Integrand size = 29, antiderivative size = 202 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {a^2 (3 A b-4 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-3 a B) x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (A b-2 a B) (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-a^2*(3*A*b-4*B*a)/b^5/((b*x+a)^2)^(1/2)+1/2*a^3*(A*b-B*a)/b^5/(b*x+a)/((b *x+a)^2)^(1/2)+(A*b-3*B*a)*x*(b*x+a)/b^4/((b*x+a)^2)^(1/2)+1/2*B*x^2*(b*x+ a)/b^3/((b*x+a)^2)^(1/2)-3*a*(A*b-2*B*a)*(b*x+a)*ln(b*x+a)/b^5/((b*x+a)^2) ^(1/2)
Time = 1.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.58 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {7 a^4 B+4 a b^3 x^2 (A-B x)+b^4 x^3 (2 A+B x)-a^2 b^2 x (4 A+11 B x)+a^3 (-5 A b+2 b B x)+6 a (-A b+2 a B) (a+b x)^2 \log (a+b x)}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \]
(7*a^4*B + 4*a*b^3*x^2*(A - B*x) + b^4*x^3*(2*A + B*x) - a^2*b^2*x*(4*A + 11*B*x) + a^3*(-5*A*b + 2*b*B*x) + 6*a*(-(A*b) + 2*a*B)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])
Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.59, 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.
\(\displaystyle \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^3 (a+b x) \int \frac {x^3 (A+B x)}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {x^3 (A+B x)}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {(a B-A b) a^3}{b^4 (a+b x)^3}-\frac {(4 a B-3 A b) a^2}{b^4 (a+b x)^2}+\frac {3 (2 a B-A b) a}{b^4 (a+b x)}+\frac {A b-3 a B}{b^4}+\frac {B x}{b^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (\frac {a^3 (A b-a B)}{2 b^5 (a+b x)^2}-\frac {a^2 (3 A b-4 a B)}{b^5 (a+b x)}-\frac {3 a (A b-2 a B) \log (a+b x)}{b^5}+\frac {x (A b-3 a B)}{b^4}+\frac {B x^2}{2 b^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(((A*b - 3*a*B)*x)/b^4 + (B*x^2)/(2*b^3) + (a^3*(A*b - a*B))/(2 *b^5*(a + b*x)^2) - (a^2*(3*A*b - 4*a*B))/(b^5*(a + b*x)) - (3*a*(A*b - 2* a*B)*Log[a + b*x])/b^5))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.8.15.3.1 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 = 0.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{2} B b \,x^{2}+A b x -3 a B x \right )}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-3 A \,a^{2} b +4 B \,a^{3}\right ) x -\frac {a^{3} \left (5 A b -7 B a \right )}{2 b}\right )}{\left (b x +a \right )^{3} b^{4}}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, a \left (A b -2 B a \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\) | \(129\) |
default | \(-\frac {\left (-b^{4} B \,x^{4}+6 A \ln \left (b x +a \right ) x^{2} a \,b^{3}-2 A \,b^{4} x^{3}-12 B \ln \left (b x +a \right ) x^{2} a^{2} b^{2}+4 B a \,b^{3} x^{3}+12 A \ln \left (b x +a \right ) x \,a^{2} b^{2}-4 A a \,b^{3} x^{2}-24 B \ln \left (b x +a \right ) x \,a^{3} b +11 B \,a^{2} b^{2} x^{2}+6 A \ln \left (b x +a \right ) a^{3} b +4 A \,a^{2} b^{2} x -12 B \ln \left (b x +a \right ) a^{4}-2 B \,a^{3} b x +5 A \,a^{3} b -7 B \,a^{4}\right ) \left (b x +a \right )}{2 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(191\) |
((b*x+a)^2)^(1/2)/(b*x+a)/b^4*(1/2*B*b*x^2+A*b*x-3*a*B*x)+((b*x+a)^2)^(1/2 )/(b*x+a)^3*((-3*A*a^2*b+4*B*a^3)*x-1/2*a^3*(5*A*b-7*B*a)/b)/b^4-3*((b*x+a )^2)^(1/2)/(b*x+a)*a/b^5*(A*b-2*B*a)*ln(b*x+a)
Time = 0.42 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.85 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B b^{4} x^{4} + 7 \, B a^{4} - 5 \, A a^{3} b - 2 \, {\left (2 \, B a b^{3} - A b^{4}\right )} x^{3} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x + 6 \, {\left (2 \, B a^{4} - A a^{3} b + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 2 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]
1/2*(B*b^4*x^4 + 7*B*a^4 - 5*A*a^3*b - 2*(2*B*a*b^3 - A*b^4)*x^3 - (11*B*a ^2*b^2 - 4*A*a*b^3)*x^2 + 2*(B*a^3*b - 2*A*a^2*b^2)*x + 6*(2*B*a^4 - A*a^3 *b + (2*B*a^2*b^2 - A*a*b^3)*x^2 + 2*(2*B*a^3*b - A*a^2*b^2)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)
\[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.20 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, B a x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {A x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {6 \, B a^{2} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {3 \, A a \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {5 \, B a^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {2 \, A a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac {12 \, B a^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {6 \, A a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, B a^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, A a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} \]
1/2*B*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*B*a*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + A*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 6*B*a^ 2*log(x + a/b)/b^5 - 3*A*a*log(x + a/b)/b^4 - 5*B*a^3/(sqrt(b^2*x^2 + 2*a* b*x + a^2)*b^5) + 2*A*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) + 12*B*a^3*x /(b^6*(x + a/b)^2) - 6*A*a^2*x/(b^5*(x + a/b)^2) + 23/2*B*a^4/(b^7*(x + a/ b)^2) - 11/2*A*a^3/(b^6*(x + a/b)^2)
Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.66 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, {\left (2 \, B a^{2} - A a b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {B b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, B a b^{2} x \mathrm {sgn}\left (b x + a\right ) + 2 \, A b^{3} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{6}} + \frac {7 \, B a^{4} - 5 \, A a^{3} b + 2 \, {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x + a\right )} \]
3*(2*B*a^2 - A*a*b)*log(abs(b*x + a))/(b^5*sgn(b*x + a)) + 1/2*(B*b^3*x^2* sgn(b*x + a) - 6*B*a*b^2*x*sgn(b*x + a) + 2*A*b^3*x*sgn(b*x + a))/b^6 + 1/ 2*(7*B*a^4 - 5*A*a^3*b + 2*(4*B*a^3*b - 3*A*a^2*b^2)*x)/((b*x + a)^2*b^5*s gn(b*x + a))
Timed out. \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]