\(\int \frac {(c+d x)^3 (A+B x+C x^2+D x^3)}{(a+b x^2)^2} \, dx\) [33]

Optimal result

Integrand size = 32, antiderivative size = 388 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {d \left (2 a d^2 D-b \left (3 c C d+B d^2+3 c^2 D\right )\right ) x}{b^3}+\frac {d^2 (C d+3 c D) x^2}{2 b^2}+\frac {d^3 D x^3}{3 b^2}-\frac {b^2 c^2 (B c+3 A d)+a^2 d^2 (C d+3 c D)-a b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )}{2 b^3 \left (a+b x^2\right )}+\frac {\left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (b^2 c^2 (c C+3 B d)+a^2 d^3 D-a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) x}{2 a b^3 \left (a+b x^2\right )}+\frac {\left (A b^2 c \left (b c^2+3 a d^2\right )+a \left (b^2 c^2 (c C+3 B d)+5 a^2 d^3 D-3 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{7/2}}-\frac {\left (2 a d^2 (C d+3 c D)-b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )\right ) \log \left (a+b x^2\right )}{2 b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {d \left (-2 a d^2 D+b \left (3 c C d+B d^2+3 c^2 D\right )\right ) x}{b^3}+\frac {d^2 (C d+3 c D) x^2}{2 b^2}+\frac {d^3 D x^3}{3 b^2}+\frac {A b^3 c^3 x-a^3 d^2 (C d+3 c D+d D x)-a b^2 c \left (c^2 C x+3 A d (c+d x)+B c (c+3 d x)\right )+a^2 b \left (c^3 D+d^3 (A+B x)+3 c d^2 (B+C x)+3 c^2 d (C+D x)\right )}{2 a b^3 \left (a+b x^2\right )}+\frac {\left (A b^2 c \left (b c^2+3 a d^2\right )+a \left (b^2 c^2 (c C+3 B d)+5 a^2 d^3 D-3 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{7/2}}+\frac {\left (-2 a d^2 (C d+3 c D)+b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )\right ) \log \left (a+b x^2\right )}{2 b^3} \] Input:

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

Output:

(d*(-2*a*d^2*D + b*(3*c*C*d + B*d^2 + 3*c^2*D))*x)/b^3 + (d^2*(C*d + 3*c*D 
)*x^2)/(2*b^2) + (d^3*D*x^3)/(3*b^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)))/(2*a*b^3* 
(a + b*x^2)) + ((A*b^2*c*(b*c^2 + 3*a*d^2) + a*(b^2*c^2*(c*C + 3*B*d) + 5* 
a^2*d^3*D - 3*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^(7/2)) + ((-2*a*d^2*(C*d + 3*c*D) + b*(3*c^2*C*d + 3*B*c 
*d^2 + A*d^3 + c^3*D))*Log[a + b*x^2])/(2*b^3)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2176, 25, 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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2160

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

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]
 

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 = 1.00 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {d \left (\frac {1}{3} D b \,x^{3} d^{2}+\frac {1}{2} C b \,x^{2} d^{2}+\frac {3}{2} D b c d \,x^{2}+B b \,d^{2} x +3 C b c d x -2 a \,d^{2} D x +3 D b \,c^{2} x \right )}{b^{3}}+\frac {\frac {-\frac {\left (3 A a \,b^{2} c \,d^{2}-A \,b^{3} c^{3}-B \,a^{2} b \,d^{3}+3 B a \,b^{2} c^{2} d -3 C \,a^{2} b c \,d^{2}+C a \,b^{2} c^{3}+D a^{3} d^{3}-3 D a^{2} b \,c^{2} d \right ) x}{2 a}+\frac {A a b \,d^{3}}{2}-\frac {3 A \,b^{2} c^{2} d}{2}+\frac {3 B a b c \,d^{2}}{2}-\frac {B \,b^{2} c^{3}}{2}-\frac {C \,a^{2} d^{3}}{2}+\frac {3 C a b \,c^{2} d}{2}-\frac {3 D a^{2} c \,d^{2}}{2}+\frac {D a b \,c^{3}}{2}}{b \,x^{2}+a}+\frac {\frac {\left (2 A \,d^{3} a \,b^{2}+6 B a \,b^{2} c \,d^{2}-4 C \,a^{2} b \,d^{3}+6 C a \,b^{2} c^{2} d -12 D a^{2} b c \,d^{2}+2 D a \,b^{2} c^{3}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (3 A a \,b^{2} c \,d^{2}+A \,b^{3} c^{3}-3 B \,a^{2} b \,d^{3}+3 B a \,b^{2} c^{2} d -9 C \,a^{2} b c \,d^{2}+C a \,b^{2} c^{3}+5 D a^{3} d^{3}-9 D a^{2} b \,c^{2} d \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{2 a}}{b^{3}}\)	\(421\)

Input:

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

Output:

d/b^3*(1/3*D*b*x^3*d^2+1/2*C*b*x^2*d^2+3/2*D*b*c*d*x^2+B*b*d^2*x+3*C*b*c*d 
*x-2*a*d^2*D*x+3*D*b*c^2*x)+1/b^3*((-1/2*(3*A*a*b^2*c*d^2-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+D*a^3*d^3-3*D*a^2*b*c^2* 
d)/a*x+1/2*A*a*b*d^3-3/2*A*b^2*c^2*d+3/2*B*a*b*c*d^2-1/2*B*b^2*c^3-1/2*C*a 
^2*d^3+3/2*C*a*b*c^2*d-3/2*D*a^2*c*d^2+1/2*D*a*b*c^3)/(b*x^2+a)+1/2/a*(1/2 
*(2*A*a*b^2*d^3+6*B*a*b^2*c*d^2-4*C*a^2*b*d^3+6*C*a*b^2*c^2*d-12*D*a^2*b*c 
*d^2+2*D*a*b^2*c^3)/b*ln(b*x^2+a)+(3*A*a*b^2*c*d^2+A*b^3*c^3-3*B*a^2*b*d^3 
+3*B*a*b^2*c^2*d-9*C*a^2*b*c*d^2+C*a*b^2*c^3+5*D*a^3*d^3-9*D*a^2*b*c^2*d)/ 
(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 1394, normalized size of antiderivative = 3.59 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/12*(4*D*a^2*b^3*d^3*x^5 + 6*(3*D*a^2*b^3*c*d^2 + C*a^2*b^3*d^3)*x^4 + 6 
*(D*a^3*b^2 - B*a^2*b^3)*c^3 + 18*(C*a^3*b^2 - A*a^2*b^3)*c^2*d - 18*(D*a^ 
4*b - B*a^3*b^2)*c*d^2 - 6*(C*a^4*b - A*a^3*b^2)*d^3 + 4*(9*D*a^2*b^3*c^2* 
d + 9*C*a^2*b^3*c*d^2 - (5*D*a^3*b^2 - 3*B*a^2*b^3)*d^3)*x^3 + 6*(3*D*a^3* 
b^2*c*d^2 + C*a^3*b^2*d^3)*x^2 + 3*((C*a^2*b^2 + A*a*b^3)*c^3 - 3*(3*D*a^3 
*b - B*a^2*b^2)*c^2*d - 3*(3*C*a^3*b - A*a^2*b^2)*c*d^2 + (5*D*a^4 - 3*B*a 
^3*b)*d^3 + ((C*a*b^3 + A*b^4)*c^3 - 3*(3*D*a^2*b^2 - B*a*b^3)*c^2*d - 3*( 
3*C*a^2*b^2 - A*a*b^3)*c*d^2 + (5*D*a^3*b - 3*B*a^2*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^2*b^3 - A*a*b^ 
4)*c^3 - 3*(3*D*a^3*b^2 - B*a^2*b^3)*c^2*d - 3*(3*C*a^3*b^2 - A*a^2*b^3)*c 
*d^2 + (5*D*a^4*b - 3*B*a^3*b^2)*d^3)*x + 6*(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 + (D* 
a^2*b^3*c^3 + 3*C*a^2*b^3*c^2*d - 3*(2*D*a^3*b^2 - B*a^2*b^3)*c*d^2 - (2*C 
*a^3*b^2 - A*a^2*b^3)*d^3)*x^2)*log(b*x^2 + a))/(a^2*b^5*x^2 + a^3*b^4), 1 
/6*(2*D*a^2*b^3*d^3*x^5 + 3*(3*D*a^2*b^3*c*d^2 + C*a^2*b^3*d^3)*x^4 + 3*(D 
*a^3*b^2 - B*a^2*b^3)*c^3 + 9*(C*a^3*b^2 - A*a^2*b^3)*c^2*d - 9*(D*a^4*b - 
 B*a^3*b^2)*c*d^2 - 3*(C*a^4*b - A*a^3*b^2)*d^3 + 2*(9*D*a^2*b^3*c^2*d + 9 
*C*a^2*b^3*c*d^2 - (5*D*a^3*b^2 - 3*B*a^2*b^3)*d^3)*x^3 + 3*(3*D*a^3*b^2*c 
*d^2 + C*a^3*b^2*d^3)*x^2 + 3*((C*a^2*b^2 + A*a*b^3)*c^3 - 3*(3*D*a^3*b - 
B*a^2*b^2)*c^2*d - 3*(3*C*a^3*b - A*a^2*b^2)*c*d^2 + (5*D*a^4 - 3*B*a^3...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1346 vs. \(2 (384) = 768\).

Time = 55.30 (sec) , antiderivative size = 1346, normalized size of antiderivative = 3.47 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

D*d**3*x**3/(3*b**2) + x**2*(C*d**3/(2*b**2) + 3*D*c*d**2/(2*b**2)) + x*(B 
*d**3/b**2 + 3*C*c*d**2/b**2 - 2*D*a*d**3/b**3 + 3*D*c**2*d/b**2) + (-(-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)/(2*b**3) - sqrt(-a**3*b**7)*(3*A*a*b**2*c*d**2 + A*b**3*c**3 - 3*B*a**2 
*b*d**3 + 3*B*a*b**2*c**2*d - 9*C*a**2*b*c*d**2 + C*a*b**2*c**3 + 5*D*a**3 
*d**3 - 9*D*a**2*b*c**2*d)/(4*a**3*b**7))*log(x + (-2*A*a**2*b*d**3 - 6*B* 
a**2*b*c*d**2 + 4*C*a**3*d**3 - 6*C*a**2*b*c**2*d + 12*D*a**3*c*d**2 - 2*D 
*a**2*b*c**3 + 4*a**2*b**3*(-(-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)/(2*b**3) - sqrt(-a**3*b**7)*(3*A*a*b** 
2*c*d**2 + A*b**3*c**3 - 3*B*a**2*b*d**3 + 3*B*a*b**2*c**2*d - 9*C*a**2*b* 
c*d**2 + C*a*b**2*c**3 + 5*D*a**3*d**3 - 9*D*a**2*b*c**2*d)/(4*a**3*b**7)) 
)/(3*A*a*b**2*c*d**2 + A*b**3*c**3 - 3*B*a**2*b*d**3 + 3*B*a*b**2*c**2*d - 
 9*C*a**2*b*c*d**2 + C*a*b**2*c**3 + 5*D*a**3*d**3 - 9*D*a**2*b*c**2*d)) + 
 (-(-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)/(2*b**3) + sqrt(-a**3*b**7)*(3*A*a*b**2*c*d**2 + A*b**3*c**3 - 3 
*B*a**2*b*d**3 + 3*B*a*b**2*c**2*d - 9*C*a**2*b*c*d**2 + C*a*b**2*c**3 + 5 
*D*a**3*d**3 - 9*D*a**2*b*c**2*d)/(4*a**3*b**7))*log(x + (-2*A*a**2*b*d**3 
 - 6*B*a**2*b*c*d**2 + 4*C*a**3*d**3 - 6*C*a**2*b*c**2*d + 12*D*a**3*c*d** 
2 - 2*D*a**2*b*c**3 + 4*a**2*b**3*(-(-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)/(2*b**3) + sqrt(-a**3*b**7)*...
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.03 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (D a^{2} b - B a b^{2}\right )} c^{3} + 3 \, {\left (C a^{2} b - A a b^{2}\right )} c^{2} d - 3 \, {\left (D a^{3} - B a^{2} b\right )} c d^{2} - {\left (C a^{3} - A a^{2} b\right )} d^{3} - {\left ({\left (C a b^{2} - A b^{3}\right )} c^{3} - 3 \, {\left (D a^{2} b - B a b^{2}\right )} c^{2} d - 3 \, {\left (C a^{2} b - A a b^{2}\right )} c d^{2} + {\left (D a^{3} - B a^{2} b\right )} d^{3}\right )} x}{2 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {{\left (D b c^{3} + 3 \, C b c^{2} d - 3 \, {\left (2 \, D a - B b\right )} c d^{2} - {\left (2 \, C a - A b\right )} d^{3}\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {2 \, D b d^{3} x^{3} + 3 \, {\left (3 \, D b c d^{2} + C b d^{3}\right )} x^{2} + 6 \, {\left (3 \, D b c^{2} d + 3 \, C b c d^{2} - {\left (2 \, D a - B b\right )} d^{3}\right )} x}{6 \, b^{3}} + \frac {{\left ({\left (C a b^{2} + A b^{3}\right )} c^{3} - 3 \, {\left (3 \, D a^{2} b - B a b^{2}\right )} c^{2} d - 3 \, {\left (3 \, C a^{2} b - A a b^{2}\right )} c d^{2} + {\left (5 \, D a^{3} - 3 \, B a^{2} b\right )} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (D b c^{3} + 3 \, C b c^{2} d - 6 \, D a c d^{2} + 3 \, B b c d^{2} - 2 \, C a d^{3} + A b d^{3}\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (C a b^{2} c^{3} + A b^{3} c^{3} - 9 \, D a^{2} b c^{2} d + 3 \, B a b^{2} c^{2} d - 9 \, C a^{2} b c d^{2} + 3 \, A a b^{2} c d^{2} + 5 \, D a^{3} d^{3} - 3 \, B a^{2} b d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b^{3}} + \frac {D a^{2} b c^{3} - B a b^{2} c^{3} + 3 \, C a^{2} b c^{2} d - 3 \, A a b^{2} c^{2} d - 3 \, D a^{3} c d^{2} + 3 \, B a^{2} b c d^{2} - C a^{3} d^{3} + A a^{2} b d^{3} - {\left (C a b^{2} c^{3} - A b^{3} c^{3} - 3 \, D a^{2} b c^{2} d + 3 \, B a b^{2} c^{2} d - 3 \, C a^{2} b c d^{2} + 3 \, A a b^{2} c d^{2} + D a^{3} d^{3} - B a^{2} b d^{3}\right )} x}{2 \, {\left (b x^{2} + a\right )} a b^{3}} + \frac {2 \, D b^{4} d^{3} x^{3} + 9 \, D b^{4} c d^{2} x^{2} + 3 \, C b^{4} d^{3} x^{2} + 18 \, D b^{4} c^{2} d x + 18 \, C b^{4} c d^{2} x - 12 \, D a b^{3} d^{3} x + 6 \, B b^{4} d^{3} x}{6 \, b^{6}} \] Input:

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

Output:

1/2*(D*b*c^3 + 3*C*b*c^2*d - 6*D*a*c*d^2 + 3*B*b*c*d^2 - 2*C*a*d^3 + A*b*d 
^3)*log(b*x^2 + a)/b^3 + 1/2*(C*a*b^2*c^3 + A*b^3*c^3 - 9*D*a^2*b*c^2*d + 
3*B*a*b^2*c^2*d - 9*C*a^2*b*c*d^2 + 3*A*a*b^2*c*d^2 + 5*D*a^3*d^3 - 3*B*a^ 
2*b*d^3)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^3) + 1/2*(D*a^2*b*c^3 - B*a* 
b^2*c^3 + 3*C*a^2*b*c^2*d - 3*A*a*b^2*c^2*d - 3*D*a^3*c*d^2 + 3*B*a^2*b*c* 
d^2 - C*a^3*d^3 + A*a^2*b*d^3 - (C*a*b^2*c^3 - 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 + D*a^3*d^3 - B*a^2 
*b*d^3)*x)/((b*x^2 + a)*a*b^3) + 1/6*(2*D*b^4*d^3*x^3 + 9*D*b^4*c*d^2*x^2 
+ 3*C*b^4*d^3*x^2 + 18*D*b^4*c^2*d*x + 18*C*b^4*c*d^2*x - 12*D*a*b^3*d^3*x 
 + 6*B*b^4*d^3*x)/b^6
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 789, normalized size of antiderivative = 2.03 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*d**4 + 9*sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c*d**2 - 9*sqrt(b)*sqrt(a)*a 
tan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*d**3 - 54*sqrt(b)*sqrt(a)*atan((b*x 
)/(sqrt(b)*sqrt(a)))*a**2*b*c**2*d**2 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqr 
t(b)*sqrt(a)))*a**2*b*d**4*x**2 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sq 
rt(a)))*a*b**3*c**3 + 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b* 
*3*c**2*d + 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**3*c*d**2* 
x**2 - 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**3*d**3*x**2 + 
3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**4 - 54*sqrt(b)*s 
qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**2*d**2*x**2 + 3*sqrt(b)*sqr 
t(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**4*c**3*x**2 + 9*sqrt(b)*sqrt(a)*atan 
((b*x)/(sqrt(b)*sqrt(a)))*b**4*c**2*d*x**2 + 3*sqrt(b)*sqrt(a)*atan((b*x)/ 
(sqrt(b)*sqrt(a)))*b**3*c**4*x**2 + 3*log(a + b*x**2)*a**3*b**2*d**3 - 24* 
log(a + b*x**2)*a**3*b*c*d**3 + 9*log(a + b*x**2)*a**2*b**3*c*d**2 + 3*log 
(a + b*x**2)*a**2*b**3*d**3*x**2 + 12*log(a + b*x**2)*a**2*b**2*c**3*d - 2 
4*log(a + b*x**2)*a**2*b**2*c*d**3*x**2 + 9*log(a + b*x**2)*a*b**4*c*d**2* 
x**2 + 12*log(a + b*x**2)*a*b**3*c**3*d*x**2 - 15*a**3*b*d**4*x - 9*a**2*b 
**3*c*d**2*x - 3*a**2*b**3*d**3*x**2 + 9*a**2*b**3*d**3*x + 54*a**2*b**2*c 
**2*d**2*x + 24*a**2*b**2*c*d**3*x**2 - 10*a**2*b**2*d**4*x**3 + 3*a*b**4* 
c**3*x + 9*a*b**4*c**2*d*x**2 - 9*a*b**4*c**2*d*x - 9*a*b**4*c*d**2*x**...