Integrand size = 22, antiderivative size = 86 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=-\frac {1}{8 a b (a+b x)^4}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{16 a^3 b (a+b x)^2}-\frac {1}{16 a^4 b (a+b x)}+\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{16 a^5 b} \] Output:
-1/8/a/b/(b*x+a)^4-1/12/a^2/b/(b*x+a)^3-1/16/a^3/b/(b*x+a)^2-1/16/a^4/b/(b *x+a)+1/16*arctanh(b*x/a)/a^5/b
Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=\frac {-2 a \left (16 a^3+19 a^2 b x+12 a b^2 x^2+3 b^3 x^3\right )-3 (a+b x)^4 \log (a-b x)+3 (a+b x)^4 \log (a+b x)}{96 a^5 b (a+b x)^4} \] Input:
Integrate[1/((a + b*x)^4*(a^2 - b^2*x^2)),x]
Output:
(-2*a*(16*a^3 + 19*a^2*b*x + 12*a*b^2*x^2 + 3*b^3*x^3) - 3*(a + b*x)^4*Log [a - b*x] + 3*(a + b*x)^4*Log[a + b*x])/(96*a^5*b*(a + b*x)^4)
Time = 0.37 (sec) , antiderivative size = 86, 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.
\(\displaystyle \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {1}{(a-b x) (a+b x)^5}dx\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \int \left (\frac {1}{16 a^4 (a+b x)^2}+\frac {1}{8 a^3 (a+b x)^3}+\frac {1}{4 a^2 (a+b x)^4}+\frac {1}{16 a^4 \left (a^2-b^2 x^2\right )}+\frac {1}{2 a (a+b x)^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {b x}{a}\right )}{16 a^5 b}-\frac {1}{16 a^4 b (a+b x)}-\frac {1}{16 a^3 b (a+b x)^2}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{8 a b (a+b x)^4}\) |
Input:
Int[1/((a + b*x)^4*(a^2 - b^2*x^2)),x]
Output:
-1/8*1/(a*b*(a + b*x)^4) - 1/(12*a^2*b*(a + b*x)^3) - 1/(16*a^3*b*(a + b*x )^2) - 1/(16*a^4*b*(a + b*x)) + ArcTanh[(b*x)/a]/(16*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 = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86
method | result | size |
norman | \(\frac {-\frac {1}{3 b a}-\frac {b \,x^{2}}{4 a^{3}}-\frac {b^{2} x^{3}}{16 a^{4}}-\frac {19 x}{48 a^{2}}}{\left (b x +a \right )^{4}}-\frac {\ln \left (-b x +a \right )}{32 a^{5} b}+\frac {\ln \left (b x +a \right )}{32 a^{5} b}\) | \(74\) |
risch | \(\frac {-\frac {1}{3 b a}-\frac {b \,x^{2}}{4 a^{3}}-\frac {b^{2} x^{3}}{16 a^{4}}-\frac {19 x}{48 a^{2}}}{\left (b x +a \right )^{4}}-\frac {\ln \left (-b x +a \right )}{32 a^{5} b}+\frac {\ln \left (b x +a \right )}{32 a^{5} b}\) | \(74\) |
default | \(\frac {\ln \left (b x +a \right )}{32 a^{5} b}-\frac {1}{16 a^{4} b \left (b x +a \right )}-\frac {1}{16 a^{3} b \left (b x +a \right )^{2}}-\frac {1}{12 a^{2} b \left (b x +a \right )^{3}}-\frac {1}{8 a b \left (b x +a \right )^{4}}-\frac {\ln \left (-b x +a \right )}{32 a^{5} b}\) | \(91\) |
parallelrisch | \(-\frac {3 \ln \left (b x -a \right ) x^{4} b^{7}-3 \ln \left (b x +a \right ) x^{4} b^{7}+12 \ln \left (b x -a \right ) x^{3} a \,b^{6}-12 \ln \left (b x +a \right ) x^{3} a \,b^{6}+18 \ln \left (b x -a \right ) x^{2} a^{2} b^{5}-18 \ln \left (b x +a \right ) x^{2} a^{2} b^{5}+6 x^{3} a \,b^{6}+12 \ln \left (b x -a \right ) x \,a^{3} b^{4}-12 \ln \left (b x +a \right ) x \,a^{3} b^{4}+24 x^{2} a^{2} b^{5}+3 \ln \left (b x -a \right ) a^{4} b^{3}-3 \ln \left (b x +a \right ) a^{4} b^{3}+38 x \,a^{3} b^{4}+32 a^{4} b^{3}}{96 a^{5} b^{4} \left (b x +a \right )^{4}}\) | \(214\) |
Input:
int(1/(b*x+a)^4/(-b^2*x^2+a^2),x,method=_RETURNVERBOSE)
Output:
(-1/3/b/a-1/4*b/a^3*x^2-1/16*b^2/a^4*x^3-19/48/a^2*x)/(b*x+a)^4-1/32/a^5/b *ln(-b*x+a)+1/32/a^5/b*ln(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (76) = 152\).
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.07 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=-\frac {6 \, a b^{3} x^{3} + 24 \, a^{2} b^{2} x^{2} + 38 \, a^{3} b x + 32 \, a^{4} - 3 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right ) + 3 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x - a\right )}{96 \, {\left (a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{3} + 6 \, a^{7} b^{3} x^{2} + 4 \, a^{8} b^{2} x + a^{9} b\right )}} \] Input:
integrate(1/(b*x+a)^4/(-b^2*x^2+a^2),x, algorithm="fricas")
Output:
-1/96*(6*a*b^3*x^3 + 24*a^2*b^2*x^2 + 38*a^3*b*x + 32*a^4 - 3*(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) + 3*(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))/(a^5*b^5*x^4 + 4*a^6*b^4*x^3 + 6*a^7*b^3*x^2 + 4*a^8*b^2*x + a^9*b)
Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=- \frac {16 a^{3} + 19 a^{2} b x + 12 a b^{2} x^{2} + 3 b^{3} x^{3}}{48 a^{8} b + 192 a^{7} b^{2} x + 288 a^{6} b^{3} x^{2} + 192 a^{5} b^{4} x^{3} + 48 a^{4} b^{5} x^{4}} - \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{32} - \frac {\log {\left (\frac {a}{b} + x \right )}}{32}}{a^{5} b} \] Input:
integrate(1/(b*x+a)**4/(-b**2*x**2+a**2),x)
Output:
-(16*a**3 + 19*a**2*b*x + 12*a*b**2*x**2 + 3*b**3*x**3)/(48*a**8*b + 192*a **7*b**2*x + 288*a**6*b**3*x**2 + 192*a**5*b**4*x**3 + 48*a**4*b**5*x**4) - (log(-a/b + x)/32 - log(a/b + x)/32)/(a**5*b)
Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=-\frac {3 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 19 \, a^{2} b x + 16 \, a^{3}}{48 \, {\left (a^{4} b^{5} x^{4} + 4 \, a^{5} b^{4} x^{3} + 6 \, a^{6} b^{3} x^{2} + 4 \, a^{7} b^{2} x + a^{8} b\right )}} + \frac {\log \left (b x + a\right )}{32 \, a^{5} b} - \frac {\log \left (b x - a\right )}{32 \, a^{5} b} \] Input:
integrate(1/(b*x+a)^4/(-b^2*x^2+a^2),x, algorithm="maxima")
Output:
-1/48*(3*b^3*x^3 + 12*a*b^2*x^2 + 19*a^2*b*x + 16*a^3)/(a^4*b^5*x^4 + 4*a^ 5*b^4*x^3 + 6*a^6*b^3*x^2 + 4*a^7*b^2*x + a^8*b) + 1/32*log(b*x + a)/(a^5* b) - 1/32*log(b*x - a)/(a^5*b)
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{32 \, a^{5} b} - \frac {\log \left ({\left | b x - a \right |}\right )}{32 \, a^{5} b} - \frac {3 \, a b^{3} x^{3} + 12 \, a^{2} b^{2} x^{2} + 19 \, a^{3} b x + 16 \, a^{4}}{48 \, {\left (b x + a\right )}^{4} a^{5} b} \] Input:
integrate(1/(b*x+a)^4/(-b^2*x^2+a^2),x, algorithm="giac")
Output:
1/32*log(abs(b*x + a))/(a^5*b) - 1/32*log(abs(b*x - a))/(a^5*b) - 1/48*(3* a*b^3*x^3 + 12*a^2*b^2*x^2 + 19*a^3*b*x + 16*a^4)/((b*x + a)^4*a^5*b)
Time = 6.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{16\,a^5\,b}-\frac {\frac {19\,x}{48\,a^2}+\frac {1}{3\,a\,b}+\frac {b\,x^2}{4\,a^3}+\frac {b^2\,x^3}{16\,a^4}}{a^4+4\,a^3\,b\,x+6\,a^2\,b^2\,x^2+4\,a\,b^3\,x^3+b^4\,x^4} \] Input:
int(1/((a^2 - b^2*x^2)*(a + b*x)^4),x)
Output:
atanh((b*x)/a)/(16*a^5*b) - ((19*x)/(48*a^2) + 1/(3*a*b) + (b*x^2)/(4*a^3) + (b^2*x^3)/(16*a^4))/(a^4 + b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^ 3*b*x)
Time = 0.25 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.62 \[ \int \frac {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx=\frac {-6 \,\mathrm {log}\left (-b x +a \right ) a^{4}-24 \,\mathrm {log}\left (-b x +a \right ) a^{3} b x -36 \,\mathrm {log}\left (-b x +a \right ) a^{2} b^{2} x^{2}-24 \,\mathrm {log}\left (-b x +a \right ) a \,b^{3} x^{3}-6 \,\mathrm {log}\left (-b x +a \right ) b^{4} x^{4}+6 \,\mathrm {log}\left (b x +a \right ) a^{4}+24 \,\mathrm {log}\left (b x +a \right ) a^{3} b x +36 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} x^{2}+24 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}+6 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}-61 a^{4}-64 a^{3} b x -30 a^{2} b^{2} x^{2}+3 b^{4} x^{4}}{192 a^{5} b \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int(1/(b*x+a)^4/(-b^2*x^2+a^2),x)
Output:
( - 6*log(a - b*x)*a**4 - 24*log(a - b*x)*a**3*b*x - 36*log(a - b*x)*a**2* b**2*x**2 - 24*log(a - b*x)*a*b**3*x**3 - 6*log(a - b*x)*b**4*x**4 + 6*log (a + b*x)*a**4 + 24*log(a + b*x)*a**3*b*x + 36*log(a + b*x)*a**2*b**2*x**2 + 24*log(a + b*x)*a*b**3*x**3 + 6*log(a + b*x)*b**4*x**4 - 61*a**4 - 64*a **3*b*x - 30*a**2*b**2*x**2 + 3*b**4*x**4)/(192*a**5*b*(a**4 + 4*a**3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3*x**3 + b**4*x**4))