Integrand size = 27, antiderivative size = 114 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {a A x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {(A b+a B) x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {b B x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \] Output:
a*A*x^2*((b*x+a)^2)^(1/2)/(2*b*x+2*a)+(A*b+B*a)*x^3*((b*x+a)^2)^(1/2)/(3*b *x+3*a)+b*B*x^4*((b*x+a)^2)^(1/2)/(4*b*x+4*a)
Time = 0.85 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x^2 (b x (4 A+3 B x)+a (6 A+4 B x)) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{12 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \] Input:
Integrate[x*(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(x^2*(b*x*(4*A + 3*B*x) + a*(6*A + 4*B*x))*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(12*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]))
Time = 0.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.
\(\displaystyle \int x \sqrt {a^2+2 a b x+b^2 x^2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b x (a+b x) (A+B x)dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x (a+b x) (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b B x^3+(A b+a B) x^2+a A x\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{3} x^3 (a B+A b)+\frac {1}{2} a A x^2+\frac {1}{4} b B x^4\right )}{a+b x}\) |
Input:
Int[x*(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((a*A*x^2)/2 + ((A*b + a*B)*x^3)/3 + (b*B*x ^4)/4))/(a + b*x)
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 = 0.54 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.39
method | result | size |
gosper | \(\frac {x^{2} \left (3 B b \,x^{2}+4 A b x +4 B a x +6 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{12 b x +12 a}\) | \(44\) |
orering | \(\frac {x^{2} \left (3 B b \,x^{2}+4 A b x +4 B a x +6 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{12 b x +12 a}\) | \(44\) |
default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (-3 x^{2} B \,b^{2}-4 x \,b^{2} A +2 x a b B +2 a b A -a^{2} B \right )}{12 b^{3}}\) | \(53\) |
risch | \(\frac {a A \,x^{2} \sqrt {\left (b x +a \right )^{2}}}{2 b x +2 a}+\frac {\left (A b +B a \right ) x^{3} \sqrt {\left (b x +a \right )^{2}}}{3 b x +3 a}+\frac {b B \,x^{4} \sqrt {\left (b x +a \right )^{2}}}{4 b x +4 a}\) | \(76\) |
Input:
int(x*(B*x+A)*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
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)
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.24 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, B b x^{4} + \frac {1}{2} \, A a x^{2} + \frac {1}{3} \, {\left (B a + A b\right )} x^{3} \] Input:
integrate(x*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="fricas")
Output:
1/4*B*b*x^4 + 1/2*A*a*x^2 + 1/3*(B*a + A*b)*x^3
Time = 1.42 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.00 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=A \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{2}}{6 b^{2}} + \frac {a x}{6 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \sqrt {a^{2}}}{2} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} x}{12 b^{2}} + \frac {a x^{2}}{12 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases}\right ) \] Input:
integrate(x*(B*x+A)*((b*x+a)**2)**(1/2),x)
Output:
A*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**2/(6*b**2) + a*x/(6*b) + x**2/3), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a**2 + 2*a*b *x)**(5/2)/5)/(2*a**2*b**2), Ne(a*b, 0)), (x**2*sqrt(a**2)/2, True)) + B*P iecewise((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))
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (75) = 150\).
Time = 0.03 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.61 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} x}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a x}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3}}{2 \, b^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2}}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a}{12 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{3 \, b^{2}} \] Input:
integrate(x*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="maxima")
Output:
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
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, B b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (B a^{4} - 2 \, A a^{3} b\right )} \mathrm {sgn}\left (b x + a\right )}{12 \, b^{3}} \] Input:
integrate(x*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="giac")
Output:
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
Time = 10.59 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.54 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {A\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^4}+\frac {B\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {5\,B\,a\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^5}-\frac {B\,a^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2} \] Input:
int(x*((a + b*x)^2)^(1/2)*(A + B*x),x)
Output:
(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)
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.21 \[ \int x (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x^{2} \left (3 b^{2} x^{2}+8 a b x +6 a^{2}\right )}{12} \] Input:
int(x*(B*x+A)*((b*x+a)^2)^(1/2),x)
Output:
(x**2*(6*a**2 + 8*a*b*x + 3*b**2*x**2))/12