3.1574 \(\int (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=32 \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b} \]

[Out]

((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(6*b)

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Rubi [A]  time = 0.0055318, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {609} \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(6*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 )^{5/2} \, dx &=\frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.0128349, size = 23, normalized size = 0.72 \[ \frac{(a+b x) \left ((a+b x)^2\right )^{5/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((a + b*x)*((a + b*x)^2)^(5/2))/(6*b)

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Maple [B]  time = 0.042, size = 71, normalized size = 2.2 \begin{align*}{\frac{x \left ({b}^{5}{x}^{5}+6\,a{b}^{4}{x}^{4}+15\,{a}^{2}{b}^{3}{x}^{3}+20\,{a}^{3}{b}^{2}{x}^{2}+15\,{a}^{4}bx+6\,{a}^{5} \right ) }{6\, \left ( bx+a \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/6*x*(b^5*x^5+6*a*b^4*x^4+15*a^2*b^3*x^3+20*a^3*b^2*x^2+15*a^4*b*x+6*a^5)*((b*x+a)^2)^(5/2)/(b*x+a)^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53539, size = 116, normalized size = 3.62 \begin{align*} \frac{1}{6} \, b^{5} x^{6} + a b^{4} x^{5} + \frac{5}{2} \, a^{2} b^{3} x^{4} + \frac{10}{3} \, a^{3} b^{2} x^{3} + \frac{5}{2} \, a^{4} b x^{2} + a^{5} 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)^(5/2),x, algorithm="fricas")

[Out]

1/6*b^5*x^6 + a*b^4*x^5 + 5/2*a^2*b^3*x^4 + 10/3*a^3*b^2*x^3 + 5/2*a^4*b*x^2 + a^5*x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((a**2 + 2*a*b*x + b**2*x**2)**(5/2), x)

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Giac [B]  time = 1.17674, size = 139, normalized size = 4.34 \begin{align*} \frac{1}{6} \, b^{5} x^{6} \mathrm{sgn}\left (b x + a\right ) + a b^{4} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a^{2} b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, a^{3} b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a^{4} b x^{2} \mathrm{sgn}\left (b x + a\right ) + a^{5} x \mathrm{sgn}\left (b x + a\right ) + \frac{a^{6} \mathrm{sgn}\left (b x + a\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/6*b^5*x^6*sgn(b*x + a) + a*b^4*x^5*sgn(b*x + a) + 5/2*a^2*b^3*x^4*sgn(b*x + a) + 10/3*a^3*b^2*x^3*sgn(b*x +
a) + 5/2*a^4*b*x^2*sgn(b*x + a) + a^5*x*sgn(b*x + a) + 1/6*a^6*sgn(b*x + a)/b