3.1725 \(\int (A+B x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=69 \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A b-a B)}{4 b^2}+\frac{B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]

[Out]

((A*b - a*B)*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(4*b^2) + (B*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(5*b^2)

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Rubi [A]  time = 0.0242733, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {640, 609} \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A b-a B)}{4 b^2}+\frac{B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(4*b^2) + (B*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(5*b^2)

Rule 640

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

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 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac{B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}+\frac{\left (2 A b^2-2 a b B\right ) \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx}{2 b^2}\\ &=\frac{(A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2}+\frac{B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0348285, size = 83, normalized size = 1.2 \[ \frac{x \sqrt{(a+b x)^2} \left (10 a^2 b x (3 A+2 B x)+10 a^3 (2 A+B x)+5 a b^2 x^2 (4 A+3 B x)+b^3 x^3 (5 A+4 B x)\right )}{20 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(10*a^3*(2*A + B*x) + 10*a^2*b*x*(3*A + 2*B*x) + 5*a*b^2*x^2*(4*A + 3*B*x) + b^3*x^3*(5*A
+ 4*B*x)))/(20*(a + b*x))

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Maple [A]  time = 0.004, size = 90, normalized size = 1.3 \begin{align*}{\frac{x \left ( 4\,B{x}^{4}{b}^{3}+5\,A{b}^{3}{x}^{3}+15\,B{x}^{3}a{b}^{2}+20\,A{x}^{2}a{b}^{2}+20\,B{x}^{2}{a}^{2}b+30\,A{a}^{2}bx+10\,{a}^{3}Bx+20\,A{a}^{3} \right ) }{20\, \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]

1/20*x*(4*B*b^3*x^4+5*A*b^3*x^3+15*B*a*b^2*x^3+20*A*a*b^2*x^2+20*B*a^2*b*x^2+30*A*a^2*b*x+10*B*a^3*x+20*A*a^3)
*((b*x+a)^2)^(3/2)/(b*x+a)^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53427, size = 150, normalized size = 2.17 \begin{align*} \frac{1}{5} \, B b^{3} x^{5} + A a^{3} x + \frac{1}{4} \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} +{\left (B a^{2} b + A a b^{2}\right )} x^{3} + \frac{1}{2} \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2} \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*b^3*x^5 + A*a^3*x + 1/4*(3*B*a*b^2 + A*b^3)*x^4 + (B*a^2*b + A*a*b^2)*x^3 + 1/2*(B*a^3 + 3*A*a^2*b)*x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}\, dx \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]

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

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Giac [B]  time = 1.09187, size = 194, normalized size = 2.81 \begin{align*} \frac{1}{5} \, B b^{3} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, B a b^{2} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, A b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + B a^{2} b x^{3} \mathrm{sgn}\left (b x + a\right ) + A a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, A a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) + A a^{3} x \mathrm{sgn}\left (b x + a\right ) - \frac{{\left (B a^{5} - 5 \, A a^{4} b\right )} \mathrm{sgn}\left (b x + a\right )}{20 \, b^{2}} \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*b^3*x^5*sgn(b*x + a) + 3/4*B*a*b^2*x^4*sgn(b*x + a) + 1/4*A*b^3*x^4*sgn(b*x + a) + B*a^2*b*x^3*sgn(b*x +
a) + A*a*b^2*x^3*sgn(b*x + a) + 1/2*B*a^3*x^2*sgn(b*x + a) + 3/2*A*a^2*b*x^2*sgn(b*x + a) + A*a^3*x*sgn(b*x +
a) - 1/20*(B*a^5 - 5*A*a^4*b)*sgn(b*x + a)/b^2