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

Optimal result

Integrand size = 37, antiderivative size = 222 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{b^2 (b c-a d)^2 (a+b x)}-\frac {c^2 C d-B c d^2+A d^3-c^3 D}{d^2 (b c-a d)^2 (c+d x)}+\frac {\left (b^3 (B c-2 A d)-a b^2 (2 c C-B d)+3 a^2 b c D-a^3 d D\right ) \log (a+b x)}{b^2 (b c-a d)^3}+\frac {\left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (B c d^2-2 A d^3-c^3 D\right )\right ) \log (c+d x)}{d^2 (b c-a d)^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {-A b^3+a \left (b^2 B-a b C+a^2 D\right )}{b^2 (b c-a d)^2 (a+b x)}+\frac {-c^2 C d+B c d^2-A d^3+c^3 D}{d^2 (b c-a d)^2 (c+d x)}+\frac {\left (b^3 (B c-2 A d)+a b^2 (-2 c C+B d)+3 a^2 b c D-a^3 d D\right ) \log (a+b x)}{b^2 (b c-a d)^3}+\frac {\left (a d \left (-2 c C d+B d^2+3 c^2 D\right )+b \left (B c d^2-2 A d^3-c^3 D\right )\right ) \log (c+d x)}{d^2 (-b c+a d)^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.33, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {2191, 25, 27, 1141, 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.

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

\(\Big \downarrow \) 2191

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1141

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1141

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ 
Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p   Int[ExpandIntegrand[ 
(d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 
1] ||  !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 
0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
 

rule 2009

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

rule 2191

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = 
PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P 
q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + 
c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ 
(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c))   Int 
[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* 
(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 
2 - 4*a*c, 0] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 1.73 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {A \,b^{3}-B a \,b^{2}+C \,a^{2} b -D a^{3}}{b^{2} \left (a d -b c \right )^{2} \left (b x +a \right )}+\frac {\left (2 A d \,b^{3}-B a \,b^{2} d -B \,b^{3} c +2 C a \,b^{2} c +a^{3} d D-3 a^{2} b c D\right ) \ln \left (b x +a \right )}{\left (a d -b c \right )^{3} b^{2}}-\frac {A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}}{d^{2} \left (a d -b c \right )^{2} \left (d x +c \right )}+\frac {\left (-2 A b \,d^{3}+B a \,d^{3}+B b c \,d^{2}-2 C a c \,d^{2}+3 D a \,c^{2} d -D b \,c^{3}\right ) \ln \left (d x +c \right )}{\left (a d -b c \right )^{3} d^{2}}\)	\(225\)
norman	\(\frac {-\frac {A \,d^{3} a \,b^{2}+A \,b^{3} c \,d^{2}-2 B a \,b^{2} c \,d^{2}+C \,a^{2} b c \,d^{2}+C a \,b^{2} c^{2} d -D a^{3} c \,d^{2}-D a \,b^{2} c^{3}}{d^{2} b^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {\left (2 A \,d^{3} b^{3}-B a \,b^{2} d^{3}-B \,b^{3} c \,d^{2}+C \,a^{2} b \,d^{3}+C \,b^{3} c^{2} d -D a^{3} d^{3}-D b^{3} c^{3}\right ) x}{d^{2} b^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}}{\left (b x +a \right ) \left (d x +c \right )}+\frac {\left (2 A d \,b^{3}-B a \,b^{2} d -B \,b^{3} c +2 C a \,b^{2} c +a^{3} d D-3 a^{2} b c D\right ) \ln \left (b x +a \right )}{\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) b^{2}}-\frac {\left (2 A b \,d^{3}-B a \,d^{3}-B b c \,d^{2}+2 C a c \,d^{2}-3 D a \,c^{2} d +D b \,c^{3}\right ) \ln \left (d x +c \right )}{d^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(404\)
parallelrisch	\(\text {Expression too large to display}\)	\(1137\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 814 vs. \(2 (221) = 442\).

Time = 0.13 (sec) , antiderivative size = 814, normalized size of antiderivative = 3.67 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1525 vs. \(2 (201) = 402\).

Time = 153.27 (sec) , antiderivative size = 1525, normalized size of antiderivative = 6.87 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.94 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {{\left ({\left (3 \, D a^{2} b - 2 \, C a b^{2} + B b^{3}\right )} c - {\left (D a^{3} - B a b^{2} + 2 \, A b^{3}\right )} d\right )} \log \left (b x + a\right )}{b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}} + \frac {{\left (D b c^{3} - 3 \, D a c^{2} d + {\left (2 \, C a - B b\right )} c d^{2} - {\left (B a - 2 \, A b\right )} d^{3}\right )} \log \left (d x + c\right )}{b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}} + \frac {D a b^{2} c^{3} - C a b^{2} c^{2} d - A a b^{2} d^{3} + {\left (D a^{3} - C a^{2} b + 2 \, B a b^{2} - A b^{3}\right )} c d^{2} + {\left (D b^{3} c^{3} - C b^{3} c^{2} d + B b^{3} c d^{2} + {\left (D a^{3} - C a^{2} b + B a b^{2} - 2 \, A b^{3}\right )} d^{3}\right )} x}{a b^{4} c^{3} d^{2} - 2 \, a^{2} b^{3} c^{2} d^{3} + a^{3} b^{2} c d^{4} + {\left (b^{5} c^{2} d^{3} - 2 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} + {\left (b^{5} c^{3} d^{2} - a b^{4} c^{2} d^{3} - a^{2} b^{3} c d^{4} + a^{3} b^{2} d^{5}\right )} x} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.63 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {{\left (3 \, D a^{2} b c - 2 \, C a b^{2} c + B b^{3} c - D a^{3} d + B a b^{2} d - 2 \, A b^{3} d\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}} + \frac {{\left (D b c^{3} - 3 \, D a c^{2} d + 2 \, C a c d^{2} - B b c d^{2} - B a d^{3} + 2 \, A b d^{3}\right )} \log \left ({\left | d x + c \right |}\right )}{b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}} + \frac {D a b^{2} c^{3} - C a b^{2} c^{2} d + D a^{3} c d^{2} - C a^{2} b c d^{2} + 2 \, B a b^{2} c d^{2} - A b^{3} c d^{2} - A a b^{2} d^{3} + {\left (D b^{3} c^{3} - C b^{3} c^{2} d + B b^{3} c d^{2} + D a^{3} d^{3} - C a^{2} b d^{3} + B a b^{2} d^{3} - 2 \, A b^{3} d^{3}\right )} x}{{\left (b c - a d\right )}^{2} {\left (b x + a\right )} {\left (d x + c\right )} b^{2} d^{2}} \] Input:

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

Output:

(3*D*a^2*b*c - 2*C*a*b^2*c + B*b^3*c - D*a^3*d + B*a*b^2*d - 2*A*b^3*d)*lo 
g(abs(b*x + a))/(b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b^3*c*d^2 - a^3*b^2*d^3) 
+ (D*b*c^3 - 3*D*a*c^2*d + 2*C*a*c*d^2 - B*b*c*d^2 - B*a*d^3 + 2*A*b*d^3)* 
log(abs(d*x + c))/(b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4 - a^3*d^5 
) + (D*a*b^2*c^3 - C*a*b^2*c^2*d + D*a^3*c*d^2 - C*a^2*b*c*d^2 + 2*B*a*b^2 
*c*d^2 - A*b^3*c*d^2 - A*a*b^2*d^3 + (D*b^3*c^3 - C*b^3*c^2*d + B*b^3*c*d^ 
2 + D*a^3*d^3 - C*a^2*b*d^3 + B*a*b^2*d^3 - 2*A*b^3*d^3)*x)/((b*c - a*d)^2 
*(b*x + a)*(d*x + c)*b^2*d^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 744, normalized size of antiderivative = 3.35 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(log(a + b*x)*a**4*c*d**3 + log(a + b*x)*a**4*d**4*x - log(a + b*x)*a**3*b 
*c**2*d**2 + log(a + b*x)*a**3*b*d**4*x**2 + log(a + b*x)*a**2*b**3*c*d**2 
 + log(a + b*x)*a**2*b**3*d**3*x - 2*log(a + b*x)*a**2*b**2*c**3*d - 3*log 
(a + b*x)*a**2*b**2*c**2*d**2*x - log(a + b*x)*a**2*b**2*c*d**3*x**2 + log 
(a + b*x)*a*b**4*c**2*d + 2*log(a + b*x)*a*b**4*c*d**2*x + log(a + b*x)*a* 
b**4*d**3*x**2 - 2*log(a + b*x)*a*b**3*c**3*d*x - 2*log(a + b*x)*a*b**3*c* 
*2*d**2*x**2 + log(a + b*x)*b**5*c**2*d*x + log(a + b*x)*b**5*c*d**2*x**2 
- log(c + d*x)*a**2*b**3*c*d**2 - log(c + d*x)*a**2*b**3*d**3*x + log(c + 
d*x)*a**2*b**2*c**3*d + log(c + d*x)*a**2*b**2*c**2*d**2*x - log(c + d*x)* 
a*b**4*c**2*d - 2*log(c + d*x)*a*b**4*c*d**2*x - log(c + d*x)*a*b**4*d**3* 
x**2 + log(c + d*x)*a*b**3*c**4 + 2*log(c + d*x)*a*b**3*c**3*d*x + log(c + 
 d*x)*a*b**3*c**2*d**2*x**2 - log(c + d*x)*b**5*c**2*d*x - log(c + d*x)*b* 
*5*c*d**2*x**2 + log(c + d*x)*b**4*c**4*x + log(c + d*x)*b**4*c**3*d*x**2 
- a**3*b**2*d**3 + a**3*b*c**2*d**2 - a**3*b*d**4*x**2 + a**2*b**3*c*d**2 
- a**2*b**2*c**3*d + a**2*b**2*c*d**3*x**2 + a*b**4*d**3*x**2 - b**5*c*d** 
2*x**2)/(b**2*d*(a**4*c*d**3 + a**4*d**4*x - a**3*b*c**2*d**2 + a**3*b*d** 
4*x**2 - a**2*b**2*c**3*d - 2*a**2*b**2*c**2*d**2*x - a**2*b**2*c*d**3*x** 
2 + a*b**3*c**4 - a*b**3*c**2*d**2*x**2 + b**4*c**4*x + b**4*c**3*d*x**2))