Integrand size = 22, antiderivative size = 55 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=\frac {x}{b^4 c^2}+\frac {a^2 x}{2 b^4 c^2 \left (a^2-b^2 x^2\right )}-\frac {3 a \text {arctanh}\left (\frac {b x}{a}\right )}{2 b^5 c^2} \] Output:
x/b^4/c^2+1/2*a^2*x/b^4/c^2/(-b^2*x^2+a^2)-3/2*a*arctanh(b*x/a)/b^5/c^2
Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.56 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=\frac {x}{b^4 c^2}-\frac {a^2}{4 b^5 c^2 (-a+b x)}-\frac {a^2}{4 b^5 c^2 (a+b x)}+\frac {3 a \log (a-b x)}{4 b^5 c^2}-\frac {3 a \log (a+b x)}{4 b^5 c^2} \] Input:
Integrate[x^4/((a + b*x)^2*(a*c - b*c*x)^2),x]
Output:
x/(b^4*c^2) - a^2/(4*b^5*c^2*(-a + b*x)) - a^2/(4*b^5*c^2*(a + b*x)) + (3* a*Log[a - b*x])/(4*b^5*c^2) - (3*a*Log[a + b*x])/(4*b^5*c^2)
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04, 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, 262, 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.
\(\displaystyle \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \int \frac {x^4}{\left (a^2 c-b^2 c x^2\right )^2}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {x^3}{2 b^2 c^2 \left (a^2-b^2 x^2\right )}-\frac {3 \int \frac {x^2}{c \left (a^2-b^2 x^2\right )}dx}{2 b^2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3}{2 b^2 c^2 \left (a^2-b^2 x^2\right )}-\frac {3 \int \frac {x^2}{a^2-b^2 x^2}dx}{2 b^2 c^2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {x^3}{2 b^2 c^2 \left (a^2-b^2 x^2\right )}-\frac {3 \left (\frac {a^2 \int \frac {1}{a^2-b^2 x^2}dx}{b^2}-\frac {x}{b^2}\right )}{2 b^2 c^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {x^3}{2 b^2 c^2 \left (a^2-b^2 x^2\right )}-\frac {3 \left (\frac {a \text {arctanh}\left (\frac {b x}{a}\right )}{b^3}-\frac {x}{b^2}\right )}{2 b^2 c^2}\) |
Input:
Int[x^4/((a + b*x)^2*(a*c - b*c*x)^2),x]
Output:
x^3/(2*b^2*c^2*(a^2 - b^2*x^2)) - (3*(-(x/b^2) + (a*ArcTanh[(b*x)/a])/b^3) )/(2*b^2*c^2)
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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Time = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22
method | result | size |
default | \(\frac {\frac {x}{b^{4}}-\frac {3 a \ln \left (b x +a \right )}{4 b^{5}}-\frac {a^{2}}{4 b^{5} \left (b x +a \right )}+\frac {3 a \ln \left (-b x +a \right )}{4 b^{5}}+\frac {a^{2}}{4 b^{5} \left (-b x +a \right )}}{c^{2}}\) | \(67\) |
risch | \(\frac {x}{b^{4} c^{2}}+\frac {a^{2} x}{2 b^{4} \left (b x +a \right ) c^{2} \left (-b x +a \right )}+\frac {3 a \ln \left (b x -a \right )}{4 b^{5} c^{2}}-\frac {3 a \ln \left (-b x -a \right )}{4 b^{5} c^{2}}\) | \(72\) |
norman | \(\frac {-\frac {x^{3}}{b^{2} c}+\frac {3 a^{2} x}{2 b^{4} c}}{\left (b x +a \right ) c \left (-b x +a \right )}+\frac {3 a \ln \left (-b x +a \right )}{4 b^{5} c^{2}}-\frac {3 a \ln \left (b x +a \right )}{4 b^{5} c^{2}}\) | \(76\) |
parallelrisch | \(\frac {3 \ln \left (b x -a \right ) x^{2} a \,b^{2}-3 \ln \left (b x +a \right ) x^{2} a \,b^{2}+4 b^{3} x^{3}-3 a^{3} \ln \left (b x -a \right )+3 a^{3} \ln \left (b x +a \right )-6 a^{2} b x}{4 b^{5} c^{2} \left (b x +a \right ) \left (b x -a \right )}\) | \(97\) |
Input:
int(x^4/(b*x+a)^2/(-b*c*x+a*c)^2,x,method=_RETURNVERBOSE)
Output:
1/c^2*(x/b^4-3/4*a*ln(b*x+a)/b^5-1/4*a^2/b^5/(b*x+a)+3/4/b^5*a*ln(-b*x+a)+ 1/4*a^2/b^5/(-b*x+a))
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.60 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=\frac {4 \, b^{3} x^{3} - 6 \, a^{2} b x - 3 \, {\left (a b^{2} x^{2} - a^{3}\right )} \log \left (b x + a\right ) + 3 \, {\left (a b^{2} x^{2} - a^{3}\right )} \log \left (b x - a\right )}{4 \, {\left (b^{7} c^{2} x^{2} - a^{2} b^{5} c^{2}\right )}} \] Input:
integrate(x^4/(b*x+a)^2/(-b*c*x+a*c)^2,x, algorithm="fricas")
Output:
1/4*(4*b^3*x^3 - 6*a^2*b*x - 3*(a*b^2*x^2 - a^3)*log(b*x + a) + 3*(a*b^2*x ^2 - a^3)*log(b*x - a))/(b^7*c^2*x^2 - a^2*b^5*c^2)
Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.18 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=- \frac {a^{2} x}{- 2 a^{2} b^{4} c^{2} + 2 b^{6} c^{2} x^{2}} + \frac {a \left (\frac {3 \log {\left (- \frac {a}{b} + x \right )}}{4} - \frac {3 \log {\left (\frac {a}{b} + x \right )}}{4}\right )}{b^{5} c^{2}} + \frac {x}{b^{4} c^{2}} \] Input:
integrate(x**4/(b*x+a)**2/(-b*c*x+a*c)**2,x)
Output:
-a**2*x/(-2*a**2*b**4*c**2 + 2*b**6*c**2*x**2) + a*(3*log(-a/b + x)/4 - 3* log(a/b + x)/4)/(b**5*c**2) + x/(b**4*c**2)
Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=-\frac {a^{2} x}{2 \, {\left (b^{6} c^{2} x^{2} - a^{2} b^{4} c^{2}\right )}} + \frac {x}{b^{4} c^{2}} - \frac {3 \, a \log \left (b x + a\right )}{4 \, b^{5} c^{2}} + \frac {3 \, a \log \left (b x - a\right )}{4 \, b^{5} c^{2}} \] Input:
integrate(x^4/(b*x+a)^2/(-b*c*x+a*c)^2,x, algorithm="maxima")
Output:
-1/2*a^2*x/(b^6*c^2*x^2 - a^2*b^4*c^2) + x/(b^4*c^2) - 3/4*a*log(b*x + a)/ (b^5*c^2) + 3/4*a*log(b*x - a)/(b^5*c^2)
Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.89 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=-\frac {a^{2}}{4 \, {\left (b c x - a c\right )} b^{5} c} - \frac {3 \, a \log \left ({\left | -\frac {2 \, a c}{b c x - a c} - 1 \right |}\right )}{4 \, b^{5} c^{2}} + \frac {{\left (b c x - a c\right )} {\left (\frac {17 \, a c}{b c x - a c} + 8\right )}}{8 \, b^{5} {\left (\frac {2 \, a c}{b c x - a c} + 1\right )} c^{3}} \] Input:
integrate(x^4/(b*x+a)^2/(-b*c*x+a*c)^2,x, algorithm="giac")
Output:
-1/4*a^2/((b*c*x - a*c)*b^5*c) - 3/4*a*log(abs(-2*a*c/(b*c*x - a*c) - 1))/ (b^5*c^2) + 1/8*(b*c*x - a*c)*(17*a*c/(b*c*x - a*c) + 8)/(b^5*(2*a*c/(b*c* x - a*c) + 1)*c^3)
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=\frac {x}{b^4\,c^2}+\frac {a^2\,x}{2\,\left (a^2\,b^4\,c^2-b^6\,c^2\,x^2\right )}-\frac {3\,a\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{2\,b^5\,c^2} \] Input:
int(x^4/((a*c - b*c*x)^2*(a + b*x)^2),x)
Output:
x/(b^4*c^2) + (a^2*x)/(2*(a^2*b^4*c^2 - b^6*c^2*x^2)) - (3*a*atanh((b*x)/a ))/(2*b^5*c^2)
Time = 0.15 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.78 \[ \int \frac {x^4}{(a+b x)^2 (a c-b c x)^2} \, dx=\frac {-3 \,\mathrm {log}\left (-b x -a \right ) a^{3}+3 \,\mathrm {log}\left (-b x -a \right ) a \,b^{2} x^{2}+3 \,\mathrm {log}\left (-b x +a \right ) a^{3}-3 \,\mathrm {log}\left (-b x +a \right ) a \,b^{2} x^{2}+6 a^{2} b x -4 b^{3} x^{3}}{4 b^{5} c^{2} \left (-b^{2} x^{2}+a^{2}\right )} \] Input:
int(x^4/(b*x+a)^2/(-b*c*x+a*c)^2,x)
Output:
( - 3*log( - a - b*x)*a**3 + 3*log( - a - b*x)*a*b**2*x**2 + 3*log(a - b*x )*a**3 - 3*log(a - b*x)*a*b**2*x**2 + 6*a**2*b*x - 4*b**3*x**3)/(4*b**5*c* *2*(a**2 - b**2*x**2))