\left (b x + a\right )} {\left (b x + a\right )}^{2 \, p}}{b {\left (2 \, p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]
time = 2.96, size = 32, normalized size = 0.94 \begin {gather*} \frac {{\left (b x + a\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \, b p + b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \\\int \frac {1}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}\, dx & \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)), (Integral(1/sqrt(a**2 + 2*a*b*x + b
**2*x**2), x), 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))

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Giac [A]
time = 1.61, size = 51, normalized size = 1.50 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 0.66, size = 41, normalized size = 1.21 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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