Integrand size = 29, antiderivative size = 86 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {(b d-a e)^3}{3 e^4 (d+e x)^3}-\frac {3 b (b d-a e)^2}{2 e^4 (d+e x)^2}+\frac {3 b^2 (b d-a e)}{e^4 (d+e x)}+\frac {b^3 \log (d+e x)}{e^4} \] Output:
1/3*(-a*e+b*d)^3/e^4/(e*x+d)^3-3/2*b*(-a*e+b*d)^2/e^4/(e*x+d)^2+3*b^2*(-a* e+b*d)/e^4/(e*x+d)+b^3*ln(e*x+d)/e^4
Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {\frac {(b d-a e) \left (2 a^2 e^2+a b e (5 d+9 e x)+b^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )}{(d+e x)^3}+6 b^3 \log (d+e x)}{6 e^4} \] Input:
Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^4,x]
Output:
(((b*d - a*e)*(2*a^2*e^2 + a*b*e*(5*d + 9*e*x) + b^2*(11*d^2 + 27*d*e*x + 18*e^2*x^2)))/(d + e*x)^3 + 6*b^3*Log[d + e*x])/(6*e^4)
Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 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.
\(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^3}{(d+e x)^4}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^3}{(d+e x)^4}dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (-\frac {3 b^2 (b d-a e)}{e^3 (d+e x)^2}+\frac {3 b (b d-a e)^2}{e^3 (d+e x)^3}+\frac {(a e-b d)^3}{e^3 (d+e x)^4}+\frac {b^3}{e^3 (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 b^2 (b d-a e)}{e^4 (d+e x)}-\frac {3 b (b d-a e)^2}{2 e^4 (d+e x)^2}+\frac {(b d-a e)^3}{3 e^4 (d+e x)^3}+\frac {b^3 \log (d+e x)}{e^4}\) |
Input:
Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^4,x]
Output:
(b*d - a*e)^3/(3*e^4*(d + e*x)^3) - (3*b*(b*d - a*e)^2)/(2*e^4*(d + e*x)^2 ) + (3*b^2*(b*d - a*e))/(e^4*(d + e*x)) + (b^3*Log[d + e*x])/e^4
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_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.78 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.33
method | result | size |
risch | \(\frac {-\frac {3 b^{2} \left (a e -b d \right ) x^{2}}{e^{2}}-\frac {3 b \left (e^{2} a^{2}+2 a b d e -3 b^{2} d^{2}\right ) x}{2 e^{3}}-\frac {2 e^{3} a^{3}+3 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -11 b^{3} d^{3}}{6 e^{4}}}{\left (e x +d \right )^{3}}+\frac {b^{3} \ln \left (e x +d \right )}{e^{4}}\) | \(114\) |
norman | \(\frac {-\frac {2 e^{3} a^{3}+3 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -11 b^{3} d^{3}}{6 e^{4}}-\frac {3 \left (e a \,b^{2}-b^{3} d \right ) x^{2}}{e^{2}}-\frac {3 \left (a^{2} b \,e^{2}+2 a \,b^{2} d e -3 b^{3} d^{2}\right ) x}{2 e^{3}}}{\left (e x +d \right )^{3}}+\frac {b^{3} \ln \left (e x +d \right )}{e^{4}}\) | \(118\) |
default | \(-\frac {e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}}{3 e^{4} \left (e x +d \right )^{3}}+\frac {b^{3} \ln \left (e x +d \right )}{e^{4}}-\frac {3 b^{2} \left (a e -b d \right )}{e^{4} \left (e x +d \right )}-\frac {3 b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{2 e^{4} \left (e x +d \right )^{2}}\) | \(120\) |
parallelrisch | \(\frac {6 \ln \left (e x +d \right ) b^{3} e^{3} x^{3}+18 \ln \left (e x +d \right ) b^{3} d \,e^{2} x^{2}+18 \ln \left (e x +d \right ) b^{3} d^{2} e x -18 x^{2} a \,b^{2} e^{3}+18 x^{2} b^{3} d \,e^{2}+6 \ln \left (e x +d \right ) b^{3} d^{3}-9 a^{2} b \,e^{3} x -18 x a \,b^{2} d \,e^{2}+27 b^{3} d^{2} e x -2 e^{3} a^{3}-3 a^{2} b d \,e^{2}-6 a \,b^{2} d^{2} e +11 b^{3} d^{3}}{6 e^{4} \left (e x +d \right )^{3}}\) | \(170\) |
Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^4,x,method=_RETURNVERBOSE)
Output:
(-3*b^2*(a*e-b*d)/e^2*x^2-3/2*b*(a^2*e^2+2*a*b*d*e-3*b^2*d^2)/e^3*x-1/6*(2 *a^3*e^3+3*a^2*b*d*e^2+6*a*b^2*d^2*e-11*b^3*d^3)/e^4)/(e*x+d)^3+b^3*ln(e*x +d)/e^4
Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (82) = 164\).
Time = 0.07 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.06 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {11 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3} + 18 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (3 \, b^{3} d^{2} e - 2 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, b^{3} d e^{2} x^{2} + 3 \, b^{3} d^{2} e x + b^{3} d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^4,x, algorithm="fricas")
Output:
1/6*(11*b^3*d^3 - 6*a*b^2*d^2*e - 3*a^2*b*d*e^2 - 2*a^3*e^3 + 18*(b^3*d*e^ 2 - a*b^2*e^3)*x^2 + 9*(3*b^3*d^2*e - 2*a*b^2*d*e^2 - a^2*b*e^3)*x + 6*(b^ 3*e^3*x^3 + 3*b^3*d*e^2*x^2 + 3*b^3*d^2*e*x + b^3*d^3)*log(e*x + d))/(e^7* x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
Time = 1.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.72 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {b^{3} \log {\left (d + e x \right )}}{e^{4}} + \frac {- 2 a^{3} e^{3} - 3 a^{2} b d e^{2} - 6 a b^{2} d^{2} e + 11 b^{3} d^{3} + x^{2} \left (- 18 a b^{2} e^{3} + 18 b^{3} d e^{2}\right ) + x \left (- 9 a^{2} b e^{3} - 18 a b^{2} d e^{2} + 27 b^{3} d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \] Input:
integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)/(e*x+d)**4,x)
Output:
b**3*log(d + e*x)/e**4 + (-2*a**3*e**3 - 3*a**2*b*d*e**2 - 6*a*b**2*d**2*e + 11*b**3*d**3 + x**2*(-18*a*b**2*e**3 + 18*b**3*d*e**2) + x*(-9*a**2*b*e **3 - 18*a*b**2*d*e**2 + 27*b**3*d**2*e))/(6*d**3*e**4 + 18*d**2*e**5*x + 18*d*e**6*x**2 + 6*e**7*x**3)
Time = 0.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {11 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3} + 18 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (3 \, b^{3} d^{2} e - 2 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac {b^{3} \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^4,x, algorithm="maxima")
Output:
1/6*(11*b^3*d^3 - 6*a*b^2*d^2*e - 3*a^2*b*d*e^2 - 2*a^3*e^3 + 18*(b^3*d*e^ 2 - a*b^2*e^3)*x^2 + 9*(3*b^3*d^2*e - 2*a*b^2*d*e^2 - a^2*b*e^3)*x)/(e^7*x ^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) + b^3*log(e*x + d)/e^4
Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {b^{3} \log \left ({\left | e x + d \right |}\right )}{e^{4}} + \frac {18 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x^{2} + 9 \, {\left (3 \, b^{3} d^{2} - 2 \, a b^{2} d e - a^{2} b e^{2}\right )} x + \frac {11 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3}}{e}}{6 \, {\left (e x + d\right )}^{3} e^{3}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^4,x, algorithm="giac")
Output:
b^3*log(abs(e*x + d))/e^4 + 1/6*(18*(b^3*d*e - a*b^2*e^2)*x^2 + 9*(3*b^3*d ^2 - 2*a*b^2*d*e - a^2*b*e^2)*x + (11*b^3*d^3 - 6*a*b^2*d^2*e - 3*a^2*b*d* e^2 - 2*a^3*e^3)/e)/((e*x + d)^3*e^3)
Time = 11.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.60 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {b^3\,\ln \left (d+e\,x\right )}{e^4}-\frac {\frac {2\,a^3\,e^3+3\,a^2\,b\,d\,e^2+6\,a\,b^2\,d^2\,e-11\,b^3\,d^3}{6\,e^4}+\frac {3\,x\,\left (a^2\,b\,e^2+2\,a\,b^2\,d\,e-3\,b^3\,d^2\right )}{2\,e^3}+\frac {3\,b^2\,x^2\,\left (a\,e-b\,d\right )}{e^2}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \] Input:
int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x))/(d + e*x)^4,x)
Output:
(b^3*log(d + e*x))/e^4 - ((2*a^3*e^3 - 11*b^3*d^3 + 6*a*b^2*d^2*e + 3*a^2* b*d*e^2)/(6*e^4) + (3*x*(a^2*b*e^2 - 3*b^3*d^2 + 2*a*b^2*d*e))/(2*e^3) + ( 3*b^2*x^2*(a*e - b*d))/e^2)/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x)
Time = 0.22 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.09 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^4} \, dx=\frac {6 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{4}+18 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{3} e x +18 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{2} e^{2} x^{2}+6 \,\mathrm {log}\left (e x +d \right ) b^{3} d \,e^{3} x^{3}-2 a^{3} d \,e^{3}-3 a^{2} b \,d^{2} e^{2}-9 a^{2} b d \,e^{3} x +6 a \,b^{2} e^{4} x^{3}+5 b^{3} d^{4}+9 b^{3} d^{3} e x -6 b^{3} d \,e^{3} x^{3}}{6 d \,e^{4} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )} \] Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^4,x)
Output:
(6*log(d + e*x)*b**3*d**4 + 18*log(d + e*x)*b**3*d**3*e*x + 18*log(d + e*x )*b**3*d**2*e**2*x**2 + 6*log(d + e*x)*b**3*d*e**3*x**3 - 2*a**3*d*e**3 - 3*a**2*b*d**2*e**2 - 9*a**2*b*d*e**3*x + 6*a*b**2*e**4*x**3 + 5*b**3*d**4 + 9*b**3*d**3*e*x - 6*b**3*d*e**3*x**3)/(6*d*e**4*(d**3 + 3*d**2*e*x + 3*d *e**2*x**2 + e**3*x**3))