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

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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)]

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]

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 = 0.03, size = 83, normalized size = 1.20 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (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)\right )}{20 (a+b x)} \end {gather*}

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))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 69, normalized size = 1.00 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

giac [B]  time = 0.19, size = 144, normalized size = 2.09 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 90, normalized size = 1.30 \begin {gather*} \frac {\left (4 b^{3} B \,x^{4}+5 A \,b^{3} x^{3}+15 x^{3} B a \,b^{2}+20 A a \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+30 x A \,a^{2} b +10 B \,a^{3} x +20 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{20 \left (b x +a \right )^{3}} \end {gather*}

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

________________________________________________________________________________________

maxima [B]  time = 0.53, size = 125, normalized size = 1.81 \begin {gather*} \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a x}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{5 \, b^{2}} \end {gather*}

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]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*x - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a*x/b - 1/4*(b^2*x^2 + 2*a*b*x
+ a^2)^(3/2)*B*a^2/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a/b + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/b^
2

________________________________________________________________________________________

mupad [B]  time = 2.21, size = 42, normalized size = 0.61 \begin {gather*} \frac {\left (a+b\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}\,\left (5\,A\,b-B\,a+4\,B\,b\,x\right )}{20\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)*(5*A*b - B*a + 4*B*b*x))/(20*b^2)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________