Integrand size = 22, antiderivative size = 73 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {x^3}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {3 x}{8 b^4 c^3 \left (a^2-b^2 x^2\right )}+\frac {3 \text {arctanh}\left (\frac {b x}{a}\right )}{8 a b^5 c^3} \] Output:
1/4*x^3/b^2/c^3/(-b^2*x^2+a^2)^2-3/8*x/b^4/c^3/(-b^2*x^2+a^2)+3/8*arctanh( b*x/a)/a/b^5/c^3
Time = 0.02 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {-6 a^3 b x+10 a b^3 x^3-3 \left (a^2-b^2 x^2\right )^2 \log (a-b x)+3 \left (a^2-b^2 x^2\right )^2 \log (a+b x)}{16 a b^5 c^3 (a-b x)^2 (a+b x)^2} \] Input:
Integrate[x^4/((a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
(-6*a^3*b*x + 10*a*b^3*x^3 - 3*(a^2 - b^2*x^2)^2*Log[a - b*x] + 3*(a^2 - b ^2*x^2)^2*Log[a + b*x])/(16*a*b^5*c^3*(a - b*x)^2*(a + b*x)^2)
Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {82, 252, 27, 252, 221}
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.
\(\displaystyle \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \int \frac {x^4}{\left (a^2 c-b^2 c x^2\right )^3}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {x^3}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {3 \int \frac {x^2}{c^2 \left (a^2-b^2 x^2\right )^2}dx}{4 b^2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {3 \int \frac {x^2}{\left (a^2-b^2 x^2\right )^2}dx}{4 b^2 c^3}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {x^3}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {3 \left (\frac {x}{2 b^2 \left (a^2-b^2 x^2\right )}-\frac {\int \frac {1}{a^2-b^2 x^2}dx}{2 b^2}\right )}{4 b^2 c^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {x^3}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {3 \left (\frac {x}{2 b^2 \left (a^2-b^2 x^2\right )}-\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{2 a b^3}\right )}{4 b^2 c^3}\) |
Input:
Int[x^4/((a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
x^3/(4*b^2*c^3*(a^2 - b^2*x^2)^2) - (3*(x/(2*b^2*(a^2 - b^2*x^2)) - ArcTan h[(b*x)/a]/(2*a*b^3)))/(4*b^2*c^3)
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_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01
method | result | size |
risch | \(\frac {\frac {5 x^{3}}{8 b^{2}}-\frac {3 a^{2} x}{8 b^{4}}}{\left (b x +a \right )^{2} c^{3} \left (-b x +a \right )^{2}}-\frac {3 \ln \left (-b x +a \right )}{16 a \,b^{5} c^{3}}+\frac {3 \ln \left (b x +a \right )}{16 a \,b^{5} c^{3}}\) | \(74\) |
norman | \(\frac {\frac {5 x^{3}}{8 b^{2} c}-\frac {3 a^{2} x}{8 b^{4} c}}{\left (b x +a \right )^{2} c^{2} \left (-b x +a \right )^{2}}-\frac {3 \ln \left (-b x +a \right )}{16 a \,b^{5} c^{3}}+\frac {3 \ln \left (b x +a \right )}{16 a \,b^{5} c^{3}}\) | \(80\) |
default | \(\frac {\frac {5}{16 b^{5} \left (b x +a \right )}+\frac {3 \ln \left (b x +a \right )}{16 a \,b^{5}}-\frac {a}{16 b^{5} \left (b x +a \right )^{2}}-\frac {5}{16 b^{5} \left (-b x +a \right )}-\frac {3 \ln \left (-b x +a \right )}{16 a \,b^{5}}+\frac {a}{16 b^{5} \left (-b x +a \right )^{2}}}{c^{3}}\) | \(87\) |
parallelrisch | \(\frac {-3 \ln \left (b x -a \right ) x^{4} b^{4}+3 \ln \left (b x +a \right ) x^{4} b^{4}+6 \ln \left (b x -a \right ) x^{2} a^{2} b^{2}-6 \ln \left (b x +a \right ) x^{2} a^{2} b^{2}+10 a \,x^{3} b^{3}-3 \ln \left (b x -a \right ) a^{4}+3 \ln \left (b x +a \right ) a^{4}-6 a^{3} b x}{16 b^{5} c^{3} \left (b x +a \right )^{2} \left (b x -a \right )^{2} a}\) | \(135\) |
Input:
int(x^4/(b*x+a)^3/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)
Output:
(5/8/b^2*x^3-3/8*a^2/b^4*x)/(b*x+a)^2/c^3/(-b*x+a)^2-3/16/a/b^5/c^3*ln(-b* x+a)+3/16/a/b^5/c^3*ln(b*x+a)
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.63 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {10 \, a b^{3} x^{3} - 6 \, a^{3} b x + 3 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \log \left (b x + a\right ) - 3 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \log \left (b x - a\right )}{16 \, {\left (a b^{9} c^{3} x^{4} - 2 \, a^{3} b^{7} c^{3} x^{2} + a^{5} b^{5} c^{3}\right )}} \] Input:
integrate(x^4/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="fricas")
Output:
1/16*(10*a*b^3*x^3 - 6*a^3*b*x + 3*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*log(b*x + a) - 3*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*log(b*x - a))/(a*b^9*c^3*x^4 - 2 *a^3*b^7*c^3*x^2 + a^5*b^5*c^3)
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.14 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=- \frac {3 a^{2} x - 5 b^{2} x^{3}}{8 a^{4} b^{4} c^{3} - 16 a^{2} b^{6} c^{3} x^{2} + 8 b^{8} c^{3} x^{4}} - \frac {\frac {3 \log {\left (- \frac {a}{b} + x \right )}}{16} - \frac {3 \log {\left (\frac {a}{b} + x \right )}}{16}}{a b^{5} c^{3}} \] Input:
integrate(x**4/(b*x+a)**3/(-b*c*x+a*c)**3,x)
Output:
-(3*a**2*x - 5*b**2*x**3)/(8*a**4*b**4*c**3 - 16*a**2*b**6*c**3*x**2 + 8*b **8*c**3*x**4) - (3*log(-a/b + x)/16 - 3*log(a/b + x)/16)/(a*b**5*c**3)
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {5 \, b^{2} x^{3} - 3 \, a^{2} x}{8 \, {\left (b^{8} c^{3} x^{4} - 2 \, a^{2} b^{6} c^{3} x^{2} + a^{4} b^{4} c^{3}\right )}} + \frac {3 \, \log \left (b x + a\right )}{16 \, a b^{5} c^{3}} - \frac {3 \, \log \left (b x - a\right )}{16 \, a b^{5} c^{3}} \] Input:
integrate(x^4/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="maxima")
Output:
1/8*(5*b^2*x^3 - 3*a^2*x)/(b^8*c^3*x^4 - 2*a^2*b^6*c^3*x^2 + a^4*b^4*c^3) + 3/16*log(b*x + a)/(a*b^5*c^3) - 3/16*log(b*x - a)/(a*b^5*c^3)
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {5 \, b^{2} x^{3} - 3 \, a^{2} x}{8 \, {\left (b^{2} x^{2} - a^{2}\right )}^{2} b^{4} c^{3}} + \frac {3 \, \log \left ({\left | b x + a \right |}\right )}{16 \, a b^{5} c^{3}} - \frac {3 \, \log \left ({\left | b x - a \right |}\right )}{16 \, a b^{5} c^{3}} \] Input:
integrate(x^4/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="giac")
Output:
1/8*(5*b^2*x^3 - 3*a^2*x)/((b^2*x^2 - a^2)^2*b^4*c^3) + 3/16*log(abs(b*x + a))/(a*b^5*c^3) - 3/16*log(abs(b*x - a))/(a*b^5*c^3)
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\frac {5\,x^3}{8\,b^2}-\frac {3\,a^2\,x}{8\,b^4}}{a^4\,c^3-2\,a^2\,b^2\,c^3\,x^2+b^4\,c^3\,x^4}+\frac {3\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{8\,a\,b^5\,c^3} \] Input:
int(x^4/((a*c - b*c*x)^3*(a + b*x)^3),x)
Output:
((5*x^3)/(8*b^2) - (3*a^2*x)/(8*b^4))/(a^4*c^3 + b^4*c^3*x^4 - 2*a^2*b^2*c ^3*x^2) + (3*atanh((b*x)/a))/(8*a*b^5*c^3)
Time = 0.16 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.03 \[ \int \frac {x^4}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {3 \,\mathrm {log}\left (-b x -a \right ) a^{4}-6 \,\mathrm {log}\left (-b x -a \right ) a^{2} b^{2} x^{2}+3 \,\mathrm {log}\left (-b x -a \right ) b^{4} x^{4}-3 \,\mathrm {log}\left (-b x +a \right ) a^{4}+6 \,\mathrm {log}\left (-b x +a \right ) a^{2} b^{2} x^{2}-3 \,\mathrm {log}\left (-b x +a \right ) b^{4} x^{4}-6 a^{3} b x +10 a \,b^{3} x^{3}}{16 a \,b^{5} c^{3} \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:
int(x^4/(b*x+a)^3/(-b*c*x+a*c)^3,x)
Output:
(3*log( - a - b*x)*a**4 - 6*log( - a - b*x)*a**2*b**2*x**2 + 3*log( - a - b*x)*b**4*x**4 - 3*log(a - b*x)*a**4 + 6*log(a - b*x)*a**2*b**2*x**2 - 3*l og(a - b*x)*b**4*x**4 - 6*a**3*b*x + 10*a*b**3*x**3)/(16*a*b**5*c**3*(a**4 - 2*a**2*b**2*x**2 + b**4*x**4))