Integrand size = 29, antiderivative size = 114 \[ \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {a A x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {(A b+a B) x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b B x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]
1/3*a*A*x^3*((b*x+a)^2)^(1/2)/(b*x+a)+1/4*(A*b+B*a)*x^4*((b*x+a)^2)^(1/2)/ (b*x+a)+1/5*b*B*x^5*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.65 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.85 \[ \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x^3 (5 a (4 A+3 B x)+3 b x (5 A+4 B x)) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{60 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]
(x^3*(5*a*(4*A + 3*B*x) + 3*b*x*(5*A + 4*B*x))*(Sqrt[a^2]*b*x + a*(Sqrt[a^ 2] - Sqrt[(a + b*x)^2])))/(60*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]) )
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 27, 85, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b x^2 (a+b x) (A+B x)dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^2 (a+b x) (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b B x^4+(A b+a B) x^3+a A x^2\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{4} x^4 (a B+A b)+\frac {1}{3} a A x^3+\frac {1}{5} b B x^5\right )}{a+b x}\) |
3.7.55.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > 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]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(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)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.39
method | result | size |
gosper | \(\frac {x^{3} \left (12 B b \,x^{2}+15 A b x +15 a B x +20 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{60 b x +60 a}\) | \(44\) |
risch | \(\frac {a A \,x^{3} \sqrt {\left (b x +a \right )^{2}}}{3 b x +3 a}+\frac {\left (A b +B a \right ) x^{4} \sqrt {\left (b x +a \right )^{2}}}{4 b x +4 a}+\frac {b B \,x^{5} \sqrt {\left (b x +a \right )^{2}}}{5 b x +5 a}\) | \(76\) |
default | \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (12 x^{3} B \,b^{3}+15 A \,b^{3} x^{2}-9 B \,x^{2} a \,b^{2}-10 A x a \,b^{2}+6 B x \,a^{2} b +5 A \,a^{2} b -3 B \,a^{3}\right )}{60 b^{4}}\) | \(77\) |
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.24 \[ \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{5} \, B b x^{5} + \frac {1}{3} \, A a x^{3} + \frac {1}{4} \, {\left (B a + A b\right )} x^{4} \]
Time = 1.70 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.56 \[ \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=A \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} x}{12 b^{2}} + \frac {a x^{2}}{12 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{4}}{20 b^{4}} + \frac {a^{3} x}{20 b^{3}} - \frac {a^{2} x^{2}}{20 b^{2}} + \frac {a x^{3}}{20 b} + \frac {x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{6} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {3 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {3 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4} \sqrt {a^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]
A*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(a**3/(12*b**3) - a**2*x/(12 *b**2) + a*x**2/(12*b) + x**3/4), Ne(b**2, 0)), ((a**4*(a**2 + 2*a*b*x)**( 3/2)/3 - 2*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/(4* a**3*b**3), Ne(a*b, 0)), (x**3*sqrt(a**2)/3, True)) + B*Piecewise((sqrt(a* *2 + 2*a*b*x + b**2*x**2)*(-a**4/(20*b**4) + a**3*x/(20*b**3) - a**2*x**2/ (20*b**2) + a*x**3/(20*b) + x**4/5), Ne(b**2, 0)), ((-a**6*(a**2 + 2*a*b*x )**(3/2)/3 + 3*a**4*(a**2 + 2*a*b*x)**(5/2)/5 - 3*a**2*(a**2 + 2*a*b*x)**( 7/2)/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(8*a**4*b**4), Ne(a*b, 0)), (x**4*sqrt (a**2)/4, True))
Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (75) = 150\).
Time = 0.34 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.11 \[ \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} x}{2 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2} x}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B x^{2}}{5 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4}}{2 \, b^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3}}{2 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a x}{20 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A x}{4 \, b^{2}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2}}{20 \, b^{4}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a}{12 \, b^{3}} \]
-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
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{5} \, B b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, B a x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A a x^{3} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (3 \, B a^{5} - 5 \, A a^{4} b\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{4}} \]
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
Time = 10.05 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.38 \[ \int x^2 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {B\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}+\frac {A\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {7\,B\,a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{60\,b^4}-\frac {5\,A\,a\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^5}-\frac {B\,a^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{60\,b^6}-\frac {A\,a^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2} \]
(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)