\(\int \frac {(c+d x)^3 (A+B x+C x^2+D x^3)}{(a+b x^2)^4} \, dx\) [45]

Optimal result

Integrand size = 32, antiderivative size = 523 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=-\frac {b^2 c^2 (B c+3 A d)+a^2 d^2 (C d+3 c D)-a b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )}{6 b^3 \left (a+b x^2\right )^3}+\frac {\left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (b^2 c^2 (c C+3 B d)+a^2 d^3 D-a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) x}{6 a b^3 \left (a+b x^2\right )^3}+\frac {2 a d^2 (C d+3 c D)-b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )}{4 b^3 \left (a+b x^2\right )^2}+\frac {\left (A b^2 c \left (5 b c^2+3 a d^2\right )+a \left (b^2 c^2 (c C+3 B d)+13 a^2 d^3 D-7 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) x}{24 a^2 b^3 \left (a+b x^2\right )^2}-\frac {d^2 (C d+3 c D)}{2 b^3 \left (a+b x^2\right )}+\frac {\left (A b^2 c \left (5 b c^2+3 a d^2\right )+a \left (b^2 c^2 (c C+3 B d)-11 a^2 d^3 D+a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) x}{16 a^3 b^3 \left (a+b x^2\right )}+\frac {\left (A b^2 c \left (5 b c^2+3 a d^2\right )+a \left (b^2 c^2 (c C+3 B d)+5 a^2 d^3 D+a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} b^{7/2}} \] Output:

-1/6*(b^2*c^2*(3*A*d+B*c)+a^2*d^2*(C*d+3*D*c)-a*b*(A*d^3+3*B*c*d^2+3*C*c^2 
*d+D*c^3))/b^3/(b*x^2+a)^3+1/6*(A*b^2*c*(-3*a*d^2+b*c^2)-a*(b^2*c^2*(3*B*d 
+C*c)+a^2*d^3*D-a*b*d*(B*d^2+3*C*c*d+3*D*c^2)))*x/a/b^3/(b*x^2+a)^3+1/4*(2 
*a*d^2*(C*d+3*D*c)-b*(A*d^3+3*B*c*d^2+3*C*c^2*d+D*c^3))/b^3/(b*x^2+a)^2+1/ 
24*(A*b^2*c*(3*a*d^2+5*b*c^2)+a*(b^2*c^2*(3*B*d+C*c)+13*a^2*d^3*D-7*a*b*d* 
(B*d^2+3*C*c*d+3*D*c^2)))*x/a^2/b^3/(b*x^2+a)^2-1/2*d^2*(C*d+3*D*c)/b^3/(b 
*x^2+a)+1/16*(A*b^2*c*(3*a*d^2+5*b*c^2)+a*(b^2*c^2*(3*B*d+C*c)-11*a^2*d^3* 
D+a*b*d*(B*d^2+3*C*c*d+3*D*c^2)))*x/a^3/b^3/(b*x^2+a)+1/16*(A*b^2*c*(3*a*d 
^2+5*b*c^2)+a*(b^2*c^2*(3*B*d+C*c)+5*a^2*d^3*D+a*b*d*(B*d^2+3*C*c*d+3*D*c^ 
2)))*arctan(b^(1/2)*x/a^(1/2))/a^(7/2)/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 447, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\frac {5 A b^3 c^3 x+a b^2 c \left (c^2 C+3 B c d+3 A d^2\right ) x+a^2 b d \left (3 c C d+B d^2+3 c^2 D\right ) x-a^3 d^2 (8 C d+24 c D+11 d D x)}{16 a^3 b^3 \left (a+b x^2\right )}+\frac {A b^3 c^3 x-a^3 d^2 (C d+3 c D+d D x)-a b^2 c \left (c^2 C x+3 A d (c+d x)+B c (c+3 d x)\right )+a^2 b \left (c^3 D+d^3 (A+B x)+3 c d^2 (B+C x)+3 c^2 d (C+D x)\right )}{6 a b^3 \left (a+b x^2\right )^3}+\frac {5 A b^3 c^3 x+a b^2 c \left (c^2 C+3 B c d+3 A d^2\right ) x+a^3 d^2 (12 C d+36 c D+13 d D x)-a^2 b \left (6 c^3 D+d^3 (6 A+7 B x)+3 c d^2 (6 B+7 C x)+3 c^2 d (6 C+7 D x)\right )}{24 a^2 b^3 \left (a+b x^2\right )^2}+\frac {\left (A b^2 c \left (5 b c^2+3 a d^2\right )+a \left (b^2 c^2 (c C+3 B d)+5 a^2 d^3 D+a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} b^{7/2}} \] Input:

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

Output:

(5*A*b^3*c^3*x + a*b^2*c*(c^2*C + 3*B*c*d + 3*A*d^2)*x + a^2*b*d*(3*c*C*d 
+ B*d^2 + 3*c^2*D)*x - a^3*d^2*(8*C*d + 24*c*D + 11*d*D*x))/(16*a^3*b^3*(a 
 + b*x^2)) + (A*b^3*c^3*x - a^3*d^2*(C*d + 3*c*D + d*D*x) - a*b^2*c*(c^2*C 
*x + 3*A*d*(c + d*x) + B*c*(c + 3*d*x)) + a^2*b*(c^3*D + d^3*(A + B*x) + 3 
*c*d^2*(B + C*x) + 3*c^2*d*(C + D*x)))/(6*a*b^3*(a + b*x^2)^3) + (5*A*b^3* 
c^3*x + a*b^2*c*(c^2*C + 3*B*c*d + 3*A*d^2)*x + a^3*d^2*(12*C*d + 36*c*D + 
 13*d*D*x) - a^2*b*(6*c^3*D + d^3*(6*A + 7*B*x) + 3*c*d^2*(6*B + 7*C*x) + 
3*c^2*d*(6*C + 7*D*x)))/(24*a^2*b^3*(a + b*x^2)^2) + ((A*b^2*c*(5*b*c^2 + 
3*a*d^2) + a*(b^2*c^2*(c*C + 3*B*d) + 5*a^2*d^3*D + a*b*d*(3*c*C*d + B*d^2 
 + 3*c^2*D)))*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*a^(7/2)*b^(7/2))
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 381, normalized size of antiderivative = 0.73, 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, 675, 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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2176

\(\displaystyle \frac {-\frac {\int -\frac {(c+d x) \left (A b \left (15 b c^2+4 a d^2\right )+a (3 b c (c C+3 B d)+a d (8 C d+9 c D))+d \left (5 A c b^2+a (b c C+3 b B d+15 a d D)\right ) x\right )}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {(c+d x)^2 \left (2 a (3 a c D+2 a C d+A b d)-x \left (a (-9 a d D+3 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {(c+d x) \left (A b \left (15 b c^2+4 a d^2\right )+a (3 b c (c C+3 B d)+a d (8 C d+9 c D))+d \left (5 A c b^2+a (b c C+3 b B d+15 a d D)\right ) x\right )}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {(c+d x)^2 \left (2 a (3 a c D+2 a C d+A b d)-x \left (a (-9 a d D+3 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 675

\(\displaystyle \frac {\frac {\frac {3 \left (a \left (5 a^2 d^3 D+a b d \left (B d^2+3 c^2 D+3 c C d\right )+b^2 c^2 (3 B d+c C)\right )+A b^2 c \left (3 a d^2+5 b c^2\right )\right ) \int \frac {1}{b x^2+a}dx}{2 a b}+\frac {x \left (-15 a^2 d^3 D+\frac {15 A b^3 c^3}{a}+a b d \left (-3 B d^2+9 c^2 D+7 c C d\right )+b^2 c \left (-A d^2+9 B c d+3 c^2 C\right )\right )}{2 b \left (a+b x^2\right )}-\frac {2 d \left (A b \left (a d^2+5 b c^2\right )+a (2 a d (3 c D+C d)+b c (3 B d+c C))\right )}{b \left (a+b x^2\right )}}{4 a b}-\frac {(c+d x)^2 \left (2 a (3 a c D+2 a C d+A b d)-x \left (a (-9 a d D+3 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {x \left (-15 a^2 d^3 D+\frac {15 A b^3 c^3}{a}+a b d \left (-3 B d^2+9 c^2 D+7 c C d\right )+b^2 c \left (-A d^2+9 B c d+3 c^2 C\right )\right )}{2 b \left (a+b x^2\right )}+\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (a \left (5 a^2 d^3 D+a b d \left (B d^2+3 c^2 D+3 c C d\right )+b^2 c^2 (3 B d+c C)\right )+A b^2 c \left (3 a d^2+5 b c^2\right )\right )}{2 a^{3/2} b^{3/2}}-\frac {2 d \left (A b \left (a d^2+5 b c^2\right )+a (2 a d (3 c D+C d)+b c (3 B d+c C))\right )}{b \left (a+b x^2\right )}}{4 a b}-\frac {(c+d x)^2 \left (2 a (3 a c D+2 a C d+A b d)-x \left (a (-9 a d D+3 b B d+b c C)+5 A b^2 c\right )\right )}{4 a b \left (a+b x^2\right )^2}}{6 a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{6 a b \left (a+b x^2\right )^3}\)

Input:

Int[((c + d*x)^3*(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)^3)/(a*b*(a + b*x^2)^3) + 
 (-1/4*((c + d*x)^2*(2*a*(A*b*d + 2*a*C*d + 3*a*c*D) - (5*A*b^2*c + a*(b*c 
*C + 3*b*B*d - 9*a*d*D))*x))/(a*b*(a + b*x^2)^2) + ((-2*d*(A*b*(5*b*c^2 + 
a*d^2) + a*(b*c*(c*C + 3*B*d) + 2*a*d*(C*d + 3*c*D))))/(b*(a + b*x^2)) + ( 
((15*A*b^3*c^3)/a + b^2*c*(3*c^2*C + 9*B*c*d - A*d^2) - 15*a^2*d^3*D + a*b 
*d*(7*c*C*d - 3*B*d^2 + 9*c^2*D))*x)/(2*b*(a + b*x^2)) + (3*(A*b^2*c*(5*b* 
c^2 + 3*a*d^2) + a*(b^2*c^2*(c*C + 3*B*d) + 5*a^2*d^3*D + a*b*d*(3*c*C*d + 
 B*d^2 + 3*c^2*D)))*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(3/2)*b^(3/2)))/(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 675

Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- 
Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* 
e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1))   Int[(a + c*x^2)^(p + 1), x], x]) / 
; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] &&  !(IntegerQ[p] && NiceSqrtQ 
[(-a)*c])
 

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 = 538, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {\frac {\left (3 A a \,b^{2} c \,d^{2}+5 A \,b^{3} c^{3}+B \,a^{2} b \,d^{3}+3 B a \,b^{2} c^{2} d +3 C \,a^{2} b c \,d^{2}+C a \,b^{2} c^{3}-11 D a^{3} d^{3}+3 D a^{2} b \,c^{2} d \right ) x^{5}}{16 a^{3} b}-\frac {d^{2} \left (C d +3 D c \right ) x^{4}}{2 b}+\frac {\left (3 A a \,b^{2} c \,d^{2}+5 A \,b^{3} c^{3}-B \,a^{2} b \,d^{3}+3 B a \,b^{2} c^{2} d -3 C \,a^{2} b c \,d^{2}+C a \,b^{2} c^{3}-5 D a^{3} d^{3}-3 D a^{2} b \,c^{2} d \right ) x^{3}}{6 a^{2} b^{2}}-\frac {\left (A b \,d^{3}+3 B b c \,d^{2}+2 C a \,d^{3}+3 C b \,c^{2} d +6 D a c \,d^{2}+D b \,c^{3}\right ) x^{2}}{4 b^{2}}-\frac {\left (3 A a \,b^{2} c \,d^{2}-11 A \,b^{3} c^{3}+B \,a^{2} b \,d^{3}+3 B a \,b^{2} c^{2} d +3 C \,a^{2} b c \,d^{2}+C a \,b^{2} c^{3}+5 D a^{3} d^{3}+3 D a^{2} b \,c^{2} d \right ) x}{16 a \,b^{3}}-\frac {A a b \,d^{3}+6 A \,b^{2} c^{2} d +3 B a b c \,d^{2}+2 B \,b^{2} c^{3}+2 C \,a^{2} d^{3}+3 C a b \,c^{2} d +6 D a^{2} c \,d^{2}+D a b \,c^{3}}{12 b^{3}}}{\left (b \,x^{2}+a \right )^{3}}+\frac {\left (3 A a \,b^{2} c \,d^{2}+5 A \,b^{3} c^{3}+B \,a^{2} b \,d^{3}+3 B a \,b^{2} c^{2} d +3 C \,a^{2} b c \,d^{2}+C a \,b^{2} c^{3}+5 D a^{3} d^{3}+3 D a^{2} b \,c^{2} d \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 a^{3} b^{3} \sqrt {a b}}\)	\(538\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 1824, normalized size of antiderivative = 3.49 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

integrate((d*x+c)**3*(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 = 592, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\frac {3 \, {\left ({\left (C a b^{4} + 5 \, A b^{5}\right )} c^{3} + 3 \, {\left (D a^{2} b^{3} + B a b^{4}\right )} c^{2} d + 3 \, {\left (C a^{2} b^{3} + A a b^{4}\right )} c d^{2} - {\left (11 \, D a^{3} b^{2} - B a^{2} b^{3}\right )} d^{3}\right )} x^{5} - 24 \, {\left (3 \, D a^{3} b^{2} c d^{2} + C a^{3} b^{2} d^{3}\right )} x^{4} - 4 \, {\left (D a^{4} b + 2 \, B a^{3} b^{2}\right )} c^{3} - 12 \, {\left (C a^{4} b + 2 \, A a^{3} b^{2}\right )} c^{2} d - 12 \, {\left (2 \, D a^{5} + B a^{4} b\right )} c d^{2} - 4 \, {\left (2 \, C a^{5} + A a^{4} b\right )} d^{3} + 8 \, {\left ({\left (C a^{2} b^{3} + 5 \, A a b^{4}\right )} c^{3} - 3 \, {\left (D a^{3} b^{2} - B a^{2} b^{3}\right )} c^{2} d - 3 \, {\left (C a^{3} b^{2} - A a^{2} b^{3}\right )} c d^{2} - {\left (5 \, D a^{4} b + B a^{3} b^{2}\right )} d^{3}\right )} x^{3} - 12 \, {\left (D a^{3} b^{2} c^{3} + 3 \, C a^{3} b^{2} c^{2} d + 3 \, {\left (2 \, D a^{4} b + B a^{3} b^{2}\right )} c d^{2} + {\left (2 \, C a^{4} b + A a^{3} b^{2}\right )} d^{3}\right )} x^{2} - 3 \, {\left ({\left (C a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} c^{3} + 3 \, {\left (D a^{4} b + B a^{3} b^{2}\right )} c^{2} d + 3 \, {\left (C a^{4} b + A a^{3} b^{2}\right )} c d^{2} + {\left (5 \, D a^{5} + B a^{4} b\right )} d^{3}\right )} x}{48 \, {\left (a^{3} b^{6} x^{6} + 3 \, a^{4} b^{5} x^{4} + 3 \, a^{5} b^{4} x^{2} + a^{6} b^{3}\right )}} + \frac {{\left ({\left (C a b^{2} + 5 \, A b^{3}\right )} c^{3} + 3 \, {\left (D a^{2} b + B a b^{2}\right )} c^{2} d + 3 \, {\left (C a^{2} b + A a b^{2}\right )} c d^{2} + {\left (5 \, D a^{3} + B a^{2} b\right )} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3} b^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 662, normalized size of antiderivative = 1.27 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^4} \, dx=\frac {{\left (C a b^{2} c^{3} + 5 \, A b^{3} c^{3} + 3 \, D a^{2} b c^{2} d + 3 \, B a b^{2} c^{2} d + 3 \, C a^{2} b c d^{2} + 3 \, A a b^{2} c d^{2} + 5 \, D a^{3} d^{3} + B a^{2} b d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3} b^{3}} + \frac {3 \, C a b^{4} c^{3} x^{5} + 15 \, A b^{5} c^{3} x^{5} + 9 \, D a^{2} b^{3} c^{2} d x^{5} + 9 \, B a b^{4} c^{2} d x^{5} + 9 \, C a^{2} b^{3} c d^{2} x^{5} + 9 \, A a b^{4} c d^{2} x^{5} - 33 \, D a^{3} b^{2} d^{3} x^{5} + 3 \, B a^{2} b^{3} d^{3} x^{5} - 72 \, D a^{3} b^{2} c d^{2} x^{4} - 24 \, C a^{3} b^{2} d^{3} x^{4} + 8 \, C a^{2} b^{3} c^{3} x^{3} + 40 \, A a b^{4} c^{3} x^{3} - 24 \, D a^{3} b^{2} c^{2} d x^{3} + 24 \, B a^{2} b^{3} c^{2} d x^{3} - 24 \, C a^{3} b^{2} c d^{2} x^{3} + 24 \, A a^{2} b^{3} c d^{2} x^{3} - 40 \, D a^{4} b d^{3} x^{3} - 8 \, B a^{3} b^{2} d^{3} x^{3} - 12 \, D a^{3} b^{2} c^{3} x^{2} - 36 \, C a^{3} b^{2} c^{2} d x^{2} - 72 \, D a^{4} b c d^{2} x^{2} - 36 \, B a^{3} b^{2} c d^{2} x^{2} - 24 \, C a^{4} b d^{3} x^{2} - 12 \, A a^{3} b^{2} d^{3} x^{2} - 3 \, C a^{3} b^{2} c^{3} x + 33 \, A a^{2} b^{3} c^{3} x - 9 \, D a^{4} b c^{2} d x - 9 \, B a^{3} b^{2} c^{2} d x - 9 \, C a^{4} b c d^{2} x - 9 \, A a^{3} b^{2} c d^{2} x - 15 \, D a^{5} d^{3} x - 3 \, B a^{4} b d^{3} x - 4 \, D a^{4} b c^{3} - 8 \, B a^{3} b^{2} c^{3} - 12 \, C a^{4} b c^{2} d - 24 \, A a^{3} b^{2} c^{2} d - 24 \, D a^{5} c d^{2} - 12 \, B a^{4} b c d^{2} - 8 \, C a^{5} d^{3} - 4 \, A a^{4} b d^{3}}{48 \, {\left (b x^{2} + a\right )}^{3} a^{3} b^{3}} \] Input:

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

Output:

1/16*(C*a*b^2*c^3 + 5*A*b^3*c^3 + 3*D*a^2*b*c^2*d + 3*B*a*b^2*c^2*d + 3*C* 
a^2*b*c*d^2 + 3*A*a*b^2*c*d^2 + 5*D*a^3*d^3 + B*a^2*b*d^3)*arctan(b*x/sqrt 
(a*b))/(sqrt(a*b)*a^3*b^3) + 1/48*(3*C*a*b^4*c^3*x^5 + 15*A*b^5*c^3*x^5 + 
9*D*a^2*b^3*c^2*d*x^5 + 9*B*a*b^4*c^2*d*x^5 + 9*C*a^2*b^3*c*d^2*x^5 + 9*A* 
a*b^4*c*d^2*x^5 - 33*D*a^3*b^2*d^3*x^5 + 3*B*a^2*b^3*d^3*x^5 - 72*D*a^3*b^ 
2*c*d^2*x^4 - 24*C*a^3*b^2*d^3*x^4 + 8*C*a^2*b^3*c^3*x^3 + 40*A*a*b^4*c^3* 
x^3 - 24*D*a^3*b^2*c^2*d*x^3 + 24*B*a^2*b^3*c^2*d*x^3 - 24*C*a^3*b^2*c*d^2 
*x^3 + 24*A*a^2*b^3*c*d^2*x^3 - 40*D*a^4*b*d^3*x^3 - 8*B*a^3*b^2*d^3*x^3 - 
 12*D*a^3*b^2*c^3*x^2 - 36*C*a^3*b^2*c^2*d*x^2 - 72*D*a^4*b*c*d^2*x^2 - 36 
*B*a^3*b^2*c*d^2*x^2 - 24*C*a^4*b*d^3*x^2 - 12*A*a^3*b^2*d^3*x^2 - 3*C*a^3 
*b^2*c^3*x + 33*A*a^2*b^3*c^3*x - 9*D*a^4*b*c^2*d*x - 9*B*a^3*b^2*c^2*d*x 
- 9*C*a^4*b*c*d^2*x - 9*A*a^3*b^2*c*d^2*x - 15*D*a^5*d^3*x - 3*B*a^4*b*d^3 
*x - 4*D*a^4*b*c^3 - 8*B*a^3*b^2*c^3 - 12*C*a^4*b*c^2*d - 24*A*a^3*b^2*c^2 
*d - 24*D*a^5*c*d^2 - 12*B*a^4*b*c*d^2 - 8*C*a^5*d^3 - 4*A*a^4*b*d^3)/((b* 
x^2 + a)^3*a^3*b^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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