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

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

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Rubi [A]  time = 0.05, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {770, 76} \begin {gather*} \frac {x^3 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 (a+b x)}+\frac {a A x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b B x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a + b*x)) + ((A*b + a*B)*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a
+ b*x)) + (b*B*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(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 (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 \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+b (A b+a B) x^2+b^2 B x^3\right ) \, dx}{a b+b^2 x}\\ &=\frac {a A x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {(A b+a B) x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {b B x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 47, normalized size = 0.41 \begin {gather*} \frac {x^2 \sqrt {(a+b x)^2} (a (6 A+4 B x)+b x (4 A+3 B x))}{12 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[(a + b*x)^2]*(b*x*(4*A + 3*B*x) + a*(6*A + 4*B*x)))/(12*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.57, size = 0, normalized size = 0.00 \begin {gather*} \int x (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*(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

Defer[IntegrateAlgebraic][x*(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}{4} \, B b x^{4} + \frac {1}{2} \, A a x^{2} + \frac {1}{3} \, {\left (B a + A b\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 77, normalized size = 0.68 \begin {gather*} \frac {1}{4} \, B b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (B a^{4} - 2 \, A a^{3} b\right )} \mathrm {sgn}\left (b x + a\right )}{12 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 44, normalized size = 0.39 \begin {gather*} \frac {\left (3 B b \,x^{2}+4 A b x +4 B a x +6 A a \right ) \sqrt {\left (b x +a \right )^{2}}\, x^{2}}{12 b x +12 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^2*(3*B*b*x^2+4*A*b*x+4*B*a*x+6*A*a)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 0.51, size = 183, normalized size = 1.61 \begin {gather*} \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} x}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a x}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3}}{2 \, b^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2}}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a}{12 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.28, size = 176, normalized size = 1.54 \begin {gather*} \frac {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}}{24\,b^4}+\frac {B\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {5\,B\,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 (\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*((a + b*x)^2)^(1/2)*(A + B*x),x)

[Out]

(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))/(24*b^4) + (B*x*(a^2 + b^
2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) - (5*B*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*(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^{2}}{2} + \frac {B b x^{4}}{4} + x^{3} \left (\frac {A b}{3} + \frac {B a}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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