Integrand size = 20, antiderivative size = 94 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^3}+\frac {(B d-A e) \left (c d^2+a e^2\right )}{2 e^4 (d+e x)^2}-\frac {3 B c d^2-2 A c d e+a B e^2}{e^4 (d+e x)}-\frac {c (3 B d-A e) \log (d+e x)}{e^4} \] Output:
B*c*x/e^3+1/2*(-A*e+B*d)*(a*e^2+c*d^2)/e^4/(e*x+d)^2-(-2*A*c*d*e+B*a*e^2+3 *B*c*d^2)/e^4/(e*x+d)-c*(-A*e+3*B*d)*ln(e*x+d)/e^4
Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 B c e x+\frac {(B d-A e) \left (c d^2+a e^2\right )}{(d+e x)^2}-\frac {2 \left (3 B c d^2-2 A c d e+a B e^2\right )}{d+e x}+2 (-3 B c d+A c e) \log (d+e x)}{2 e^4} \] Input:
Integrate[((A + B*x)*(a + c*x^2))/(d + e*x)^3,x]
Output:
(2*B*c*e*x + ((B*d - A*e)*(c*d^2 + a*e^2))/(d + e*x)^2 - (2*(3*B*c*d^2 - 2 *A*c*d*e + a*B*e^2))/(d + e*x) + 2*(-3*B*c*d + A*c*e)*Log[d + e*x])/(2*e^4 )
Time = 0.26 (sec) , antiderivative size = 94, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) (A+B x)}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {a B e^2-2 A c d e+3 B c d^2}{e^3 (d+e x)^2}+\frac {\left (a e^2+c d^2\right ) (A e-B d)}{e^3 (d+e x)^3}+\frac {c (A e-3 B d)}{e^3 (d+e x)}+\frac {B c}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a e^2+c d^2\right ) (B d-A e)}{2 e^4 (d+e x)^2}-\frac {a B e^2-2 A c d e+3 B c d^2}{e^4 (d+e x)}-\frac {c (3 B d-A e) \log (d+e x)}{e^4}+\frac {B c x}{e^3}\) |
Input:
Int[((A + B*x)*(a + c*x^2))/(d + e*x)^3,x]
Output:
(B*c*x)/e^3 + ((B*d - A*e)*(c*d^2 + a*e^2))/(2*e^4*(d + e*x)^2) - (3*B*c*d ^2 - 2*A*c*d*e + a*B*e^2)/(e^4*(d + e*x)) - (c*(3*B*d - A*e)*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 = 0.58 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {\frac {\left (2 A c d e -B a \,e^{2}-6 B c \,d^{2}\right ) x}{e^{3}}+\frac {B c \,x^{3}}{e}-\frac {A a \,e^{3}-3 A c \,d^{2} e +B a d \,e^{2}+9 B c \,d^{3}}{2 e^{4}}}{\left (e x +d \right )^{2}}+\frac {c \left (A e -3 B d \right ) \ln \left (e x +d \right )}{e^{4}}\) | \(99\) |
default | \(\frac {B c x}{e^{3}}+\frac {c \left (A e -3 B d \right ) \ln \left (e x +d \right )}{e^{4}}-\frac {-2 A c d e +B a \,e^{2}+3 B c \,d^{2}}{e^{4} \left (e x +d \right )}-\frac {A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(101\) |
risch | \(\frac {B c x}{e^{3}}+\frac {\left (2 A c d e -B a \,e^{2}-3 B c \,d^{2}\right ) x -\frac {A a \,e^{3}-3 A c \,d^{2} e +B a d \,e^{2}+5 B c \,d^{3}}{2 e}}{e^{3} \left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right ) A}{e^{3}}-\frac {3 c \ln \left (e x +d \right ) B d}{e^{4}}\) | \(104\) |
parallelrisch | \(\frac {2 A \ln \left (e x +d \right ) x^{2} c \,e^{3}-6 B \ln \left (e x +d \right ) x^{2} c d \,e^{2}+2 B c \,x^{3} e^{3}+4 A \ln \left (e x +d \right ) x c d \,e^{2}-12 B \ln \left (e x +d \right ) x c \,d^{2} e +2 A \ln \left (e x +d \right ) c \,d^{2} e +4 A x c d \,e^{2}-6 B \ln \left (e x +d \right ) c \,d^{3}-2 B x a \,e^{3}-12 B x c \,d^{2} e -A a \,e^{3}+3 A c \,d^{2} e -B a d \,e^{2}-9 B c \,d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(170\) |
Input:
int((B*x+A)*(c*x^2+a)/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
((2*A*c*d*e-B*a*e^2-6*B*c*d^2)/e^3*x+B*c*x^3/e-1/2*(A*a*e^3-3*A*c*d^2*e+B* a*d*e^2+9*B*c*d^3)/e^4)/(e*x+d)^2+c/e^4*(A*e-3*B*d)*ln(e*x+d)
Time = 0.14 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.79 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 \, B c e^{3} x^{3} + 4 \, B c d e^{2} x^{2} - 5 \, B c d^{3} + 3 \, A c d^{2} e - B a d e^{2} - A a e^{3} - 2 \, {\left (2 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x - 2 \, {\left (3 \, B c d^{3} - A c d^{2} e + {\left (3 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 2 \, {\left (3 \, B c d^{2} e - A c d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^3,x, algorithm="fricas")
Output:
1/2*(2*B*c*e^3*x^3 + 4*B*c*d*e^2*x^2 - 5*B*c*d^3 + 3*A*c*d^2*e - B*a*d*e^2 - A*a*e^3 - 2*(2*B*c*d^2*e - 2*A*c*d*e^2 + B*a*e^3)*x - 2*(3*B*c*d^3 - A* c*d^2*e + (3*B*c*d*e^2 - A*c*e^3)*x^2 + 2*(3*B*c*d^2*e - A*c*d*e^2)*x)*log (e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)
Time = 0.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^{3}} - \frac {c \left (- A e + 3 B d\right ) \log {\left (d + e x \right )}}{e^{4}} + \frac {- A a e^{3} + 3 A c d^{2} e - B a d e^{2} - 5 B c d^{3} + x \left (4 A c d e^{2} - 2 B a e^{3} - 6 B c d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \] Input:
integrate((B*x+A)*(c*x**2+a)/(e*x+d)**3,x)
Output:
B*c*x/e**3 - c*(-A*e + 3*B*d)*log(d + e*x)/e**4 + (-A*a*e**3 + 3*A*c*d**2* e - B*a*d*e**2 - 5*B*c*d**3 + x*(4*A*c*d*e**2 - 2*B*a*e**3 - 6*B*c*d**2*e) )/(2*d**2*e**4 + 4*d*e**5*x + 2*e**6*x**2)
Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=-\frac {5 \, B c d^{3} - 3 \, A c d^{2} e + B a d e^{2} + A a e^{3} + 2 \, {\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - A c e\right )} \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^3,x, algorithm="maxima")
Output:
-1/2*(5*B*c*d^3 - 3*A*c*d^2*e + B*a*d*e^2 + A*a*e^3 + 2*(3*B*c*d^2*e - 2*A *c*d*e^2 + B*a*e^3)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + B*c*x/e^3 - (3*B* c*d - A*c*e)*log(e*x + d)/e^4
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - A c e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} - \frac {5 \, B c d^{3} - 3 \, A c d^{2} e + B a d e^{2} + A a e^{3} + 2 \, {\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{4}} \] Input:
integrate((B*x+A)*(c*x^2+a)/(e*x+d)^3,x, algorithm="giac")
Output:
B*c*x/e^3 - (3*B*c*d - A*c*e)*log(abs(e*x + d))/e^4 - 1/2*(5*B*c*d^3 - 3*A *c*d^2*e + B*a*d*e^2 + A*a*e^3 + 2*(3*B*c*d^2*e - 2*A*c*d*e^2 + B*a*e^3)*x )/((e*x + d)^2*e^4)
Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (A\,c\,e-3\,B\,c\,d\right )}{e^4}-\frac {\frac {5\,B\,c\,d^3-3\,A\,c\,d^2\,e+B\,a\,d\,e^2+A\,a\,e^3}{2\,e}+x\,\left (3\,B\,c\,d^2-2\,A\,c\,d\,e+B\,a\,e^2\right )}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,c\,x}{e^3} \] Input:
int(((a + c*x^2)*(A + B*x))/(d + e*x)^3,x)
Output:
(log(d + e*x)*(A*c*e - 3*B*c*d))/e^4 - ((A*a*e^3 + 5*B*c*d^3 + B*a*d*e^2 - 3*A*c*d^2*e)/(2*e) + x*(B*a*e^2 + 3*B*c*d^2 - 2*A*c*d*e))/(d^2*e^3 + e^5* x^2 + 2*d*e^4*x) + (B*c*x)/e^3
Time = 0.22 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.01 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 \,\mathrm {log}\left (e x +d \right ) a c \,d^{3} e +4 \,\mathrm {log}\left (e x +d \right ) a c \,d^{2} e^{2} x +2 \,\mathrm {log}\left (e x +d \right ) a c d \,e^{3} x^{2}-6 \,\mathrm {log}\left (e x +d \right ) b c \,d^{4}-12 \,\mathrm {log}\left (e x +d \right ) b c \,d^{3} e x -6 \,\mathrm {log}\left (e x +d \right ) b c \,d^{2} e^{2} x^{2}-a^{2} d \,e^{3}+a b \,e^{4} x^{2}+a c \,d^{3} e -2 a c d \,e^{3} x^{2}-3 b c \,d^{4}+6 b c \,d^{2} e^{2} x^{2}+2 b c d \,e^{3} x^{3}}{2 d \,e^{4} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((B*x+A)*(c*x^2+a)/(e*x+d)^3,x)
Output:
(2*log(d + e*x)*a*c*d**3*e + 4*log(d + e*x)*a*c*d**2*e**2*x + 2*log(d + e* x)*a*c*d*e**3*x**2 - 6*log(d + e*x)*b*c*d**4 - 12*log(d + e*x)*b*c*d**3*e* x - 6*log(d + e*x)*b*c*d**2*e**2*x**2 - a**2*d*e**3 + a*b*e**4*x**2 + a*c* d**3*e - 2*a*c*d*e**3*x**2 - 3*b*c*d**4 + 6*b*c*d**2*e**2*x**2 + 2*b*c*d*e **3*x**3)/(2*d*e**4*(d**2 + 2*d*e*x + e**2*x**2))