Integrand size = 26, antiderivative size = 108 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e^2 \left (6 b^2 d^2-8 a b d e+3 a^2 e^2\right ) x}{b^4}+\frac {e^3 (2 b d-a e) x^2}{b^3}+\frac {e^4 x^3}{3 b^2}-\frac {(b d-a e)^4}{b^5 (a+b x)}+\frac {4 e (b d-a e)^3 \log (a+b x)}{b^5} \] Output:
e^2*(3*a^2*e^2-8*a*b*d*e+6*b^2*d^2)*x/b^4+e^3*(-a*e+2*b*d)*x^2/b^3+1/3*e^4 *x^3/b^2-(-a*e+b*d)^4/b^5/(b*x+a)+4*e*(-a*e+b*d)^3*ln(b*x+a)/b^5
Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.54 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=\frac {-3 a^4 e^4+3 a^3 b e^3 (4 d+3 e x)+6 a^2 b^2 e^2 \left (-3 d^2-4 d e x+e^2 x^2\right )-2 a b^3 e \left (-6 d^3-9 d^2 e x+9 d e^2 x^2+e^3 x^3\right )+b^4 \left (-3 d^4+18 d^2 e^2 x^2+6 d e^3 x^3+e^4 x^4\right )-12 e (-b d+a e)^3 (a+b x) \log (a+b x)}{3 b^5 (a+b x)} \] Input:
Integrate[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
(-3*a^4*e^4 + 3*a^3*b*e^3*(4*d + 3*e*x) + 6*a^2*b^2*e^2*(-3*d^2 - 4*d*e*x + e^2*x^2) - 2*a*b^3*e*(-6*d^3 - 9*d^2*e*x + 9*d*e^2*x^2 + e^3*x^3) + b^4* (-3*d^4 + 18*d^2*e^2*x^2 + 6*d*e^3*x^3 + e^4*x^4) - 12*e*(-(b*d) + a*e)^3* (a + b*x)*Log[a + b*x])/(3*b^5*(a + b*x))
Time = 0.50 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1098, 27, 49, 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.
\(\displaystyle \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx\) |
\(\Big \downarrow \) 1098 |
\(\displaystyle b^2 \int \frac {(d+e x)^4}{b^2 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(d+e x)^4}{(a+b x)^2}dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (\frac {4 e^3 (a+b x) (b d-a e)}{b^4}+\frac {6 e^2 (b d-a e)^2}{b^4}+\frac {4 e (b d-a e)^3}{b^4 (a+b x)}+\frac {(b d-a e)^4}{b^4 (a+b x)^2}+\frac {e^4 (a+b x)^2}{b^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e^3 (a+b x)^2 (b d-a e)}{b^5}-\frac {(b d-a e)^4}{b^5 (a+b x)}+\frac {4 e (b d-a e)^3 \log (a+b x)}{b^5}+\frac {e^4 (a+b x)^3}{3 b^5}+\frac {6 e^2 x (b d-a e)^2}{b^4}\) |
Input:
Int[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
(6*e^2*(b*d - a*e)^2*x)/b^4 - (b*d - a*e)^4/(b^5*(a + b*x)) + (2*e^3*(b*d - a*e)*(a + b*x)^2)/b^5 + (e^4*(a + b*x)^3)/(3*b^5) + (4*e*(b*d - a*e)^3*L og[a + b*x])/b^5
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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ {a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.03 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.62
method | result | size |
default | \(\frac {e^{2} \left (\frac {1}{3} b^{2} e^{2} x^{3}-x^{2} a b \,e^{2}+2 b^{2} d e \,x^{2}+3 a^{2} e^{2} x -8 a b d e x +6 b^{2} d^{2} x \right )}{b^{4}}-\frac {a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}{b^{5} \left (b x +a \right )}-\frac {4 e \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \ln \left (b x +a \right )}{b^{5}}\) | \(175\) |
norman | \(\frac {\frac {\left (4 a^{4} e^{4}-12 a^{3} b d \,e^{3}+12 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) x}{a \,b^{4}}+\frac {e^{4} x^{4}}{3 b}+\frac {2 e^{2} \left (e^{2} a^{2}-3 a b d e +3 b^{2} d^{2}\right ) x^{2}}{b^{3}}-\frac {2 e^{3} \left (a e -3 b d \right ) x^{3}}{3 b^{2}}}{b x +a}-\frac {4 e \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \ln \left (b x +a \right )}{b^{5}}\) | \(180\) |
risch | \(\frac {e^{4} x^{3}}{3 b^{2}}-\frac {e^{4} x^{2} a}{b^{3}}+\frac {2 e^{3} d \,x^{2}}{b^{2}}+\frac {3 e^{4} a^{2} x}{b^{4}}-\frac {8 e^{3} a d x}{b^{3}}+\frac {6 e^{2} d^{2} x}{b^{2}}-\frac {a^{4} e^{4}}{b^{5} \left (b x +a \right )}+\frac {4 a^{3} d \,e^{3}}{b^{4} \left (b x +a \right )}-\frac {6 a^{2} d^{2} e^{2}}{b^{3} \left (b x +a \right )}+\frac {4 a \,d^{3} e}{b^{2} \left (b x +a \right )}-\frac {d^{4}}{b \left (b x +a \right )}-\frac {4 e^{4} \ln \left (b x +a \right ) a^{3}}{b^{5}}+\frac {12 e^{3} \ln \left (b x +a \right ) a^{2} d}{b^{4}}-\frac {12 e^{2} \ln \left (b x +a \right ) a \,d^{2}}{b^{3}}+\frac {4 e \ln \left (b x +a \right ) d^{3}}{b^{2}}\) | \(230\) |
parallelrisch | \(-\frac {-b^{4} x^{4} e^{4}+2 x^{3} a \,b^{3} e^{4}-6 x^{3} b^{4} d \,e^{3}+12 \ln \left (b x +a \right ) x \,a^{3} b \,e^{4}-36 \ln \left (b x +a \right ) x \,a^{2} b^{2} d \,e^{3}+36 \ln \left (b x +a \right ) x a \,b^{3} d^{2} e^{2}-12 \ln \left (b x +a \right ) x \,b^{4} d^{3} e -6 x^{2} a^{2} b^{2} e^{4}+18 x^{2} a \,b^{3} d \,e^{3}-18 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (b x +a \right ) a^{4} e^{4}-36 \ln \left (b x +a \right ) a^{3} b d \,e^{3}+36 \ln \left (b x +a \right ) a^{2} b^{2} d^{2} e^{2}-12 \ln \left (b x +a \right ) a \,b^{3} d^{3} e +12 a^{4} e^{4}-36 a^{3} b d \,e^{3}+36 a^{2} b^{2} d^{2} e^{2}-12 a \,b^{3} d^{3} e +3 b^{4} d^{4}}{3 b^{5} \left (b x +a \right )}\) | \(276\) |
Input:
int((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
e^2/b^4*(1/3*b^2*e^2*x^3-x^2*a*b*e^2+2*b^2*d*e*x^2+3*a^2*e^2*x-8*a*b*d*e*x +6*b^2*d^2*x)-1/b^5*(a^4*e^4-4*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2-4*a*b^3*d^3*e +b^4*d^4)/(b*x+a)-4/b^5*e*(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*ln (b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (106) = 212\).
Time = 0.08 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.48 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=\frac {b^{4} e^{4} x^{4} - 3 \, b^{4} d^{4} + 12 \, a b^{3} d^{3} e - 18 \, a^{2} b^{2} d^{2} e^{2} + 12 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 2 \, {\left (3 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (3 \, b^{4} d^{2} e^{2} - 3 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (6 \, a b^{3} d^{2} e^{2} - 8 \, a^{2} b^{2} d e^{3} + 3 \, a^{3} b e^{4}\right )} x + 12 \, {\left (a b^{3} d^{3} e - 3 \, a^{2} b^{2} d^{2} e^{2} + 3 \, a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{3 \, {\left (b^{6} x + a b^{5}\right )}} \] Input:
integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
1/3*(b^4*e^4*x^4 - 3*b^4*d^4 + 12*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^2 + 12*a^ 3*b*d*e^3 - 3*a^4*e^4 + 2*(3*b^4*d*e^3 - a*b^3*e^4)*x^3 + 6*(3*b^4*d^2*e^2 - 3*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 3*(6*a*b^3*d^2*e^2 - 8*a^2*b^2*d*e^3 + 3*a^3*b*e^4)*x + 12*(a*b^3*d^3*e - 3*a^2*b^2*d^2*e^2 + 3*a^3*b*d*e^3 - a^4*e^4 + (b^4*d^3*e - 3*a*b^3*d^2*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)*l og(b*x + a))/(b^6*x + a*b^5)
Time = 0.36 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=x^{2} \left (- \frac {a e^{4}}{b^{3}} + \frac {2 d e^{3}}{b^{2}}\right ) + x \left (\frac {3 a^{2} e^{4}}{b^{4}} - \frac {8 a d e^{3}}{b^{3}} + \frac {6 d^{2} e^{2}}{b^{2}}\right ) + \frac {- a^{4} e^{4} + 4 a^{3} b d e^{3} - 6 a^{2} b^{2} d^{2} e^{2} + 4 a b^{3} d^{3} e - b^{4} d^{4}}{a b^{5} + b^{6} x} + \frac {e^{4} x^{3}}{3 b^{2}} - \frac {4 e \left (a e - b d\right )^{3} \log {\left (a + b x \right )}}{b^{5}} \] Input:
integrate((e*x+d)**4/(b**2*x**2+2*a*b*x+a**2),x)
Output:
x**2*(-a*e**4/b**3 + 2*d*e**3/b**2) + x*(3*a**2*e**4/b**4 - 8*a*d*e**3/b** 3 + 6*d**2*e**2/b**2) + (-a**4*e**4 + 4*a**3*b*d*e**3 - 6*a**2*b**2*d**2*e **2 + 4*a*b**3*d**3*e - b**4*d**4)/(a*b**5 + b**6*x) + e**4*x**3/(3*b**2) - 4*e*(a*e - b*d)**3*log(a + b*x)/b**5
Time = 0.04 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.70 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}}{b^{6} x + a b^{5}} + \frac {b^{2} e^{4} x^{3} + 3 \, {\left (2 \, b^{2} d e^{3} - a b e^{4}\right )} x^{2} + 3 \, {\left (6 \, b^{2} d^{2} e^{2} - 8 \, a b d e^{3} + 3 \, a^{2} e^{4}\right )} x}{3 \, b^{4}} + \frac {4 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} \log \left (b x + a\right )}{b^{5}} \] Input:
integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
-(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)/( b^6*x + a*b^5) + 1/3*(b^2*e^4*x^3 + 3*(2*b^2*d*e^3 - a*b*e^4)*x^2 + 3*(6*b ^2*d^2*e^2 - 8*a*b*d*e^3 + 3*a^2*e^4)*x)/b^4 + 4*(b^3*d^3*e - 3*a*b^2*d^2* e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*log(b*x + a)/b^5
Time = 0.20 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.74 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=\frac {4 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac {b^{4} e^{4} x^{3} + 6 \, b^{4} d e^{3} x^{2} - 3 \, a b^{3} e^{4} x^{2} + 18 \, b^{4} d^{2} e^{2} x - 24 \, a b^{3} d e^{3} x + 9 \, a^{2} b^{2} e^{4} x}{3 \, b^{6}} - \frac {b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}}{{\left (b x + a\right )} b^{5}} \] Input:
integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
4*(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*log(abs(b*x + a) )/b^5 + 1/3*(b^4*e^4*x^3 + 6*b^4*d*e^3*x^2 - 3*a*b^3*e^4*x^2 + 18*b^4*d^2* e^2*x - 24*a*b^3*d*e^3*x + 9*a^2*b^2*e^4*x)/b^6 - (b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)/((b*x + a)*b^5)
Time = 9.02 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.88 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=x\,\left (\frac {2\,a\,\left (\frac {2\,a\,e^4}{b^3}-\frac {4\,d\,e^3}{b^2}\right )}{b}-\frac {a^2\,e^4}{b^4}+\frac {6\,d^2\,e^2}{b^2}\right )-x^2\,\left (\frac {a\,e^4}{b^3}-\frac {2\,d\,e^3}{b^2}\right )+\frac {e^4\,x^3}{3\,b^2}-\frac {\ln \left (a+b\,x\right )\,\left (4\,a^3\,e^4-12\,a^2\,b\,d\,e^3+12\,a\,b^2\,d^2\,e^2-4\,b^3\,d^3\,e\right )}{b^5}-\frac {a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4}{b\,\left (x\,b^5+a\,b^4\right )} \] Input:
int((d + e*x)^4/(a^2 + b^2*x^2 + 2*a*b*x),x)
Output:
x*((2*a*((2*a*e^4)/b^3 - (4*d*e^3)/b^2))/b - (a^2*e^4)/b^4 + (6*d^2*e^2)/b ^2) - x^2*((a*e^4)/b^3 - (2*d*e^3)/b^2) + (e^4*x^3)/(3*b^2) - (log(a + b*x )*(4*a^3*e^4 - 4*b^3*d^3*e + 12*a*b^2*d^2*e^2 - 12*a^2*b*d*e^3))/b^5 - (a^ 4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3)/(b*(a *b^4 + b^5*x))
Time = 0.24 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.75 \[ \int \frac {(d+e x)^4}{a^2+2 a b x+b^2 x^2} \, dx=\frac {-12 \,\mathrm {log}\left (b x +a \right ) a^{5} e^{4}+36 \,\mathrm {log}\left (b x +a \right ) a^{4} b d \,e^{3}-12 \,\mathrm {log}\left (b x +a \right ) a^{4} b \,e^{4} x -36 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{2} d^{2} e^{2}+36 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{2} d \,e^{3} x +12 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{3} d^{3} e -36 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{3} d^{2} e^{2} x +12 \,\mathrm {log}\left (b x +a \right ) a \,b^{4} d^{3} e x +12 a^{4} b \,e^{4} x -36 a^{3} b^{2} d \,e^{3} x +6 a^{3} b^{2} e^{4} x^{2}+36 a^{2} b^{3} d^{2} e^{2} x -18 a^{2} b^{3} d \,e^{3} x^{2}-2 a^{2} b^{3} e^{4} x^{3}-12 a \,b^{4} d^{3} e x +18 a \,b^{4} d^{2} e^{2} x^{2}+6 a \,b^{4} d \,e^{3} x^{3}+a \,b^{4} e^{4} x^{4}+3 b^{5} d^{4} x}{3 a \,b^{5} \left (b x +a \right )} \] Input:
int((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2),x)
Output:
( - 12*log(a + b*x)*a**5*e**4 + 36*log(a + b*x)*a**4*b*d*e**3 - 12*log(a + b*x)*a**4*b*e**4*x - 36*log(a + b*x)*a**3*b**2*d**2*e**2 + 36*log(a + b*x )*a**3*b**2*d*e**3*x + 12*log(a + b*x)*a**2*b**3*d**3*e - 36*log(a + b*x)* a**2*b**3*d**2*e**2*x + 12*log(a + b*x)*a*b**4*d**3*e*x + 12*a**4*b*e**4*x - 36*a**3*b**2*d*e**3*x + 6*a**3*b**2*e**4*x**2 + 36*a**2*b**3*d**2*e**2* x - 18*a**2*b**3*d*e**3*x**2 - 2*a**2*b**3*e**4*x**3 - 12*a*b**4*d**3*e*x + 18*a*b**4*d**2*e**2*x**2 + 6*a*b**4*d*e**3*x**3 + a*b**4*e**4*x**4 + 3*b **5*d**4*x)/(3*a*b**5*(a + b*x))