3.1975 \(\int (a+b x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=27 \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b} \]

[Out]

(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(5*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0076531, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {629} \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(5*b)

Rule 629

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

Rubi steps

\begin{align*} \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0069656, size = 25, normalized size = 0.93 \[ \frac{(a+b x)^4 \sqrt{(a+b x)^2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^4*Sqrt[(a + b*x)^2])/(5*b)

________________________________________________________________________________________

Maple [B]  time = 0.003, size = 60, normalized size = 2.2 \begin{align*}{\frac{x \left ({b}^{4}{x}^{4}+5\,a{b}^{3}{x}^{3}+10\,{a}^{2}{b}^{2}{x}^{2}+10\,{a}^{3}bx+5\,{a}^{4} \right ) }{5\, \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*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.956461, size = 31, normalized size = 1.15 \begin{align*} \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{2}}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)/b

________________________________________________________________________________________

Fricas [A]  time = 1.46884, size = 85, normalized size = 3.15 \begin{align*} \frac{1}{5} \, b^{4} x^{5} + a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{3} + 2 \, a^{3} b x^{2} + a^{4} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.884506, size = 158, normalized size = 5.85 \begin{align*} \begin{cases} \frac{a^{4} \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{5 b} + \frac{4 a^{3} x \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{5} + \frac{6 a^{2} b x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{5} + \frac{4 a b^{2} x^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{5} + \frac{b^{3} x^{4} \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{5} & \text{for}\: b \neq 0 \\a x \left (a^{2}\right )^{\frac{3}{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*sqrt(a**2 + 2*a*b*x + b**2*x**2)/(5*b) + 4*a**3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2)/5 + 6*a**2*
b*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2)/5 + 4*a*b**2*x**3*sqrt(a**2 + 2*a*b*x + b**2*x**2)/5 + b**3*x**4*sqrt(
a**2 + 2*a*b*x + b**2*x**2)/5, Ne(b, 0)), (a*x*(a**2)**(3/2), True))

________________________________________________________________________________________

Giac [B]  time = 1.19815, size = 116, normalized size = 4.3 \begin{align*} \frac{1}{5} \, b^{4} x^{5} \mathrm{sgn}\left (b x + a\right ) + a b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{2} b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{3} b x^{2} \mathrm{sgn}\left (b x + a\right ) + a^{4} x \mathrm{sgn}\left (b x + a\right ) + \frac{a^{5} \mathrm{sgn}\left (b x + a\right )}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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