Integrand size = 27, antiderivative size = 121 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {a (A b-a B) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^3}+\frac {(A b-2 a B) (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^3}+\frac {B (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3} \] Output:
-1/4*a*(A*b-B*a)*(b*x+a)^3*((b*x+a)^2)^(1/2)/b^3+1/5*(A*b-2*B*a)*(b*x+a)^4 *((b*x+a)^2)^(1/2)/b^3+1/6*B*(b*x+a)^5*((b*x+a)^2)^(1/2)/b^3
Time = 1.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^2 \sqrt {(a+b x)^2} \left (10 a^3 (3 A+2 B x)+15 a^2 b x (4 A+3 B x)+9 a b^2 x^2 (5 A+4 B x)+2 b^3 x^3 (6 A+5 B x)\right )}{60 (a+b x)} \] Input:
Integrate[x*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
(x^2*Sqrt[(a + b*x)^2]*(10*a^3*(3*A + 2*B*x) + 15*a^2*b*x*(4*A + 3*B*x) + 9*a*b^2*x^2*(5*A + 4*B*x) + 2*b^3*x^3*(6*A + 5*B*x)))/(60*(a + b*x))
Time = 0.42 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74, 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.
\(\displaystyle \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 x (a+b x)^3 (A+B x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x (a+b x)^3 (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B (a+b x)^5}{b^2}+\frac {(A b-2 a B) (a+b x)^4}{b^2}+\frac {a (a B-A b) (a+b x)^3}{b^2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {(a+b x)^5 (A b-2 a B)}{5 b^3}-\frac {a (a+b x)^4 (A b-a B)}{4 b^3}+\frac {B (a+b x)^6}{6 b^3}\right )}{a+b x}\) |
Input:
Int[x*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/4*(a*(A*b - a*B)*(a + b*x)^4)/b^3 + ((A *b - 2*a*B)*(a + b*x)^5)/(5*b^3) + (B*(a + b*x)^6)/(6*b^3)))/(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 = 0.98 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76
method | result | size |
gosper | \(\frac {x^{2} \left (10 B \,b^{3} x^{4}+12 A \,b^{3} x^{3}+36 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+60 A \,a^{2} b x +20 B \,a^{3} x +30 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (b x +a \right )^{3}}\) | \(92\) |
default | \(\frac {x^{2} \left (10 B \,b^{3} x^{4}+12 A \,b^{3} x^{3}+36 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+60 A \,a^{2} b x +20 B \,a^{3} x +30 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (b x +a \right )^{3}}\) | \(92\) |
orering | \(\frac {x^{2} \left (10 B \,b^{3} x^{4}+12 A \,b^{3} x^{3}+36 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+60 A \,a^{2} b x +20 B \,a^{3} x +30 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{60 \left (b x +a \right )^{3}}\) | \(101\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, B \,b^{3} x^{6}}{6 b x +6 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{3}+3 B a \,b^{2}\right ) x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A a \,b^{2}+3 B \,a^{2} b \right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} A \,x^{2}}{2 b x +2 a}\) | \(156\) |
Input:
int(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/60*x^2*(10*B*b^3*x^4+12*A*b^3*x^3+36*B*a*b^2*x^3+45*A*a*b^2*x^2+45*B*a^2 *b*x^2+60*A*a^2*b*x+20*B*a^3*x+30*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{6} \, B b^{3} x^{6} + \frac {1}{2} \, A a^{3} x^{2} + \frac {1}{5} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{5} + \frac {3}{4} \, {\left (B a^{2} b + A a b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3} \] Input:
integrate(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
Output:
1/6*B*b^3*x^6 + 1/2*A*a^3*x^2 + 1/5*(3*B*a*b^2 + A*b^3)*x^5 + 3/4*(B*a^2*b + A*a*b^2)*x^4 + 1/3*(B*a^3 + 3*A*a^2*b)*x^3
Leaf count of result is larger than twice the leaf count of optimal. 1258 vs. \(2 (85) = 170\).
Time = 1.11 (sec) , antiderivative size = 1258, normalized size of antiderivative = 10.40 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\text {Too large to display} \] Input:
integrate(x*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
Output:
Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*b**2*x**5/6 + x**4*(A*b**4 + 13*B*a*b**3/6)/(5*b**2) + x**3*(4*A*a*b**3 + 31*B*a**2*b**2/6 - 9*a*(A*b **4 + 13*B*a*b**3/6)/(5*b))/(4*b**2) + x**2*(6*A*a**2*b**2 + 4*B*a**3*b - 4*a**2*(A*b**4 + 13*B*a*b**3/6)/(5*b**2) - 7*a*(4*A*a*b**3 + 31*B*a**2*b** 2/6 - 9*a*(A*b**4 + 13*B*a*b**3/6)/(5*b))/(4*b))/(3*b**2) + x*(4*A*a**3*b + B*a**4 - 3*a**2*(4*A*a*b**3 + 31*B*a**2*b**2/6 - 9*a*(A*b**4 + 13*B*a*b* *3/6)/(5*b))/(4*b**2) - 5*a*(6*A*a**2*b**2 + 4*B*a**3*b - 4*a**2*(A*b**4 + 13*B*a*b**3/6)/(5*b**2) - 7*a*(4*A*a*b**3 + 31*B*a**2*b**2/6 - 9*a*(A*b** 4 + 13*B*a*b**3/6)/(5*b))/(4*b))/(3*b))/(2*b**2) + (A*a**4 - 2*a**2*(6*A*a **2*b**2 + 4*B*a**3*b - 4*a**2*(A*b**4 + 13*B*a*b**3/6)/(5*b**2) - 7*a*(4* A*a*b**3 + 31*B*a**2*b**2/6 - 9*a*(A*b**4 + 13*B*a*b**3/6)/(5*b))/(4*b))/( 3*b**2) - 3*a*(4*A*a**3*b + B*a**4 - 3*a**2*(4*A*a*b**3 + 31*B*a**2*b**2/6 - 9*a*(A*b**4 + 13*B*a*b**3/6)/(5*b))/(4*b**2) - 5*a*(6*A*a**2*b**2 + 4*B *a**3*b - 4*a**2*(A*b**4 + 13*B*a*b**3/6)/(5*b**2) - 7*a*(4*A*a*b**3 + 31* B*a**2*b**2/6 - 9*a*(A*b**4 + 13*B*a*b**3/6)/(5*b))/(4*b))/(3*b))/(2*b))/b **2) + (a/b + x)*(-a**2*(4*A*a**3*b + B*a**4 - 3*a**2*(4*A*a*b**3 + 31*B*a **2*b**2/6 - 9*a*(A*b**4 + 13*B*a*b**3/6)/(5*b))/(4*b**2) - 5*a*(6*A*a**2* b**2 + 4*B*a**3*b - 4*a**2*(A*b**4 + 13*B*a*b**3/6)/(5*b**2) - 7*a*(4*A*a* b**3 + 31*B*a**2*b**2/6 - 9*a*(A*b**4 + 13*B*a*b**3/6)/(5*b))/(4*b))/(3*b) )/(2*b**2) - a*(A*a**4 - 2*a**2*(6*A*a**2*b**2 + 4*B*a**3*b - 4*a**2*(A...
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (83) = 166\).
Time = 0.04 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.51 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} x}{4 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3}}{4 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x}{6 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a}{30 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{5 \, b^{2}} \] Input:
integrate(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
Output:
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^2*x/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a*x/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^3/b^3 - 1/4* (b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^2/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^ (5/2)*B*x/b^2 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a/b^3 + 1/5*(b^2*x^ 2 + 2*a*b*x + a^2)^(5/2)*A/b^2
Time = 0.21 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.22 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{6} \, B b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, B a b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a^{2} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, A a b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + A a^{2} b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (B a^{6} - 3 \, A a^{5} b\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{3}} \] Input:
integrate(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
Output:
1/6*B*b^3*x^6*sgn(b*x + a) + 3/5*B*a*b^2*x^5*sgn(b*x + a) + 1/5*A*b^3*x^5* sgn(b*x + a) + 3/4*B*a^2*b*x^4*sgn(b*x + a) + 3/4*A*a*b^2*x^4*sgn(b*x + a) + 1/3*B*a^3*x^3*sgn(b*x + a) + A*a^2*b*x^3*sgn(b*x + a) + 1/2*A*a^3*x^2*s gn(b*x + a) + 1/60*(B*a^6 - 3*A*a^5*b)*sgn(b*x + a)/b^3
Timed out. \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \] Input:
int(x*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
Output:
int(x*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.38 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^{2} \left (5 b^{4} x^{4}+24 a \,b^{3} x^{3}+45 a^{2} b^{2} x^{2}+40 a^{3} b x +15 a^{4}\right )}{30} \] Input:
int(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
Output:
(x**2*(15*a**4 + 40*a**3*b*x + 45*a**2*b**2*x**2 + 24*a*b**3*x**3 + 5*b**4 *x**4))/30