\(\int \frac {(A+B x) (d+e x)^2}{(a+c x^2)^3} \, dx\) [98]

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=-\frac {(d+e x) (a (B d+A e)-(A c d-a B e) x)}{4 a c \left (a+c x^2\right )^2}-\frac {2 a e (A c d+a B e)-c \left (3 A c d^2+2 a B d e+a A e^2\right ) x}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (3 A c d^2+2 a B d e+a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}} \] Output:

-1/4*(e*x+d)*(a*(A*e+B*d)-(A*c*d-B*a*e)*x)/a/c/(c*x^2+a)^2-1/8*(2*a*e*(A*c 
*d+B*a*e)-c*(A*a*e^2+3*A*c*d^2+2*B*a*d*e)*x)/a^2/c^2/(c*x^2+a)+1/8*(A*a*e^ 
2+3*A*c*d^2+2*B*a*d*e)*arctan(c^(1/2)*x/a^(1/2))/a^(5/2)/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\frac {\frac {\sqrt {a} \left (-4 a^2 B e^2+3 A c^2 d^2 x+a c e (2 B d+A e) x\right )}{a+c x^2}+\frac {2 a^{3/2} \left (a^2 B e^2+A c^2 d^2 x-a c (A e (2 d+e x)+B d (d+2 e x))\right )}{\left (a+c x^2\right )^2}+\sqrt {c} \left (3 A c d^2+2 a B d e+a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^2} \] Input:

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

Output:

((Sqrt[a]*(-4*a^2*B*e^2 + 3*A*c^2*d^2*x + a*c*e*(2*B*d + A*e)*x))/(a + c*x 
^2) + (2*a^(3/2)*(a^2*B*e^2 + A*c^2*d^2*x - a*c*(A*e*(2*d + e*x) + B*d*(d 
+ 2*e*x))))/(a + c*x^2)^2 + Sqrt[c]*(3*A*c*d^2 + 2*a*B*d*e + a*A*e^2)*ArcT 
an[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^2)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {685, 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.

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

\(\Big \downarrow \) 685

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

\(\Big \downarrow \) 675

\(\displaystyle \frac {\frac {\left (a A e^2+2 a B d e+3 A c d^2\right ) \int \frac {1}{c x^2+a}dx}{2 a}+\frac {x \left (-a A e^2+2 a B d e+3 A c d^2\right )}{2 a \left (a+c x^2\right )}-\frac {e (a B e+2 A c d)}{c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^2 (a B-A c x)}{4 a c \left (a+c x^2\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a A e^2+2 a B d e+3 A c d^2\right )}{2 a^{3/2} \sqrt {c}}+\frac {x \left (-a A e^2+2 a B d e+3 A c d^2\right )}{2 a \left (a+c x^2\right )}-\frac {e (a B e+2 A c d)}{c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^2 (a B-A c x)}{4 a c \left (a+c x^2\right )^2}\)

Input:

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

Output:

-1/4*((a*B - A*c*x)*(d + e*x)^2)/(a*c*(a + c*x^2)^2) + (-((e*(2*A*c*d + a* 
B*e))/(c*(a + c*x^2))) + ((3*A*c*d^2 + 2*a*B*d*e - a*A*e^2)*x)/(2*a*(a + c 
*x^2)) + ((3*A*c*d^2 + 2*a*B*d*e + a*A*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/( 
2*a^(3/2)*Sqrt[c]))/(4*a*c)
 

Defintions of rubi rules used

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 685

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c 
*(p + 1))), x] - Simp[1/(2*a*c*(p + 1))   Int[(d + e*x)^(m - 1)*(a + c*x^2) 
^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] 
 /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] 
 || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\frac {\left (A a \,e^{2}+3 A c \,d^{2}+2 B a d e \right ) x^{3}}{8 a^{2}}-\frac {B \,e^{2} x^{2}}{2 c}-\frac {\left (A a \,e^{2}-5 A c \,d^{2}+2 B a d e \right ) x}{8 a c}-\frac {2 A c d e +B a \,e^{2}+B c \,d^{2}}{4 c^{2}}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\left (A a \,e^{2}+3 A c \,d^{2}+2 B a d e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 a^{2} c \sqrt {a c}}\)	\(148\)
risch	\(\frac {\frac {\left (A a \,e^{2}+3 A c \,d^{2}+2 B a d e \right ) x^{3}}{8 a^{2}}-\frac {B \,e^{2} x^{2}}{2 c}-\frac {\left (A a \,e^{2}-5 A c \,d^{2}+2 B a d e \right ) x}{8 a c}-\frac {2 A c d e +B a \,e^{2}+B c \,d^{2}}{4 c^{2}}}{\left (c \,x^{2}+a \right )^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) A \,e^{2}}{16 \sqrt {-a c}\, c a}-\frac {3 \ln \left (c x +\sqrt {-a c}\right ) A \,d^{2}}{16 \sqrt {-a c}\, a^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) B d e}{8 \sqrt {-a c}\, c a}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) A \,e^{2}}{16 \sqrt {-a c}\, c a}+\frac {3 \ln \left (-c x +\sqrt {-a c}\right ) A \,d^{2}}{16 \sqrt {-a c}\, a^{2}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) B d e}{8 \sqrt {-a c}\, c a}\)	\(275\)

Input:

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

Output:

(1/8*(A*a*e^2+3*A*c*d^2+2*B*a*d*e)/a^2*x^3-1/2*B*e^2*x^2/c-1/8*(A*a*e^2-5* 
A*c*d^2+2*B*a*d*e)/a/c*x-1/4*(2*A*c*d*e+B*a*e^2+B*c*d^2)/c^2)/(c*x^2+a)^2+ 
1/8*(A*a*e^2+3*A*c*d^2+2*B*a*d*e)/a^2/c/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2) 
)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 537, normalized size of antiderivative = 3.53 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\left [-\frac {8 \, B a^{3} c e^{2} x^{2} + 4 \, B a^{3} c d^{2} + 8 \, A a^{3} c d e + 4 \, B a^{4} e^{2} - 2 \, {\left (3 \, A a c^{3} d^{2} + 2 \, B a^{2} c^{2} d e + A a^{2} c^{2} e^{2}\right )} x^{3} + {\left (3 \, A a^{2} c d^{2} + 2 \, B a^{3} d e + A a^{3} e^{2} + {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d^{2} + 2 \, B a^{2} c d e + A a^{2} c e^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (5 \, A a^{2} c^{2} d^{2} - 2 \, B a^{3} c d e - A a^{3} c e^{2}\right )} x}{16 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac {4 \, B a^{3} c e^{2} x^{2} + 2 \, B a^{3} c d^{2} + 4 \, A a^{3} c d e + 2 \, B a^{4} e^{2} - {\left (3 \, A a c^{3} d^{2} + 2 \, B a^{2} c^{2} d e + A a^{2} c^{2} e^{2}\right )} x^{3} - {\left (3 \, A a^{2} c d^{2} + 2 \, B a^{3} d e + A a^{3} e^{2} + {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d^{2} + 2 \, B a^{2} c d e + A a^{2} c e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, A a^{2} c^{2} d^{2} - 2 \, B a^{3} c d e - A a^{3} c e^{2}\right )} x}{8 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}\right ] \] Input:

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

Output:

[-1/16*(8*B*a^3*c*e^2*x^2 + 4*B*a^3*c*d^2 + 8*A*a^3*c*d*e + 4*B*a^4*e^2 - 
2*(3*A*a*c^3*d^2 + 2*B*a^2*c^2*d*e + A*a^2*c^2*e^2)*x^3 + (3*A*a^2*c*d^2 + 
 2*B*a^3*d*e + A*a^3*e^2 + (3*A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x^4 
 + 2*(3*A*a*c^2*d^2 + 2*B*a^2*c*d*e + A*a^2*c*e^2)*x^2)*sqrt(-a*c)*log((c* 
x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*(5*A*a^2*c^2*d^2 - 2*B*a^3*c*d* 
e - A*a^3*c*e^2)*x)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/8*(4*B*a^3 
*c*e^2*x^2 + 2*B*a^3*c*d^2 + 4*A*a^3*c*d*e + 2*B*a^4*e^2 - (3*A*a*c^3*d^2 
+ 2*B*a^2*c^2*d*e + A*a^2*c^2*e^2)*x^3 - (3*A*a^2*c*d^2 + 2*B*a^3*d*e + A* 
a^3*e^2 + (3*A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x^4 + 2*(3*A*a*c^2*d 
^2 + 2*B*a^2*c*d*e + A*a^2*c*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - ( 
5*A*a^2*c^2*d^2 - 2*B*a^3*c*d*e - A*a^3*c*e^2)*x)/(a^3*c^4*x^4 + 2*a^4*c^3 
*x^2 + a^5*c^2)]
 

Sympy [A] (verification not implemented)

Time = 3.99 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.80 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \left (A a e^{2} + 3 A c d^{2} + 2 B a d e\right ) \log {\left (- a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \left (A a e^{2} + 3 A c d^{2} + 2 B a d e\right ) \log {\left (a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {- 4 A a^{2} c d e - 2 B a^{3} e^{2} - 2 B a^{2} c d^{2} - 4 B a^{2} c e^{2} x^{2} + x^{3} \left (A a c^{2} e^{2} + 3 A c^{3} d^{2} + 2 B a c^{2} d e\right ) + x \left (- A a^{2} c e^{2} + 5 A a c^{2} d^{2} - 2 B a^{2} c d e\right )}{8 a^{4} c^{2} + 16 a^{3} c^{3} x^{2} + 8 a^{2} c^{4} x^{4}} \] Input:

integrate((B*x+A)*(e*x+d)**2/(c*x**2+a)**3,x)
 

Output:

-sqrt(-1/(a**5*c**3))*(A*a*e**2 + 3*A*c*d**2 + 2*B*a*d*e)*log(-a**3*c*sqrt 
(-1/(a**5*c**3)) + x)/16 + sqrt(-1/(a**5*c**3))*(A*a*e**2 + 3*A*c*d**2 + 2 
*B*a*d*e)*log(a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + (-4*A*a**2*c*d*e - 2*B 
*a**3*e**2 - 2*B*a**2*c*d**2 - 4*B*a**2*c*e**2*x**2 + x**3*(A*a*c**2*e**2 
+ 3*A*c**3*d**2 + 2*B*a*c**2*d*e) + x*(-A*a**2*c*e**2 + 5*A*a*c**2*d**2 - 
2*B*a**2*c*d*e))/(8*a**4*c**2 + 16*a**3*c**3*x**2 + 8*a**2*c**4*x**4)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=-\frac {4 \, B a^{2} c e^{2} x^{2} + 2 \, B a^{2} c d^{2} + 4 \, A a^{2} c d e + 2 \, B a^{3} e^{2} - {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{3} - {\left (5 \, A a c^{2} d^{2} - 2 \, B a^{2} c d e - A a^{2} c e^{2}\right )} x}{8 \, {\left (a^{2} c^{4} x^{4} + 2 \, a^{3} c^{3} x^{2} + a^{4} c^{2}\right )}} + \frac {{\left (3 \, A c d^{2} + 2 \, B a d e + A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} \] Input:

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

Output:

-1/8*(4*B*a^2*c*e^2*x^2 + 2*B*a^2*c*d^2 + 4*A*a^2*c*d*e + 2*B*a^3*e^2 - (3 
*A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x^3 - (5*A*a*c^2*d^2 - 2*B*a^2*c 
*d*e - A*a^2*c*e^2)*x)/(a^2*c^4*x^4 + 2*a^3*c^3*x^2 + a^4*c^2) + 1/8*(3*A* 
c*d^2 + 2*B*a*d*e + A*a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\frac {{\left (3 \, A c d^{2} + 2 \, B a d e + A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} + \frac {3 \, A c^{3} d^{2} x^{3} + 2 \, B a c^{2} d e x^{3} + A a c^{2} e^{2} x^{3} - 4 \, B a^{2} c e^{2} x^{2} + 5 \, A a c^{2} d^{2} x - 2 \, B a^{2} c d e x - A a^{2} c e^{2} x - 2 \, B a^{2} c d^{2} - 4 \, A a^{2} c d e - 2 \, B a^{3} e^{2}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \] Input:

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

Output:

1/8*(3*A*c*d^2 + 2*B*a*d*e + A*a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2 
*c) + 1/8*(3*A*c^3*d^2*x^3 + 2*B*a*c^2*d*e*x^3 + A*a*c^2*e^2*x^3 - 4*B*a^2 
*c*e^2*x^2 + 5*A*a*c^2*d^2*x - 2*B*a^2*c*d*e*x - A*a^2*c*e^2*x - 2*B*a^2*c 
*d^2 - 4*A*a^2*c*d*e - 2*B*a^3*e^2)/((c*x^2 + a)^2*a^2*c^2)
 

Mupad [B] (verification not implemented)

Time = 6.65 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a^{5/2}\,c^{3/2}}-\frac {\frac {B\,c\,d^2+2\,A\,c\,d\,e+B\,a\,e^2}{4\,c^2}-\frac {x^3\,\left (3\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a^2}+\frac {x\,\left (-5\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a\,c}+\frac {B\,e^2\,x^2}{2\,c}}{a^2+2\,a\,c\,x^2+c^2\,x^4} \] Input:

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

Output:

(atan((c^(1/2)*x)/a^(1/2))*(A*a*e^2 + 3*A*c*d^2 + 2*B*a*d*e))/(8*a^(5/2)*c 
^(3/2)) - ((B*a*e^2 + B*c*d^2 + 2*A*c*d*e)/(4*c^2) - (x^3*(A*a*e^2 + 3*A*c 
*d^2 + 2*B*a*d*e))/(8*a^2) + (x*(A*a*e^2 - 5*A*c*d^2 + 2*B*a*d*e))/(8*a*c) 
 + (B*e^2*x^2)/(2*c))/(a^2 + c^2*x^4 + 2*a*c*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.39 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a^{3} e^{2}+2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a^{2} b d e +3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a^{2} c \,d^{2}+2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a^{2} c \,e^{2} x^{2}+4 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a b c d e \,x^{2}+6 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a \,c^{2} d^{2} x^{2}+\sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a \,c^{2} e^{2} x^{4}+2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) b \,c^{2} d e \,x^{4}+3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) c^{3} d^{2} x^{4}-4 a^{3} c d e -a^{3} c \,e^{2} x -2 a^{2} b c \,d^{2}-2 a^{2} b c d e x +5 a^{2} c^{2} d^{2} x +a^{2} c^{2} e^{2} x^{3}+2 a b \,c^{2} d e \,x^{3}+2 a b \,c^{2} e^{2} x^{4}+3 a \,c^{3} d^{2} x^{3}}{8 a^{2} c^{2} \left (c^{2} x^{4}+2 a c \,x^{2}+a^{2}\right )} \] Input:

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

Output:

(sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**3*e**2 + 2*sqrt(c)*sqrt( 
a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**2*b*d*e + 3*sqrt(c)*sqrt(a)*atan((c*x) 
/(sqrt(c)*sqrt(a)))*a**2*c*d**2 + 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sq 
rt(a)))*a**2*c*e**2*x**2 + 4*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a))) 
*a*b*c*d*e*x**2 + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*c**2*d 
**2*x**2 + sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*c**2*e**2*x**4 
+ 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*b*c**2*d*e*x**4 + 3*sqrt 
(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*c**3*d**2*x**4 - 4*a**3*c*d*e - 
a**3*c*e**2*x - 2*a**2*b*c*d**2 - 2*a**2*b*c*d*e*x + 5*a**2*c**2*d**2*x + 
a**2*c**2*e**2*x**3 + 2*a*b*c**2*d*e*x**3 + 2*a*b*c**2*e**2*x**4 + 3*a*c** 
3*d**2*x**3)/(8*a**2*c**2*(a**2 + 2*a*c*x**2 + c**2*x**4))