\(\int \frac {(A+B x) (a+c x^2)^2}{(d+e x)^2} \, dx\) [53]

Optimal result

Integrand size = 22, antiderivative size = 180 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=-\frac {c \left (4 B c d^3-3 A c d^2 e+4 a B d e^2-2 a A e^3\right ) x}{e^5}+\frac {c \left (3 B c d^2-2 A c d e+2 a B e^2\right ) x^2}{2 e^4}-\frac {c^2 (2 B d-A e) x^3}{3 e^3}+\frac {B c^2 x^4}{4 e^2}+\frac {(B d-A e) \left (c d^2+a e^2\right )^2}{e^6 (d+e x)}+\frac {\left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right ) \log (d+e x)}{e^6} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\frac {12 c e \left (A e \left (3 c d^2+2 a e^2\right )-4 B \left (c d^3+a d e^2\right )\right ) x+6 c e^2 \left (3 B c d^2-2 A c d e+2 a B e^2\right ) x^2+4 c^2 e^3 (-2 B d+A e) x^3+3 B c^2 e^4 x^4+\frac {12 (B d-A e) \left (c d^2+a e^2\right )^2}{d+e x}+12 \left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right ) \log (d+e x)}{12 e^6} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 180, 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.

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

\(\Big \downarrow \) 652

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((c*(4*B*c*d^3 - 3*A*c*d^2*e + 4*a*B*d*e^2 - 2*a*A*e^3)*x)/e^5) + (c*(3*B 
*c*d^2 - 2*A*c*d*e + 2*a*B*e^2)*x^2)/(2*e^4) - (c^2*(2*B*d - A*e)*x^3)/(3* 
e^3) + (B*c^2*x^4)/(4*e^2) + ((B*d - A*e)*(c*d^2 + a*e^2)^2)/(e^6*(d + e*x 
)) + ((c*d^2 + a*e^2)*(5*B*c*d^2 - 4*A*c*d*e + a*B*e^2)*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 = 0.67 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.29

method	result	size
default	\(\frac {c \left (\frac {1}{4} B c \,x^{4} e^{3}+\frac {1}{3} A \,x^{3} c \,e^{3}-\frac {2}{3} B \,x^{3} c d \,e^{2}-A \,x^{2} c d \,e^{2}+B \,x^{2} a \,e^{3}+\frac {3}{2} B \,x^{2} c \,d^{2} e +2 A a \,e^{3} x +3 A c \,d^{2} e x -4 B a d \,e^{2} x -4 B c \,d^{3} x \right )}{e^{5}}+\frac {\left (-4 A a c d \,e^{3}-4 A \,c^{2} d^{3} e +B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}+5 B \,c^{2} d^{4}\right ) \ln \left (e x +d \right )}{e^{6}}-\frac {A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}}{e^{6} \left (e x +d \right )}\)	\(233\)
norman	\(\frac {\frac {\left (A \,a^{2} e^{5}+4 A a c \,d^{2} e^{3}+4 A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-6 B a c \,d^{3} e^{2}-5 B \,c^{2} d^{5}\right ) x}{d \,e^{5}}+\frac {B \,c^{2} x^{5}}{4 e}-\frac {c \left (4 A c d e -6 B a \,e^{2}-5 B c \,d^{2}\right ) x^{3}}{6 e^{3}}+\frac {c \left (4 A a \,e^{3}+4 A c \,d^{2} e -6 B a d \,e^{2}-5 B c \,d^{3}\right ) x^{2}}{2 e^{4}}+\frac {c^{2} \left (4 A e -5 B d \right ) x^{4}}{12 e^{2}}}{e x +d}-\frac {\left (4 A a c d \,e^{3}+4 A \,c^{2} d^{3} e -B \,e^{4} a^{2}-6 B a c \,d^{2} e^{2}-5 B \,c^{2} d^{4}\right ) \ln \left (e x +d \right )}{e^{6}}\)	\(241\)
risch	\(\frac {B \,c^{2} x^{4}}{4 e^{2}}+\frac {c^{2} A \,x^{3}}{3 e^{2}}-\frac {2 c^{2} B \,x^{3} d}{3 e^{3}}-\frac {c^{2} A \,x^{2} d}{e^{3}}+\frac {c B \,x^{2} a}{e^{2}}+\frac {3 c^{2} B \,x^{2} d^{2}}{2 e^{4}}+\frac {2 c A a x}{e^{2}}+\frac {3 c^{2} A \,d^{2} x}{e^{4}}-\frac {4 c B a d x}{e^{3}}-\frac {4 c^{2} B \,d^{3} x}{e^{5}}-\frac {A \,a^{2}}{e \left (e x +d \right )}-\frac {2 A a c \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {A \,c^{2} d^{4}}{e^{5} \left (e x +d \right )}+\frac {B \,a^{2} d}{e^{2} \left (e x +d \right )}+\frac {2 B a c \,d^{3}}{e^{4} \left (e x +d \right )}+\frac {B \,c^{2} d^{5}}{e^{6} \left (e x +d \right )}-\frac {4 \ln \left (e x +d \right ) A a c d}{e^{3}}-\frac {4 \ln \left (e x +d \right ) A \,c^{2} d^{3}}{e^{5}}+\frac {\ln \left (e x +d \right ) B \,a^{2}}{e^{2}}+\frac {6 \ln \left (e x +d \right ) B a c \,d^{2}}{e^{4}}+\frac {5 \ln \left (e x +d \right ) B \,c^{2} d^{4}}{e^{6}}\)	\(309\)
parallelrisch	\(-\frac {-4 A \,x^{4} c^{2} e^{5}-60 B \ln \left (e x +d \right ) c^{2} d^{5}+48 A \ln \left (e x +d \right ) x \,c^{2} d^{3} e^{2}-60 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e +48 A a c \,d^{2} e^{3}-72 B a c \,d^{3} e^{2}+5 B \,x^{4} c^{2} d \,e^{4}+8 A \,x^{3} c^{2} d \,e^{4}-12 B \,x^{3} a c \,e^{5}-72 B \ln \left (e x +d \right ) x a c \,d^{2} e^{3}+48 A \ln \left (e x +d \right ) x a c d \,e^{4}-12 B \ln \left (e x +d \right ) a^{2} d \,e^{4}+36 B \,x^{2} a c d \,e^{4}+48 A \ln \left (e x +d \right ) a c \,d^{2} e^{3}-12 B \,a^{2} d \,e^{4}+30 B \,x^{2} c^{2} d^{3} e^{2}+48 A \ln \left (e x +d \right ) c^{2} d^{4} e +48 A \,c^{2} d^{4} e -12 B \ln \left (e x +d \right ) x \,a^{2} e^{5}-10 B \,x^{3} c^{2} d^{2} e^{3}-24 A \,x^{2} a c \,e^{5}-24 A \,x^{2} c^{2} d^{2} e^{3}-72 B \ln \left (e x +d \right ) a c \,d^{3} e^{2}-3 B \,x^{5} c^{2} e^{5}-60 B \,c^{2} d^{5}+12 A \,a^{2} e^{5}}{12 e^{6} \left (e x +d \right )}\)	\(370\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (174) = 348\).

Time = 0.09 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.97 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\frac {3 \, B c^{2} e^{5} x^{5} + 12 \, B c^{2} d^{5} - 12 \, A c^{2} d^{4} e + 24 \, B a c d^{3} e^{2} - 24 \, A a c d^{2} e^{3} + 12 \, B a^{2} d e^{4} - 12 \, A a^{2} e^{5} - {\left (5 \, B c^{2} d e^{4} - 4 \, A c^{2} e^{5}\right )} x^{4} + 2 \, {\left (5 \, B c^{2} d^{2} e^{3} - 4 \, A c^{2} d e^{4} + 6 \, B a c e^{5}\right )} x^{3} - 6 \, {\left (5 \, B c^{2} d^{3} e^{2} - 4 \, A c^{2} d^{2} e^{3} + 6 \, B a c d e^{4} - 4 \, A a c e^{5}\right )} x^{2} - 12 \, {\left (4 \, B c^{2} d^{4} e - 3 \, A c^{2} d^{3} e^{2} + 4 \, B a c d^{2} e^{3} - 2 \, A a c d e^{4}\right )} x + 12 \, {\left (5 \, B c^{2} d^{5} - 4 \, A c^{2} d^{4} e + 6 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + {\left (5 \, B c^{2} d^{4} e - 4 \, A c^{2} d^{3} e^{2} + 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} + B a^{2} e^{5}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{7} x + d e^{6}\right )}} \] Input:

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

Output:

1/12*(3*B*c^2*e^5*x^5 + 12*B*c^2*d^5 - 12*A*c^2*d^4*e + 24*B*a*c*d^3*e^2 - 
 24*A*a*c*d^2*e^3 + 12*B*a^2*d*e^4 - 12*A*a^2*e^5 - (5*B*c^2*d*e^4 - 4*A*c 
^2*e^5)*x^4 + 2*(5*B*c^2*d^2*e^3 - 4*A*c^2*d*e^4 + 6*B*a*c*e^5)*x^3 - 6*(5 
*B*c^2*d^3*e^2 - 4*A*c^2*d^2*e^3 + 6*B*a*c*d*e^4 - 4*A*a*c*e^5)*x^2 - 12*( 
4*B*c^2*d^4*e - 3*A*c^2*d^3*e^2 + 4*B*a*c*d^2*e^3 - 2*A*a*c*d*e^4)*x + 12* 
(5*B*c^2*d^5 - 4*A*c^2*d^4*e + 6*B*a*c*d^3*e^2 - 4*A*a*c*d^2*e^3 + B*a^2*d 
*e^4 + (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)*log(e*x + d))/(e^7*x + d*e^6)
 

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\frac {B c^{2} x^{4}}{4 e^{2}} + x^{3} \left (\frac {A c^{2}}{3 e^{2}} - \frac {2 B c^{2} d}{3 e^{3}}\right ) + x^{2} \left (- \frac {A c^{2} d}{e^{3}} + \frac {B a c}{e^{2}} + \frac {3 B c^{2} d^{2}}{2 e^{4}}\right ) + x \left (\frac {2 A a c}{e^{2}} + \frac {3 A c^{2} d^{2}}{e^{4}} - \frac {4 B a c d}{e^{3}} - \frac {4 B c^{2} d^{3}}{e^{5}}\right ) + \frac {- A a^{2} e^{5} - 2 A a c d^{2} e^{3} - A c^{2} d^{4} e + B a^{2} d e^{4} + 2 B a c d^{3} e^{2} + B c^{2} d^{5}}{d e^{6} + e^{7} x} + \frac {\left (a e^{2} + c d^{2}\right ) \left (- 4 A c d e + B a e^{2} + 5 B c d^{2}\right ) \log {\left (d + e x \right )}}{e^{6}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\frac {B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}}{e^{7} x + d e^{6}} + \frac {3 \, B c^{2} e^{3} x^{4} - 4 \, {\left (2 \, B c^{2} d e^{2} - A c^{2} e^{3}\right )} x^{3} + 6 \, {\left (3 \, B c^{2} d^{2} e - 2 \, A c^{2} d e^{2} + 2 \, B a c e^{3}\right )} x^{2} - 12 \, {\left (4 \, B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 4 \, B a c d e^{2} - 2 \, A a c e^{3}\right )} x}{12 \, e^{5}} + \frac {{\left (5 \, B c^{2} d^{4} - 4 \, A c^{2} d^{3} e + 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} + B a^{2} e^{4}\right )} \log \left (e x + d\right )}{e^{6}} \] Input:

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

Output:

(B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3 + B*a^2*d*e^4 
 - A*a^2*e^5)/(e^7*x + d*e^6) + 1/12*(3*B*c^2*e^3*x^4 - 4*(2*B*c^2*d*e^2 - 
 A*c^2*e^3)*x^3 + 6*(3*B*c^2*d^2*e - 2*A*c^2*d*e^2 + 2*B*a*c*e^3)*x^2 - 12 
*(4*B*c^2*d^3 - 3*A*c^2*d^2*e + 4*B*a*c*d*e^2 - 2*A*a*c*e^3)*x)/e^5 + (5*B 
*c^2*d^4 - 4*A*c^2*d^3*e + 6*B*a*c*d^2*e^2 - 4*A*a*c*d*e^3 + B*a^2*e^4)*lo 
g(e*x + d)/e^6
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\frac {{\left (3 \, B c^{2} - \frac {4 \, {\left (5 \, B c^{2} d e - A c^{2} e^{2}\right )}}{{\left (e x + d\right )} e} + \frac {12 \, {\left (5 \, B c^{2} d^{2} e^{2} - 2 \, A c^{2} d e^{3} + B a c e^{4}\right )}}{{\left (e x + d\right )}^{2} e^{2}} - \frac {24 \, {\left (5 \, B c^{2} d^{3} e^{3} - 3 \, A c^{2} d^{2} e^{4} + 3 \, B a c d e^{5} - A a c e^{6}\right )}}{{\left (e x + d\right )}^{3} e^{3}}\right )} {\left (e x + d\right )}^{4}}{12 \, e^{6}} - \frac {{\left (5 \, B c^{2} d^{4} - 4 \, A c^{2} d^{3} e + 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} + B a^{2} e^{4}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{6}} + \frac {\frac {B c^{2} d^{5} e^{4}}{e x + d} - \frac {A c^{2} d^{4} e^{5}}{e x + d} + \frac {2 \, B a c d^{3} e^{6}}{e x + d} - \frac {2 \, A a c d^{2} e^{7}}{e x + d} + \frac {B a^{2} d e^{8}}{e x + d} - \frac {A a^{2} e^{9}}{e x + d}}{e^{10}} \] Input:

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

Output:

1/12*(3*B*c^2 - 4*(5*B*c^2*d*e - A*c^2*e^2)/((e*x + d)*e) + 12*(5*B*c^2*d^ 
2*e^2 - 2*A*c^2*d*e^3 + B*a*c*e^4)/((e*x + d)^2*e^2) - 24*(5*B*c^2*d^3*e^3 
 - 3*A*c^2*d^2*e^4 + 3*B*a*c*d*e^5 - A*a*c*e^6)/((e*x + d)^3*e^3))*(e*x + 
d)^4/e^6 - (5*B*c^2*d^4 - 4*A*c^2*d^3*e + 6*B*a*c*d^2*e^2 - 4*A*a*c*d*e^3 
+ B*a^2*e^4)*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^6 + (B*c^2*d^5*e^4/( 
e*x + d) - A*c^2*d^4*e^5/(e*x + d) + 2*B*a*c*d^3*e^6/(e*x + d) - 2*A*a*c*d 
^2*e^7/(e*x + d) + B*a^2*d*e^8/(e*x + d) - A*a^2*e^9/(e*x + d))/e^10
 

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.73 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=x^3\,\left (\frac {A\,c^2}{3\,e^2}-\frac {2\,B\,c^2\,d}{3\,e^3}\right )-x^2\,\left (\frac {d\,\left (\frac {A\,c^2}{e^2}-\frac {2\,B\,c^2\,d}{e^3}\right )}{e}-\frac {B\,a\,c}{e^2}+\frac {B\,c^2\,d^2}{2\,e^4}\right )+x\,\left (\frac {2\,d\,\left (\frac {2\,d\,\left (\frac {A\,c^2}{e^2}-\frac {2\,B\,c^2\,d}{e^3}\right )}{e}-\frac {2\,B\,a\,c}{e^2}+\frac {B\,c^2\,d^2}{e^4}\right )}{e}-\frac {d^2\,\left (\frac {A\,c^2}{e^2}-\frac {2\,B\,c^2\,d}{e^3}\right )}{e^2}+\frac {2\,A\,a\,c}{e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (B\,a^2\,e^4+6\,B\,a\,c\,d^2\,e^2-4\,A\,a\,c\,d\,e^3+5\,B\,c^2\,d^4-4\,A\,c^2\,d^3\,e\right )}{e^6}-\frac {-B\,a^2\,d\,e^4+A\,a^2\,e^5-2\,B\,a\,c\,d^3\,e^2+2\,A\,a\,c\,d^2\,e^3-B\,c^2\,d^5+A\,c^2\,d^4\,e}{e\,\left (x\,e^6+d\,e^5\right )}+\frac {B\,c^2\,x^4}{4\,e^2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.22 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\frac {12 a^{3} e^{6} x -36 a b c \,d^{2} e^{4} x^{2}+12 a b c d \,e^{5} x^{3}+3 b \,c^{2} d \,e^{5} x^{5}-5 b \,c^{2} d^{2} e^{4} x^{4}+10 b \,c^{2} d^{3} e^{3} x^{3}-60 b \,c^{2} d^{5} e x -30 b \,c^{2} d^{4} e^{2} x^{2}+4 a \,c^{2} d \,e^{5} x^{4}+24 a \,c^{2} d^{3} e^{3} x^{2}-8 a \,c^{2} d^{2} e^{4} x^{3}-12 a^{2} b d \,e^{5} x +48 a^{2} c \,d^{2} e^{4} x +24 a^{2} c d \,e^{5} x^{2}+48 a \,c^{2} d^{4} e^{2} x +12 \,\mathrm {log}\left (e x +d \right ) a^{2} b \,d^{2} e^{4}-48 \,\mathrm {log}\left (e x +d \right ) a^{2} c \,d^{3} e^{3}-48 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{5} e +60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{6}+72 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{3} e^{3} x +12 \,\mathrm {log}\left (e x +d \right ) a^{2} b d \,e^{5} x -48 \,\mathrm {log}\left (e x +d \right ) a^{2} c \,d^{2} e^{4} x +72 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{4} e^{2}-48 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{4} e^{2} x +60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{5} e x -72 a b c \,d^{3} e^{3} x}{12 d \,e^{6} \left (e x +d \right )} \] Input:

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

Output:

(12*log(d + e*x)*a**2*b*d**2*e**4 + 12*log(d + e*x)*a**2*b*d*e**5*x - 48*l 
og(d + e*x)*a**2*c*d**3*e**3 - 48*log(d + e*x)*a**2*c*d**2*e**4*x + 72*log 
(d + e*x)*a*b*c*d**4*e**2 + 72*log(d + e*x)*a*b*c*d**3*e**3*x - 48*log(d + 
 e*x)*a*c**2*d**5*e - 48*log(d + e*x)*a*c**2*d**4*e**2*x + 60*log(d + e*x) 
*b*c**2*d**6 + 60*log(d + e*x)*b*c**2*d**5*e*x + 12*a**3*e**6*x - 12*a**2* 
b*d*e**5*x + 48*a**2*c*d**2*e**4*x + 24*a**2*c*d*e**5*x**2 - 72*a*b*c*d**3 
*e**3*x - 36*a*b*c*d**2*e**4*x**2 + 12*a*b*c*d*e**5*x**3 + 48*a*c**2*d**4* 
e**2*x + 24*a*c**2*d**3*e**3*x**2 - 8*a*c**2*d**2*e**4*x**3 + 4*a*c**2*d*e 
**5*x**4 - 60*b*c**2*d**5*e*x - 30*b*c**2*d**4*e**2*x**2 + 10*b*c**2*d**3* 
e**3*x**3 - 5*b*c**2*d**2*e**4*x**4 + 3*b*c**2*d*e**5*x**5)/(12*d*e**6*(d 
+ e*x))