\(\int \frac {(A+B x) (a+c x^2)^2}{(d+e x)^3} \, dx\) [54]

Optimal result

Integrand size = 22, antiderivative size = 185 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {c \left (6 B c d^2-3 A c d e+2 a B e^2\right ) x}{e^5}-\frac {c^2 (3 B d-A e) x^2}{2 e^4}+\frac {B c^2 x^3}{3 e^3}+\frac {(B d-A e) \left (c d^2+a e^2\right )^2}{2 e^6 (d+e x)^2}-\frac {\left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right )}{e^6 (d+e x)}-\frac {2 c \left (5 B c d^3-3 A c d^2 e+3 a B d e^2-a A e^3\right ) \log (d+e x)}{e^6} \] Output:

c*(-3*A*c*d*e+2*B*a*e^2+6*B*c*d^2)*x/e^5-1/2*c^2*(-A*e+3*B*d)*x^2/e^4+1/3* 
B*c^2*x^3/e^3+1/2*(-A*e+B*d)*(a*e^2+c*d^2)^2/e^6/(e*x+d)^2-(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)-2*c*(-A*a*e^3-3*A*c*d^2*e+3*B*a* 
d*e^2+5*B*c*d^3)*ln(e*x+d)/e^6
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {6 c e \left (6 B c d^2-3 A c d e+2 a B e^2\right ) x+3 c^2 e^2 (-3 B d+A e) x^2+2 B c^2 e^3 x^3+\frac {3 (B d-A e) \left (c d^2+a e^2\right )^2}{(d+e x)^2}-\frac {6 \left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right )}{d+e x}+12 c \left (-5 B c d^3+3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log (d+e x)}{6 e^6} \] Input:

Integrate[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^3,x]
 

Output:

(6*c*e*(6*B*c*d^2 - 3*A*c*d*e + 2*a*B*e^2)*x + 3*c^2*e^2*(-3*B*d + A*e)*x^ 
2 + 2*B*c^2*e^3*x^3 + (3*(B*d - A*e)*(c*d^2 + a*e^2)^2)/(d + e*x)^2 - (6*( 
c*d^2 + a*e^2)*(5*B*c*d^2 - 4*A*c*d*e + a*B*e^2))/(d + e*x) + 12*c*(-5*B*c 
*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2 + a*A*e^3)*Log[d + e*x])/(6*e^6)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 185, 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.

\(\displaystyle \int \frac {\left (a+c x^2\right )^2 (A+B x)}{(d+e x)^3} \, dx\)

\(\Big \downarrow \) 652

\(\displaystyle \int \left (\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^5 (d+e x)^2}+\frac {\left (a e^2+c d^2\right )^2 (A e-B d)}{e^5 (d+e x)^3}-\frac {c \left (-2 a B e^2+3 A c d e-6 B c d^2\right )}{e^5}+\frac {2 c \left (a A e^3-3 a B d e^2+3 A c d^2 e-5 B c d^3\right )}{e^5 (d+e x)}+\frac {c^2 x (A e-3 B d)}{e^4}+\frac {B c^2 x^2}{e^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^6 (d+e x)}+\frac {\left (a e^2+c d^2\right )^2 (B d-A e)}{2 e^6 (d+e x)^2}+\frac {c x \left (2 a B e^2-3 A c d e+6 B c d^2\right )}{e^5}-\frac {2 c \log (d+e x) \left (-a A e^3+3 a B d e^2-3 A c d^2 e+5 B c d^3\right )}{e^6}-\frac {c^2 x^2 (3 B d-A e)}{2 e^4}+\frac {B c^2 x^3}{3 e^3}\)

Input:

Int[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^3,x]
 

Output:

(c*(6*B*c*d^2 - 3*A*c*d*e + 2*a*B*e^2)*x)/e^5 - (c^2*(3*B*d - A*e)*x^2)/(2 
*e^4) + (B*c^2*x^3)/(3*e^3) + ((B*d - A*e)*(c*d^2 + a*e^2)^2)/(2*e^6*(d + 
e*x)^2) - ((c*d^2 + a*e^2)*(5*B*c*d^2 - 4*A*c*d*e + a*B*e^2))/(e^6*(d + e* 
x)) - (2*c*(5*B*c*d^3 - 3*A*c*d^2*e + 3*a*B*d*e^2 - a*A*e^3)*Log[d + e*x]) 
/e^6
 

Defintions of rubi rules used

rule 652

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {\frac {\left (4 A a c d \,e^{3}+12 A \,c^{2} d^{3} e -B \,e^{4} a^{2}-12 B a c \,d^{2} e^{2}-20 B \,c^{2} d^{4}\right ) x}{e^{5}}-\frac {A \,a^{2} e^{5}-6 A a c \,d^{2} e^{3}-18 A \,c^{2} d^{4} e +B \,a^{2} d \,e^{4}+18 B a c \,d^{3} e^{2}+30 B \,c^{2} d^{5}}{2 e^{6}}+\frac {B \,c^{2} x^{5}}{3 e}-\frac {2 c \left (3 A c d e -3 B a \,e^{2}-5 B c \,d^{2}\right ) x^{3}}{3 e^{3}}+\frac {c^{2} \left (3 A e -5 B d \right ) x^{4}}{6 e^{2}}}{\left (e x +d \right )^{2}}+\frac {2 c \left (A a \,e^{3}+3 A c \,d^{2} e -3 B a d \,e^{2}-5 B c \,d^{3}\right ) \ln \left (e x +d \right )}{e^{6}}\)	\(233\)
default	\(-\frac {c \left (-\frac {1}{3} B c \,x^{3} e^{2}-\frac {1}{2} A c \,e^{2} x^{2}+\frac {3}{2} B c d e \,x^{2}+3 A c d e x -2 B a \,e^{2} x -6 B c \,d^{2} x \right )}{e^{5}}+\frac {2 c \left (A a \,e^{3}+3 A c \,d^{2} e -3 B a d \,e^{2}-5 B c \,d^{3}\right ) \ln \left (e x +d \right )}{e^{6}}-\frac {-4 A a c d \,e^{3}-4 A \,c^{2} d^{3} e +B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}+5 B \,c^{2} d^{4}}{e^{6} \left (e x +d \right )}-\frac {A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}}{2 e^{6} \left (e x +d \right )^{2}}\)	\(234\)
risch	\(\frac {B \,c^{2} x^{3}}{3 e^{3}}+\frac {c^{2} A \,x^{2}}{2 e^{3}}-\frac {3 c^{2} B d \,x^{2}}{2 e^{4}}-\frac {3 c^{2} A d x}{e^{4}}+\frac {2 c B a x}{e^{3}}+\frac {6 c^{2} B \,d^{2} x}{e^{5}}+\frac {\left (4 A a c d \,e^{3}+4 A \,c^{2} d^{3} e -B \,e^{4} a^{2}-6 B a c \,d^{2} e^{2}-5 B \,c^{2} d^{4}\right ) x -\frac {A \,a^{2} e^{5}-6 A a c \,d^{2} e^{3}-7 A \,c^{2} d^{4} e +B \,a^{2} d \,e^{4}+10 B a c \,d^{3} e^{2}+9 B \,c^{2} d^{5}}{2 e}}{e^{5} \left (e x +d \right )^{2}}+\frac {2 c \ln \left (e x +d \right ) A a}{e^{3}}+\frac {6 c^{2} \ln \left (e x +d \right ) A \,d^{2}}{e^{5}}-\frac {6 c \ln \left (e x +d \right ) B a d}{e^{4}}-\frac {10 c^{2} \ln \left (e x +d \right ) B \,d^{3}}{e^{6}}\)	\(264\)
parallelrisch	\(\frac {3 A \,x^{4} c^{2} e^{5}-60 B \ln \left (e x +d \right ) c^{2} d^{5}-6 B x \,a^{2} e^{5}+72 A \ln \left (e x +d \right ) x \,c^{2} d^{3} e^{2}-120 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e +72 A x \,c^{2} d^{3} e^{2}+18 A a c \,d^{2} e^{3}-54 B a c \,d^{3} e^{2}-5 B \,x^{4} c^{2} d \,e^{4}-12 A \,x^{3} c^{2} d \,e^{4}+12 B \,x^{3} a c \,e^{5}-36 B \ln \left (e x +d \right ) x^{2} a c d \,e^{4}-72 B \ln \left (e x +d \right ) x a c \,d^{2} e^{3}+24 A \ln \left (e x +d \right ) x a c d \,e^{4}-120 B x \,c^{2} d^{4} e +12 A \ln \left (e x +d \right ) a c \,d^{2} e^{3}+24 A x a c d \,e^{4}-3 B \,a^{2} d \,e^{4}+36 A \ln \left (e x +d \right ) c^{2} d^{4} e +54 A \,c^{2} d^{4} e +20 B \,x^{3} c^{2} d^{2} e^{3}-36 B \ln \left (e x +d \right ) a c \,d^{3} e^{2}-72 B x a c \,d^{2} e^{3}+12 A \ln \left (e x +d \right ) x^{2} a c \,e^{5}+36 A \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{3}-60 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}+2 B \,x^{5} c^{2} e^{5}-90 B \,c^{2} d^{5}-3 A \,a^{2} e^{5}}{6 e^{6} \left (e x +d \right )^{2}}\)	\(418\)

Input:

int((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x,method=_RETURNVERBOSE)
 

Output:

((4*A*a*c*d*e^3+12*A*c^2*d^3*e-B*a^2*e^4-12*B*a*c*d^2*e^2-20*B*c^2*d^4)/e^ 
5*x-1/2*(A*a^2*e^5-6*A*a*c*d^2*e^3-18*A*c^2*d^4*e+B*a^2*d*e^4+18*B*a*c*d^3 
*e^2+30*B*c^2*d^5)/e^6+1/3*B*c^2*x^5/e-2/3*c*(3*A*c*d*e-3*B*a*e^2-5*B*c*d^ 
2)/e^3*x^3+1/6*c^2*(3*A*e-5*B*d)/e^2*x^4)/(e*x+d)^2+2*c/e^6*(A*a*e^3+3*A*c 
*d^2*e-3*B*a*d*e^2-5*B*c*d^3)*ln(e*x+d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (179) = 358\).

Time = 0.08 (sec) , antiderivative size = 394, normalized size of antiderivative = 2.13 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {2 \, B c^{2} e^{5} x^{5} - 27 \, B c^{2} d^{5} + 21 \, A c^{2} d^{4} e - 30 \, B a c d^{3} e^{2} + 18 \, A a c d^{2} e^{3} - 3 \, B a^{2} d e^{4} - 3 \, A a^{2} e^{5} - {\left (5 \, B c^{2} d e^{4} - 3 \, A c^{2} e^{5}\right )} x^{4} + 4 \, {\left (5 \, B c^{2} d^{2} e^{3} - 3 \, A c^{2} d e^{4} + 3 \, B a c e^{5}\right )} x^{3} + 3 \, {\left (21 \, B c^{2} d^{3} e^{2} - 11 \, A c^{2} d^{2} e^{3} + 8 \, B a c d e^{4}\right )} x^{2} + 6 \, {\left (B c^{2} d^{4} e + A c^{2} d^{3} e^{2} - 4 \, B a c d^{2} e^{3} + 4 \, A a c d e^{4} - B a^{2} e^{5}\right )} x - 12 \, {\left (5 \, B c^{2} d^{5} - 3 \, A c^{2} d^{4} e + 3 \, B a c d^{3} e^{2} - A a c d^{2} e^{3} + {\left (5 \, B c^{2} d^{3} e^{2} - 3 \, A c^{2} d^{2} e^{3} + 3 \, B a c d e^{4} - A a c e^{5}\right )} x^{2} + 2 \, {\left (5 \, B c^{2} d^{4} e - 3 \, A c^{2} d^{3} e^{2} + 3 \, B a c d^{2} e^{3} - A a c d e^{4}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} \] Input:

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="fricas")
 

Output:

1/6*(2*B*c^2*e^5*x^5 - 27*B*c^2*d^5 + 21*A*c^2*d^4*e - 30*B*a*c*d^3*e^2 + 
18*A*a*c*d^2*e^3 - 3*B*a^2*d*e^4 - 3*A*a^2*e^5 - (5*B*c^2*d*e^4 - 3*A*c^2* 
e^5)*x^4 + 4*(5*B*c^2*d^2*e^3 - 3*A*c^2*d*e^4 + 3*B*a*c*e^5)*x^3 + 3*(21*B 
*c^2*d^3*e^2 - 11*A*c^2*d^2*e^3 + 8*B*a*c*d*e^4)*x^2 + 6*(B*c^2*d^4*e + A* 
c^2*d^3*e^2 - 4*B*a*c*d^2*e^3 + 4*A*a*c*d*e^4 - B*a^2*e^5)*x - 12*(5*B*c^2 
*d^5 - 3*A*c^2*d^4*e + 3*B*a*c*d^3*e^2 - A*a*c*d^2*e^3 + (5*B*c^2*d^3*e^2 
- 3*A*c^2*d^2*e^3 + 3*B*a*c*d*e^4 - A*a*c*e^5)*x^2 + 2*(5*B*c^2*d^4*e - 3* 
A*c^2*d^3*e^2 + 3*B*a*c*d^2*e^3 - A*a*c*d*e^4)*x)*log(e*x + d))/(e^8*x^2 + 
 2*d*e^7*x + d^2*e^6)
 

Sympy [A] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.52 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {B c^{2} x^{3}}{3 e^{3}} - \frac {2 c \left (- A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} + 5 B c d^{3}\right ) \log {\left (d + e x \right )}}{e^{6}} + x^{2} \left (\frac {A c^{2}}{2 e^{3}} - \frac {3 B c^{2} d}{2 e^{4}}\right ) + x \left (- \frac {3 A c^{2} d}{e^{4}} + \frac {2 B a c}{e^{3}} + \frac {6 B c^{2} d^{2}}{e^{5}}\right ) + \frac {- A a^{2} e^{5} + 6 A a c d^{2} e^{3} + 7 A c^{2} d^{4} e - B a^{2} d e^{4} - 10 B a c d^{3} e^{2} - 9 B c^{2} d^{5} + x \left (8 A a c d e^{4} + 8 A c^{2} d^{3} e^{2} - 2 B a^{2} e^{5} - 12 B a c d^{2} e^{3} - 10 B c^{2} d^{4} e\right )}{2 d^{2} e^{6} + 4 d e^{7} x + 2 e^{8} x^{2}} \] Input:

integrate((B*x+A)*(c*x**2+a)**2/(e*x+d)**3,x)
 

Output:

B*c**2*x**3/(3*e**3) - 2*c*(-A*a*e**3 - 3*A*c*d**2*e + 3*B*a*d*e**2 + 5*B* 
c*d**3)*log(d + e*x)/e**6 + x**2*(A*c**2/(2*e**3) - 3*B*c**2*d/(2*e**4)) + 
 x*(-3*A*c**2*d/e**4 + 2*B*a*c/e**3 + 6*B*c**2*d**2/e**5) + (-A*a**2*e**5 
+ 6*A*a*c*d**2*e**3 + 7*A*c**2*d**4*e - B*a**2*d*e**4 - 10*B*a*c*d**3*e**2 
 - 9*B*c**2*d**5 + x*(8*A*a*c*d*e**4 + 8*A*c**2*d**3*e**2 - 2*B*a**2*e**5 
- 12*B*a*c*d**2*e**3 - 10*B*c**2*d**4*e))/(2*d**2*e**6 + 4*d*e**7*x + 2*e* 
*8*x**2)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=-\frac {9 \, B c^{2} d^{5} - 7 \, A c^{2} d^{4} e + 10 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + A a^{2} e^{5} + 2 \, {\left (5 \, B c^{2} d^{4} e - 4 \, A c^{2} d^{3} e^{2} + 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} + B a^{2} e^{5}\right )} x}{2 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} + \frac {2 \, B c^{2} e^{2} x^{3} - 3 \, {\left (3 \, B c^{2} d e - A c^{2} e^{2}\right )} x^{2} + 6 \, {\left (6 \, B c^{2} d^{2} - 3 \, A c^{2} d e + 2 \, B a c e^{2}\right )} x}{6 \, e^{5}} - \frac {2 \, {\left (5 \, B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, B a c d e^{2} - A a c e^{3}\right )} \log \left (e x + d\right )}{e^{6}} \] Input:

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="maxima")
 

Output:

-1/2*(9*B*c^2*d^5 - 7*A*c^2*d^4*e + 10*B*a*c*d^3*e^2 - 6*A*a*c*d^2*e^3 + B 
*a^2*d*e^4 + A*a^2*e^5 + 2*(5*B*c^2*d^4*e - 4*A*c^2*d^3*e^2 + 6*B*a*c*d^2* 
e^3 - 4*A*a*c*d*e^4 + B*a^2*e^5)*x)/(e^8*x^2 + 2*d*e^7*x + d^2*e^6) + 1/6* 
(2*B*c^2*e^2*x^3 - 3*(3*B*c^2*d*e - A*c^2*e^2)*x^2 + 6*(6*B*c^2*d^2 - 3*A* 
c^2*d*e + 2*B*a*c*e^2)*x)/e^5 - 2*(5*B*c^2*d^3 - 3*A*c^2*d^2*e + 3*B*a*c*d 
*e^2 - A*a*c*e^3)*log(e*x + d)/e^6
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=-\frac {2 \, {\left (5 \, B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, B a c d e^{2} - A a c e^{3}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} - \frac {9 \, B c^{2} d^{5} - 7 \, A c^{2} d^{4} e + 10 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + A a^{2} e^{5} + 2 \, {\left (5 \, B c^{2} d^{4} e - 4 \, A c^{2} d^{3} e^{2} + 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} + B a^{2} e^{5}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{6}} + \frac {2 \, B c^{2} e^{6} x^{3} - 9 \, B c^{2} d e^{5} x^{2} + 3 \, A c^{2} e^{6} x^{2} + 36 \, B c^{2} d^{2} e^{4} x - 18 \, A c^{2} d e^{5} x + 12 \, B a c e^{6} x}{6 \, e^{9}} \] Input:

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="giac")
 

Output:

-2*(5*B*c^2*d^3 - 3*A*c^2*d^2*e + 3*B*a*c*d*e^2 - A*a*c*e^3)*log(abs(e*x + 
 d))/e^6 - 1/2*(9*B*c^2*d^5 - 7*A*c^2*d^4*e + 10*B*a*c*d^3*e^2 - 6*A*a*c*d 
^2*e^3 + B*a^2*d*e^4 + A*a^2*e^5 + 2*(5*B*c^2*d^4*e - 4*A*c^2*d^3*e^2 + 6* 
B*a*c*d^2*e^3 - 4*A*a*c*d*e^4 + B*a^2*e^5)*x)/((e*x + d)^2*e^6) + 1/6*(2*B 
*c^2*e^6*x^3 - 9*B*c^2*d*e^5*x^2 + 3*A*c^2*e^6*x^2 + 36*B*c^2*d^2*e^4*x - 
18*A*c^2*d*e^5*x + 12*B*a*c*e^6*x)/e^9
 

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.49 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=x^2\,\left (\frac {A\,c^2}{2\,e^3}-\frac {3\,B\,c^2\,d}{2\,e^4}\right )-\frac {x\,\left (B\,a^2\,e^4+6\,B\,a\,c\,d^2\,e^2-4\,A\,a\,c\,d\,e^3+5\,B\,c^2\,d^4-4\,A\,c^2\,d^3\,e\right )+\frac {B\,a^2\,d\,e^4+A\,a^2\,e^5+10\,B\,a\,c\,d^3\,e^2-6\,A\,a\,c\,d^2\,e^3+9\,B\,c^2\,d^5-7\,A\,c^2\,d^4\,e}{2\,e}}{d^2\,e^5+2\,d\,e^6\,x+e^7\,x^2}-x\,\left (\frac {3\,d\,\left (\frac {A\,c^2}{e^3}-\frac {3\,B\,c^2\,d}{e^4}\right )}{e}-\frac {2\,B\,a\,c}{e^3}+\frac {3\,B\,c^2\,d^2}{e^5}\right )-\frac {\ln \left (d+e\,x\right )\,\left (10\,B\,c^2\,d^3-6\,A\,c^2\,d^2\,e+6\,B\,a\,c\,d\,e^2-2\,A\,a\,c\,e^3\right )}{e^6}+\frac {B\,c^2\,x^3}{3\,e^3} \] Input:

int(((a + c*x^2)^2*(A + B*x))/(d + e*x)^3,x)
 

Output:

x^2*((A*c^2)/(2*e^3) - (3*B*c^2*d)/(2*e^4)) - (x*(B*a^2*e^4 + 5*B*c^2*d^4 
- 4*A*c^2*d^3*e - 4*A*a*c*d*e^3 + 6*B*a*c*d^2*e^2) + (A*a^2*e^5 + 9*B*c^2* 
d^5 + B*a^2*d*e^4 - 7*A*c^2*d^4*e - 6*A*a*c*d^2*e^3 + 10*B*a*c*d^3*e^2)/(2 
*e))/(d^2*e^5 + e^7*x^2 + 2*d*e^6*x) - x*((3*d*((A*c^2)/e^3 - (3*B*c^2*d)/ 
e^4))/e - (2*B*a*c)/e^3 + (3*B*c^2*d^2)/e^5) - (log(d + e*x)*(10*B*c^2*d^3 
 - 2*A*a*c*e^3 - 6*A*c^2*d^2*e + 6*B*a*c*d*e^2))/e^6 + (B*c^2*x^3)/(3*e^3)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.43 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {36 a b c \,d^{2} e^{4} x^{2}+12 a b c d \,e^{5} x^{3}-3 a^{3} d \,e^{5}-30 b \,c^{2} d^{6}+2 b \,c^{2} d \,e^{5} x^{5}-5 b \,c^{2} d^{2} e^{4} x^{4}+20 b \,c^{2} d^{3} e^{3} x^{3}+60 b \,c^{2} d^{4} e^{2} x^{2}+3 a \,c^{2} d \,e^{5} x^{4}-36 a \,c^{2} d^{3} e^{3} x^{2}-12 a \,c^{2} d^{2} e^{4} x^{3}-12 a^{2} c d \,e^{5} x^{2}+12 \,\mathrm {log}\left (e x +d \right ) a^{2} c \,d^{3} e^{3}+36 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{5} e +12 \,\mathrm {log}\left (e x +d \right ) a^{2} c d \,e^{5} x^{2}+36 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{3} e^{3} x^{2}-60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{4} e^{2} x^{2}-60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{6}+3 a^{2} b \,e^{6} x^{2}+6 a^{2} c \,d^{3} e^{3}+18 a \,c^{2} d^{5} e -18 a b c \,d^{4} e^{2}-72 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{3} e^{3} x +24 \,\mathrm {log}\left (e x +d \right ) a^{2} c \,d^{2} e^{4} x -36 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{4} e^{2}+72 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{4} e^{2} x -120 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{5} e x -36 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{2} e^{4} x^{2}}{6 d \,e^{6} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:

int((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x)
 

Output:

(12*log(d + e*x)*a**2*c*d**3*e**3 + 24*log(d + e*x)*a**2*c*d**2*e**4*x + 1 
2*log(d + e*x)*a**2*c*d*e**5*x**2 - 36*log(d + e*x)*a*b*c*d**4*e**2 - 72*l 
og(d + e*x)*a*b*c*d**3*e**3*x - 36*log(d + e*x)*a*b*c*d**2*e**4*x**2 + 36* 
log(d + e*x)*a*c**2*d**5*e + 72*log(d + e*x)*a*c**2*d**4*e**2*x + 36*log(d 
 + e*x)*a*c**2*d**3*e**3*x**2 - 60*log(d + e*x)*b*c**2*d**6 - 120*log(d + 
e*x)*b*c**2*d**5*e*x - 60*log(d + e*x)*b*c**2*d**4*e**2*x**2 - 3*a**3*d*e* 
*5 + 3*a**2*b*e**6*x**2 + 6*a**2*c*d**3*e**3 - 12*a**2*c*d*e**5*x**2 - 18* 
a*b*c*d**4*e**2 + 36*a*b*c*d**2*e**4*x**2 + 12*a*b*c*d*e**5*x**3 + 18*a*c* 
*2*d**5*e - 36*a*c**2*d**3*e**3*x**2 - 12*a*c**2*d**2*e**4*x**3 + 3*a*c**2 
*d*e**5*x**4 - 30*b*c**2*d**6 + 60*b*c**2*d**4*e**2*x**2 + 20*b*c**2*d**3* 
e**3*x**3 - 5*b*c**2*d**2*e**4*x**4 + 2*b*c**2*d*e**5*x**5)/(6*d*e**6*(d** 
2 + 2*d*e*x + e**2*x**2))