Integrand size = 20, antiderivative size = 108 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=\frac {(B d-A e) \left (c d^2+a e^2\right )}{5 e^4 (d+e x)^5}-\frac {3 B c d^2-2 A c d e+a B e^2}{4 e^4 (d+e x)^4}+\frac {c (3 B d-A e)}{3 e^4 (d+e x)^3}-\frac {B c}{2 e^4 (d+e x)^2} \] Output:
1/5*(-A*e+B*d)*(a*e^2+c*d^2)/e^4/(e*x+d)^5-1/4*(-2*A*c*d*e+B*a*e^2+3*B*c*d ^2)/e^4/(e*x+d)^4+1/3*c*(-A*e+3*B*d)/e^4/(e*x+d)^3-1/2*B*c/e^4/(e*x+d)^2
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=-\frac {2 A e \left (6 a e^2+c \left (d^2+5 d e x+10 e^2 x^2\right )\right )+3 B \left (a e^2 (d+5 e x)+c \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )\right )}{60 e^4 (d+e x)^5} \] Input:
Integrate[((A + B*x)*(a + c*x^2))/(d + e*x)^6,x]
Output:
-1/60*(2*A*e*(6*a*e^2 + c*(d^2 + 5*d*e*x + 10*e^2*x^2)) + 3*B*(a*e^2*(d + 5*e*x) + c*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3)))/(e^4*(d + e*x)^ 5)
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) (A+B x)}{(d+e x)^6} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {a B e^2-2 A c d e+3 B c d^2}{e^3 (d+e x)^5}+\frac {\left (a e^2+c d^2\right ) (A e-B d)}{e^3 (d+e x)^6}+\frac {c (A e-3 B d)}{e^3 (d+e x)^4}+\frac {B c}{e^3 (d+e x)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a B e^2-2 A c d e+3 B c d^2}{4 e^4 (d+e x)^4}+\frac {\left (a e^2+c d^2\right ) (B d-A e)}{5 e^4 (d+e x)^5}+\frac {c (3 B d-A e)}{3 e^4 (d+e x)^3}-\frac {B c}{2 e^4 (d+e x)^2}\) |
Input:
Int[((A + B*x)*(a + c*x^2))/(d + e*x)^6,x]
Output:
((B*d - A*e)*(c*d^2 + a*e^2))/(5*e^4*(d + e*x)^5) - (3*B*c*d^2 - 2*A*c*d*e + a*B*e^2)/(4*e^4*(d + e*x)^4) + (c*(3*B*d - A*e))/(3*e^4*(d + e*x)^3) - (B*c)/(2*e^4*(d + e*x)^2)
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 = 0.60 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94
method | result | size |
gosper | \(-\frac {30 B c \,x^{3} e^{3}+20 A \,x^{2} c \,e^{3}+30 B \,x^{2} c d \,e^{2}+10 A x c d \,e^{2}+15 B x a \,e^{3}+15 B x c \,d^{2} e +12 A a \,e^{3}+2 A c \,d^{2} e +3 B a d \,e^{2}+3 B c \,d^{3}}{60 e^{4} \left (e x +d \right )^{5}}\) | \(101\) |
risch | \(\frac {-\frac {B c \,x^{3}}{2 e}-\frac {c \left (2 A e +3 B d \right ) x^{2}}{6 e^{2}}-\frac {\left (2 A c d e +3 B a \,e^{2}+3 B c \,d^{2}\right ) x}{12 e^{3}}-\frac {12 A a \,e^{3}+2 A c \,d^{2} e +3 B a d \,e^{2}+3 B c \,d^{3}}{60 e^{4}}}{\left (e x +d \right )^{5}}\) | \(101\) |
orering | \(-\frac {30 B c \,x^{3} e^{3}+20 A \,x^{2} c \,e^{3}+30 B \,x^{2} c d \,e^{2}+10 A x c d \,e^{2}+15 B x a \,e^{3}+15 B x c \,d^{2} e +12 A a \,e^{3}+2 A c \,d^{2} e +3 B a d \,e^{2}+3 B c \,d^{3}}{60 e^{4} \left (e x +d \right )^{5}}\) | \(101\) |
parallelrisch | \(-\frac {30 B c \,x^{3} e^{4}+20 A c \,e^{4} x^{2}+30 B c d \,e^{3} x^{2}+10 A c d \,e^{3} x +15 B a \,e^{4} x +15 B c \,d^{2} e^{2} x +12 A a \,e^{4}+2 A c \,d^{2} e^{2}+3 B a d \,e^{3}+3 B c \,d^{3} e}{60 e^{5} \left (e x +d \right )^{5}}\) | \(106\) |
default | \(-\frac {c \left (A e -3 B d \right )}{3 e^{4} \left (e x +d \right )^{3}}-\frac {-2 A c d e +B a \,e^{2}+3 B c \,d^{2}}{4 e^{4} \left (e x +d \right )^{4}}-\frac {A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}}{5 e^{4} \left (e x +d \right )^{5}}-\frac {B c}{2 e^{4} \left (e x +d \right )^{2}}\) | \(110\) |
norman | \(\frac {-\frac {B c \,x^{3}}{2 e}-\frac {\left (2 A c \,e^{2}+3 B c d e \right ) x^{2}}{6 e^{3}}-\frac {\left (2 A c d \,e^{2}+3 B \,e^{3} a +3 B c \,d^{2} e \right ) x}{12 e^{4}}-\frac {12 A a \,e^{4}+2 A c \,d^{2} e^{2}+3 B a d \,e^{3}+3 B c \,d^{3} e}{60 e^{5}}}{\left (e x +d \right )^{5}}\) | \(111\) |
Input:
int((B*x+A)*(c*x^2+a)/(e*x+d)^6,x,method=_RETURNVERBOSE)
Output:
-1/60/e^4*(30*B*c*e^3*x^3+20*A*c*e^3*x^2+30*B*c*d*e^2*x^2+10*A*c*d*e^2*x+1 5*B*a*e^3*x+15*B*c*d^2*e*x+12*A*a*e^3+2*A*c*d^2*e+3*B*a*d*e^2+3*B*c*d^3)/( e*x+d)^5
Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B c e^{3} x^{3} + 3 \, B c d^{3} + 2 \, A c d^{2} e + 3 \, B a d e^{2} + 12 \, A a e^{3} + 10 \, {\left (3 \, B c d e^{2} + 2 \, A c e^{3}\right )} x^{2} + 5 \, {\left (3 \, B c d^{2} e + 2 \, A c d e^{2} + 3 \, B a e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^6,x, algorithm="fricas")
Output:
-1/60*(30*B*c*e^3*x^3 + 3*B*c*d^3 + 2*A*c*d^2*e + 3*B*a*d*e^2 + 12*A*a*e^3 + 10*(3*B*c*d*e^2 + 2*A*c*e^3)*x^2 + 5*(3*B*c*d^2*e + 2*A*c*d*e^2 + 3*B*a *e^3)*x)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^7*x^3 + 10*d^3*e^6*x^2 + 5*d^4* e^5*x + d^5*e^4)
Time = 3.46 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.53 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=\frac {- 12 A a e^{3} - 2 A c d^{2} e - 3 B a d e^{2} - 3 B c d^{3} - 30 B c e^{3} x^{3} + x^{2} \left (- 20 A c e^{3} - 30 B c d e^{2}\right ) + x \left (- 10 A c d e^{2} - 15 B a e^{3} - 15 B c d^{2} e\right )}{60 d^{5} e^{4} + 300 d^{4} e^{5} x + 600 d^{3} e^{6} x^{2} + 600 d^{2} e^{7} x^{3} + 300 d e^{8} x^{4} + 60 e^{9} x^{5}} \] Input:
integrate((B*x+A)*(c*x**2+a)/(e*x+d)**6,x)
Output:
(-12*A*a*e**3 - 2*A*c*d**2*e - 3*B*a*d*e**2 - 3*B*c*d**3 - 30*B*c*e**3*x** 3 + x**2*(-20*A*c*e**3 - 30*B*c*d*e**2) + x*(-10*A*c*d*e**2 - 15*B*a*e**3 - 15*B*c*d**2*e))/(60*d**5*e**4 + 300*d**4*e**5*x + 600*d**3*e**6*x**2 + 6 00*d**2*e**7*x**3 + 300*d*e**8*x**4 + 60*e**9*x**5)
Time = 0.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B c e^{3} x^{3} + 3 \, B c d^{3} + 2 \, A c d^{2} e + 3 \, B a d e^{2} + 12 \, A a e^{3} + 10 \, {\left (3 \, B c d e^{2} + 2 \, A c e^{3}\right )} x^{2} + 5 \, {\left (3 \, B c d^{2} e + 2 \, A c d e^{2} + 3 \, B a e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^6,x, algorithm="maxima")
Output:
-1/60*(30*B*c*e^3*x^3 + 3*B*c*d^3 + 2*A*c*d^2*e + 3*B*a*d*e^2 + 12*A*a*e^3 + 10*(3*B*c*d*e^2 + 2*A*c*e^3)*x^2 + 5*(3*B*c*d^2*e + 2*A*c*d*e^2 + 3*B*a *e^3)*x)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^7*x^3 + 10*d^3*e^6*x^2 + 5*d^4* e^5*x + d^5*e^4)
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B c e^{3} x^{3} + 30 \, B c d e^{2} x^{2} + 20 \, A c e^{3} x^{2} + 15 \, B c d^{2} e x + 10 \, A c d e^{2} x + 15 \, B a e^{3} x + 3 \, B c d^{3} + 2 \, A c d^{2} e + 3 \, B a d e^{2} + 12 \, A a e^{3}}{60 \, {\left (e x + d\right )}^{5} e^{4}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^6,x, algorithm="giac")
Output:
-1/60*(30*B*c*e^3*x^3 + 30*B*c*d*e^2*x^2 + 20*A*c*e^3*x^2 + 15*B*c*d^2*e*x + 10*A*c*d*e^2*x + 15*B*a*e^3*x + 3*B*c*d^3 + 2*A*c*d^2*e + 3*B*a*d*e^2 + 12*A*a*e^3)/((e*x + d)^5*e^4)
Time = 0.04 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=-\frac {\frac {3\,B\,c\,d^3+2\,A\,c\,d^2\,e+3\,B\,a\,d\,e^2+12\,A\,a\,e^3}{60\,e^4}+\frac {x\,\left (3\,B\,c\,d^2+2\,A\,c\,d\,e+3\,B\,a\,e^2\right )}{12\,e^3}+\frac {B\,c\,x^3}{2\,e}+\frac {c\,x^2\,\left (2\,A\,e+3\,B\,d\right )}{6\,e^2}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \] Input:
int(((a + c*x^2)*(A + B*x))/(d + e*x)^6,x)
Output:
-((12*A*a*e^3 + 3*B*c*d^3 + 3*B*a*d*e^2 + 2*A*c*d^2*e)/(60*e^4) + (x*(3*B* a*e^2 + 3*B*c*d^2 + 2*A*c*d*e))/(12*e^3) + (B*c*x^3)/(2*e) + (c*x^2*(2*A*e + 3*B*d))/(6*e^2))/(d^5 + e^5*x^5 + 5*d*e^4*x^4 + 10*d^3*e^2*x^2 + 10*d^2 *e^3*x^3 + 5*d^4*e*x)
Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^6} \, dx=\frac {-30 b c \,e^{3} x^{3}-20 a c \,e^{3} x^{2}-30 b c d \,e^{2} x^{2}-15 a b \,e^{3} x -10 a c d \,e^{2} x -15 b c \,d^{2} e x -12 a^{2} e^{3}-3 a b d \,e^{2}-2 a c \,d^{2} e -3 b c \,d^{3}}{60 e^{4} \left (e^{5} x^{5}+5 d \,e^{4} x^{4}+10 d^{2} e^{3} x^{3}+10 d^{3} e^{2} x^{2}+5 d^{4} e x +d^{5}\right )} \] Input:
int((B*x+A)*(c*x^2+a)/(e*x+d)^6,x)
Output:
( - 12*a**2*e**3 - 3*a*b*d*e**2 - 15*a*b*e**3*x - 2*a*c*d**2*e - 10*a*c*d* e**2*x - 20*a*c*e**3*x**2 - 3*b*c*d**3 - 15*b*c*d**2*e*x - 30*b*c*d*e**2*x **2 - 30*b*c*e**3*x**3)/(60*e**4*(d**5 + 5*d**4*e*x + 10*d**3*e**2*x**2 + 10*d**2*e**3*x**3 + 5*d*e**4*x**4 + e**5*x**5))