3.655 \(\int x^2 (A+B x) \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=114 \[ \frac{x^4 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{4 (a+b x)}+\frac{a A x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{b B x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0571252, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {770, 76} \[ \frac{x^4 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{4 (a+b x)}+\frac{a A x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{b B x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

Mathematica [A]  time = 0.0166503, size = 49, normalized size = 0.43 \[ \frac{x^3 \sqrt{(a+b x)^2} (5 a (4 A+3 B x)+3 b x (5 A+4 B x))}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x)^2]*(5*a*(4*A + 3*B*x) + 3*b*x*(5*A + 4*B*x)))/(60*(a + b*x))

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Maple [A]  time = 0.003, size = 44, normalized size = 0.4 \begin{align*}{\frac{{x}^{3} \left ( 12\,Bb{x}^{2}+15\,Abx+15\,aBx+20\,aA \right ) }{60\,bx+60\,a}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x+A)*((b*x+a)^2)^(1/2),x)

[Out]

1/60*x^3*(12*B*b*x^2+15*A*b*x+15*B*a*x+20*A*a)*((b*x+a)^2)^(1/2)/(b*x+a)

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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(x^2*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27198, size = 66, normalized size = 0.58 \begin{align*} \frac{1}{5} \, B b x^{5} + \frac{1}{3} \, A a x^{3} + \frac{1}{4} \,{\left (B a + A b\right )} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.130704, size = 29, normalized size = 0.25 \begin{align*} \frac{A a x^{3}}{3} + \frac{B b x^{5}}{5} + x^{4} \left (\frac{A b}{4} + \frac{B a}{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x+A)*((b*x+a)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.15922, size = 105, normalized size = 0.92 \begin{align*} \frac{1}{5} \, B b x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, B a x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, A b x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, A a x^{3} \mathrm{sgn}\left (b x + a\right ) - \frac{{\left (3 \, B a^{5} - 5 \, A a^{4} b\right )} \mathrm{sgn}\left (b x + a\right )}{60 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/5*B*b*x^5*sgn(b*x + a) + 1/4*B*a*x^4*sgn(b*x + a) + 1/4*A*b*x^4*sgn(b*x + a) + 1/3*A*a*x^3*sgn(b*x + a) - 1/
60*(3*B*a^5 - 5*A*a^4*b)*sgn(b*x + a)/b^4