Integrand size = 29, antiderivative size = 154 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {a (2 A b-3 a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2 (A b-a B)}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-3 a B) (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]
a*(2*A*b-3*B*a)/b^4/((b*x+a)^2)^(1/2)-1/2*a^2*(A*b-B*a)/b^4/(b*x+a)/((b*x+ a)^2)^(1/2)+B*x*(b*x+a)/b^3/((b*x+a)^2)^(1/2)+(A*b-3*B*a)*(b*x+a)*ln(b*x+a )/b^4/((b*x+a)^2)^(1/2)
Time = 0.96 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.51 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\frac {b \left (a x \sqrt {(a+b x)^2} \left (6 a^3 B+A b^3 x^2-a b^2 x (A+B x)+a^2 (-2 A b+3 b B x)\right )-\sqrt {a^2} x \left (6 a^4 B-A b^4 x^3+a b^3 B x^3+a^2 b^2 x (-3 A+2 B x)+a^3 (-2 A b+9 b B x)\right )\right )}{a^2 (a+b x) \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )}-4 A b \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )+12 a B \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )}{2 b^4} \]
((b*(a*x*Sqrt[(a + b*x)^2]*(6*a^3*B + A*b^3*x^2 - a*b^2*x*(A + B*x) + a^2* (-2*A*b + 3*b*B*x)) - Sqrt[a^2]*x*(6*a^4*B - A*b^4*x^3 + a*b^3*B*x^3 + a^2 *b^2*x*(-3*A + 2*B*x) + a^3*(-2*A*b + 9*b*B*x))))/(a^2*(a + b*x)*(a^2 + a* b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2])) - 4*A*b*ArcTanh[(b*x)/(Sqrt[a^2] - Sqr t[(a + b*x)^2])] + 12*a*B*ArcTanh[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])])/ (2*b^4)
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, 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.
\(\displaystyle \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^3 (a+b x) \int \frac {x^2 (A+B x)}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {x^2 (A+B x)}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (-\frac {(a B-A b) a^2}{b^3 (a+b x)^3}+\frac {(3 a B-2 A b) a}{b^3 (a+b x)^2}+\frac {B}{b^3}+\frac {A b-3 a B}{b^3 (a+b x)}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {a^2 (A b-a B)}{2 b^4 (a+b x)^2}+\frac {a (2 A b-3 a B)}{b^4 (a+b x)}+\frac {(A b-3 a B) \log (a+b x)}{b^4}+\frac {B x}{b^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((B*x)/b^3 - (a^2*(A*b - a*B))/(2*b^4*(a + b*x)^2) + (a*(2*A*b - 3*a*B))/(b^4*(a + b*x)) + ((A*b - 3*a*B)*Log[a + b*x])/b^4))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.8.16.3.1 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[((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 = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, B x}{\left (b x +a \right ) b^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (2 A b a -3 B \,a^{2}\right ) x +\frac {a^{2} \left (3 A b -5 B a \right )}{2 b}\right )}{\left (b x +a \right )^{3} b^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -3 B a \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\) | \(110\) |
default | \(\frac {\left (2 A \ln \left (b x +a \right ) x^{2} b^{3}-6 B \ln \left (b x +a \right ) x^{2} a \,b^{2}+2 x^{3} B \,b^{3}+4 A \ln \left (b x +a \right ) x a \,b^{2}-12 B \ln \left (b x +a \right ) x \,a^{2} b +4 B \,x^{2} a \,b^{2}+2 A \ln \left (b x +a \right ) a^{2} b +4 A x a \,b^{2}-6 B \ln \left (b x +a \right ) a^{3}-4 B x \,a^{2} b +3 A \,a^{2} b -5 B \,a^{3}\right ) \left (b x +a \right )}{2 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(153\) |
((b*x+a)^2)^(1/2)/(b*x+a)*B/b^3*x+((b*x+a)^2)^(1/2)/(b*x+a)^3*((2*A*a*b-3* B*a^2)*x+1/2*a^2*(3*A*b-5*B*a)/b)/b^3+((b*x+a)^2)^(1/2)/(b*x+a)/b^4*(A*b-3 *B*a)*ln(b*x+a)
Time = 0.40 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, B b^{3} x^{3} + 4 \, B a b^{2} x^{2} - 5 \, B a^{3} + 3 \, A a^{2} b - 4 \, {\left (B a^{2} b - A a b^{2}\right )} x - 2 \, {\left (3 \, B a^{3} - A a^{2} b + {\left (3 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]
1/2*(2*B*b^3*x^3 + 4*B*a*b^2*x^2 - 5*B*a^3 + 3*A*a^2*b - 4*(B*a^2*b - A*a* b^2)*x - 2*(3*B*a^3 - A*a^2*b + (3*B*a*b^2 - A*b^3)*x^2 + 2*(3*B*a^2*b - A *a*b^2)*x)*log(b*x + a))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)
\[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {3 \, B a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {A \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, B a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {6 \, B a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, A a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, B a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, A a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} \]
B*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 3*B*a*log(x + a/b)/b^4 + A*log (x + a/b)/b^3 + 2*B*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 6*B*a^2*x/(b ^5*(x + a/b)^2) + 2*A*a*x/(b^4*(x + a/b)^2) - 11/2*B*a^3/(b^6*(x + a/b)^2) + 3/2*A*a^2/(b^5*(x + a/b)^2)
Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B x}{b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {{\left (3 \, B a - A b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {5 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x + a\right )} \]
B*x/(b^3*sgn(b*x + a)) - (3*B*a - A*b)*log(abs(b*x + a))/(b^4*sgn(b*x + a) ) - 1/2*(5*B*a^3 - 3*A*a^2*b + 2*(3*B*a^2*b - 2*A*a*b^2)*x)/((b*x + a)^2*b ^4*sgn(b*x + a))
Timed out. \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]