Integrand size = 26, antiderivative size = 111 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx=-\frac {(b d-a e)^4}{4 e^5 (d+e x)^4}+\frac {4 b (b d-a e)^3}{3 e^5 (d+e x)^3}-\frac {3 b^2 (b d-a e)^2}{e^5 (d+e x)^2}+\frac {4 b^3 (b d-a e)}{e^5 (d+e x)}+\frac {b^4 \log (d+e x)}{e^5} \]
-1/4*(-a*e+b*d)^4/e^5/(e*x+d)^4+4/3*b*(-a*e+b*d)^3/e^5/(e*x+d)^3-3*b^2*(-a *e+b*d)^2/e^5/(e*x+d)^2+4*b^3*(-a*e+b*d)/e^5/(e*x+d)+b^4*ln(e*x+d)/e^5
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {\frac {(b d-a e) \left (3 a^3 e^3+a^2 b e^2 (7 d+16 e x)+a b^2 e \left (13 d^2+40 d e x+36 e^2 x^2\right )+b^3 \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )}{(d+e x)^4}+12 b^4 \log (d+e x)}{12 e^5} \]
(((b*d - a*e)*(3*a^3*e^3 + a^2*b*e^2*(7*d + 16*e*x) + a*b^2*e*(13*d^2 + 40 *d*e*x + 36*e^2*x^2) + b^3*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x ^3)))/(d + e*x)^4 + 12*b^4*Log[d + e*x])/(12*e^5)
Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx\) |
\(\Big \downarrow \) 1098 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4}{(d+e x)^5}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4}{(d+e x)^5}dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (-\frac {4 b^3 (b d-a e)}{e^4 (d+e x)^2}+\frac {6 b^2 (b d-a e)^2}{e^4 (d+e x)^3}-\frac {4 b (b d-a e)^3}{e^4 (d+e x)^4}+\frac {(a e-b d)^4}{e^4 (d+e x)^5}+\frac {b^4}{e^4 (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 b^3 (b d-a e)}{e^5 (d+e x)}-\frac {3 b^2 (b d-a e)^2}{e^5 (d+e x)^2}+\frac {4 b (b d-a e)^3}{3 e^5 (d+e x)^3}-\frac {(b d-a e)^4}{4 e^5 (d+e x)^4}+\frac {b^4 \log (d+e x)}{e^5}\) |
-1/4*(b*d - a*e)^4/(e^5*(d + e*x)^4) + (4*b*(b*d - a*e)^3)/(3*e^5*(d + e*x )^3) - (3*b^2*(b*d - a*e)^2)/(e^5*(d + e*x)^2) + (4*b^3*(b*d - a*e))/(e^5* (d + e*x)) + (b^4*Log[d + e*x])/e^5
3.15.74.3.1 Defintions of rubi rules used
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 = 2.23 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.59
method | result | size |
risch | \(\frac {-\frac {4 b^{3} \left (a e -b d \right ) x^{3}}{e^{2}}-\frac {3 b^{2} \left (a^{2} e^{2}+2 a b d e -3 b^{2} d^{2}\right ) x^{2}}{e^{3}}-\frac {2 b \left (2 a^{3} e^{3}+3 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -11 b^{3} d^{3}\right ) x}{3 e^{4}}-\frac {3 e^{4} a^{4}+4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}+12 a \,b^{3} d^{3} e -25 b^{4} d^{4}}{12 e^{5}}}{\left (e x +d \right )^{4}}+\frac {b^{4} \ln \left (e x +d \right )}{e^{5}}\) | \(176\) |
norman | \(\frac {-\frac {3 e^{4} a^{4}+4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}+12 a \,b^{3} d^{3} e -25 b^{4} d^{4}}{12 e^{5}}-\frac {4 \left (e a \,b^{3}-d \,b^{4}\right ) x^{3}}{e^{2}}-\frac {3 \left (a^{2} b^{2} e^{2}+2 a \,b^{3} d e -3 b^{4} d^{2}\right ) x^{2}}{e^{3}}-\frac {2 \left (2 a^{3} b \,e^{3}+3 a^{2} b^{2} d \,e^{2}+6 d^{2} e a \,b^{3}-11 b^{4} d^{3}\right ) x}{3 e^{4}}}{\left (e x +d \right )^{4}}+\frac {b^{4} \ln \left (e x +d \right )}{e^{5}}\) | \(182\) |
default | \(-\frac {4 b^{3} \left (a e -b d \right )}{e^{5} \left (e x +d \right )}-\frac {4 b \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{3 e^{5} \left (e x +d \right )^{3}}-\frac {e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}{4 e^{5} \left (e x +d \right )^{4}}-\frac {3 b^{2} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}{e^{5} \left (e x +d \right )^{2}}+\frac {b^{4} \ln \left (e x +d \right )}{e^{5}}\) | \(184\) |
parallelrisch | \(\frac {12 \ln \left (e x +d \right ) x^{4} b^{4} e^{4}+48 \ln \left (e x +d \right ) x^{3} b^{4} d \,e^{3}+72 \ln \left (e x +d \right ) x^{2} b^{4} d^{2} e^{2}-48 x^{3} a \,b^{3} e^{4}+48 x^{3} b^{4} d \,e^{3}+48 \ln \left (e x +d \right ) x \,b^{4} d^{3} e -36 x^{2} a^{2} b^{2} e^{4}-72 x^{2} a \,b^{3} d \,e^{3}+108 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (e x +d \right ) b^{4} d^{4}-16 x \,a^{3} b \,e^{4}-24 x \,a^{2} b^{2} d \,e^{3}-48 x a \,b^{3} d^{2} e^{2}+88 x \,b^{4} d^{3} e -3 e^{4} a^{4}-4 b \,e^{3} d \,a^{3}-6 b^{2} e^{2} d^{2} a^{2}-12 a \,b^{3} d^{3} e +25 b^{4} d^{4}}{12 e^{5} \left (e x +d \right )^{4}}\) | \(260\) |
(-4*b^3*(a*e-b*d)/e^2*x^3-3*b^2*(a^2*e^2+2*a*b*d*e-3*b^2*d^2)/e^3*x^2-2/3* b*(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*x-1/12*(3*a^4*e^4 +4*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2+12*a*b^3*d^3*e-25*b^4*d^4)/e^5)/(e*x+d)^4 +b^4*ln(e*x+d)/e^5
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (107) = 214\).
Time = 0.27 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.41 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {25 \, b^{4} d^{4} - 12 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 48 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 36 \, {\left (3 \, b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 8 \, {\left (11 \, b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} - 3 \, a^{2} b^{2} d e^{3} - 2 \, a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} e^{4} x^{4} + 4 \, b^{4} d e^{3} x^{3} + 6 \, b^{4} d^{2} e^{2} x^{2} + 4 \, b^{4} d^{3} e x + b^{4} d^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \]
1/12*(25*b^4*d^4 - 12*a*b^3*d^3*e - 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - 3* a^4*e^4 + 48*(b^4*d*e^3 - a*b^3*e^4)*x^3 + 36*(3*b^4*d^2*e^2 - 2*a*b^3*d*e ^3 - a^2*b^2*e^4)*x^2 + 8*(11*b^4*d^3*e - 6*a*b^3*d^2*e^2 - 3*a^2*b^2*d*e^ 3 - 2*a^3*b*e^4)*x + 12*(b^4*e^4*x^4 + 4*b^4*d*e^3*x^3 + 6*b^4*d^2*e^2*x^2 + 4*b^4*d^3*e*x + b^4*d^4)*log(e*x + d))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e ^7*x^2 + 4*d^3*e^6*x + d^4*e^5)
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (100) = 200\).
Time = 1.62 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.07 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {b^{4} \log {\left (d + e x \right )}}{e^{5}} + \frac {- 3 a^{4} e^{4} - 4 a^{3} b d e^{3} - 6 a^{2} b^{2} d^{2} e^{2} - 12 a b^{3} d^{3} e + 25 b^{4} d^{4} + x^{3} \left (- 48 a b^{3} e^{4} + 48 b^{4} d e^{3}\right ) + x^{2} \left (- 36 a^{2} b^{2} e^{4} - 72 a b^{3} d e^{3} + 108 b^{4} d^{2} e^{2}\right ) + x \left (- 16 a^{3} b e^{4} - 24 a^{2} b^{2} d e^{3} - 48 a b^{3} d^{2} e^{2} + 88 b^{4} d^{3} e\right )}{12 d^{4} e^{5} + 48 d^{3} e^{6} x + 72 d^{2} e^{7} x^{2} + 48 d e^{8} x^{3} + 12 e^{9} x^{4}} \]
b**4*log(d + e*x)/e**5 + (-3*a**4*e**4 - 4*a**3*b*d*e**3 - 6*a**2*b**2*d** 2*e**2 - 12*a*b**3*d**3*e + 25*b**4*d**4 + x**3*(-48*a*b**3*e**4 + 48*b**4 *d*e**3) + x**2*(-36*a**2*b**2*e**4 - 72*a*b**3*d*e**3 + 108*b**4*d**2*e** 2) + x*(-16*a**3*b*e**4 - 24*a**2*b**2*d*e**3 - 48*a*b**3*d**2*e**2 + 88*b **4*d**3*e))/(12*d**4*e**5 + 48*d**3*e**6*x + 72*d**2*e**7*x**2 + 48*d*e** 8*x**3 + 12*e**9*x**4)
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (107) = 214\).
Time = 0.20 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {25 \, b^{4} d^{4} - 12 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 48 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 36 \, {\left (3 \, b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 8 \, {\left (11 \, b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} - 3 \, a^{2} b^{2} d e^{3} - 2 \, a^{3} b e^{4}\right )} x}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} + \frac {b^{4} \log \left (e x + d\right )}{e^{5}} \]
1/12*(25*b^4*d^4 - 12*a*b^3*d^3*e - 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - 3* a^4*e^4 + 48*(b^4*d*e^3 - a*b^3*e^4)*x^3 + 36*(3*b^4*d^2*e^2 - 2*a*b^3*d*e ^3 - a^2*b^2*e^4)*x^2 + 8*(11*b^4*d^3*e - 6*a*b^3*d^2*e^2 - 3*a^2*b^2*d*e^ 3 - 2*a^3*b*e^4)*x)/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*e^5) + b^4*log(e*x + d)/e^5
Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (107) = 214\).
Time = 0.26 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.53 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx=-\frac {b^{4} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} + \frac {\frac {48 \, b^{4} d e^{15}}{e x + d} - \frac {36 \, b^{4} d^{2} e^{15}}{{\left (e x + d\right )}^{2}} + \frac {16 \, b^{4} d^{3} e^{15}}{{\left (e x + d\right )}^{3}} - \frac {3 \, b^{4} d^{4} e^{15}}{{\left (e x + d\right )}^{4}} - \frac {48 \, a b^{3} e^{16}}{e x + d} + \frac {72 \, a b^{3} d e^{16}}{{\left (e x + d\right )}^{2}} - \frac {48 \, a b^{3} d^{2} e^{16}}{{\left (e x + d\right )}^{3}} + \frac {12 \, a b^{3} d^{3} e^{16}}{{\left (e x + d\right )}^{4}} - \frac {36 \, a^{2} b^{2} e^{17}}{{\left (e x + d\right )}^{2}} + \frac {48 \, a^{2} b^{2} d e^{17}}{{\left (e x + d\right )}^{3}} - \frac {18 \, a^{2} b^{2} d^{2} e^{17}}{{\left (e x + d\right )}^{4}} - \frac {16 \, a^{3} b e^{18}}{{\left (e x + d\right )}^{3}} + \frac {12 \, a^{3} b d e^{18}}{{\left (e x + d\right )}^{4}} - \frac {3 \, a^{4} e^{19}}{{\left (e x + d\right )}^{4}}}{12 \, e^{20}} \]
-b^4*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^5 + 1/12*(48*b^4*d*e^15/(e*x + d) - 36*b^4*d^2*e^15/(e*x + d)^2 + 16*b^4*d^3*e^15/(e*x + d)^3 - 3*b^4* d^4*e^15/(e*x + d)^4 - 48*a*b^3*e^16/(e*x + d) + 72*a*b^3*d*e^16/(e*x + d) ^2 - 48*a*b^3*d^2*e^16/(e*x + d)^3 + 12*a*b^3*d^3*e^16/(e*x + d)^4 - 36*a^ 2*b^2*e^17/(e*x + d)^2 + 48*a^2*b^2*d*e^17/(e*x + d)^3 - 18*a^2*b^2*d^2*e^ 17/(e*x + d)^4 - 16*a^3*b*e^18/(e*x + d)^3 + 12*a^3*b*d*e^18/(e*x + d)^4 - 3*a^4*e^19/(e*x + d)^4)/e^20
Time = 9.77 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.92 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {b^4\,\ln \left (d+e\,x\right )}{e^5}-\frac {\frac {3\,a^4\,e^4+4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2+12\,a\,b^3\,d^3\,e-25\,b^4\,d^4}{12\,e^5}+\frac {3\,x^2\,\left (a^2\,b^2\,e^2+2\,a\,b^3\,d\,e-3\,b^4\,d^2\right )}{e^3}+\frac {2\,x\,\left (2\,a^3\,b\,e^3+3\,a^2\,b^2\,d\,e^2+6\,a\,b^3\,d^2\,e-11\,b^4\,d^3\right )}{3\,e^4}+\frac {4\,b^3\,x^3\,\left (a\,e-b\,d\right )}{e^2}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]
(b^4*log(d + e*x))/e^5 - ((3*a^4*e^4 - 25*b^4*d^4 + 6*a^2*b^2*d^2*e^2 + 12 *a*b^3*d^3*e + 4*a^3*b*d*e^3)/(12*e^5) + (3*x^2*(a^2*b^2*e^2 - 3*b^4*d^2 + 2*a*b^3*d*e))/e^3 + (2*x*(2*a^3*b*e^3 - 11*b^4*d^3 + 3*a^2*b^2*d*e^2 + 6* a*b^3*d^2*e))/(3*e^4) + (4*b^3*x^3*(a*e - b*d))/e^2)/(d^4 + e^4*x^4 + 4*d* e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x)