Integrand size = 27, antiderivative size = 121 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a (A b-a B)}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-2 a B}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
1/4*a*(A*b-B*a)/b^3/(b*x+a)^3/((b*x+a)^2)^(1/2)-1/3*(A*b-2*B*a)/b^3/(b*x+a )^2/((b*x+a)^2)^(1/2)-1/2*B/b^3/(b*x+a)/((b*x+a)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(121)=242\).
Time = 1.31 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.64 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {x^2 \left (-3 a^2 b^5 B x^6-3 a^3 \sqrt {a^2} b^2 x^2 \sqrt {(a+b x)^2} (A+B x)+a^6 b x (4 A+B x)+a^4 \sqrt {a^2} b x \sqrt {(a+b x)^2} (2 A+3 B x)+a^7 (6 A+4 B x)-3 a b^4 x^4 \left (-A b^2 x^2+\sqrt {a^2} A \sqrt {(a+b x)^2}+\sqrt {a^2} B x \sqrt {(a+b x)^2}\right )-a^5 \left (-A b^2 x^2+6 \sqrt {a^2} A \sqrt {(a+b x)^2}+4 \sqrt {a^2} B x \sqrt {(a+b x)^2}\right )+3 \sqrt {a^2} b^3 x^3 \sqrt {(a+b x)^2} \left (A b^2 x^2+a^2 (A+B x)\right )\right )}{12 a^7 (a+b x)^3 \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )} \] Input:
Integrate[(x*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
-1/12*(x^2*(-3*a^2*b^5*B*x^6 - 3*a^3*Sqrt[a^2]*b^2*x^2*Sqrt[(a + b*x)^2]*( A + B*x) + a^6*b*x*(4*A + B*x) + a^4*Sqrt[a^2]*b*x*Sqrt[(a + b*x)^2]*(2*A + 3*B*x) + a^7*(6*A + 4*B*x) - 3*a*b^4*x^4*(-(A*b^2*x^2) + Sqrt[a^2]*A*Sqr t[(a + b*x)^2] + Sqrt[a^2]*B*x*Sqrt[(a + b*x)^2]) - a^5*(-(A*b^2*x^2) + 6* Sqrt[a^2]*A*Sqrt[(a + b*x)^2] + 4*Sqrt[a^2]*B*x*Sqrt[(a + b*x)^2]) + 3*Sqr t[a^2]*b^3*x^3*Sqrt[(a + b*x)^2]*(A*b^2*x^2 + a^2*(A + B*x))))/(a^7*(a + b *x)^3*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))
Time = 0.42 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1187, 27, 86, 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.
\(\displaystyle \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {x (A+B x)}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {x (A+B x)}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {B}{b^2 (a+b x)^3}+\frac {A b-2 a B}{b^2 (a+b x)^4}+\frac {a (a B-A b)}{b^2 (a+b x)^5}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {A b-2 a B}{3 b^3 (a+b x)^3}+\frac {a (A b-a B)}{4 b^3 (a+b x)^4}-\frac {B}{2 b^3 (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(x*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
((a + b*x)*((a*(A*b - a*B))/(4*b^3*(a + b*x)^4) - (A*b - 2*a*B)/(3*b^3*(a + b*x)^3) - B/(2*b^3*(a + b*x)^2)))/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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 1.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43
method | result | size |
gosper | \(-\frac {\left (b x +a \right ) \left (6 x^{2} B \,b^{2}+4 x \,b^{2} A +4 x a b B +a b A +a^{2} B \right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(52\) |
default | \(-\frac {\left (b x +a \right ) \left (6 x^{2} B \,b^{2}+4 x \,b^{2} A +4 x a b B +a b A +a^{2} B \right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(52\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,x^{2}}{2 b}-\frac {\left (A b +B a \right ) x}{3 b^{2}}-\frac {a \left (A b +B a \right )}{12 b^{3}}\right )}{\left (b x +a \right )^{5}}\) | \(54\) |
orering | \(-\frac {\left (6 x^{2} B \,b^{2}+4 x \,b^{2} A +4 x a b B +a b A +a^{2} B \right ) \left (b x +a \right )}{12 b^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}\) | \(61\) |
Input:
int(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/12*(b*x+a)/b^3*(6*B*b^2*x^2+4*A*b^2*x+4*B*a*b*x+A*a*b+B*a^2)/((b*x+a)^2 )^(5/2)
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, B b^{2} x^{2} + B a^{2} + A a b + 4 \, {\left (B a b + A b^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \] Input:
integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
Output:
-1/12*(6*B*b^2*x^2 + B*a^2 + A*a*b + 4*(B*a*b + A*b^2)*x)/(b^7*x^4 + 4*a*b ^6*x^3 + 6*a^2*b^5*x^2 + 4*a^3*b^4*x + a^4*b^3)
\[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
Output:
Integral(x*(A + B*x)/((a + b*x)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.74 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {A}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {B}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, B a}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {B a^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {A a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:
integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
Output:
-1/3*A/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 1/2*B/(b^5*(x + a/b)^2) + 2 /3*B*a/(b^6*(x + a/b)^3) - 1/4*B*a^2/(b^7*(x + a/b)^4) + 1/4*A*a/(b^6*(x + a/b)^4)
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, B b^{2} x^{2} + 4 \, B a b x + 4 \, A b^{2} x + B a^{2} + A a b}{12 \, {\left (b x + a\right )}^{4} b^{3} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
Output:
-1/12*(6*B*b^2*x^2 + 4*B*a*b*x + 4*A*b^2*x + B*a^2 + A*a*b)/((b*x + a)^4*b ^3*sgn(b*x + a))
Time = 10.80 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.51 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (B\,a^2+4\,B\,a\,b\,x+A\,a\,b+6\,B\,b^2\,x^2+4\,A\,b^2\,x\right )}{12\,b^3\,{\left (a+b\,x\right )}^5} \] Input:
int((x*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
Output:
-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(B*a^2 + 6*B*b^2*x^2 + A*a*b + 4*A*b^2*x + 4*B*a*b*x))/(12*b^3*(a + b*x)^5)
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-3 b x -a}{6 b^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:
int(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
Output:
( - a - 3*b*x)/(6*b**2*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))