Integrand size = 29, antiderivative size = 256 \[ \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {A (a+b x)}{4 a x^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (a+b x)}{3 a^2 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (A b-a B) (a+b x)}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (A b-a B) (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 (A b-a B) (a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 (A b-a B) (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/4*A*(b*x+a)/a/x^4/((b*x+a)^2)^(1/2)+1/3*(A*b-B*a)*(b*x+a)/a^2/x^3/((b*x +a)^2)^(1/2)-1/2*b*(A*b-B*a)*(b*x+a)/a^3/x^2/((b*x+a)^2)^(1/2)+b^2*(A*b-B* a)*(b*x+a)/a^4/x/((b*x+a)^2)^(1/2)+b^3*(A*b-B*a)*(b*x+a)*ln(x)/a^5/((b*x+a )^2)^(1/2)-b^3*(A*b-B*a)*(b*x+a)*ln(b*x+a)/a^5/((b*x+a)^2)^(1/2)
Time = 0.60 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\frac {a^3 \left (-12 A b^3 x^3+6 a b^2 x^2 (A+2 B x)-2 a^2 b x (2 A+3 B x)+a^3 (3 A+4 B x)\right )}{\sqrt {a^2} x^4}-\frac {a \sqrt {(a+b x)^2} \left (-25 A b^3 x^3+a^3 (3 A+4 B x)-a^2 b x (7 A+10 B x)+a b^2 x^2 (13 A+22 B x)\right )}{x^4}+24 \sqrt {a^2} b^3 (-A b+a B) \log (x)-12 \left (-a+\sqrt {a^2}\right ) b^3 (-A b+a B) \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )-12 \left (a+\sqrt {a^2}\right ) b^3 (-A b+a B) \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )}{24 a^6} \]
((a^3*(-12*A*b^3*x^3 + 6*a*b^2*x^2*(A + 2*B*x) - 2*a^2*b*x*(2*A + 3*B*x) + a^3*(3*A + 4*B*x)))/(Sqrt[a^2]*x^4) - (a*Sqrt[(a + b*x)^2]*(-25*A*b^3*x^3 + a^3*(3*A + 4*B*x) - a^2*b*x*(7*A + 10*B*x) + a*b^2*x^2*(13*A + 22*B*x)) )/x^4 + 24*Sqrt[a^2]*b^3*(-(A*b) + a*B)*Log[x] - 12*(-a + Sqrt[a^2])*b^3*( -(A*b) + a*B)*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2]] - 12*(a + Sqrt[a^2] )*b^3*(-(A*b) + a*B)*Log[Sqrt[a^2] + b*x - Sqrt[(a + b*x)^2]])/(24*a^6)
Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.52, 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.
| \(\displaystyle \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b x^5 (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^5 (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {(a B-A b) b^4}{a^5 (a+b x)}-\frac {(a B-A b) b^3}{a^5 x}+\frac {(a B-A b) b^2}{a^4 x^2}-\frac {(a B-A b) b}{a^3 x^3}+\frac {a B-A b}{a^2 x^4}+\frac {A}{a x^5}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (\frac {b^3 \log (x) (A b-a B)}{a^5}-\frac {b^3 (A b-a B) \log (a+b x)}{a^5}+\frac {b^2 (A b-a B)}{a^4 x}-\frac {b (A b-a B)}{2 a^3 x^2}+\frac {A b-a B}{3 a^2 x^3}-\frac {A}{4 a x^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/4*A/(a*x^4) + (A*b - a*B)/(3*a^2*x^3) - (b*(A*b - a*B))/(2* a^3*x^2) + (b^2*(A*b - a*B))/(a^4*x) + (b^3*(A*b - a*B)*Log[x])/a^5 - (b^3 *(A*b - a*B)*Log[a + b*x])/a^5))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.8.13.3.1 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 = 0.41 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.56
| method | result | size |
| default | \(\frac {\left (b x +a \right ) \left (12 A \ln \left (x \right ) x^{4} b^{4}-12 A \ln \left (b x +a \right ) x^{4} b^{4}-12 B \ln \left (x \right ) x^{4} a \,b^{3}+12 B \ln \left (b x +a \right ) x^{4} a \,b^{3}+12 a A \,b^{3} x^{3}-12 x^{3} B \,a^{2} b^{2}-6 a^{2} A \,b^{2} x^{2}+6 x^{2} B \,a^{3} b +4 a^{3} A b x -4 a^{4} B x -3 A \,a^{4}\right )}{12 \sqrt {\left (b x +a \right )^{2}}\, a^{5} x^{4}}\) | \(143\) |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {\left (A b -B a \right ) b^{2} x^{3}}{a^{4}}-\frac {\left (A b -B a \right ) b \,x^{2}}{2 a^{3}}+\frac {\left (A b -B a \right ) x}{3 a^{2}}-\frac {A}{4 a}\right )}{\left (b x +a \right ) x^{4}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -B a \right ) b^{3} \ln \left (b x +a \right )}{\left (b x +a \right ) a^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -B a \right ) b^{3} \ln \left (-x \right )}{\left (b x +a \right ) a^{5}}\) | \(151\) |
1/12*(b*x+a)*(12*A*ln(x)*x^4*b^4-12*A*ln(b*x+a)*x^4*b^4-12*B*ln(x)*x^4*a*b ^3+12*B*ln(b*x+a)*x^4*a*b^3+12*a*A*b^3*x^3-12*x^3*B*a^2*b^2-6*a^2*A*b^2*x^ 2+6*x^2*B*a^3*b+4*a^3*A*b*x-4*a^4*B*x-3*A*a^4)/((b*x+a)^2)^(1/2)/a^5/x^4
Time = 0.41 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.46 \[ \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {12 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} \log \left (b x + a\right ) - 12 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} \log \left (x\right ) - 3 \, A a^{4} - 12 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} + 6 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} - 4 \, {\left (B a^{4} - A a^{3} b\right )} x}{12 \, a^{5} x^{4}} \]
1/12*(12*(B*a*b^3 - A*b^4)*x^4*log(b*x + a) - 12*(B*a*b^3 - A*b^4)*x^4*log (x) - 3*A*a^4 - 12*(B*a^2*b^2 - A*a*b^3)*x^3 + 6*(B*a^3*b - A*a^2*b^2)*x^2 - 4*(B*a^4 - A*a^3*b)*x)/(a^5*x^4)
\[ \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A + B x}{x^{5} \sqrt {\left (a + b x\right )^{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} - \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{4} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} - \frac {11 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{2}}{6 \, a^{4} x} + \frac {25 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{3}}{12 \, a^{5} x} + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b}{6 \, a^{3} x^{2}} - \frac {13 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2}}{12 \, a^{4} x^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{3 \, a^{2} x^{3}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{12 \, a^{3} x^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A}{4 \, a^{2} x^{4}} \]
(-1)^(2*a*b*x + 2*a^2)*B*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^4 - (-1) ^(2*a*b*x + 2*a^2)*A*b^4*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^5 - 11/6*sqr t(b^2*x^2 + 2*a*b*x + a^2)*B*b^2/(a^4*x) + 25/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^3/(a^5*x) + 5/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b/(a^3*x^2) - 13/ 12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^2/(a^4*x^2) - 1/3*sqrt(b^2*x^2 + 2*a* b*x + a^2)*B/(a^2*x^3) + 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b/(a^3*x^3) - 1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A/(a^2*x^4)
Time = 0.28 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {{\left (B a b^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right )}{a^{5}} + \frac {{\left (B a b^{4} \mathrm {sgn}\left (b x + a\right ) - A b^{5} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{a^{5} b} - \frac {3 \, A a^{4} \mathrm {sgn}\left (b x + a\right ) + 12 \, {\left (B a^{2} b^{2} \mathrm {sgn}\left (b x + a\right ) - A a b^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} - 6 \, {\left (B a^{3} b \mathrm {sgn}\left (b x + a\right ) - A a^{2} b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 4 \, {\left (B a^{4} \mathrm {sgn}\left (b x + a\right ) - A a^{3} b \mathrm {sgn}\left (b x + a\right )\right )} x}{12 \, a^{5} x^{4}} \]
-(B*a*b^3*sgn(b*x + a) - A*b^4*sgn(b*x + a))*log(abs(x))/a^5 + (B*a*b^4*sg n(b*x + a) - A*b^5*sgn(b*x + a))*log(abs(b*x + a))/(a^5*b) - 1/12*(3*A*a^4 *sgn(b*x + a) + 12*(B*a^2*b^2*sgn(b*x + a) - A*a*b^3*sgn(b*x + a))*x^3 - 6 *(B*a^3*b*sgn(b*x + a) - A*a^2*b^2*sgn(b*x + a))*x^2 + 4*(B*a^4*sgn(b*x + a) - A*a^3*b*sgn(b*x + a))*x)/(a^5*x^4)
Timed out. \[ \int \frac {A+B x}{x^5 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{x^5\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]