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

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

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Rubi [A]  time = 0.06, antiderivative size = 114, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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])

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]

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*}

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.43 \begin {gather*} \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)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.68, size = 0, normalized size = 0.00 \begin {gather*} \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 27, normalized size = 0.24 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.22, size = 78, normalized size = 0.68 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.05, size = 44, normalized size = 0.39 \begin {gather*} \frac {\left (12 B b \,x^{2}+15 A b x +15 B a x +20 A a \right ) \sqrt {\left (b x +a \right )^{2}}\, x^{3}}{60 b x +60 a} \end {gather*}

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 [B]  time = 0.57, size = 241, normalized size = 2.11 \begin {gather*} -\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} x}{2 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2} x}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B x^{2}}{5 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4}}{2 \, b^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3}}{2 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a x}{20 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A x}{4 \, b^{2}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2}}{20 \, b^{4}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a}{12 \, b^{3}} \end {gather*}

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]

-1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^3*x/b^3 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^2*x/b^2 + 1/5*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*B*x^2/b^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^4/b^4 + 1/2*sqrt(b^2*x^2 + 2*a*b*x +
a^2)*A*a^3/b^3 - 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a*x/b^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*x/b^2
+ 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^2/b^4 - 5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a/b^3

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mupad [B]  time = 1.30, size = 271, normalized size = 2.38 \begin {gather*} \frac {B\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}+\frac {A\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {7\,B\,a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{60\,b^4}-\frac {5\,A\,a\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^5}-\frac {B\,a^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{60\,b^6}-\frac {A\,a^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b^2) + (A*x*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) - (7*B*a*(a^2
+ b^2*x^2 + 2*a*b*x)^(1/2)*(a^3 - 5*a*b^2*x^2 + 3*b*x*(a^2 + b^2*x^2 + 2*a*b*x) - 4*a^2*b*x))/(60*b^4) - (5*A*
a*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(96*b^5) - (B*a^2*(8*b^2*(
a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(60*b^6) - (A*a^2*(x/2 + a/(2*b))*(a
^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b^2)

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sympy [A]  time = 0.10, size = 29, normalized size = 0.25 \begin {gather*} \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 {gather*}

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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