### 3.1559 $$\int (a^2+2 a b x+b^2 x^2)^{3/2} \, dx$$

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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0049789, 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} $\frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b}$

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*}

Mathematica [A]  time = 0.0088064, 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)

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 49, normalized size = 1.5 \begin{align*}{\frac{x \left ({b}^{3}{x}^{3}+4\,a{b}^{2}{x}^{2}+6\,{a}^{2}bx+4\,{a}^{3} \right ) }{4\, \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{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)^(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

________________________________________________________________________________________

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.42429, size = 66, normalized size = 2.06 \begin{align*} \frac{1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac{3}{2} \, a^{2} b x^{2} + a^{3} 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)^(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

________________________________________________________________________________________

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{3}{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)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.11532, size = 93, normalized size = 2.91 \begin{align*} \frac{1}{4} \, b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) + a^{3} x \mathrm{sgn}\left (b x + a\right ) + \frac{a^{4} \mathrm{sgn}\left (b x + a\right )}{4 \, 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)^(3/2),x, algorithm="giac")

[Out]

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