\(\int \frac {(a+b x^2) (A+B x+C x^2+D x^3)}{(c+d x)^2} \, dx\) [6]

Optimal result

Integrand size = 30, antiderivative size = 230 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {\left (a d^2 (C d-2 c D)+b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )\right ) x}{d^5}+\frac {\left (a d^2 D-b \left (2 c C d-B d^2-3 c^2 D\right )\right ) x^2}{2 d^4}+\frac {b (C d-2 c D) x^3}{3 d^3}+\frac {b D x^4}{4 d^2}-\frac {\left (b c^2+a d^2\right ) \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{d^6 (c+d x)}-\frac {\left (a d^2 \left (2 c C d-B d^2-3 c^2 D\right )+b c \left (4 c^2 C d-3 B c d^2+2 A d^3-5 c^3 D\right )\right ) \log (c+d x)}{d^6} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {12 d \left (a d^2 (C d-2 c D)+b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )\right ) x+6 d^2 \left (a d^2 D+b \left (-2 c C d+B d^2+3 c^2 D\right )\right ) x^2+4 b d^3 (C d-2 c D) x^3+3 b d^4 D x^4+\frac {12 \left (b c^2+a d^2\right ) \left (-c^2 C d+B c d^2-A d^3+c^3 D\right )}{c+d x}+12 \left (a d^2 \left (-2 c C d+B d^2+3 c^2 D\right )+b c \left (-4 c^2 C d+3 B c d^2-2 A d^3+5 c^3 D\right )\right ) \log (c+d x)}{12 d^6} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2160, 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.

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

\(\Big \downarrow \) 2160

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2160

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, 
 d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.20

method	result	size
default	\(\frac {\frac {1}{4} D b \,x^{4} d^{3}+\frac {1}{3} C b \,d^{3} x^{3}-\frac {2}{3} D b c \,d^{2} x^{3}+\frac {1}{2} B b \,d^{3} x^{2}-C b c \,d^{2} x^{2}+\frac {1}{2} D a \,d^{3} x^{2}+\frac {3}{2} D b \,c^{2} d \,x^{2}+A b \,d^{3} x -2 B b c \,d^{2} x +C a \,d^{3} x +3 C b \,c^{2} d x -2 D a c \,d^{2} x -4 D b \,c^{3} x}{d^{5}}-\frac {A a \,d^{5}+A \,d^{3} b \,c^{2}-B a c \,d^{4}-B b \,c^{3} d^{2}+C a \,c^{2} d^{3}+C b \,c^{4} d -D a \,c^{3} d^{2}-D b \,c^{5}}{d^{6} \left (d x +c \right )}+\frac {\left (-2 A \,d^{3} b c +B a \,d^{4}+3 B b \,c^{2} d^{2}-2 C a c \,d^{3}-4 C b \,c^{3} d +3 D a \,c^{2} d^{2}+5 D b \,c^{4}\right ) \ln \left (d x +c \right )}{d^{6}}\)	\(276\)
norman	\(\frac {\frac {\left (A a \,d^{5}+2 A \,d^{3} b \,c^{2}-B a c \,d^{4}-3 B b \,c^{3} d^{2}+2 C a \,c^{2} d^{3}+4 C b \,c^{4} d -3 D a \,c^{3} d^{2}-5 D b \,c^{5}\right ) x}{d^{5} c}+\frac {\left (3 B b \,d^{2}-4 C b c d +3 a \,d^{2} D+5 D b \,c^{2}\right ) x^{3}}{6 d^{3}}+\frac {\left (2 A b \,d^{3}-3 B b c \,d^{2}+2 C a \,d^{3}+4 C b \,c^{2} d -3 D a c \,d^{2}-5 D b \,c^{3}\right ) x^{2}}{2 d^{4}}+\frac {b D x^{5}}{4 d}+\frac {b \left (4 C d -5 D c \right ) x^{4}}{12 d^{2}}}{d x +c}-\frac {\left (2 A \,d^{3} b c -B a \,d^{4}-3 B b \,c^{2} d^{2}+2 C a c \,d^{3}+4 C b \,c^{3} d -3 D a \,c^{2} d^{2}-5 D b \,c^{4}\right ) \ln \left (d x +c \right )}{d^{6}}\)	\(277\)
parallelrisch	\(-\frac {24 C \ln \left (d x +c \right ) a \,c^{2} d^{3}+5 D x^{4} b c \,d^{4}+24 C \ln \left (d x +c \right ) x a c \,d^{4}+24 A \,d^{3} b \,c^{2}-12 B \ln \left (d x +c \right ) x a \,d^{5}+48 C \ln \left (d x +c \right ) b \,c^{4} d +18 D x^{2} a c \,d^{4}+30 D x^{2} b \,c^{3} d^{2}+24 C a \,c^{2} d^{3}-12 B a c \,d^{4}-36 B b \,c^{3} d^{2}-36 D \ln \left (d x +c \right ) a \,c^{3} d^{2}-36 D a \,c^{3} d^{2}+8 C \,x^{3} b c \,d^{4}-10 D x^{3} b \,c^{2} d^{3}+18 B \,x^{2} b c \,d^{4}-24 C \,x^{2} b \,c^{2} d^{3}+24 A \ln \left (d x +c \right ) b \,c^{2} d^{3}-12 B \ln \left (d x +c \right ) a c \,d^{4}-36 B \ln \left (d x +c \right ) b \,c^{3} d^{2}+48 C b \,c^{4} d +24 A \ln \left (d x +c \right ) x b c \,d^{4}-36 B \ln \left (d x +c \right ) x b \,c^{2} d^{3}-3 D x^{5} b \,d^{5}-4 C \,x^{4} b \,d^{5}-6 D x^{3} a \,d^{5}-12 A \,x^{2} b \,d^{5}-12 C \,x^{2} a \,d^{5}-60 D \ln \left (d x +c \right ) b \,c^{5}-6 B \,x^{3} b \,d^{5}+48 C \ln \left (d x +c \right ) x b \,c^{3} d^{2}-36 D \ln \left (d x +c \right ) x a \,c^{2} d^{3}-60 D \ln \left (d x +c \right ) x b \,c^{4} d -60 D b \,c^{5}+12 A a \,d^{5}}{12 d^{6} \left (d x +c \right )}\)	\(442\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {3 \, D b d^{5} x^{5} + 12 \, D b c^{5} - 12 \, C b c^{4} d + 12 \, B a c d^{4} - 12 \, A a d^{5} + 12 \, {\left (D a + B b\right )} c^{3} d^{2} - 12 \, {\left (C a + A b\right )} c^{2} d^{3} - {\left (5 \, D b c d^{4} - 4 \, C b d^{5}\right )} x^{4} + 2 \, {\left (5 \, D b c^{2} d^{3} - 4 \, C b c d^{4} + 3 \, {\left (D a + B b\right )} d^{5}\right )} x^{3} - 6 \, {\left (5 \, D b c^{3} d^{2} - 4 \, C b c^{2} d^{3} + 3 \, {\left (D a + B b\right )} c d^{4} - 2 \, {\left (C a + A b\right )} d^{5}\right )} x^{2} - 12 \, {\left (4 \, D b c^{4} d - 3 \, C b c^{3} d^{2} + 2 \, {\left (D a + B b\right )} c^{2} d^{3} - {\left (C a + A b\right )} c d^{4}\right )} x + 12 \, {\left (5 \, D b c^{5} - 4 \, C b c^{4} d + B a c d^{4} + 3 \, {\left (D a + B b\right )} c^{3} d^{2} - 2 \, {\left (C a + A b\right )} c^{2} d^{3} + {\left (5 \, D b c^{4} d - 4 \, C b c^{3} d^{2} + B a d^{5} + 3 \, {\left (D a + B b\right )} c^{2} d^{3} - 2 \, {\left (C a + A b\right )} c d^{4}\right )} x\right )} \log \left (d x + c\right )}{12 \, {\left (d^{7} x + c d^{6}\right )}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {D b x^{4}}{4 d^{2}} + x^{3} \left (\frac {C b}{3 d^{2}} - \frac {2 D b c}{3 d^{3}}\right ) + x^{2} \left (\frac {B b}{2 d^{2}} - \frac {C b c}{d^{3}} + \frac {D a}{2 d^{2}} + \frac {3 D b c^{2}}{2 d^{4}}\right ) + x \left (\frac {A b}{d^{2}} - \frac {2 B b c}{d^{3}} + \frac {C a}{d^{2}} + \frac {3 C b c^{2}}{d^{4}} - \frac {2 D a c}{d^{3}} - \frac {4 D b c^{3}}{d^{5}}\right ) + \frac {- A a d^{5} - A b c^{2} d^{3} + B a c d^{4} + B b c^{3} d^{2} - C a c^{2} d^{3} - C b c^{4} d + D a c^{3} d^{2} + D b c^{5}}{c d^{6} + d^{7} x} + \frac {\left (- 2 A b c d^{3} + B a d^{4} + 3 B b c^{2} d^{2} - 2 C a c d^{3} - 4 C b c^{3} d + 3 D a c^{2} d^{2} + 5 D b c^{4}\right ) \log {\left (c + d x \right )}}{d^{6}} \] Input:

integrate((b*x**2+a)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**2,x)
                                                                                    
                                                                                    
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {D b c^{5} - C b c^{4} d + B a c d^{4} - A a d^{5} + {\left (D a + B b\right )} c^{3} d^{2} - {\left (C a + A b\right )} c^{2} d^{3}}{d^{7} x + c d^{6}} + \frac {3 \, D b d^{3} x^{4} - 4 \, {\left (2 \, D b c d^{2} - C b d^{3}\right )} x^{3} + 6 \, {\left (3 \, D b c^{2} d - 2 \, C b c d^{2} + {\left (D a + B b\right )} d^{3}\right )} x^{2} - 12 \, {\left (4 \, D b c^{3} - 3 \, C b c^{2} d + 2 \, {\left (D a + B b\right )} c d^{2} - {\left (C a + A b\right )} d^{3}\right )} x}{12 \, d^{5}} + \frac {{\left (5 \, D b c^{4} - 4 \, C b c^{3} d + B a d^{4} + 3 \, {\left (D a + B b\right )} c^{2} d^{2} - 2 \, {\left (C a + A b\right )} c d^{3}\right )} \log \left (d x + c\right )}{d^{6}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.58 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {{\left (3 \, D b - \frac {4 \, {\left (5 \, D b c d - C b d^{2}\right )}}{{\left (d x + c\right )} d} + \frac {6 \, {\left (10 \, D b c^{2} d^{2} - 4 \, C b c d^{3} + D a d^{4} + B b d^{4}\right )}}{{\left (d x + c\right )}^{2} d^{2}} - \frac {12 \, {\left (10 \, D b c^{3} d^{3} - 6 \, C b c^{2} d^{4} + 3 \, D a c d^{5} + 3 \, B b c d^{5} - C a d^{6} - A b d^{6}\right )}}{{\left (d x + c\right )}^{3} d^{3}}\right )} {\left (d x + c\right )}^{4}}{12 \, d^{6}} - \frac {{\left (5 \, D b c^{4} - 4 \, C b c^{3} d + 3 \, D a c^{2} d^{2} + 3 \, B b c^{2} d^{2} - 2 \, C a c d^{3} - 2 \, A b c d^{3} + B a d^{4}\right )} \log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{d^{6}} + \frac {\frac {D b c^{5} d^{4}}{d x + c} - \frac {C b c^{4} d^{5}}{d x + c} + \frac {D a c^{3} d^{6}}{d x + c} + \frac {B b c^{3} d^{6}}{d x + c} - \frac {C a c^{2} d^{7}}{d x + c} - \frac {A b c^{2} d^{7}}{d x + c} + \frac {B a c d^{8}}{d x + c} - \frac {A a d^{9}}{d x + c}}{d^{10}} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\int \frac {\left (b\,x^2+a\right )\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {-b \,c^{2} d^{4} x^{4}+12 \,\mathrm {log}\left (d x +c \right ) a \,c^{4} d^{2}+36 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{4} d -12 a \,c^{3} d^{3} x -6 a \,c^{2} d^{4} x^{2}+6 a c \,d^{5} x^{3}-36 b^{2} c^{3} d^{2} x -18 b^{2} c^{2} d^{3} x^{2}+6 b^{2} c \,d^{4} x^{3}-12 b \,c^{5} d x -6 b \,c^{4} d^{2} x^{2}+2 b \,c^{3} d^{3} x^{3}+3 b c \,d^{5} x^{5}-24 \,\mathrm {log}\left (d x +c \right ) a b \,c^{3} d^{2}+12 \,\mathrm {log}\left (d x +c \right ) a b \,c^{2} d^{3}+12 \,\mathrm {log}\left (d x +c \right ) a \,c^{3} d^{3} x +36 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{3} d^{2} x +12 \,\mathrm {log}\left (d x +c \right ) b \,c^{5} d x +24 a b \,c^{2} d^{3} x +12 a b c \,d^{4} x^{2}-12 a b c \,d^{4} x +12 \,\mathrm {log}\left (d x +c \right ) b \,c^{6}+12 a^{2} d^{5} x -24 \,\mathrm {log}\left (d x +c \right ) a b \,c^{2} d^{3} x +12 \,\mathrm {log}\left (d x +c \right ) a b c \,d^{4} x}{12 c \,d^{5} \left (d x +c \right )} \] Input:

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

Output:

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