Integrand size = 24, antiderivative size = 278 \[ \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {10 b^2}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{4 a^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{a^4 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2}{a^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b (a+b x)}{a^6 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 b^2 (a+b x) \log (x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 b^2 (a+b x) \log (a+b x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}} \]
10*b^2/a^6/((b*x+a)^2)^(1/2)+1/4*b^2/a^3/(b*x+a)^3/((b*x+a)^2)^(1/2)+b^2/a ^4/(b*x+a)^2/((b*x+a)^2)^(1/2)+3*b^2/a^5/(b*x+a)/((b*x+a)^2)^(1/2)+1/2*(-b *x-a)/a^5/x^2/((b*x+a)^2)^(1/2)+5*b*(b*x+a)/a^6/x/((b*x+a)^2)^(1/2)+15*b^2 *(b*x+a)*ln(x)/a^7/((b*x+a)^2)^(1/2)-15*b^2*(b*x+a)*ln(b*x+a)/a^7/((b*x+a) ^2)^(1/2)
Time = 1.05 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a \left (-2 a^5+12 a^4 b x+125 a^3 b^2 x^2+260 a^2 b^3 x^3+210 a b^4 x^4+60 b^5 x^5\right )+60 b^2 x^2 (a+b x)^4 \log (x)-60 b^2 x^2 (a+b x)^4 \log (a+b x)}{4 a^7 x^2 (a+b x)^3 \sqrt {(a+b x)^2}} \]
(a*(-2*a^5 + 12*a^4*b*x + 125*a^3*b^2*x^2 + 260*a^2*b^3*x^3 + 210*a*b^4*x^ 4 + 60*b^5*x^5) + 60*b^2*x^2*(a + b*x)^4*Log[x] - 60*b^2*x^2*(a + b*x)^4*L og[a + b*x])/(4*a^7*x^2*(a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.47, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1102, 27, 54, 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.
\(\displaystyle \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {1}{b^5 x^3 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {1}{x^3 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {(a+b x) \int \left (-\frac {15 b^3}{a^7 (a+b x)}-\frac {10 b^3}{a^6 (a+b x)^2}-\frac {6 b^3}{a^5 (a+b x)^3}-\frac {3 b^3}{a^4 (a+b x)^4}-\frac {b^3}{a^3 (a+b x)^5}+\frac {15 b^2}{a^7 x}-\frac {5 b}{a^6 x^2}+\frac {1}{a^5 x^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (\frac {15 b^2 \log (x)}{a^7}-\frac {15 b^2 \log (a+b x)}{a^7}+\frac {10 b^2}{a^6 (a+b x)}+\frac {5 b}{a^6 x}+\frac {3 b^2}{a^5 (a+b x)^2}-\frac {1}{2 a^5 x^2}+\frac {b^2}{a^4 (a+b x)^3}+\frac {b^2}{4 a^3 (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*1/(a^5*x^2) + (5*b)/(a^6*x) + b^2/(4*a^3*(a + b*x)^4) + b ^2/(a^4*(a + b*x)^3) + (3*b^2)/(a^5*(a + b*x)^2) + (10*b^2)/(a^6*(a + b*x) ) + (15*b^2*Log[x])/a^7 - (15*b^2*Log[a + b*x])/a^7))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.3.7.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.49
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {15 b^{5} x^{5}}{a^{6}}+\frac {105 b^{4} x^{4}}{2 a^{5}}+\frac {65 b^{3} x^{3}}{a^{4}}+\frac {125 b^{2} x^{2}}{4 a^{3}}+\frac {3 b x}{a^{2}}-\frac {1}{2 a}\right )}{\left (b x +a \right )^{5} x^{2}}-\frac {15 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) a^{7}}+\frac {15 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (-x \right )}{\left (b x +a \right ) a^{7}}\) | \(137\) |
default | \(\frac {\left (60 \ln \left (x \right ) b^{6} x^{6}-60 \ln \left (b x +a \right ) b^{6} x^{6}+240 \ln \left (x \right ) a \,b^{5} x^{5}-240 \ln \left (b x +a \right ) a \,b^{5} x^{5}+360 \ln \left (x \right ) a^{2} b^{4} x^{4}-360 \ln \left (b x +a \right ) a^{2} b^{4} x^{4}+60 a \,x^{5} b^{5}+240 \ln \left (x \right ) a^{3} b^{3} x^{3}-240 \ln \left (b x +a \right ) a^{3} b^{3} x^{3}+210 a^{2} x^{4} b^{4}+60 \ln \left (x \right ) a^{4} b^{2} x^{2}-60 \ln \left (b x +a \right ) a^{4} b^{2} x^{2}+260 a^{3} x^{3} b^{3}+125 a^{4} x^{2} b^{2}+12 a^{5} x b -2 a^{6}\right ) \left (b x +a \right )}{4 x^{2} a^{7} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(218\) |
((b*x+a)^2)^(1/2)/(b*x+a)^5*(15*b^5/a^6*x^5+105/2*b^4/a^5*x^4+65*b^3/a^4*x ^3+125/4*b^2/a^3*x^2+3*b/a^2*x-1/2/a)/x^2-15*((b*x+a)^2)^(1/2)/(b*x+a)/a^7 *b^2*ln(b*x+a)+15*((b*x+a)^2)^(1/2)/(b*x+a)/a^7*b^2*ln(-x)
Time = 0.25 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {60 \, a b^{5} x^{5} + 210 \, a^{2} b^{4} x^{4} + 260 \, a^{3} b^{3} x^{3} + 125 \, a^{4} b^{2} x^{2} + 12 \, a^{5} b x - 2 \, a^{6} - 60 \, {\left (b^{6} x^{6} + 4 \, a b^{5} x^{5} + 6 \, a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{3} + a^{4} b^{2} x^{2}\right )} \log \left (b x + a\right ) + 60 \, {\left (b^{6} x^{6} + 4 \, a b^{5} x^{5} + 6 \, a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{3} + a^{4} b^{2} x^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{7} b^{4} x^{6} + 4 \, a^{8} b^{3} x^{5} + 6 \, a^{9} b^{2} x^{4} + 4 \, a^{10} b x^{3} + a^{11} x^{2}\right )}} \]
1/4*(60*a*b^5*x^5 + 210*a^2*b^4*x^4 + 260*a^3*b^3*x^3 + 125*a^4*b^2*x^2 + 12*a^5*b*x - 2*a^6 - 60*(b^6*x^6 + 4*a*b^5*x^5 + 6*a^2*b^4*x^4 + 4*a^3*b^3 *x^3 + a^4*b^2*x^2)*log(b*x + a) + 60*(b^6*x^6 + 4*a*b^5*x^5 + 6*a^2*b^4*x ^4 + 4*a^3*b^3*x^3 + a^4*b^2*x^2)*log(x))/(a^7*b^4*x^6 + 4*a^8*b^3*x^5 + 6 *a^9*b^2*x^4 + 4*a^10*b*x^3 + a^11*x^2)
\[ \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{x^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {15 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{7}} + \frac {5 \, b^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{4}} + \frac {15 \, b^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{6}} + \frac {7 \, b}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} x} - \frac {1}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{2}} + \frac {15}{2 \, a^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {1}{4 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{4}} \]
-15*(-1)^(2*a*b*x + 2*a^2)*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^7 + 5* b^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^4) + 15*b^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^6) + 7/2*b/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*x) - 1/2/((b^2*x ^2 + 2*a*b*x + a^2)^(3/2)*a^2*x^2) + 15/2/(a^5*(x + a/b)^2) + 1/4/(a^3*b^2 *(x + a/b)^4)
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {15 \, b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{7} \mathrm {sgn}\left (b x + a\right )} + \frac {15 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{7} \mathrm {sgn}\left (b x + a\right )} + \frac {60 \, a b^{5} x^{5} + 210 \, a^{2} b^{4} x^{4} + 260 \, a^{3} b^{3} x^{3} + 125 \, a^{4} b^{2} x^{2} + 12 \, a^{5} b x - 2 \, a^{6}}{4 \, {\left (b x + a\right )}^{4} a^{7} x^{2} \mathrm {sgn}\left (b x + a\right )} \]
-15*b^2*log(abs(b*x + a))/(a^7*sgn(b*x + a)) + 15*b^2*log(abs(x))/(a^7*sgn (b*x + a)) + 1/4*(60*a*b^5*x^5 + 210*a^2*b^4*x^4 + 260*a^3*b^3*x^3 + 125*a ^4*b^2*x^2 + 12*a^5*b*x - 2*a^6)/((b*x + a)^4*a^7*x^2*sgn(b*x + a))
Timed out. \[ \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]