Integrand size = 29, antiderivative size = 75 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(A b-a B) x^4 (a+b x)}{4 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}+\frac {A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \] Output:
-1/4*(A*b-B*a)*x^4*(b*x+a)/a^2/(b^2*x^2+2*a*b*x+a^2)^(5/2)+1/3*A*x^3/a^2/( b^2*x^2+2*a*b*x+a^2)^(3/2)
Time = 1.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-3 a^3 B-6 b^3 x^2 (A+2 B x)-2 a b^2 x (2 A+9 B x)-a^2 b (A+12 B x)}{12 b^4 (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:
Integrate[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
(-3*a^3*B - 6*b^3*x^2*(A + 2*B*x) - 2*a*b^2*x*(2*A + 9*B*x) - a^2*b*(A + 1 2*B*x))/(12*b^4*(a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.39 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1186, 1102, 27, 48}
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.
\(\displaystyle \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1186 |
\(\displaystyle \frac {A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {(A b-a B) \int \frac {x^3}{\left (a^2+2 b x a+b^2 x^2\right )^{5/2}}dx}{a}\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {b^5 (a+b x) (A b-a B) \int \frac {x^3}{b^5 (a+b x)^5}dx}{a \sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {(a+b x) (A b-a B) \int \frac {x^3}{(a+b x)^5}dx}{a \sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {x^4 (A b-a B)}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
(A*x^3)/(3*a^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - ((A*b - a*B)*x^4)/(4*a^2 *(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e)^2)), x] + Simp[(2*c*f - b*g)/ (2*c*d - b*e) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[ {a, b, c, d, e, f, g, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]
Time = 1.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,x^{3}}{b}-\frac {\left (A b +3 B a \right ) x^{2}}{2 b^{2}}-\frac {a \left (A b +3 B a \right ) x}{3 b^{3}}-\frac {a^{2} \left (A b +3 B a \right )}{12 b^{4}}\right )}{\left (b x +a \right )^{5}}\) | \(75\) |
gosper | \(-\frac {\left (b x +a \right ) \left (12 x^{3} B \,b^{3}+6 A \,b^{3} x^{2}+18 B a \,b^{2} x^{2}+4 A a \,b^{2} x +12 B \,a^{2} b x +A \,a^{2} b +3 B \,a^{3}\right )}{12 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(77\) |
default | \(-\frac {\left (b x +a \right ) \left (12 x^{3} B \,b^{3}+6 A \,b^{3} x^{2}+18 B a \,b^{2} x^{2}+4 A a \,b^{2} x +12 B \,a^{2} b x +A \,a^{2} b +3 B \,a^{3}\right )}{12 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(77\) |
orering | \(-\frac {\left (12 x^{3} B \,b^{3}+6 A \,b^{3} x^{2}+18 B a \,b^{2} x^{2}+4 A a \,b^{2} x +12 B \,a^{2} b x +A \,a^{2} b +3 B \,a^{3}\right ) \left (b x +a \right )}{12 b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}\) | \(86\) |
Input:
int(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)^5*(-B*x^3/b-1/2*(A*b+3*B*a)/b^2*x^2-1/3*a*(A*b+3 *B*a)/b^3*x-1/12*a^2*(A*b+3*B*a)/b^4)
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.40 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {12 \, B b^{3} x^{3} + 3 \, B a^{3} + A a^{2} b + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 4 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x}{12 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
Output:
-1/12*(12*B*b^3*x^3 + 3*B*a^3 + A*a^2*b + 6*(3*B*a*b^2 + A*b^3)*x^2 + 4*(3 *B*a^2*b + A*a*b^2)*x)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5* x + a^4*b^4)
\[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{2} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**2*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
Output:
Integral(x**2*(A + B*x)/((a + b*x)**2)**(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (67) = 134\).
Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.08 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {B x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, B a^{2}}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} - \frac {B a}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, B a^{2}}{3 \, b^{7} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {2 \, A a}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {B a^{3}}{4 \, b^{8} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {A a^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
Output:
-B*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 2/3*B*a^2/((b^2*x^2 + 2*a*b *x + a^2)^(3/2)*b^4) - 1/2*B*a/(b^6*(x + a/b)^2) - 1/2*A/(b^5*(x + a/b)^2) + 2/3*B*a^2/(b^7*(x + a/b)^3) + 2/3*A*a/(b^6*(x + a/b)^3) + 1/4*B*a^3/(b^ 8*(x + a/b)^4) - 1/4*A*a^2/(b^7*(x + a/b)^4)
Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {12 \, B b^{3} x^{3} + 18 \, B a b^{2} x^{2} + 6 \, A b^{3} x^{2} + 12 \, B a^{2} b x + 4 \, A a b^{2} x + 3 \, B a^{3} + A a^{2} b}{12 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
Output:
-1/12*(12*B*b^3*x^3 + 18*B*a*b^2*x^2 + 6*A*b^3*x^2 + 12*B*a^2*b*x + 4*A*a* b^2*x + 3*B*a^3 + A*a^2*b)/((b*x + a)^4*b^4*sgn(b*x + a))
Time = 10.82 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.68 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\left (\frac {B\,a^2-A\,a\,b}{3\,b^4}-\frac {a\,\left (\frac {A\,b^2-B\,a\,b}{3\,b^4}-\frac {B\,a}{3\,b^3}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^4}-\frac {\left (\frac {A\,b-2\,B\,a}{2\,b^4}-\frac {B\,a}{2\,b^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^3}-\frac {B\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^2}-\frac {a^2\,\left (\frac {A}{4\,b}-\frac {B\,a}{4\,b^2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^2\,{\left (a+b\,x\right )}^5} \] Input:
int((x^2*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
Output:
- (((B*a^2 - A*a*b)/(3*b^4) - (a*((A*b^2 - B*a*b)/(3*b^4) - (B*a)/(3*b^3)) )/b)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(a + b*x)^4 - (((A*b - 2*B*a)/(2*b^4 ) - (B*a)/(2*b^4))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(a + b*x)^3 - (B*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(b^4*(a + b*x)^2) - (a^2*(A/(4*b) - (B*a)/(4*b ^2))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(b^2*(a + b*x)^5)
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {x^{3}}{3 a \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:
int(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
Output:
x**3/(3*a*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))