3.153 \(\int (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=32 \[ \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {609} \[ \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Rule 609

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 )^{3/2} \, dx &=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.72 \[ \frac {(a+b x) \left ((a+b x)^2\right )^{3/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((a + b*x)*((a + b*x)^2)^(3/2))/(4*b)

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fricas [A]  time = 0.97, size = 31, normalized size = 0.97 \[ \frac {1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} + a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.23, size = 57, normalized size = 1.78 \[ \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} a^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{4} \mathrm {sgn}\left (b x + a\right )}{4 \, b} + \frac {1}{4} \, {\left (b x^{2} + 2 \, a x\right )}^{2} b \mathrm {sgn}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 49, normalized size = 1.53 \[ \frac {\left (b^{3} x^{3}+4 a \,b^{2} x^{2}+6 a^{2} b x +4 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{4 \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/4*x*(b^3*x^3+4*a*b^2*x^2+6*a^2*b*x+4*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [A]  time = 1.35, size = 46, normalized size = 1.44 \[ \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.16, size = 32, normalized size = 1.00 \[ \frac {\left (x\,b^2+a\,b\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((a**2 + 2*a*b*x + b**2*x**2)**(3/2), x)

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