Integrand size = 22, antiderivative size = 105 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=\frac {1}{32 a^4 b (a-b x)^2}+\frac {1}{8 a^5 b (a-b x)}-\frac {1}{24 a^3 b (a+b x)^3}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {3}{16 a^5 b (a+b x)}+\frac {5 \text {arctanh}\left (\frac {b x}{a}\right )}{16 a^6 b} \] Output:
1/32/a^4/b/(-b*x+a)^2+1/8/a^5/b/(-b*x+a)-1/24/a^3/b/(b*x+a)^3-3/32/a^4/b/( b*x+a)^2-3/16/a^5/b/(b*x+a)+5/16*arctanh(b*x/a)/a^6/b
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=\frac {-\frac {2 a \left (8 a^4-25 a^3 b x-25 a^2 b^2 x^2+15 a b^3 x^3+15 b^4 x^4\right )}{(a-b x)^2 (a+b x)^3}-15 \log (a-b x)+15 \log (a+b x)}{96 a^6 b} \] Input:
Integrate[1/((a + b*x)*(a^2 - b^2*x^2)^3),x]
Output:
((-2*a*(8*a^4 - 25*a^3*b*x - 25*a^2*b^2*x^2 + 15*a*b^3*x^3 + 15*b^4*x^4))/ ((a - b*x)^2*(a + b*x)^3) - 15*Log[a - b*x] + 15*Log[a + b*x])/(96*a^6*b)
Time = 0.44 (sec) , antiderivative size = 105, 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.
\(\displaystyle \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {1}{(a-b x)^3 (a+b x)^4}dx\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \int \left (\frac {1}{8 a^5 (a-b x)^2}+\frac {3}{16 a^5 (a+b x)^2}+\frac {1}{16 a^4 (a-b x)^3}+\frac {3}{16 a^4 (a+b x)^3}+\frac {1}{8 a^3 (a+b x)^4}+\frac {5}{16 a^5 \left (a^2-b^2 x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 \text {arctanh}\left (\frac {b x}{a}\right )}{16 a^6 b}+\frac {1}{8 a^5 b (a-b x)}-\frac {3}{16 a^5 b (a+b x)}+\frac {1}{32 a^4 b (a-b x)^2}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {1}{24 a^3 b (a+b x)^3}\) |
Input:
Int[1/((a + b*x)*(a^2 - b^2*x^2)^3),x]
Output:
1/(32*a^4*b*(a - b*x)^2) + 1/(8*a^5*b*(a - b*x)) - 1/(24*a^3*b*(a + b*x)^3 ) - 3/(32*a^4*b*(a + b*x)^2) - 3/(16*a^5*b*(a + b*x)) + (5*ArcTanh[(b*x)/a ])/(16*a^6*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 = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89
method | result | size |
norman | \(\frac {\frac {7}{48 b a}-\frac {15 b^{2} x^{3}}{16 a^{4}}+\frac {5 b^{4} x^{5}}{16 a^{6}}+\frac {5 x}{6 a^{2}}-\frac {5 b \,x^{2}}{48 a^{3}}}{\left (b x +a \right )^{3} \left (-b x +a \right )^{2}}-\frac {5 \ln \left (-b x +a \right )}{32 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{32 a^{6} b}\) | \(93\) |
risch | \(\frac {-\frac {5 b^{3} x^{4}}{16 a^{5}}-\frac {5 b^{2} x^{3}}{16 a^{4}}+\frac {25 b \,x^{2}}{48 a^{3}}+\frac {25 x}{48 a^{2}}-\frac {1}{6 b a}}{\left (b x +a \right ) \left (-b^{2} x^{2}+a^{2}\right )^{2}}-\frac {5 \ln \left (-b x +a \right )}{32 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{32 a^{6} b}\) | \(99\) |
default | \(\frac {5 \ln \left (b x +a \right )}{32 a^{6} b}-\frac {3}{16 a^{5} b \left (b x +a \right )}-\frac {3}{32 a^{4} b \left (b x +a \right )^{2}}-\frac {1}{24 a^{3} b \left (b x +a \right )^{3}}-\frac {5 \ln \left (-b x +a \right )}{32 a^{6} b}+\frac {1}{8 a^{5} b \left (-b x +a \right )}+\frac {1}{32 a^{4} b \left (-b x +a \right )^{2}}\) | \(108\) |
parallelrisch | \(-\frac {15 \ln \left (b x -a \right ) x^{5} b^{8}-15 \ln \left (b x +a \right ) x^{5} b^{8}+15 \ln \left (b x -a \right ) x^{4} a \,b^{7}-15 \ln \left (b x +a \right ) x^{4} a \,b^{7}-30 x^{5} b^{8}-30 \ln \left (b x -a \right ) x^{3} a^{2} b^{6}+30 \ln \left (b x +a \right ) x^{3} a^{2} b^{6}-30 \ln \left (b x -a \right ) x^{2} a^{3} b^{5}+30 \ln \left (b x +a \right ) x^{2} a^{3} b^{5}+90 x^{3} a^{2} b^{6}+15 \ln \left (b x -a \right ) x \,a^{4} b^{4}-15 \ln \left (b x +a \right ) x \,a^{4} b^{4}+10 x^{2} a^{3} b^{5}+15 \ln \left (b x -a \right ) a^{5} b^{3}-15 \ln \left (b x +a \right ) a^{5} b^{3}-80 x \,a^{4} b^{4}-14 a^{5} b^{3}}{96 b^{4} a^{6} \left (b x +a \right ) \left (b^{2} x^{2}-a^{2}\right )^{2}}\) | \(275\) |
Input:
int(1/(b*x+a)/(-b^2*x^2+a^2)^3,x,method=_RETURNVERBOSE)
Output:
(7/48/b/a-15/16*b^2/a^4*x^3+5/16*b^4/a^6*x^5+5/6/a^2*x-5/48*b/a^3*x^2)/(b* x+a)^3/(-b*x+a)^2-5/32/a^6/b*ln(-b*x+a)+5/32/a^6/b*ln(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (95) = 190\).
Time = 0.07 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.06 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=-\frac {30 \, a b^{4} x^{4} + 30 \, a^{2} b^{3} x^{3} - 50 \, a^{3} b^{2} x^{2} - 50 \, a^{4} b x + 16 \, a^{5} - 15 \, {\left (b^{5} x^{5} + a b^{4} x^{4} - 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} + a^{4} b x + a^{5}\right )} \log \left (b x + a\right ) + 15 \, {\left (b^{5} x^{5} + a b^{4} x^{4} - 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} + a^{4} b x + a^{5}\right )} \log \left (b x - a\right )}{96 \, {\left (a^{6} b^{6} x^{5} + a^{7} b^{5} x^{4} - 2 \, a^{8} b^{4} x^{3} - 2 \, a^{9} b^{3} x^{2} + a^{10} b^{2} x + a^{11} b\right )}} \] Input:
integrate(1/(b*x+a)/(-b^2*x^2+a^2)^3,x, algorithm="fricas")
Output:
-1/96*(30*a*b^4*x^4 + 30*a^2*b^3*x^3 - 50*a^3*b^2*x^2 - 50*a^4*b*x + 16*a^ 5 - 15*(b^5*x^5 + a*b^4*x^4 - 2*a^2*b^3*x^3 - 2*a^3*b^2*x^2 + a^4*b*x + a^ 5)*log(b*x + a) + 15*(b^5*x^5 + a*b^4*x^4 - 2*a^2*b^3*x^3 - 2*a^3*b^2*x^2 + a^4*b*x + a^5)*log(b*x - a))/(a^6*b^6*x^5 + a^7*b^5*x^4 - 2*a^8*b^4*x^3 - 2*a^9*b^3*x^2 + a^10*b^2*x + a^11*b)
Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=- \frac {8 a^{4} - 25 a^{3} b x - 25 a^{2} b^{2} x^{2} + 15 a b^{3} x^{3} + 15 b^{4} x^{4}}{48 a^{10} b + 48 a^{9} b^{2} x - 96 a^{8} b^{3} x^{2} - 96 a^{7} b^{4} x^{3} + 48 a^{6} b^{5} x^{4} + 48 a^{5} b^{6} x^{5}} - \frac {\frac {5 \log {\left (- \frac {a}{b} + x \right )}}{32} - \frac {5 \log {\left (\frac {a}{b} + x \right )}}{32}}{a^{6} b} \] Input:
integrate(1/(b*x+a)/(-b**2*x**2+a**2)**3,x)
Output:
-(8*a**4 - 25*a**3*b*x - 25*a**2*b**2*x**2 + 15*a*b**3*x**3 + 15*b**4*x**4 )/(48*a**10*b + 48*a**9*b**2*x - 96*a**8*b**3*x**2 - 96*a**7*b**4*x**3 + 4 8*a**6*b**5*x**4 + 48*a**5*b**6*x**5) - (5*log(-a/b + x)/32 - 5*log(a/b + x)/32)/(a**6*b)
Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=-\frac {15 \, b^{4} x^{4} + 15 \, a b^{3} x^{3} - 25 \, a^{2} b^{2} x^{2} - 25 \, a^{3} b x + 8 \, a^{4}}{48 \, {\left (a^{5} b^{6} x^{5} + a^{6} b^{5} x^{4} - 2 \, a^{7} b^{4} x^{3} - 2 \, a^{8} b^{3} x^{2} + a^{9} b^{2} x + a^{10} b\right )}} + \frac {5 \, \log \left (b x + a\right )}{32 \, a^{6} b} - \frac {5 \, \log \left (b x - a\right )}{32 \, a^{6} b} \] Input:
integrate(1/(b*x+a)/(-b^2*x^2+a^2)^3,x, algorithm="maxima")
Output:
-1/48*(15*b^4*x^4 + 15*a*b^3*x^3 - 25*a^2*b^2*x^2 - 25*a^3*b*x + 8*a^4)/(a ^5*b^6*x^5 + a^6*b^5*x^4 - 2*a^7*b^4*x^3 - 2*a^8*b^3*x^2 + a^9*b^2*x + a^1 0*b) + 5/32*log(b*x + a)/(a^6*b) - 5/32*log(b*x - a)/(a^6*b)
Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=\frac {5 \, \log \left ({\left | b x + a \right |}\right )}{32 \, a^{6} b} - \frac {5 \, \log \left ({\left | b x - a \right |}\right )}{32 \, a^{6} b} - \frac {15 \, a b^{4} x^{4} + 15 \, a^{2} b^{3} x^{3} - 25 \, a^{3} b^{2} x^{2} - 25 \, a^{4} b x + 8 \, a^{5}}{48 \, {\left (b x + a\right )}^{3} {\left (b x - a\right )}^{2} a^{6} b} \] Input:
integrate(1/(b*x+a)/(-b^2*x^2+a^2)^3,x, algorithm="giac")
Output:
5/32*log(abs(b*x + a))/(a^6*b) - 5/32*log(abs(b*x - a))/(a^6*b) - 1/48*(15 *a*b^4*x^4 + 15*a^2*b^3*x^3 - 25*a^3*b^2*x^2 - 25*a^4*b*x + 8*a^5)/((b*x + a)^3*(b*x - a)^2*a^6*b)
Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=\frac {5\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{16\,a^6\,b}-\frac {\frac {1}{6\,a\,b}-\frac {25\,x}{48\,a^2}-\frac {25\,b\,x^2}{48\,a^3}+\frac {5\,b^2\,x^3}{16\,a^4}+\frac {5\,b^3\,x^4}{16\,a^5}}{a^5+a^4\,b\,x-2\,a^3\,b^2\,x^2-2\,a^2\,b^3\,x^3+a\,b^4\,x^4+b^5\,x^5} \] Input:
int(1/((a^2 - b^2*x^2)^3*(a + b*x)),x)
Output:
(5*atanh((b*x)/a))/(16*a^6*b) - (1/(6*a*b) - (25*x)/(48*a^2) - (25*b*x^2)/ (48*a^3) + (5*b^2*x^3)/(16*a^4) + (5*b^3*x^4)/(16*a^5))/(a^5 + b^5*x^5 + a *b^4*x^4 - 2*a^3*b^2*x^2 - 2*a^2*b^3*x^3 + a^4*b*x)
Time = 0.19 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.67 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )^3} \, dx=\frac {-15 \,\mathrm {log}\left (-b x +a \right ) a^{5}-15 \,\mathrm {log}\left (-b x +a \right ) a^{4} b x +30 \,\mathrm {log}\left (-b x +a \right ) a^{3} b^{2} x^{2}+30 \,\mathrm {log}\left (-b x +a \right ) a^{2} b^{3} x^{3}-15 \,\mathrm {log}\left (-b x +a \right ) a \,b^{4} x^{4}-15 \,\mathrm {log}\left (-b x +a \right ) b^{5} x^{5}+15 \,\mathrm {log}\left (b x +a \right ) a^{5}+15 \,\mathrm {log}\left (b x +a \right ) a^{4} b x -30 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{2} x^{2}-30 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{3} x^{3}+15 \,\mathrm {log}\left (b x +a \right ) a \,b^{4} x^{4}+15 \,\mathrm {log}\left (b x +a \right ) b^{5} x^{5}+14 a^{5}+80 a^{4} b x -10 a^{3} b^{2} x^{2}-90 a^{2} b^{3} x^{3}+30 b^{5} x^{5}}{96 a^{6} b \left (b^{5} x^{5}+a \,b^{4} x^{4}-2 a^{2} b^{3} x^{3}-2 a^{3} b^{2} x^{2}+a^{4} b x +a^{5}\right )} \] Input:
int(1/(b*x+a)/(-b^2*x^2+a^2)^3,x)
Output:
( - 15*log(a - b*x)*a**5 - 15*log(a - b*x)*a**4*b*x + 30*log(a - b*x)*a**3 *b**2*x**2 + 30*log(a - b*x)*a**2*b**3*x**3 - 15*log(a - b*x)*a*b**4*x**4 - 15*log(a - b*x)*b**5*x**5 + 15*log(a + b*x)*a**5 + 15*log(a + b*x)*a**4* b*x - 30*log(a + b*x)*a**3*b**2*x**2 - 30*log(a + b*x)*a**2*b**3*x**3 + 15 *log(a + b*x)*a*b**4*x**4 + 15*log(a + b*x)*b**5*x**5 + 14*a**5 + 80*a**4* b*x - 10*a**3*b**2*x**2 - 90*a**2*b**3*x**3 + 30*b**5*x**5)/(96*a**6*b*(a* *5 + a**4*b*x - 2*a**3*b**2*x**2 - 2*a**2*b**3*x**3 + a*b**4*x**4 + b**5*x **5))