\(\int \frac {(c+d x)^2 (A+B x+C x^2+D x^3)}{(a+b x^2)^4} \, dx\) [46]

Optimal result

Integrand size = 32, antiderivative size = 376 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=-\frac {b^2 c (B c+2 A d)+a^2 d^2 D-a b \left (2 c C d+B d^2+c^2 D\right )}{6 b^3 \left (a+b x^2\right )^3}+\frac {\left (A b \left (b c^2-a d^2\right )-a (b c (c C+2 B d)-a d (C d+2 c D))\right ) x}{6 a b^2 \left (a+b x^2\right )^3}+\frac {2 a d^2 D-b \left (2 c C d+B d^2+c^2 D\right )}{4 b^3 \left (a+b x^2\right )^2}+\frac {\left (A b \left (5 b c^2+a d^2\right )+a (b c (c C+2 B d)-7 a d (C d+2 c D))\right ) x}{24 a^2 b^2 \left (a+b x^2\right )^2}-\frac {d^2 D}{2 b^3 \left (a+b x^2\right )}+\frac {\left (A b \left (5 b c^2+a d^2\right )+a (b c (c C+2 B d)+a d (C d+2 c D))\right ) x}{16 a^3 b^2 \left (a+b x^2\right )}+\frac {\left (A b \left (5 b c^2+a d^2\right )+a (b c (c C+2 B d)+a d (C d+2 c D))\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} b^{5/2}} \] Output:

-1/6*(b^2*c*(2*A*d+B*c)+a^2*d^2*D-a*b*(B*d^2+2*C*c*d+D*c^2))/b^3/(b*x^2+a) 
^3+1/6*(A*b*(-a*d^2+b*c^2)-a*(b*c*(2*B*d+C*c)-a*d*(C*d+2*D*c)))*x/a/b^2/(b 
*x^2+a)^3+1/4*(2*a*d^2*D-b*(B*d^2+2*C*c*d+D*c^2))/b^3/(b*x^2+a)^2+1/24*(A* 
b*(a*d^2+5*b*c^2)+a*(b*c*(2*B*d+C*c)-7*a*d*(C*d+2*D*c)))*x/a^2/b^2/(b*x^2+ 
a)^2-1/2*d^2*D/b^3/(b*x^2+a)+1/16*(A*b*(a*d^2+5*b*c^2)+a*(b*c*(2*B*d+C*c)+ 
a*d*(C*d+2*D*c)))*x/a^3/b^2/(b*x^2+a)+1/16*(A*b*(a*d^2+5*b*c^2)+a*(b*c*(2* 
B*d+C*c)+a*d*(C*d+2*D*c)))*arctan(b^(1/2)*x/a^(1/2))/a^(7/2)/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 340, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\frac {-\frac {3 \sqrt {a} \left (8 a^3 d^2 D-5 A b^3 c^2 x-a b^2 \left (c^2 C+2 B c d+A d^2\right ) x-a^2 b d (C d+2 c D) x\right )}{a+b x^2}-\frac {8 a^{5/2} \left (a^3 d^2 D-A b^3 c^2 x+a b^2 \left (c^2 C x+A d (2 c+d x)+B c (c+2 d x)\right )-a^2 b \left (c^2 D+d^2 (B+C x)+2 c d (C+D x)\right )\right )}{\left (a+b x^2\right )^3}+\frac {2 a^{3/2} \left (12 a^3 d^2 D+5 A b^3 c^2 x+a b^2 \left (c^2 C+2 B c d+A d^2\right ) x-a^2 b \left (6 c^2 D+d^2 (6 B+7 C x)+2 c d (6 C+7 D x)\right )\right )}{\left (a+b x^2\right )^2}+3 \sqrt {b} \left (A b \left (5 b c^2+a d^2\right )+a (b c (c C+2 B d)+a d (C d+2 c D))\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{48 a^{7/2} b^3} \] Input:

Integrate[((c + d*x)^2*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^4,x]
 

Output:

((-3*Sqrt[a]*(8*a^3*d^2*D - 5*A*b^3*c^2*x - a*b^2*(c^2*C + 2*B*c*d + A*d^2 
)*x - a^2*b*d*(C*d + 2*c*D)*x))/(a + b*x^2) - (8*a^(5/2)*(a^3*d^2*D - A*b^ 
3*c^2*x + a*b^2*(c^2*C*x + A*d*(2*c + d*x) + B*c*(c + 2*d*x)) - a^2*b*(c^2 
*D + d^2*(B + C*x) + 2*c*d*(C + D*x))))/(a + b*x^2)^3 + (2*a^(3/2)*(12*a^3 
*d^2*D + 5*A*b^3*c^2*x + a*b^2*(c^2*C + 2*B*c*d + A*d^2)*x - a^2*b*(6*c^2* 
D + d^2*(6*B + 7*C*x) + 2*c*d*(6*C + 7*D*x))))/(a + b*x^2)^2 + 3*Sqrt[b]*( 
A*b*(5*b*c^2 + a*d^2) + a*(b*c*(c*C + 2*B*d) + a*d*(C*d + 2*c*D)))*ArcTan[ 
(Sqrt[b]*x)/Sqrt[a]])/(48*a^(7/2)*b^3)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2176, 25, 2176, 25, 454, 218}

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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2176

\(\displaystyle \frac {-\frac {\int -\frac {3 \left (A b \left (5 b c^2+a d^2\right )+a (b c (c C+2 B d)+a d (C d+2 c D))\right )+2 d \left (5 A c b^2+a (b c C+2 b B d+4 a d D)\right ) x}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {(c+d x) \left (3 a (2 a c D+a C d+A b d)-x \left (a (-8 a d D+2 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {3 \left (A b \left (5 b c^2+a d^2\right )+a (b c (c C+2 B d)+a d (C d+2 c D))\right )+2 d \left (5 A c b^2+a (b c C+2 b B d+4 a d D)\right ) x}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {(c+d x) \left (3 a (2 a c D+a C d+A b d)-x \left (a (-8 a d D+2 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 454

\(\displaystyle \frac {\frac {\frac {3 \left (A b \left (a d^2+5 b c^2\right )+a (a d (2 c D+C d)+b c (2 B d+c C))\right ) \int \frac {1}{b x^2+a}dx}{2 a}-\frac {2 a d \left (a (4 a d D+2 b B d+b c C)+5 A b^2 c\right )-3 b x \left (A b \left (a d^2+5 b c^2\right )+a (a d (2 c D+C d)+b c (2 B d+c C))\right )}{2 a b \left (a+b x^2\right )}}{4 a b}-\frac {(c+d x) \left (3 a (2 a c D+a C d+A b d)-x \left (a (-8 a d D+2 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (A b \left (a d^2+5 b c^2\right )+a (a d (2 c D+C d)+b c (2 B d+c C))\right )}{2 a^{3/2} \sqrt {b}}-\frac {2 a d \left (a (4 a d D+2 b B d+b c C)+5 A b^2 c\right )-3 b x \left (A b \left (a d^2+5 b c^2\right )+a (a d (2 c D+C d)+b c (2 B d+c C))\right )}{2 a b \left (a+b x^2\right )}}{4 a b}-\frac {(c+d x) \left (3 a (2 a c D+a C d+A b d)-x \left (a (-8 a d D+2 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

Input:

Int[((c + d*x)^2*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^4,x]
 

Output:

-1/6*((a*(B - (a*D)/b) - (A*b - a*C)*x)*(c + d*x)^2)/(a*b*(a + b*x^2)^3) + 
 (-1/4*((c + d*x)*(3*a*(A*b*d + a*C*d + 2*a*c*D) - (5*A*b^2*c + a*(b*c*C + 
 2*b*B*d - 8*a*d*D))*x))/(a*b*(a + b*x^2)^2) + (-1/2*(2*a*d*(5*A*b^2*c + a 
*(b*c*C + 2*b*B*d + 4*a*d*D)) - 3*b*(A*b*(5*b*c^2 + a*d^2) + a*(b*c*(c*C + 
 2*B*d) + a*d*(C*d + 2*c*D)))*x)/(a*b*(a + b*x^2)) + (3*(A*b*(5*b*c^2 + a* 
d^2) + a*(b*c*(c*C + 2*B*d) + a*d*(C*d + 2*c*D)))*ArcTan[(Sqrt[b]*x)/Sqrt[ 
a]])/(2*a^(3/2)*Sqrt[b]))/(4*a*b))/(6*a*b)
 

Defintions of rubi rules used

rule 25

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

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 454

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d 
 - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a 
*(p + 1)))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L 
tQ[p, -1] && NeQ[p, -3/2]
 

rule 2176

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema 
inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 
, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p 
 + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p 
 + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + 
b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && R 
ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 351, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\frac {\left (A a b \,d^{2}+5 A \,b^{2} c^{2}+2 a b B c d +a^{2} C \,d^{2}+C a b \,c^{2}+2 a^{2} c d D\right ) x^{5}}{16 a^{3}}-\frac {d^{2} D x^{4}}{2 b}+\frac {\left (A a b \,d^{2}+5 A \,b^{2} c^{2}+2 a b B c d -a^{2} C \,d^{2}+C a b \,c^{2}-2 a^{2} c d D\right ) x^{3}}{6 a^{2} b}-\frac {\left (B b \,d^{2}+2 C b c d +2 a \,d^{2} D+D b \,c^{2}\right ) x^{2}}{4 b^{2}}-\frac {\left (A a b \,d^{2}-11 A \,b^{2} c^{2}+2 a b B c d +a^{2} C \,d^{2}+C a b \,c^{2}+2 a^{2} c d D\right ) x}{16 a \,b^{2}}-\frac {4 A \,b^{2} c d +B a b \,d^{2}+2 B \,b^{2} c^{2}+2 a b c d C +2 a^{2} d^{2} D+D a b \,c^{2}}{12 b^{3}}}{\left (b \,x^{2}+a \right )^{3}}+\frac {\left (A a b \,d^{2}+5 A \,b^{2} c^{2}+2 a b B c d +a^{2} C \,d^{2}+C a b \,c^{2}+2 a^{2} c d D\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 a^{3} b^{2} \sqrt {a b}}\)	\(351\)

Input:

int((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^4,x,method=_RETURNVERBOSE)
 

Output:

(1/16*(A*a*b*d^2+5*A*b^2*c^2+2*B*a*b*c*d+C*a^2*d^2+C*a*b*c^2+2*D*a^2*c*d)/ 
a^3*x^5-1/2*d^2*D*x^4/b+1/6*(A*a*b*d^2+5*A*b^2*c^2+2*B*a*b*c*d-C*a^2*d^2+C 
*a*b*c^2-2*D*a^2*c*d)/a^2/b*x^3-1/4*(B*b*d^2+2*C*b*c*d+2*D*a*d^2+D*b*c^2)/ 
b^2*x^2-1/16*(A*a*b*d^2-11*A*b^2*c^2+2*B*a*b*c*d+C*a^2*d^2+C*a*b*c^2+2*D*a 
^2*c*d)/a/b^2*x-1/12*(4*A*b^2*c*d+B*a*b*d^2+2*B*b^2*c^2+2*C*a*b*c*d+2*D*a^ 
2*d^2+D*a*b*c^2)/b^3)/(b*x^2+a)^3+1/16*(A*a*b*d^2+5*A*b^2*c^2+2*B*a*b*c*d+ 
C*a^2*d^2+C*a*b*c^2+2*D*a^2*c*d)/a^3/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2 
))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 1294, normalized size of antiderivative = 3.44 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^4,x, algorithm="fricas")
 

Output:

[-1/96*(48*D*a^4*b^2*d^2*x^4 - 6*((C*a^2*b^4 + 5*A*a*b^5)*c^2 + 2*(D*a^3*b 
^3 + B*a^2*b^4)*c*d + (C*a^3*b^3 + A*a^2*b^4)*d^2)*x^5 - 16*((C*a^3*b^3 + 
5*A*a^2*b^4)*c^2 - 2*(D*a^4*b^2 - B*a^3*b^3)*c*d - (C*a^4*b^2 - A*a^3*b^3) 
*d^2)*x^3 + 8*(D*a^5*b + 2*B*a^4*b^2)*c^2 + 16*(C*a^5*b + 2*A*a^4*b^2)*c*d 
 + 8*(2*D*a^6 + B*a^5*b)*d^2 + 24*(D*a^4*b^2*c^2 + 2*C*a^4*b^2*c*d + (2*D* 
a^5*b + B*a^4*b^2)*d^2)*x^2 + 3*(((C*a*b^4 + 5*A*b^5)*c^2 + 2*(D*a^2*b^3 + 
 B*a*b^4)*c*d + (C*a^2*b^3 + A*a*b^4)*d^2)*x^6 + 3*((C*a^2*b^3 + 5*A*a*b^4 
)*c^2 + 2*(D*a^3*b^2 + B*a^2*b^3)*c*d + (C*a^3*b^2 + A*a^2*b^3)*d^2)*x^4 + 
 (C*a^4*b + 5*A*a^3*b^2)*c^2 + 2*(D*a^5 + B*a^4*b)*c*d + (C*a^5 + A*a^4*b) 
*d^2 + 3*((C*a^3*b^2 + 5*A*a^2*b^3)*c^2 + 2*(D*a^4*b + B*a^3*b^2)*c*d + (C 
*a^4*b + A*a^3*b^2)*d^2)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/ 
(b*x^2 + a)) + 6*((C*a^4*b^2 - 11*A*a^3*b^3)*c^2 + 2*(D*a^5*b + B*a^4*b^2) 
*c*d + (C*a^5*b + A*a^4*b^2)*d^2)*x)/(a^4*b^6*x^6 + 3*a^5*b^5*x^4 + 3*a^6* 
b^4*x^2 + a^7*b^3), -1/48*(24*D*a^4*b^2*d^2*x^4 - 3*((C*a^2*b^4 + 5*A*a*b^ 
5)*c^2 + 2*(D*a^3*b^3 + B*a^2*b^4)*c*d + (C*a^3*b^3 + A*a^2*b^4)*d^2)*x^5 
- 8*((C*a^3*b^3 + 5*A*a^2*b^4)*c^2 - 2*(D*a^4*b^2 - B*a^3*b^3)*c*d - (C*a^ 
4*b^2 - A*a^3*b^3)*d^2)*x^3 + 4*(D*a^5*b + 2*B*a^4*b^2)*c^2 + 8*(C*a^5*b + 
 2*A*a^4*b^2)*c*d + 4*(2*D*a^6 + B*a^5*b)*d^2 + 12*(D*a^4*b^2*c^2 + 2*C*a^ 
4*b^2*c*d + (2*D*a^5*b + B*a^4*b^2)*d^2)*x^2 - 3*(((C*a*b^4 + 5*A*b^5)*c^2 
 + 2*(D*a^2*b^3 + B*a*b^4)*c*d + (C*a^2*b^3 + A*a*b^4)*d^2)*x^6 + 3*((C...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**2*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**4,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.14 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=-\frac {24 \, D a^{3} b^{2} d^{2} x^{4} - 3 \, {\left ({\left (C a b^{4} + 5 \, A b^{5}\right )} c^{2} + 2 \, {\left (D a^{2} b^{3} + B a b^{4}\right )} c d + {\left (C a^{2} b^{3} + A a b^{4}\right )} d^{2}\right )} x^{5} - 8 \, {\left ({\left (C a^{2} b^{3} + 5 \, A a b^{4}\right )} c^{2} - 2 \, {\left (D a^{3} b^{2} - B a^{2} b^{3}\right )} c d - {\left (C a^{3} b^{2} - A a^{2} b^{3}\right )} d^{2}\right )} x^{3} + 4 \, {\left (D a^{4} b + 2 \, B a^{3} b^{2}\right )} c^{2} + 8 \, {\left (C a^{4} b + 2 \, A a^{3} b^{2}\right )} c d + 4 \, {\left (2 \, D a^{5} + B a^{4} b\right )} d^{2} + 12 \, {\left (D a^{3} b^{2} c^{2} + 2 \, C a^{3} b^{2} c d + {\left (2 \, D a^{4} b + B a^{3} b^{2}\right )} d^{2}\right )} x^{2} + 3 \, {\left ({\left (C a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} c^{2} + 2 \, {\left (D a^{4} b + B a^{3} b^{2}\right )} c d + {\left (C a^{4} b + A a^{3} b^{2}\right )} d^{2}\right )} x}{48 \, {\left (a^{3} b^{6} x^{6} + 3 \, a^{4} b^{5} x^{4} + 3 \, a^{5} b^{4} x^{2} + a^{6} b^{3}\right )}} + \frac {{\left ({\left (C a b + 5 \, A b^{2}\right )} c^{2} + 2 \, {\left (D a^{2} + B a b\right )} c d + {\left (C a^{2} + A a b\right )} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3} b^{2}} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^4,x, algorithm="maxima")
 

Output:

-1/48*(24*D*a^3*b^2*d^2*x^4 - 3*((C*a*b^4 + 5*A*b^5)*c^2 + 2*(D*a^2*b^3 + 
B*a*b^4)*c*d + (C*a^2*b^3 + A*a*b^4)*d^2)*x^5 - 8*((C*a^2*b^3 + 5*A*a*b^4) 
*c^2 - 2*(D*a^3*b^2 - B*a^2*b^3)*c*d - (C*a^3*b^2 - A*a^2*b^3)*d^2)*x^3 + 
4*(D*a^4*b + 2*B*a^3*b^2)*c^2 + 8*(C*a^4*b + 2*A*a^3*b^2)*c*d + 4*(2*D*a^5 
 + B*a^4*b)*d^2 + 12*(D*a^3*b^2*c^2 + 2*C*a^3*b^2*c*d + (2*D*a^4*b + B*a^3 
*b^2)*d^2)*x^2 + 3*((C*a^3*b^2 - 11*A*a^2*b^3)*c^2 + 2*(D*a^4*b + B*a^3*b^ 
2)*c*d + (C*a^4*b + A*a^3*b^2)*d^2)*x)/(a^3*b^6*x^6 + 3*a^4*b^5*x^4 + 3*a^ 
5*b^4*x^2 + a^6*b^3) + 1/16*((C*a*b + 5*A*b^2)*c^2 + 2*(D*a^2 + B*a*b)*c*d 
 + (C*a^2 + A*a*b)*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b^2)
 

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\frac {{\left (C a b c^{2} + 5 \, A b^{2} c^{2} + 2 \, D a^{2} c d + 2 \, B a b c d + C a^{2} d^{2} + A a b d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3} b^{2}} + \frac {3 \, C a b^{4} c^{2} x^{5} + 15 \, A b^{5} c^{2} x^{5} + 6 \, D a^{2} b^{3} c d x^{5} + 6 \, B a b^{4} c d x^{5} + 3 \, C a^{2} b^{3} d^{2} x^{5} + 3 \, A a b^{4} d^{2} x^{5} - 24 \, D a^{3} b^{2} d^{2} x^{4} + 8 \, C a^{2} b^{3} c^{2} x^{3} + 40 \, A a b^{4} c^{2} x^{3} - 16 \, D a^{3} b^{2} c d x^{3} + 16 \, B a^{2} b^{3} c d x^{3} - 8 \, C a^{3} b^{2} d^{2} x^{3} + 8 \, A a^{2} b^{3} d^{2} x^{3} - 12 \, D a^{3} b^{2} c^{2} x^{2} - 24 \, C a^{3} b^{2} c d x^{2} - 24 \, D a^{4} b d^{2} x^{2} - 12 \, B a^{3} b^{2} d^{2} x^{2} - 3 \, C a^{3} b^{2} c^{2} x + 33 \, A a^{2} b^{3} c^{2} x - 6 \, D a^{4} b c d x - 6 \, B a^{3} b^{2} c d x - 3 \, C a^{4} b d^{2} x - 3 \, A a^{3} b^{2} d^{2} x - 4 \, D a^{4} b c^{2} - 8 \, B a^{3} b^{2} c^{2} - 8 \, C a^{4} b c d - 16 \, A a^{3} b^{2} c d - 8 \, D a^{5} d^{2} - 4 \, B a^{4} b d^{2}}{48 \, {\left (b x^{2} + a\right )}^{3} a^{3} b^{3}} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^4,x, algorithm="giac")
 

Output:

1/16*(C*a*b*c^2 + 5*A*b^2*c^2 + 2*D*a^2*c*d + 2*B*a*b*c*d + C*a^2*d^2 + A* 
a*b*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b^2) + 1/48*(3*C*a*b^4*c^2*x 
^5 + 15*A*b^5*c^2*x^5 + 6*D*a^2*b^3*c*d*x^5 + 6*B*a*b^4*c*d*x^5 + 3*C*a^2* 
b^3*d^2*x^5 + 3*A*a*b^4*d^2*x^5 - 24*D*a^3*b^2*d^2*x^4 + 8*C*a^2*b^3*c^2*x 
^3 + 40*A*a*b^4*c^2*x^3 - 16*D*a^3*b^2*c*d*x^3 + 16*B*a^2*b^3*c*d*x^3 - 8* 
C*a^3*b^2*d^2*x^3 + 8*A*a^2*b^3*d^2*x^3 - 12*D*a^3*b^2*c^2*x^2 - 24*C*a^3* 
b^2*c*d*x^2 - 24*D*a^4*b*d^2*x^2 - 12*B*a^3*b^2*d^2*x^2 - 3*C*a^3*b^2*c^2* 
x + 33*A*a^2*b^3*c^2*x - 6*D*a^4*b*c*d*x - 6*B*a^3*b^2*c*d*x - 3*C*a^4*b*d 
^2*x - 3*A*a^3*b^2*d^2*x - 4*D*a^4*b*c^2 - 8*B*a^3*b^2*c^2 - 8*C*a^4*b*c*d 
 - 16*A*a^3*b^2*c*d - 8*D*a^5*d^2 - 4*B*a^4*b*d^2)/((b*x^2 + a)^3*a^3*b^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\int \frac {{\left (c+d\,x\right )}^2\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^4} \,d x \] Input:

int(((c + d*x)^2*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^4,x)
 

Output:

int(((c + d*x)^2*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 874, normalized size of antiderivative = 2.32 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^4,x)
 

Output:

(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*b*d**2 + 9*sqrt(b)*s 
qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*c*d**2 + 15*sqrt(b)*sqrt(a)*atan 
((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**2*c**2 + 6*sqrt(b)*sqrt(a)*atan((b*x)/(s 
qrt(b)*sqrt(a)))*a**3*b**2*c*d + 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqr 
t(a)))*a**3*b**2*d**2*x**2 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a) 
))*a**3*b*c**3 + 27*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c 
*d**2*x**2 + 45*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**3*c* 
*2*x**2 + 18*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**3*c*d*x 
**2 + 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**3*d**2*x**4 
+ 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**3*x**2 + 27 
*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c*d**2*x**4 + 45* 
sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**4*c**2*x**4 + 18*sqrt(b 
)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**4*c*d*x**4 + 3*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**4*d**2*x**6 + 9*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*a*b**3*c**3*x**4 + 9*sqrt(b)*sqrt(a)*atan((b*x)/( 
sqrt(b)*sqrt(a)))*a*b**3*c*d**2*x**6 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt 
(b)*sqrt(a)))*b**5*c**2*x**6 + 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt( 
a)))*b**5*c*d*x**6 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**4* 
c**3*x**6 - 16*a**4*b**2*c*d - 3*a**4*b**2*d**2*x - 4*a**4*b**2*d**2 - 12* 
a**4*b*c**2*d - 9*a**4*b*c*d**2*x + 33*a**3*b**3*c**2*x - 8*a**3*b**3*c...