Integrand size = 24, antiderivative size = 194 \[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
1/a^4/((b*x+a)^2)^(1/2)+1/4/a/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/3/a^2/(b*x+a)^ 2/((b*x+a)^2)^(1/2)+1/2/a^3/(b*x+a)/((b*x+a)^2)^(1/2)+(b*x+a)*ln(x)/a^5/(( b*x+a)^2)^(1/2)-(b*x+a)*ln(b*x+a)/a^5/((b*x+a)^2)^(1/2)
Time = 1.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a \left (25 a^3+52 a^2 b x+42 a b^2 x^2+12 b^3 x^3\right )+12 (a+b x)^4 \log (x)-12 (a+b x)^4 \log (a+b x)}{12 a^5 (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:
Integrate[1/(x*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
Output:
(a*(25*a^3 + 52*a^2*b*x + 42*a*b^2*x^2 + 12*b^3*x^3) + 12*(a + b*x)^4*Log[ x] - 12*(a + b*x)^4*Log[a + b*x])/(12*a^5*(a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.50, 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.
\(\displaystyle \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {1}{b^5 x (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {1}{x (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {(a+b x) \int \left (-\frac {b}{a^5 (a+b x)}-\frac {b}{a^4 (a+b x)^2}-\frac {b}{a^3 (a+b x)^3}-\frac {b}{a^2 (a+b x)^4}-\frac {b}{a (a+b x)^5}+\frac {1}{a^5 x}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {\log (a+b x)}{a^5}+\frac {\log (x)}{a^5}+\frac {1}{a^4 (a+b x)}+\frac {1}{2 a^3 (a+b x)^2}+\frac {1}{3 a^2 (a+b x)^3}+\frac {1}{4 a (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[1/(x*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
Output:
((a + b*x)*(1/(4*a*(a + b*x)^4) + 1/(3*a^2*(a + b*x)^3) + 1/(2*a^3*(a + b* x)^2) + 1/(a^4*(a + b*x)) + Log[x]/a^5 - Log[a + b*x]/a^5))/Sqrt[a^2 + 2*a *b*x + b^2*x^2]
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 = 0.64 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.54
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b^{3} x^{3}}{a^{4}}+\frac {7 b^{2} x^{2}}{2 a^{3}}+\frac {13 b x}{3 a^{2}}+\frac {25}{12 a}\right )}{\left (b x +a \right )^{5}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) a^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (-x \right )}{\left (b x +a \right ) a^{5}}\) | \(104\) |
default | \(-\frac {\left (12 \ln \left (b x +a \right ) x^{4} b^{4}-12 \ln \left (x \right ) x^{4} b^{4}+48 \ln \left (b x +a \right ) x^{3} a \,b^{3}-48 \ln \left (x \right ) x^{3} a \,b^{3}+72 \ln \left (b x +a \right ) x^{2} a^{2} b^{2}-72 a^{2} b^{2} \ln \left (x \right ) x^{2}-12 a \,b^{3} x^{3}+48 \ln \left (b x +a \right ) x \,a^{3} b -48 a^{3} b \ln \left (x \right ) x -42 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )-12 a^{4} \ln \left (x \right )-52 a^{3} b x -25 a^{4}\right ) \left (b x +a \right )}{12 a^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(173\) |
Input:
int(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)^5*(b^3/a^4*x^3+7/2*b^2/a^3*x^2+13/3*b/a^2*x+25/1 2/a)-((b*x+a)^2)^(1/2)/(b*x+a)/a^5*ln(b*x+a)+((b*x+a)^2)^(1/2)/(b*x+a)/a^5 *ln(-x)
Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {12 \, a b^{3} x^{3} + 42 \, a^{2} b^{2} x^{2} + 52 \, a^{3} b x + 25 \, a^{4} - 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (x\right )}{12 \, {\left (a^{5} b^{4} x^{4} + 4 \, a^{6} b^{3} x^{3} + 6 \, a^{7} b^{2} x^{2} + 4 \, a^{8} b x + a^{9}\right )}} \] Input:
integrate(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
Output:
1/12*(12*a*b^3*x^3 + 42*a^2*b^2*x^2 + 52*a^3*b*x + 25*a^4 - 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*log(b*x + a) + 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*log(x))/(a^5*b^4*x^4 + 4* a^6*b^3*x^3 + 6*a^7*b^2*x^2 + 4*a^8*b*x + a^9)
\[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{x \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/x/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
Output:
Integral(1/(x*((a + b*x)**2)**(5/2)), x)
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {1}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}} + \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} + \frac {1}{2 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {1}{4 \, a b^{4} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:
integrate(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
Output:
-(-1)^(2*a*b*x + 2*a^2)*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^5 + 1/3/((b^2 *x^2 + 2*a*b*x + a^2)^(3/2)*a^2) + 1/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4) + 1/2/(a^3*b^2*(x + a/b)^2) + 1/4/(a*b^4*(x + a/b)^4)
Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\log \left ({\left | b x + a \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {12 \, a b^{3} x^{3} + 42 \, a^{2} b^{2} x^{2} + 52 \, a^{3} b x + 25 \, a^{4}}{12 \, {\left (b x + a\right )}^{4} a^{5} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
Output:
-log(abs(b*x + a))/(a^5*sgn(b*x + a)) + log(abs(x))/(a^5*sgn(b*x + a)) + 1 /12*(12*a*b^3*x^3 + 42*a^2*b^2*x^2 + 52*a^3*b*x + 25*a^4)/((b*x + a)^4*a^5 *sgn(b*x + a))
Timed out. \[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:
int(1/(x*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)
Output:
int(1/(x*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)
Time = 0.38 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-12 \,\mathrm {log}\left (b x +a \right ) a^{4}-48 \,\mathrm {log}\left (b x +a \right ) a^{3} b x -72 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} x^{2}-48 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}-12 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}+12 \,\mathrm {log}\left (x \right ) a^{4}+48 \,\mathrm {log}\left (x \right ) a^{3} b x +72 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{2}+48 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{3}+12 \,\mathrm {log}\left (x \right ) b^{4} x^{4}+22 a^{4}+40 a^{3} b x +24 a^{2} b^{2} x^{2}-3 b^{4} x^{4}}{12 a^{5} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
Output:
( - 12*log(a + b*x)*a**4 - 48*log(a + b*x)*a**3*b*x - 72*log(a + b*x)*a**2 *b**2*x**2 - 48*log(a + b*x)*a*b**3*x**3 - 12*log(a + b*x)*b**4*x**4 + 12* log(x)*a**4 + 48*log(x)*a**3*b*x + 72*log(x)*a**2*b**2*x**2 + 48*log(x)*a* b**3*x**3 + 12*log(x)*b**4*x**4 + 22*a**4 + 40*a**3*b*x + 24*a**2*b**2*x** 2 - 3*b**4*x**4)/(12*a**5*(a**4 + 4*a**3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3 *x**3 + b**4*x**4))