Integrand size = 22, antiderivative size = 104 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {1}{32 a^5 b (a-b x)}-\frac {1}{16 a^2 b (a+b x)^4}-\frac {1}{12 a^3 b (a+b x)^3}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {1}{8 a^5 b (a+b x)}+\frac {5 \text {arctanh}\left (\frac {b x}{a}\right )}{32 a^6 b} \] Output:
1/32/a^5/b/(-b*x+a)-1/16/a^2/b/(b*x+a)^4-1/12/a^3/b/(b*x+a)^3-3/32/a^4/b/( b*x+a)^2-1/8/a^5/b/(b*x+a)+5/32*arctanh(b*x/a)/a^6/b
Time = 0.02 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {-64 a^5-30 a^4 b x+70 a^3 b^2 x^2+90 a^2 b^3 x^3+30 a b^4 x^4-15 (a-b x) (a+b x)^4 \log (a-b x)+15 (a-b x) (a+b x)^4 \log (a+b x)}{192 a^6 b (a-b x) (a+b x)^4} \] Input:
Integrate[1/((a + b*x)^3*(a^2 - b^2*x^2)^2),x]
Output:
(-64*a^5 - 30*a^4*b*x + 70*a^3*b^2*x^2 + 90*a^2*b^3*x^3 + 30*a*b^4*x^4 - 1 5*(a - b*x)*(a + b*x)^4*Log[a - b*x] + 15*(a - b*x)*(a + b*x)^4*Log[a + b* x])/(192*a^6*b*(a - b*x)*(a + b*x)^4)
Time = 0.41 (sec) , antiderivative size = 104, 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.
\(\displaystyle \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {1}{(a-b x)^2 (a+b x)^5}dx\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \int \left (\frac {1}{32 a^5 (a-b x)^2}+\frac {1}{8 a^5 (a+b x)^2}+\frac {3}{16 a^4 (a+b x)^3}+\frac {1}{4 a^3 (a+b x)^4}+\frac {1}{4 a^2 (a+b x)^5}+\frac {5}{32 a^5 \left (a^2-b^2 x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 \text {arctanh}\left (\frac {b x}{a}\right )}{32 a^6 b}+\frac {1}{32 a^5 b (a-b x)}-\frac {1}{8 a^5 b (a+b x)}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {1}{12 a^3 b (a+b x)^3}-\frac {1}{16 a^2 b (a+b x)^4}\) |
Input:
Int[1/((a + b*x)^3*(a^2 - b^2*x^2)^2),x]
Output:
1/(32*a^5*b*(a - b*x)) - 1/(16*a^2*b*(a + b*x)^4) - 1/(12*a^3*b*(a + b*x)^ 3) - 3/(32*a^4*b*(a + b*x)^2) - 1/(8*a^5*b*(a + b*x)) + (5*ArcTanh[(b*x)/a ])/(32*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 = 96, normalized size of antiderivative = 0.92
method | result | size |
norman | \(\frac {\frac {27 x}{32 a^{2}}+\frac {33 b \,x^{2}}{32 a^{3}}-\frac {19 b^{2} x^{3}}{96 a^{4}}-\frac {27 b^{3} x^{4}}{32 a^{5}}-\frac {b^{4} x^{5}}{3 a^{6}}}{\left (b x +a \right )^{4} \left (-b x +a \right )}-\frac {5 \ln \left (-b x +a \right )}{64 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{64 a^{6} b}\) | \(96\) |
risch | \(\frac {\frac {5 b^{3} x^{4}}{32 a^{5}}+\frac {15 b^{2} x^{3}}{32 a^{4}}+\frac {35 b \,x^{2}}{96 a^{3}}-\frac {5 x}{32 a^{2}}-\frac {1}{3 b a}}{\left (b x +a \right )^{3} \left (-b^{2} x^{2}+a^{2}\right )}-\frac {5 \ln \left (-b x +a \right )}{64 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{64 a^{6} b}\) | \(99\) |
default | \(\frac {5 \ln \left (b x +a \right )}{64 a^{6} b}-\frac {1}{8 a^{5} b \left (b x +a \right )}-\frac {3}{32 a^{4} b \left (b x +a \right )^{2}}-\frac {1}{12 a^{3} b \left (b x +a \right )^{3}}-\frac {1}{16 a^{2} b \left (b x +a \right )^{4}}-\frac {5 \ln \left (-b x +a \right )}{64 a^{6} b}+\frac {1}{32 a^{5} b \left (-b x +a \right )}\) | \(107\) |
parallelrisch | \(-\frac {15 \ln \left (b x -a \right ) x^{5} b^{5}-15 \ln \left (b x +a \right ) x^{5} b^{5}+45 \ln \left (b x -a \right ) x^{4} a \,b^{4}-45 \ln \left (b x +a \right ) x^{4} a \,b^{4}-64 b^{5} x^{5}+30 \ln \left (b x -a \right ) x^{3} a^{2} b^{3}-30 \ln \left (b x +a \right ) x^{3} a^{2} b^{3}-162 a \,b^{4} x^{4}-30 \ln \left (b x -a \right ) x^{2} a^{3} b^{2}+30 \ln \left (b x +a \right ) x^{2} a^{3} b^{2}-38 a^{2} b^{3} x^{3}-45 \ln \left (b x -a \right ) x \,a^{4} b +45 \ln \left (b x +a \right ) x \,a^{4} b +198 a^{3} b^{2} x^{2}-15 a^{5} \ln \left (b x -a \right )+15 \ln \left (b x +a \right ) a^{5}+162 a^{4} b x}{192 a^{6} \left (b x +a \right )^{3} \left (b^{2} x^{2}-a^{2}\right ) b}\) | \(264\) |
Input:
int(1/(b*x+a)^3/(-b^2*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(27/32/a^2*x+33/32*b/a^3*x^2-19/96*b^2/a^4*x^3-27/32*b^3/a^5*x^4-1/3*b^4/a ^6*x^5)/(b*x+a)^4/(-b*x+a)-5/64/a^6/b*ln(-b*x+a)+5/64/a^6/b*ln(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (93) = 186\).
Time = 0.08 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.18 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=-\frac {30 \, a b^{4} x^{4} + 90 \, a^{2} b^{3} x^{3} + 70 \, a^{3} b^{2} x^{2} - 30 \, a^{4} b x - 64 \, a^{5} - 15 \, {\left (b^{5} x^{5} + 3 \, a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} - 3 \, a^{4} b x - a^{5}\right )} \log \left (b x + a\right ) + 15 \, {\left (b^{5} x^{5} + 3 \, a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} - 3 \, a^{4} b x - a^{5}\right )} \log \left (b x - a\right )}{192 \, {\left (a^{6} b^{6} x^{5} + 3 \, a^{7} b^{5} x^{4} + 2 \, a^{8} b^{4} x^{3} - 2 \, a^{9} b^{3} x^{2} - 3 \, a^{10} b^{2} x - a^{11} b\right )}} \] Input:
integrate(1/(b*x+a)^3/(-b^2*x^2+a^2)^2,x, algorithm="fricas")
Output:
-1/192*(30*a*b^4*x^4 + 90*a^2*b^3*x^3 + 70*a^3*b^2*x^2 - 30*a^4*b*x - 64*a ^5 - 15*(b^5*x^5 + 3*a*b^4*x^4 + 2*a^2*b^3*x^3 - 2*a^3*b^2*x^2 - 3*a^4*b*x - a^5)*log(b*x + a) + 15*(b^5*x^5 + 3*a*b^4*x^4 + 2*a^2*b^3*x^3 - 2*a^3*b ^2*x^2 - 3*a^4*b*x - a^5)*log(b*x - a))/(a^6*b^6*x^5 + 3*a^7*b^5*x^4 + 2*a ^8*b^4*x^3 - 2*a^9*b^3*x^2 - 3*a^10*b^2*x - a^11*b)
Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {32 a^{4} + 15 a^{3} b x - 35 a^{2} b^{2} x^{2} - 45 a b^{3} x^{3} - 15 b^{4} x^{4}}{- 96 a^{10} b - 288 a^{9} b^{2} x - 192 a^{8} b^{3} x^{2} + 192 a^{7} b^{4} x^{3} + 288 a^{6} b^{5} x^{4} + 96 a^{5} b^{6} x^{5}} + \frac {- \frac {5 \log {\left (- \frac {a}{b} + x \right )}}{64} + \frac {5 \log {\left (\frac {a}{b} + x \right )}}{64}}{a^{6} b} \] Input:
integrate(1/(b*x+a)**3/(-b**2*x**2+a**2)**2,x)
Output:
(32*a**4 + 15*a**3*b*x - 35*a**2*b**2*x**2 - 45*a*b**3*x**3 - 15*b**4*x**4 )/(-96*a**10*b - 288*a**9*b**2*x - 192*a**8*b**3*x**2 + 192*a**7*b**4*x**3 + 288*a**6*b**5*x**4 + 96*a**5*b**6*x**5) + (-5*log(-a/b + x)/64 + 5*log( a/b + x)/64)/(a**6*b)
Time = 0.04 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=-\frac {15 \, b^{4} x^{4} + 45 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 15 \, a^{3} b x - 32 \, a^{4}}{96 \, {\left (a^{5} b^{6} x^{5} + 3 \, a^{6} b^{5} x^{4} + 2 \, a^{7} b^{4} x^{3} - 2 \, a^{8} b^{3} x^{2} - 3 \, a^{9} b^{2} x - a^{10} b\right )}} + \frac {5 \, \log \left (b x + a\right )}{64 \, a^{6} b} - \frac {5 \, \log \left (b x - a\right )}{64 \, a^{6} b} \] Input:
integrate(1/(b*x+a)^3/(-b^2*x^2+a^2)^2,x, algorithm="maxima")
Output:
-1/96*(15*b^4*x^4 + 45*a*b^3*x^3 + 35*a^2*b^2*x^2 - 15*a^3*b*x - 32*a^4)/( a^5*b^6*x^5 + 3*a^6*b^5*x^4 + 2*a^7*b^4*x^3 - 2*a^8*b^3*x^2 - 3*a^9*b^2*x - a^10*b) + 5/64*log(b*x + a)/(a^6*b) - 5/64*log(b*x - a)/(a^6*b)
Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {5 \, \log \left ({\left | b x + a \right |}\right )}{64 \, a^{6} b} - \frac {5 \, \log \left ({\left | b x - a \right |}\right )}{64 \, a^{6} b} - \frac {15 \, a b^{4} x^{4} + 45 \, a^{2} b^{3} x^{3} + 35 \, a^{3} b^{2} x^{2} - 15 \, a^{4} b x - 32 \, a^{5}}{96 \, {\left (b x + a\right )}^{4} {\left (b x - a\right )} a^{6} b} \] Input:
integrate(1/(b*x+a)^3/(-b^2*x^2+a^2)^2,x, algorithm="giac")
Output:
5/64*log(abs(b*x + a))/(a^6*b) - 5/64*log(abs(b*x - a))/(a^6*b) - 1/96*(15 *a*b^4*x^4 + 45*a^2*b^3*x^3 + 35*a^3*b^2*x^2 - 15*a^4*b*x - 32*a^5)/((b*x + a)^4*(b*x - a)*a^6*b)
Time = 6.42 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {\frac {35\,b\,x^2}{96\,a^3}-\frac {1}{3\,a\,b}-\frac {5\,x}{32\,a^2}+\frac {15\,b^2\,x^3}{32\,a^4}+\frac {5\,b^3\,x^4}{32\,a^5}}{a^5+3\,a^4\,b\,x+2\,a^3\,b^2\,x^2-2\,a^2\,b^3\,x^3-3\,a\,b^4\,x^4-b^5\,x^5}+\frac {5\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{32\,a^6\,b} \] Input:
int(1/((a^2 - b^2*x^2)^2*(a + b*x)^3),x)
Output:
((35*b*x^2)/(96*a^3) - 1/(3*a*b) - (5*x)/(32*a^2) + (15*b^2*x^3)/(32*a^4) + (5*b^3*x^4)/(32*a^5))/(a^5 - b^5*x^5 - 3*a*b^4*x^4 + 2*a^3*b^2*x^2 - 2*a ^2*b^3*x^3 + 3*a^4*b*x) + (5*atanh((b*x)/a))/(32*a^6*b)
Time = 0.22 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.65 \[ \int \frac {1}{(a+b x)^3 \left (a^2-b^2 x^2\right )^2} \, dx=\frac {-15 \,\mathrm {log}\left (-b x +a \right ) a^{5}-45 \,\mathrm {log}\left (-b x +a \right ) a^{4} b x -30 \,\mathrm {log}\left (-b x +a \right ) a^{3} b^{2} x^{2}+30 \,\mathrm {log}\left (-b x +a \right ) a^{2} b^{3} x^{3}+45 \,\mathrm {log}\left (-b x +a \right ) a \,b^{4} x^{4}+15 \,\mathrm {log}\left (-b x +a \right ) b^{5} x^{5}+15 \,\mathrm {log}\left (b x +a \right ) a^{5}+45 \,\mathrm {log}\left (b x +a \right ) a^{4} b x +30 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{2} x^{2}-30 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{3} x^{3}-45 \,\mathrm {log}\left (b x +a \right ) a \,b^{4} x^{4}-15 \,\mathrm {log}\left (b x +a \right ) b^{5} x^{5}-54 a^{5}+90 a^{3} b^{2} x^{2}+70 a^{2} b^{3} x^{3}-10 b^{5} x^{5}}{192 a^{6} b \left (-b^{5} x^{5}-3 a \,b^{4} x^{4}-2 a^{2} b^{3} x^{3}+2 a^{3} b^{2} x^{2}+3 a^{4} b x +a^{5}\right )} \] Input:
int(1/(b*x+a)^3/(-b^2*x^2+a^2)^2,x)
Output:
( - 15*log(a - b*x)*a**5 - 45*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 + 45*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 + 45*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 - 45 *log(a + b*x)*a*b**4*x**4 - 15*log(a + b*x)*b**5*x**5 - 54*a**5 + 90*a**3* b**2*x**2 + 70*a**2*b**3*x**3 - 10*b**5*x**5)/(192*a**6*b*(a**5 + 3*a**4*b *x + 2*a**3*b**2*x**2 - 2*a**2*b**3*x**3 - 3*a*b**4*x**4 - b**5*x**5))