Integrand size = 22, antiderivative size = 87 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {1}{16 a^4 b (a-b x)}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{8 a^3 b (a+b x)^2}-\frac {3}{16 a^4 b (a+b x)}+\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{4 a^5 b} \] Output:
1/16/a^4/b/(-b*x+a)-1/12/a^2/b/(b*x+a)^3-1/8/a^3/b/(b*x+a)^2-3/16/a^4/b/(b *x+a)+1/4*arctanh(b*x/a)/a^5/b
Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {\frac {2 a \left (-4 a^3+a^2 b x+6 a b^2 x^2+3 b^3 x^3\right )}{(a-b x) (a+b x)^3}-3 \log (a-b x)+3 \log (a+b x)}{24 a^5 b} \] Input:
Integrate[1/((a + b*x)^2*(a^2 - b^2*x^2)^2),x]
Output:
((2*a*(-4*a^3 + a^2*b*x + 6*a*b^2*x^2 + 3*b^3*x^3))/((a - b*x)*(a + b*x)^3 ) - 3*Log[a - b*x] + 3*Log[a + b*x])/(24*a^5*b)
Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {456, 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.
\(\displaystyle \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {1}{(a-b x)^2 (a+b x)^4}dx\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \int \left (\frac {1}{16 a^4 (a-b x)^2}+\frac {3}{16 a^4 (a+b x)^2}+\frac {1}{4 a^3 (a+b x)^3}+\frac {1}{4 a^2 (a+b x)^4}+\frac {1}{4 a^4 \left (a^2-b^2 x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {b x}{a}\right )}{4 a^5 b}+\frac {1}{16 a^4 b (a-b x)}-\frac {3}{16 a^4 b (a+b x)}-\frac {1}{8 a^3 b (a+b x)^2}-\frac {1}{12 a^2 b (a+b x)^3}\) |
Input:
Int[1/((a + b*x)^2*(a^2 - b^2*x^2)^2),x]
Output:
1/(16*a^4*b*(a - b*x)) - 1/(12*a^2*b*(a + b*x)^3) - 1/(8*a^3*b*(a + b*x)^2 ) - 3/(16*a^4*b*(a + b*x)) + ArcTanh[(b*x)/a]/(4*a^5*b)
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !Integ erQ[n]))
Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98
method | result | size |
norman | \(\frac {\frac {b \,x^{2}}{2 a^{3}}+\frac {3 x}{4 a^{2}}-\frac {5 b^{2} x^{3}}{12 a^{4}}-\frac {b^{3} x^{4}}{3 a^{5}}}{\left (b x +a \right )^{3} \left (-b x +a \right )}-\frac {\ln \left (-b x +a \right )}{8 a^{5} b}+\frac {\ln \left (b x +a \right )}{8 a^{5} b}\) | \(85\) |
risch | \(\frac {\frac {b^{2} x^{3}}{4 a^{4}}+\frac {b \,x^{2}}{2 a^{3}}+\frac {x}{12 a^{2}}-\frac {1}{3 b a}}{\left (b x +a \right )^{2} \left (-b^{2} x^{2}+a^{2}\right )}-\frac {\ln \left (-b x +a \right )}{8 a^{5} b}+\frac {\ln \left (b x +a \right )}{8 a^{5} b}\) | \(88\) |
default | \(\frac {\ln \left (b x +a \right )}{8 a^{5} b}-\frac {3}{16 a^{4} b \left (b x +a \right )}-\frac {1}{8 a^{3} b \left (b x +a \right )^{2}}-\frac {1}{12 a^{2} b \left (b x +a \right )^{3}}-\frac {\ln \left (-b x +a \right )}{8 a^{5} b}+\frac {1}{16 a^{4} b \left (-b x +a \right )}\) | \(92\) |
parallelrisch | \(-\frac {3 \ln \left (b x -a \right ) x^{4} b^{7}-3 \ln \left (b x +a \right ) x^{4} b^{7}+6 \ln \left (b x -a \right ) x^{3} a \,b^{6}-6 \ln \left (b x +a \right ) x^{3} a \,b^{6}+6 x^{3} a \,b^{6}-6 \ln \left (b x -a \right ) x \,a^{3} b^{4}+6 \ln \left (b x +a \right ) x \,a^{3} b^{4}+12 x^{2} a^{2} b^{5}-3 \ln \left (b x -a \right ) a^{4} b^{3}+3 \ln \left (b x +a \right ) a^{4} b^{3}+2 x \,a^{3} b^{4}-8 a^{4} b^{3}}{24 a^{5} b^{4} \left (b x +a \right )^{2} \left (b^{2} x^{2}-a^{2}\right )}\) | \(193\) |
Input:
int(1/(b*x+a)^2/(-b^2*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(1/2*b/a^3*x^2+3/4/a^2*x-5/12*b^2/a^4*x^3-1/3*b^3/a^5*x^4)/(b*x+a)^3/(-b*x +a)-1/8/a^5/b*ln(-b*x+a)+1/8/a^5/b*ln(b*x+a)
Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.72 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=-\frac {6 \, a b^{3} x^{3} + 12 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x - 8 \, a^{4} - 3 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} - 2 \, a^{3} b x - a^{4}\right )} \log \left (b x + a\right ) + 3 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} - 2 \, a^{3} b x - a^{4}\right )} \log \left (b x - a\right )}{24 \, {\left (a^{5} b^{5} x^{4} + 2 \, a^{6} b^{4} x^{3} - 2 \, a^{8} b^{2} x - a^{9} b\right )}} \] Input:
integrate(1/(b*x+a)^2/(-b^2*x^2+a^2)^2,x, algorithm="fricas")
Output:
-1/24*(6*a*b^3*x^3 + 12*a^2*b^2*x^2 + 2*a^3*b*x - 8*a^4 - 3*(b^4*x^4 + 2*a *b^3*x^3 - 2*a^3*b*x - a^4)*log(b*x + a) + 3*(b^4*x^4 + 2*a*b^3*x^3 - 2*a^ 3*b*x - a^4)*log(b*x - a))/(a^5*b^5*x^4 + 2*a^6*b^4*x^3 - 2*a^8*b^2*x - a^ 9*b)
Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {4 a^{3} - a^{2} b x - 6 a b^{2} x^{2} - 3 b^{3} x^{3}}{- 12 a^{8} b - 24 a^{7} b^{2} x + 24 a^{5} b^{4} x^{3} + 12 a^{4} b^{5} x^{4}} + \frac {- \frac {\log {\left (- \frac {a}{b} + x \right )}}{8} + \frac {\log {\left (\frac {a}{b} + x \right )}}{8}}{a^{5} b} \] Input:
integrate(1/(b*x+a)**2/(-b**2*x**2+a**2)**2,x)
Output:
(4*a**3 - a**2*b*x - 6*a*b**2*x**2 - 3*b**3*x**3)/(-12*a**8*b - 24*a**7*b* *2*x + 24*a**5*b**4*x**3 + 12*a**4*b**5*x**4) + (-log(-a/b + x)/8 + log(a/ b + x)/8)/(a**5*b)
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=-\frac {3 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + a^{2} b x - 4 \, a^{3}}{12 \, {\left (a^{4} b^{5} x^{4} + 2 \, a^{5} b^{4} x^{3} - 2 \, a^{7} b^{2} x - a^{8} b\right )}} + \frac {\log \left (b x + a\right )}{8 \, a^{5} b} - \frac {\log \left (b x - a\right )}{8 \, a^{5} b} \] Input:
integrate(1/(b*x+a)^2/(-b^2*x^2+a^2)^2,x, algorithm="maxima")
Output:
-1/12*(3*b^3*x^3 + 6*a*b^2*x^2 + a^2*b*x - 4*a^3)/(a^4*b^5*x^4 + 2*a^5*b^4 *x^3 - 2*a^7*b^2*x - a^8*b) + 1/8*log(b*x + a)/(a^5*b) - 1/8*log(b*x - a)/ (a^5*b)
Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=-\frac {\log \left ({\left | -\frac {2 \, a}{b x + a} + 1 \right |}\right )}{8 \, a^{5} b} + \frac {1}{32 \, a^{5} b {\left (\frac {2 \, a}{b x + a} - 1\right )}} - \frac {\frac {9 \, a^{2} b^{5}}{b x + a} + \frac {6 \, a^{3} b^{5}}{{\left (b x + a\right )}^{2}} + \frac {4 \, a^{4} b^{5}}{{\left (b x + a\right )}^{3}}}{48 \, a^{6} b^{6}} \] Input:
integrate(1/(b*x+a)^2/(-b^2*x^2+a^2)^2,x, algorithm="giac")
Output:
-1/8*log(abs(-2*a/(b*x + a) + 1))/(a^5*b) + 1/32/(a^5*b*(2*a/(b*x + a) - 1 )) - 1/48*(9*a^2*b^5/(b*x + a) + 6*a^3*b^5/(b*x + a)^2 + 4*a^4*b^5/(b*x + a)^3)/(a^6*b^6)
Time = 6.35 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {\frac {x}{12\,a^2}-\frac {1}{3\,a\,b}+\frac {b\,x^2}{2\,a^3}+\frac {b^2\,x^3}{4\,a^4}}{a^4+2\,a^3\,b\,x-2\,a\,b^3\,x^3-b^4\,x^4}+\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{4\,a^5\,b} \] Input:
int(1/((a^2 - b^2*x^2)^2*(a + b*x)^2),x)
Output:
(x/(12*a^2) - 1/(3*a*b) + (b*x^2)/(2*a^3) + (b^2*x^3)/(4*a^4))/(a^4 - b^4* x^4 - 2*a*b^3*x^3 + 2*a^3*b*x) + atanh((b*x)/a)/(4*a^5*b)
Time = 0.20 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.07 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {-3 \,\mathrm {log}\left (-b x +a \right ) a^{4}-6 \,\mathrm {log}\left (-b x +a \right ) a^{3} b x +6 \,\mathrm {log}\left (-b x +a \right ) a \,b^{3} x^{3}+3 \,\mathrm {log}\left (-b x +a \right ) b^{4} x^{4}+3 \,\mathrm {log}\left (b x +a \right ) a^{4}+6 \,\mathrm {log}\left (b x +a \right ) a^{3} b x -6 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}-3 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}-5 a^{4}+8 a^{3} b x +12 a^{2} b^{2} x^{2}-3 b^{4} x^{4}}{24 a^{5} b \left (-b^{4} x^{4}-2 a \,b^{3} x^{3}+2 a^{3} b x +a^{4}\right )} \] Input:
int(1/(b*x+a)^2/(-b^2*x^2+a^2)^2,x)
Output:
( - 3*log(a - b*x)*a**4 - 6*log(a - b*x)*a**3*b*x + 6*log(a - b*x)*a*b**3* x**3 + 3*log(a - b*x)*b**4*x**4 + 3*log(a + b*x)*a**4 + 6*log(a + b*x)*a** 3*b*x - 6*log(a + b*x)*a*b**3*x**3 - 3*log(a + b*x)*b**4*x**4 - 5*a**4 + 8 *a**3*b*x + 12*a**2*b**2*x**2 - 3*b**4*x**4)/(24*a**5*b*(a**4 + 2*a**3*b*x - 2*a*b**3*x**3 - b**4*x**4))