Integrand size = 20, antiderivative size = 101 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {(B d-A e) \left (c d^2+a e^2\right )}{3 e^4 (d+e x)^3}-\frac {3 B c d^2-2 A c d e+a B e^2}{2 e^4 (d+e x)^2}+\frac {c (3 B d-A e)}{e^4 (d+e x)}+\frac {B c \log (d+e x)}{e^4} \] Output:
1/3*(-A*e+B*d)*(a*e^2+c*d^2)/e^4/(e*x+d)^3-1/2*(-2*A*c*d*e+B*a*e^2+3*B*c*d ^2)/e^4/(e*x+d)^2+c*(-A*e+3*B*d)/e^4/(e*x+d)+B*c*ln(e*x+d)/e^4
Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {-2 A e \left (a e^2+c \left (d^2+3 d e x+3 e^2 x^2\right )\right )+B \left (-a e^2 (d+3 e x)+c d \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )+6 B c (d+e x)^3 \log (d+e x)}{6 e^4 (d+e x)^3} \] Input:
Integrate[((A + B*x)*(a + c*x^2))/(d + e*x)^4,x]
Output:
(-2*A*e*(a*e^2 + c*(d^2 + 3*d*e*x + 3*e^2*x^2)) + B*(-(a*e^2*(d + 3*e*x)) + c*d*(11*d^2 + 27*d*e*x + 18*e^2*x^2)) + 6*B*c*(d + e*x)^3*Log[d + e*x])/ (6*e^4*(d + e*x)^3)
Time = 0.26 (sec) , antiderivative size = 101, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) (A+B x)}{(d+e x)^4} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {a B e^2-2 A c d e+3 B c d^2}{e^3 (d+e x)^3}+\frac {\left (a e^2+c d^2\right ) (A e-B d)}{e^3 (d+e x)^4}+\frac {c (A e-3 B d)}{e^3 (d+e x)^2}+\frac {B c}{e^3 (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a B e^2-2 A c d e+3 B c d^2}{2 e^4 (d+e x)^2}+\frac {\left (a e^2+c d^2\right ) (B d-A e)}{3 e^4 (d+e x)^3}+\frac {c (3 B d-A e)}{e^4 (d+e x)}+\frac {B c \log (d+e x)}{e^4}\) |
Input:
Int[((A + B*x)*(a + c*x^2))/(d + e*x)^4,x]
Output:
((B*d - A*e)*(c*d^2 + a*e^2))/(3*e^4*(d + e*x)^3) - (3*B*c*d^2 - 2*A*c*d*e + a*B*e^2)/(2*e^4*(d + e*x)^2) + (c*(3*B*d - A*e))/(e^4*(d + e*x)) + (B*c *Log[d + e*x])/e^4
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 = 0.57 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {-\frac {c \left (A e -3 B d \right ) x^{2}}{e^{2}}-\frac {\left (2 A c d e +B a \,e^{2}-9 B c \,d^{2}\right ) x}{2 e^{3}}-\frac {2 A a \,e^{3}+2 A c \,d^{2} e +B a d \,e^{2}-11 B c \,d^{3}}{6 e^{4}}}{\left (e x +d \right )^{3}}+\frac {B c \ln \left (e x +d \right )}{e^{4}}\) | \(101\) |
norman | \(\frac {-\frac {2 A a \,e^{3}+2 A c \,d^{2} e +B a d \,e^{2}-11 B c \,d^{3}}{6 e^{4}}-\frac {\left (A c e -3 B c d \right ) x^{2}}{e^{2}}-\frac {\left (2 A c d e +B a \,e^{2}-9 B c \,d^{2}\right ) x}{2 e^{3}}}{\left (e x +d \right )^{3}}+\frac {B c \ln \left (e x +d \right )}{e^{4}}\) | \(102\) |
default | \(-\frac {A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}}{3 e^{4} \left (e x +d \right )^{3}}+\frac {B c \ln \left (e x +d \right )}{e^{4}}-\frac {c \left (A e -3 B d \right )}{e^{4} \left (e x +d \right )}-\frac {-2 A c d e +B a \,e^{2}+3 B c \,d^{2}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(108\) |
parallelrisch | \(-\frac {-6 B \ln \left (e x +d \right ) x^{3} c \,e^{3}-18 B \ln \left (e x +d \right ) x^{2} c d \,e^{2}+6 A \,x^{2} c \,e^{3}-18 B \ln \left (e x +d \right ) x c \,d^{2} e -18 B \,x^{2} c d \,e^{2}+6 A x c d \,e^{2}-6 B \ln \left (e x +d \right ) c \,d^{3}+3 B x a \,e^{3}-27 B x c \,d^{2} e +2 A a \,e^{3}+2 A c \,d^{2} e +B a d \,e^{2}-11 B c \,d^{3}}{6 e^{4} \left (e x +d \right )^{3}}\) | \(151\) |
Input:
int((B*x+A)*(c*x^2+a)/(e*x+d)^4,x,method=_RETURNVERBOSE)
Output:
(-c*(A*e-3*B*d)/e^2*x^2-1/2*(2*A*c*d*e+B*a*e^2-9*B*c*d^2)/e^3*x-1/6*(2*A*a *e^3+2*A*c*d^2*e+B*a*d*e^2-11*B*c*d^3)/e^4)/(e*x+d)^3+B*c*ln(e*x+d)/e^4
Time = 0.08 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.58 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {11 \, B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} - 2 \, A a e^{3} + 6 \, {\left (3 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x + 6 \, {\left (B c e^{3} x^{3} + 3 \, B c d e^{2} x^{2} + 3 \, B c d^{2} e x + B c d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^4,x, algorithm="fricas")
Output:
1/6*(11*B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 - 2*A*a*e^3 + 6*(3*B*c*d*e^2 - A *c*e^3)*x^2 + 3*(9*B*c*d^2*e - 2*A*c*d*e^2 - B*a*e^3)*x + 6*(B*c*e^3*x^3 + 3*B*c*d*e^2*x^2 + 3*B*c*d^2*e*x + B*c*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 = 1.14 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {B c \log {\left (d + e x \right )}}{e^{4}} + \frac {- 2 A a e^{3} - 2 A c d^{2} e - B a d e^{2} + 11 B c d^{3} + x^{2} \left (- 6 A c e^{3} + 18 B c d e^{2}\right ) + x \left (- 6 A c d e^{2} - 3 B a e^{3} + 27 B c d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \] Input:
integrate((B*x+A)*(c*x**2+a)/(e*x+d)**4,x)
Output:
B*c*log(d + e*x)/e**4 + (-2*A*a*e**3 - 2*A*c*d**2*e - B*a*d*e**2 + 11*B*c* d**3 + x**2*(-6*A*c*e**3 + 18*B*c*d*e**2) + x*(-6*A*c*d*e**2 - 3*B*a*e**3 + 27*B*c*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 = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {11 \, B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} - 2 \, A a e^{3} + 6 \, {\left (3 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac {B c \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^4,x, algorithm="maxima")
Output:
1/6*(11*B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 - 2*A*a*e^3 + 6*(3*B*c*d*e^2 - A *c*e^3)*x^2 + 3*(9*B*c*d^2*e - 2*A*c*d*e^2 - B*a*e^3)*x)/(e^7*x^3 + 3*d*e^ 6*x^2 + 3*d^2*e^5*x + d^3*e^4) + B*c*log(e*x + d)/e^4
Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {B c \log \left ({\left | e x + d \right |}\right )}{e^{4}} + \frac {6 \, {\left (3 \, B c d e - A c e^{2}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} - 2 \, A c d e - B a e^{2}\right )} x + \frac {11 \, B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} - 2 \, A a e^{3}}{e}}{6 \, {\left (e x + d\right )}^{3} e^{3}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^4,x, algorithm="giac")
Output:
B*c*log(abs(e*x + d))/e^4 + 1/6*(6*(3*B*c*d*e - A*c*e^2)*x^2 + 3*(9*B*c*d^ 2 - 2*A*c*d*e - B*a*e^2)*x + (11*B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 - 2*A*a *e^3)/e)/((e*x + d)^3*e^3)
Time = 5.78 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {B\,c\,\ln \left (d+e\,x\right )}{e^4}-\frac {\frac {-11\,B\,c\,d^3+2\,A\,c\,d^2\,e+B\,a\,d\,e^2+2\,A\,a\,e^3}{6\,e^4}+\frac {x\,\left (-9\,B\,c\,d^2+2\,A\,c\,d\,e+B\,a\,e^2\right )}{2\,e^3}+\frac {c\,x^2\,\left (A\,e-3\,B\,d\right )}{e^2}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \] Input:
int(((a + c*x^2)*(A + B*x))/(d + e*x)^4,x)
Output:
(B*c*log(d + e*x))/e^4 - ((2*A*a*e^3 - 11*B*c*d^3 + B*a*d*e^2 + 2*A*c*d^2* e)/(6*e^4) + (x*(B*a*e^2 - 9*B*c*d^2 + 2*A*c*d*e))/(2*e^3) + (c*x^2*(A*e - 3*B*d))/e^2)/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x)
Time = 0.24 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^4} \, dx=\frac {6 \,\mathrm {log}\left (e x +d \right ) b c \,d^{4}+18 \,\mathrm {log}\left (e x +d \right ) b c \,d^{3} e x +18 \,\mathrm {log}\left (e x +d \right ) b c \,d^{2} e^{2} x^{2}+6 \,\mathrm {log}\left (e x +d \right ) b c d \,e^{3} x^{3}-2 a^{2} d \,e^{3}-a b \,d^{2} e^{2}-3 a b d \,e^{3} x +2 a c \,e^{4} x^{3}+5 b c \,d^{4}+9 b c \,d^{3} e x -6 b c d \,e^{3} x^{3}}{6 d \,e^{4} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )} \] Input:
int((B*x+A)*(c*x^2+a)/(e*x+d)^4,x)
Output:
(6*log(d + e*x)*b*c*d**4 + 18*log(d + e*x)*b*c*d**3*e*x + 18*log(d + e*x)* b*c*d**2*e**2*x**2 + 6*log(d + e*x)*b*c*d*e**3*x**3 - 2*a**2*d*e**3 - a*b* d**2*e**2 - 3*a*b*d*e**3*x + 2*a*c*e**4*x**3 + 5*b*c*d**4 + 9*b*c*d**3*e*x - 6*b*c*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))