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

Optimal. Leaf size=114 \[ \frac {x^5 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{5 (a+b x)}+\frac {a A x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b B x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)} \]

________________________________________________________________________________________

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^5 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{5 (a+b x)}+\frac {a A x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b B x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x)) + ((A*b + a*B)*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a
+ b*x)) + (b*B*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*(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^3 (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^3 \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^3+b (A b+a B) x^4+b^2 B x^5\right ) \, dx}{a b+b^2 x}\\ &=\frac {a A x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {(A b+a B) x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {b B x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 49, normalized size = 0.43 \begin {gather*} \frac {x^4 \sqrt {(a+b x)^2} (3 a (5 A+4 B x)+2 b x (6 A+5 B x))}{60 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[(a + b*x)^2]*(3*a*(5*A + 4*B*x) + 2*b*x*(6*A + 5*B*x)))/(60*(a + b*x))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.75, size = 0, normalized size = 0.00 \begin {gather*} \int x^3 (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^3*(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 27, normalized size = 0.24 \begin {gather*} \frac {1}{6} \, B b x^{6} + \frac {1}{4} \, A a x^{4} + \frac {1}{5} \, {\left (B a + A b\right )} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*B*b*x^6 + 1/4*A*a*x^4 + 1/5*(B*a + A*b)*x^5

________________________________________________________________________________________

giac [A]  time = 0.15, size = 78, normalized size = 0.68 \begin {gather*} \frac {1}{6} \, B b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B a x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A a x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (2 \, B a^{6} - 3 \, A a^{5} b\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*B*b*x^6*sgn(b*x + a) + 1/5*B*a*x^5*sgn(b*x + a) + 1/5*A*b*x^5*sgn(b*x + a) + 1/4*A*a*x^4*sgn(b*x + a) + 1/
60*(2*B*a^6 - 3*A*a^5*b)*sgn(b*x + a)/b^5

________________________________________________________________________________________

maple [A]  time = 0.04, size = 44, normalized size = 0.39 \begin {gather*} \frac {\left (10 B b \,x^{2}+12 A b x +12 B a x +15 A a \right ) \sqrt {\left (b x +a \right )^{2}}\, x^{4}}{60 b x +60 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*x^4*(10*B*b*x^2+12*A*b*x+12*B*a*x+15*A*a)*((b*x+a)^2)^(1/2)/(b*x+a)

________________________________________________________________________________________

maxima [B]  time = 0.59, size = 301, normalized size = 2.64 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B x^{3}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4} x}{2 \, b^{4}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3} x}{2 \, b^{3}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a x^{2}}{10 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A x^{2}}{5 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{5}}{2 \, b^{5}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{4}}{2 \, b^{4}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} x}{5 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a x}{20 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3}}{15 \, b^{5}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{2}}{20 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*x^3/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^4*x/b^4 - 1/2*sqrt(b^2*x
^2 + 2*a*b*x + a^2)*A*a^3*x/b^3 - 3/10*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a*x^2/b^3 + 1/5*(b^2*x^2 + 2*a*b*x +
a^2)^(3/2)*A*x^2/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^5/b^5 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^4/b
^4 + 2/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^2*x/b^4 - 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a*x/b^3 - 7/15*(
b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^3/b^5 + 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^2/b^4

________________________________________________________________________________________

mupad [B]  time = 1.30, size = 340, normalized size = 2.98 \begin {gather*} \frac {A\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}+\frac {B\,x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{6\,b^2}-\frac {7\,A\,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 {B\,a^2\,\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 )}{24\,b^5}-\frac {A\,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 {3\,B\,a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (4\,b^2\,x^2\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-a^4+9\,a^2\,b^2\,x^2+8\,a^3\,b\,x-7\,a\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )\right )}{40\,b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b^2) + (B*x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(6*b^2) - (7*A*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) - (B*
a^2*(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))/(24*b^5
) - (A*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) - (3*B*a
*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(4*b^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x) - a^4 + 9*a^2*b^2*x^2 + 8*a^3*b*x - 7*a*
b*x*(a^2 + b^2*x^2 + 2*a*b*x)))/(40*b^5)

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 29, normalized size = 0.25 \begin {gather*} \frac {A a x^{4}}{4} + \frac {B b x^{6}}{6} + x^{5} \left (\frac {A b}{5} + \frac {B a}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)*((b*x+a)**2)**(1/2),x)

[Out]

A*a*x**4/4 + B*b*x**6/6 + x**5*(A*b/5 + B*a/5)

________________________________________________________________________________________