Integrand size = 22, antiderivative size = 197 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {(B d-A e) \left (c d^2+a e^2\right )^2}{5 e^6 (d+e x)^5}-\frac {\left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right )}{4 e^6 (d+e x)^4}+\frac {2 c \left (5 B c d^3-3 A c d^2 e+3 a B d e^2-a A e^3\right )}{3 e^6 (d+e x)^3}-\frac {c \left (5 B c d^2-2 A c d e+a B e^2\right )}{e^6 (d+e x)^2}+\frac {c^2 (5 B d-A e)}{e^6 (d+e x)}+\frac {B c^2 \log (d+e x)}{e^6} \]
1/5*(-A*e+B*d)*(a*e^2+c*d^2)^2/e^6/(e*x+d)^5-1/4*(a*e^2+c*d^2)*(-4*A*c*d*e +B*a*e^2+5*B*c*d^2)/e^6/(e*x+d)^4+2/3*c*(-A*a*e^3-3*A*c*d^2*e+3*B*a*d*e^2+ 5*B*c*d^3)/e^6/(e*x+d)^3-c*(-2*A*c*d*e+B*a*e^2+5*B*c*d^2)/e^6/(e*x+d)^2+c^ 2*(-A*e+5*B*d)/e^6/(e*x+d)+B*c^2*ln(e*x+d)/e^6
Time = 0.07 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {-4 A e \left (3 a^2 e^4+a c e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+3 c^2 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )\right )+B \left (-3 a^2 e^4 (d+5 e x)-6 a c e^2 \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+c^2 d \left (137 d^4+625 d^3 e x+1100 d^2 e^2 x^2+900 d e^3 x^3+300 e^4 x^4\right )\right )+60 B c^2 (d+e x)^5 \log (d+e x)}{60 e^6 (d+e x)^5} \]
(-4*A*e*(3*a^2*e^4 + a*c*e^2*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*c^2*(d^4 + 5 *d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 + 5*e^4*x^4)) + B*(-3*a^2*e^4*(d + 5*e*x) - 6*a*c*e^2*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + c^2*d *(137*d^4 + 625*d^3*e*x + 1100*d^2*e^2*x^2 + 900*d*e^3*x^3 + 300*e^4*x^4)) + 60*B*c^2*(d + e*x)^5*Log[d + e*x])/(60*e^6*(d + e*x)^5)
Time = 0.40 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {652, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^2 (A+B x)}{(d+e x)^6} \, dx\) |
| \(\Big \downarrow \) 652 |
| \(\displaystyle \int \left (-\frac {2 c \left (-a B e^2+2 A c d e-5 B c d^2\right )}{e^5 (d+e x)^3}+\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^5 (d+e x)^5}+\frac {\left (a e^2+c d^2\right )^2 (A e-B d)}{e^5 (d+e x)^6}+\frac {2 c \left (a A e^3-3 a B d e^2+3 A c d^2 e-5 B c d^3\right )}{e^5 (d+e x)^4}+\frac {c^2 (A e-5 B d)}{e^5 (d+e x)^2}+\frac {B c^2}{e^5 (d+e x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {c \left (a B e^2-2 A c d e+5 B c d^2\right )}{e^6 (d+e x)^2}-\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{4 e^6 (d+e x)^4}+\frac {\left (a e^2+c d^2\right )^2 (B d-A e)}{5 e^6 (d+e x)^5}+\frac {2 c \left (-a A e^3+3 a B d e^2-3 A c d^2 e+5 B c d^3\right )}{3 e^6 (d+e x)^3}+\frac {c^2 (5 B d-A e)}{e^6 (d+e x)}+\frac {B c^2 \log (d+e x)}{e^6}\) |
((B*d - A*e)*(c*d^2 + a*e^2)^2)/(5*e^6*(d + e*x)^5) - ((c*d^2 + a*e^2)*(5* B*c*d^2 - 4*A*c*d*e + a*B*e^2))/(4*e^6*(d + e*x)^4) + (2*c*(5*B*c*d^3 - 3* A*c*d^2*e + 3*a*B*d*e^2 - a*A*e^3))/(3*e^6*(d + e*x)^3) - (c*(5*B*c*d^2 - 2*A*c*d*e + a*B*e^2))/(e^6*(d + e*x)^2) + (c^2*(5*B*d - A*e))/(e^6*(d + e* x)) + (B*c^2*Log[d + e*x])/e^6
3.14.8.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
Time = 0.27 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.19
| method | result | size |
| risch | \(\frac {-\frac {c^{2} \left (A e -5 B d \right ) x^{4}}{e^{2}}-\frac {c \left (2 A c d e +B a \,e^{2}-15 B c \,d^{2}\right ) x^{3}}{e^{3}}-\frac {c \left (2 A a \,e^{3}+6 A c \,d^{2} e +3 B a d \,e^{2}-55 B c \,d^{3}\right ) x^{2}}{3 e^{4}}-\frac {\left (4 A a c d \,e^{3}+12 A \,c^{2} d^{3} e +3 B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}-125 B \,c^{2} d^{4}\right ) x}{12 e^{5}}-\frac {12 A \,a^{2} e^{5}+4 A a c \,d^{2} e^{3}+12 A \,c^{2} d^{4} e +3 B \,a^{2} d \,e^{4}+6 B a c \,d^{3} e^{2}-137 B \,c^{2} d^{5}}{60 e^{6}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\) | \(234\) |
| default | \(-\frac {A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}}{5 e^{6} \left (e x +d \right )^{5}}-\frac {2 c \left (A a \,e^{3}+3 A c \,d^{2} e -3 B a d \,e^{2}-5 B c \,d^{3}\right )}{3 e^{6} \left (e x +d \right )^{3}}-\frac {c^{2} \left (A e -5 B d \right )}{e^{6} \left (e x +d \right )}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}-\frac {-4 A a c d \,e^{3}-4 A \,c^{2} d^{3} e +B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}+5 B \,c^{2} d^{4}}{4 e^{6} \left (e x +d \right )^{4}}+\frac {c \left (2 A c d e -B a \,e^{2}-5 B c \,d^{2}\right )}{e^{6} \left (e x +d \right )^{2}}\) | \(246\) |
| norman | \(\frac {-\frac {12 A \,a^{2} e^{5}+4 A a c \,d^{2} e^{3}+12 A \,c^{2} d^{4} e +3 B \,a^{2} d \,e^{4}+6 B a c \,d^{3} e^{2}-137 B \,c^{2} d^{5}}{60 e^{6}}-\frac {\left (A \,c^{2} e -5 B \,c^{2} d \right ) x^{4}}{e^{2}}-\frac {\left (2 A \,c^{2} d e +B \,e^{2} a c -15 B \,c^{2} d^{2}\right ) x^{3}}{e^{3}}-\frac {\left (2 A a c \,e^{3}+6 A \,c^{2} d^{2} e +3 B a c d \,e^{2}-55 B \,c^{2} d^{3}\right ) x^{2}}{3 e^{4}}-\frac {\left (4 A a c d \,e^{3}+12 A \,c^{2} d^{3} e +3 B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}-125 B \,c^{2} d^{4}\right ) x}{12 e^{5}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\) | \(246\) |
| parallelrisch | \(-\frac {12 A \,a^{2} e^{5}-137 B \,c^{2} d^{5}+30 B x a c \,d^{2} e^{3}+20 A x a c d \,e^{4}+60 B \,x^{2} a c d \,e^{4}+15 B x \,a^{2} e^{5}-300 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e -600 B \ln \left (e x +d \right ) x^{3} c^{2} d^{2} e^{3}-300 B \ln \left (e x +d \right ) x^{4} c^{2} d \,e^{4}-600 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}-60 B \ln \left (e x +d \right ) c^{2} d^{5}+60 A \,x^{4} c^{2} e^{5}+3 B \,a^{2} d \,e^{4}+12 A \,c^{2} d^{4} e -60 B \ln \left (e x +d \right ) x^{5} c^{2} e^{5}+60 B \,x^{3} a c \,e^{5}+40 A \,x^{2} a c \,e^{5}+4 A a c \,d^{2} e^{3}+6 B a c \,d^{3} e^{2}+120 A \,x^{3} c^{2} d \,e^{4}-900 B \,x^{3} c^{2} d^{2} e^{3}+120 A \,x^{2} c^{2} d^{2} e^{3}-1100 B \,x^{2} c^{2} d^{3} e^{2}+60 A x \,c^{2} d^{3} e^{2}-625 B x \,c^{2} d^{4} e -300 B \,x^{4} c^{2} d \,e^{4}}{60 e^{6} \left (e x +d \right )^{5}}\) | \(358\) |
(-c^2*(A*e-5*B*d)/e^2*x^4-c*(2*A*c*d*e+B*a*e^2-15*B*c*d^2)/e^3*x^3-1/3*c*( 2*A*a*e^3+6*A*c*d^2*e+3*B*a*d*e^2-55*B*c*d^3)/e^4*x^2-1/12*(4*A*a*c*d*e^3+ 12*A*c^2*d^3*e+3*B*a^2*e^4+6*B*a*c*d^2*e^2-125*B*c^2*d^4)/e^5*x-1/60*(12*A *a^2*e^5+4*A*a*c*d^2*e^3+12*A*c^2*d^4*e+3*B*a^2*d*e^4+6*B*a*c*d^3*e^2-137* B*c^2*d^5)/e^6)/(e*x+d)^5+B*c^2*ln(e*x+d)/e^6
Time = 0.27 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.85 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 12 \, A c^{2} d^{4} e - 6 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} - 3 \, B a^{2} d e^{4} - 12 \, A a^{2} e^{5} + 60 \, {\left (5 \, B c^{2} d e^{4} - A c^{2} e^{5}\right )} x^{4} + 60 \, {\left (15 \, B c^{2} d^{2} e^{3} - 2 \, A c^{2} d e^{4} - B a c e^{5}\right )} x^{3} + 20 \, {\left (55 \, B c^{2} d^{3} e^{2} - 6 \, A c^{2} d^{2} e^{3} - 3 \, B a c d e^{4} - 2 \, A a c e^{5}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 12 \, A c^{2} d^{3} e^{2} - 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} - 3 \, B a^{2} e^{5}\right )} x + 60 \, {\left (B c^{2} e^{5} x^{5} + 5 \, B c^{2} d e^{4} x^{4} + 10 \, B c^{2} d^{2} e^{3} x^{3} + 10 \, B c^{2} d^{3} e^{2} x^{2} + 5 \, B c^{2} d^{4} e x + B c^{2} d^{5}\right )} \log \left (e x + d\right )}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} \]
1/60*(137*B*c^2*d^5 - 12*A*c^2*d^4*e - 6*B*a*c*d^3*e^2 - 4*A*a*c*d^2*e^3 - 3*B*a^2*d*e^4 - 12*A*a^2*e^5 + 60*(5*B*c^2*d*e^4 - A*c^2*e^5)*x^4 + 60*(1 5*B*c^2*d^2*e^3 - 2*A*c^2*d*e^4 - B*a*c*e^5)*x^3 + 20*(55*B*c^2*d^3*e^2 - 6*A*c^2*d^2*e^3 - 3*B*a*c*d*e^4 - 2*A*a*c*e^5)*x^2 + 5*(125*B*c^2*d^4*e - 12*A*c^2*d^3*e^2 - 6*B*a*c*d^2*e^3 - 4*A*a*c*d*e^4 - 3*B*a^2*e^5)*x + 60*( B*c^2*e^5*x^5 + 5*B*c^2*d*e^4*x^4 + 10*B*c^2*d^2*e^3*x^3 + 10*B*c^2*d^3*e^ 2*x^2 + 5*B*c^2*d^4*e*x + B*c^2*d^5)*log(e*x + d))/(e^11*x^5 + 5*d*e^10*x^ 4 + 10*d^2*e^9*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5*e^6)
Timed out. \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.51 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 12 \, A c^{2} d^{4} e - 6 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} - 3 \, B a^{2} d e^{4} - 12 \, A a^{2} e^{5} + 60 \, {\left (5 \, B c^{2} d e^{4} - A c^{2} e^{5}\right )} x^{4} + 60 \, {\left (15 \, B c^{2} d^{2} e^{3} - 2 \, A c^{2} d e^{4} - B a c e^{5}\right )} x^{3} + 20 \, {\left (55 \, B c^{2} d^{3} e^{2} - 6 \, A c^{2} d^{2} e^{3} - 3 \, B a c d e^{4} - 2 \, A a c e^{5}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 12 \, A c^{2} d^{3} e^{2} - 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} - 3 \, B a^{2} e^{5}\right )} x}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} + \frac {B c^{2} \log \left (e x + d\right )}{e^{6}} \]
1/60*(137*B*c^2*d^5 - 12*A*c^2*d^4*e - 6*B*a*c*d^3*e^2 - 4*A*a*c*d^2*e^3 - 3*B*a^2*d*e^4 - 12*A*a^2*e^5 + 60*(5*B*c^2*d*e^4 - A*c^2*e^5)*x^4 + 60*(1 5*B*c^2*d^2*e^3 - 2*A*c^2*d*e^4 - B*a*c*e^5)*x^3 + 20*(55*B*c^2*d^3*e^2 - 6*A*c^2*d^2*e^3 - 3*B*a*c*d*e^4 - 2*A*a*c*e^5)*x^2 + 5*(125*B*c^2*d^4*e - 12*A*c^2*d^3*e^2 - 6*B*a*c*d^2*e^3 - 4*A*a*c*d*e^4 - 3*B*a^2*e^5)*x)/(e^11 *x^5 + 5*d*e^10*x^4 + 10*d^2*e^9*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5* e^6) + B*c^2*log(e*x + d)/e^6
Time = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B c^{2} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {60 \, {\left (5 \, B c^{2} d e^{3} - A c^{2} e^{4}\right )} x^{4} + 60 \, {\left (15 \, B c^{2} d^{2} e^{2} - 2 \, A c^{2} d e^{3} - B a c e^{4}\right )} x^{3} + 20 \, {\left (55 \, B c^{2} d^{3} e - 6 \, A c^{2} d^{2} e^{2} - 3 \, B a c d e^{3} - 2 \, A a c e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} - 12 \, A c^{2} d^{3} e - 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} - 3 \, B a^{2} e^{4}\right )} x + \frac {137 \, B c^{2} d^{5} - 12 \, A c^{2} d^{4} e - 6 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} - 3 \, B a^{2} d e^{4} - 12 \, A a^{2} e^{5}}{e}}{60 \, {\left (e x + d\right )}^{5} e^{5}} \]
B*c^2*log(abs(e*x + d))/e^6 + 1/60*(60*(5*B*c^2*d*e^3 - A*c^2*e^4)*x^4 + 6 0*(15*B*c^2*d^2*e^2 - 2*A*c^2*d*e^3 - B*a*c*e^4)*x^3 + 20*(55*B*c^2*d^3*e - 6*A*c^2*d^2*e^2 - 3*B*a*c*d*e^3 - 2*A*a*c*e^4)*x^2 + 5*(125*B*c^2*d^4 - 12*A*c^2*d^3*e - 6*B*a*c*d^2*e^2 - 4*A*a*c*d*e^3 - 3*B*a^2*e^4)*x + (137*B *c^2*d^5 - 12*A*c^2*d^4*e - 6*B*a*c*d^3*e^2 - 4*A*a*c*d^2*e^3 - 3*B*a^2*d* e^4 - 12*A*a^2*e^5)/e)/((e*x + d)^5*e^5)
Time = 10.60 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B\,c^2\,\ln \left (d+e\,x\right )}{e^6}-\frac {x^2\,\left (-\frac {55\,B\,c^2\,d^3\,e^2}{3}+2\,A\,c^2\,d^2\,e^3+B\,a\,c\,d\,e^4+\frac {2\,A\,a\,c\,e^5}{3}\right )+x^3\,\left (-15\,B\,c^2\,d^2\,e^3+2\,A\,c^2\,d\,e^4+B\,a\,c\,e^5\right )+x^4\,\left (A\,c^2\,e^5-5\,B\,c^2\,d\,e^4\right )+x\,\left (\frac {B\,a^2\,e^5}{4}+\frac {B\,a\,c\,d^2\,e^3}{2}+\frac {A\,a\,c\,d\,e^4}{3}-\frac {125\,B\,c^2\,d^4\,e}{12}+A\,c^2\,d^3\,e^2\right )+\frac {A\,a^2\,e^5}{5}-\frac {137\,B\,c^2\,d^5}{60}+\frac {B\,a^2\,d\,e^4}{20}+\frac {A\,c^2\,d^4\,e}{5}+\frac {A\,a\,c\,d^2\,e^3}{15}+\frac {B\,a\,c\,d^3\,e^2}{10}}{e^6\,{\left (d+e\,x\right )}^5} \]
(B*c^2*log(d + e*x))/e^6 - (x^2*((2*A*a*c*e^5)/3 + 2*A*c^2*d^2*e^3 - (55*B *c^2*d^3*e^2)/3 + B*a*c*d*e^4) + x^3*(B*a*c*e^5 + 2*A*c^2*d*e^4 - 15*B*c^2 *d^2*e^3) + x^4*(A*c^2*e^5 - 5*B*c^2*d*e^4) + x*((B*a^2*e^5)/4 - (125*B*c^ 2*d^4*e)/12 + A*c^2*d^3*e^2 + (A*a*c*d*e^4)/3 + (B*a*c*d^2*e^3)/2) + (A*a^ 2*e^5)/5 - (137*B*c^2*d^5)/60 + (B*a^2*d*e^4)/20 + (A*c^2*d^4*e)/5 + (A*a* c*d^2*e^3)/15 + (B*a*c*d^3*e^2)/10)/(e^6*(d + e*x)^5)