Integrand size = 29, antiderivative size = 210 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {A}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A (a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
A/a^4/((b*x+a)^2)^(1/2)+1/4*(A*b-B*a)/a/b/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/3* A/a^2/(b*x+a)^2/((b*x+a)^2)^(1/2)+1/2*A/a^3/(b*x+a)/((b*x+a)^2)^(1/2)+A*(b *x+a)*ln(x)/a^5/((b*x+a)^2)^(1/2)-A*(b*x+a)*ln(b*x+a)/a^5/((b*x+a)^2)^(1/2 )
Time = 1.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.50 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a \left (25 a^3 A b-3 a^4 B+52 a^2 A b^2 x+42 a A b^3 x^2+12 A b^4 x^3\right )+12 A b (a+b x)^4 \log (x)-12 A b (a+b x)^4 \log (a+b x)}{12 a^5 b (a+b x)^3 \sqrt {(a+b x)^2}} \]
(a*(25*a^3*A*b - 3*a^4*B + 52*a^2*A*b^2*x + 42*a*A*b^3*x^2 + 12*A*b^4*x^3) + 12*A*b*(a + b*x)^4*Log[x] - 12*A*b*(a + b*x)^4*Log[a + b*x])/(12*a^5*b* (a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.54, 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.
\(\displaystyle \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {A+B x}{b^5 x (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {A}{a^5 x}-\frac {b A}{a^5 (a+b x)}-\frac {b A}{a^4 (a+b x)^2}-\frac {b A}{a^3 (a+b x)^3}-\frac {b A}{a^2 (a+b x)^4}+\frac {a B-A b}{a (a+b x)^5}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {A \log (a+b x)}{a^5}+\frac {A \log (x)}{a^5}+\frac {A}{a^4 (a+b x)}+\frac {A}{2 a^3 (a+b x)^2}+\frac {A}{3 a^2 (a+b x)^3}+\frac {A b-a B}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((A*b - a*B)/(4*a*b*(a + b*x)^4) + A/(3*a^2*(a + b*x)^3) + A/(2 *a^3*(a + b*x)^2) + A/(a^4*(a + b*x)) + (A*Log[x])/a^5 - (A*Log[a + b*x])/ a^5))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.8.27.3.1 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 = 0.19 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b^{3} A \,x^{3}}{a^{4}}+\frac {7 b^{2} A \,x^{2}}{2 a^{3}}+\frac {13 A b x}{3 a^{2}}+\frac {25 A b -3 B a}{12 b a}\right )}{\left (b x +a \right )^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, A \ln \left (-x \right )}{\left (b x +a \right ) a^{5}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, A \ln \left (b x +a \right )}{\left (b x +a \right ) a^{5}}\) | \(121\) |
default | \(\frac {\left (12 A \ln \left (x \right ) x^{4} b^{5}-12 A \ln \left (b x +a \right ) b^{5} x^{4}+48 A \ln \left (x \right ) x^{3} a \,b^{4}-48 A \ln \left (b x +a \right ) x^{3} a \,b^{4}+72 A \ln \left (x \right ) x^{2} a^{2} b^{3}-72 A \ln \left (b x +a \right ) x^{2} a^{2} b^{3}+12 A \,x^{3} a \,b^{4}+48 A \ln \left (x \right ) x \,a^{3} b^{2}-48 A \ln \left (b x +a \right ) x \,a^{3} b^{2}+42 A \,x^{2} a^{2} b^{3}+12 A \ln \left (x \right ) a^{4} b -12 A \ln \left (b x +a \right ) a^{4} b +52 A \,a^{3} b^{2} x +25 A \,a^{4} b -3 a^{5} B \right ) \left (b x +a \right )}{12 b \,a^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(205\) |
((b*x+a)^2)^(1/2)/(b*x+a)^5*(1/a^4*b^3*A*x^3+7/2/a^3*b^2*A*x^2+13/3/a^2*A* b*x+1/12*(25*A*b-3*B*a)/b/a)+((b*x+a)^2)^(1/2)/(b*x+a)*A/a^5*ln(-x)-((b*x+ a)^2)^(1/2)/(b*x+a)*A/a^5*ln(b*x+a)
Time = 0.42 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {12 \, A a b^{4} x^{3} + 42 \, A a^{2} b^{3} x^{2} + 52 \, A a^{3} b^{2} x - 3 \, B a^{5} + 25 \, A a^{4} b - 12 \, {\left (A b^{5} x^{4} + 4 \, A a b^{4} x^{3} + 6 \, A a^{2} b^{3} x^{2} + 4 \, A a^{3} b^{2} x + A a^{4} b\right )} \log \left (b x + a\right ) + 12 \, {\left (A b^{5} x^{4} + 4 \, A a b^{4} x^{3} + 6 \, A a^{2} b^{3} x^{2} + 4 \, A a^{3} b^{2} x + A a^{4} b\right )} \log \left (x\right )}{12 \, {\left (a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{3} + 6 \, a^{7} b^{3} x^{2} + 4 \, a^{8} b^{2} x + a^{9} b\right )}} \]
1/12*(12*A*a*b^4*x^3 + 42*A*a^2*b^3*x^2 + 52*A*a^3*b^2*x - 3*B*a^5 + 25*A* a^4*b - 12*(A*b^5*x^4 + 4*A*a*b^4*x^3 + 6*A*a^2*b^3*x^2 + 4*A*a^3*b^2*x + A*a^4*b)*log(b*x + a) + 12*(A*b^5*x^4 + 4*A*a*b^4*x^3 + 6*A*a^2*b^3*x^2 + 4*A*a^3*b^2*x + A*a^4*b)*log(x))/(a^5*b^5*x^4 + 4*a^6*b^4*x^3 + 6*a^7*b^3* x^2 + 4*a^8*b^2*x + a^9*b)
\[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {A + B x}{x \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {A}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}} + \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} + \frac {A}{2 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {B}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {A}{4 \, a b^{4} {\left (x + \frac {a}{b}\right )}^{4}} \]
-(-1)^(2*a*b*x + 2*a^2)*A*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^5 + 1/3*A/( (b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2) + A/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^ 4) + 1/2*A/(a^3*b^2*(x + a/b)^2) - 1/4*B/(b^5*(x + a/b)^4) + 1/4*A/(a*b^4* (x + a/b)^4)
Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.51 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {A \log \left ({\left | b x + a \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {A \log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {12 \, A a b^{4} x^{3} + 42 \, A a^{2} b^{3} x^{2} + 52 \, A a^{3} b^{2} x - 3 \, B a^{5} + 25 \, A a^{4} b}{12 \, {\left (b x + a\right )}^{4} a^{5} b \mathrm {sgn}\left (b x + a\right )} \]
-A*log(abs(b*x + a))/(a^5*sgn(b*x + a)) + A*log(abs(x))/(a^5*sgn(b*x + a)) + 1/12*(12*A*a*b^4*x^3 + 42*A*a^2*b^3*x^2 + 52*A*a^3*b^2*x - 3*B*a^5 + 25 *A*a^4*b)/((b*x + a)^4*a^5*b*sgn(b*x + a))
Timed out. \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {A+B\,x}{x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]