\(\int \frac {A+B x}{(d+e x)^2 (a+c x^2)} \, dx\) [85]

Optimal result

Integrand size = 22, antiderivative size = 173 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {B d-A e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {\sqrt {c} \left (A c d^2+2 a B d e-a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (c d^2+a e^2\right )^2}-\frac {\left (B c d^2-2 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac {\left (B c d^2-2 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \] Output:

(-A*e+B*d)/(a*e^2+c*d^2)/(e*x+d)+c^(1/2)*(-A*a*e^2+A*c*d^2+2*B*a*d*e)*arct 
an(c^(1/2)*x/a^(1/2))/a^(1/2)/(a*e^2+c*d^2)^2-(-2*A*c*d*e-B*a*e^2+B*c*d^2) 
*ln(e*x+d)/(a*e^2+c*d^2)^2+1/2*(-2*A*c*d*e-B*a*e^2+B*c*d^2)*ln(c*x^2+a)/(a 
*e^2+c*d^2)^2
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {\frac {2 (B d-A e) \left (c d^2+a e^2\right )}{d+e x}+\frac {2 \sqrt {c} \left (A c d^2+2 a B d e-a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a}}+\left (-2 B c d^2+4 A c d e+2 a B e^2\right ) \log (d+e x)+\left (B c d^2-2 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {657, 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.

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {A e -B d}{\left (a \,e^{2}+c \,d^{2}\right ) \left (e x +d \right )}+\frac {\left (2 A c d e +B a \,e^{2}-B c \,d^{2}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {c \left (\frac {\left (2 A c d e +B a \,e^{2}-B c \,d^{2}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (A a \,e^{2}-A c \,d^{2}-2 B a d e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}\)	\(157\)
risch	\(\text {Expression too large to display}\)	\(5625\)

Input:

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

Output:

-(A*e-B*d)/(a*e^2+c*d^2)/(e*x+d)+(2*A*c*d*e+B*a*e^2-B*c*d^2)/(a*e^2+c*d^2) 
^2*ln(e*x+d)-c/(a*e^2+c*d^2)^2*(1/2*(2*A*c*d*e+B*a*e^2-B*c*d^2)/c*ln(c*x^2 
+a)+(A*a*e^2-A*c*d^2-2*B*a*d*e)/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 3.88 (sec) , antiderivative size = 562, normalized size of antiderivative = 3.25 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\left [\frac {2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} - {\left (A c d^{3} + 2 \, B a d^{2} e - A a d e^{2} + {\left (A c d^{2} e + 2 \, B a d e^{2} - A a e^{3}\right )} x\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} - 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) + {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}, \frac {2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} + 2 \, {\left (A c d^{3} + 2 \, B a d^{2} e - A a d e^{2} + {\left (A c d^{2} e + 2 \, B a d e^{2} - A a e^{3}\right )} x\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} - \frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {{\left (A c^{2} d^{2} + 2 \, B a c d e - A a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} + \frac {B d - A e}{c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.37 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {a e^{2}}{{\left (e x + d\right )}^{2}}\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {\frac {B d e^{2}}{e x + d} - \frac {A e^{3}}{e x + d}}{c d^{2} e^{2} + a e^{4}} + \frac {{\left (A c^{2} d^{2} e^{2} + 2 \, B a c d e^{3} - A a c e^{4}\right )} \arctan \left (\frac {c d - \frac {c d^{2}}{e x + d} - \frac {a e^{2}}{e x + d}}{\sqrt {a c} e}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c} e^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 7.07 (sec) , antiderivative size = 810, normalized size of antiderivative = 4.68 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {\ln \left (3\,B\,a^3\,e^4+3\,B\,a\,c^2\,d^4+A\,c^3\,d^4\,x+A\,a^2\,e^4\,\sqrt {-a\,c}+A\,c^2\,d^4\,\sqrt {-a\,c}+14\,A\,d^2\,e^2\,{\left (-a\,c\right )}^{3/2}+A\,a^2\,c\,e^4\,x-8\,B\,a^2\,d\,e^3\,\sqrt {-a\,c}-3\,B\,a^2\,e^4\,x\,\sqrt {-a\,c}-3\,B\,c^2\,d^4\,x\,\sqrt {-a\,c}-10\,B\,a^2\,c\,d^2\,e^2+8\,A\,d\,e^3\,x\,{\left (-a\,c\right )}^{3/2}-8\,A\,a\,c^2\,d^3\,e+8\,A\,a^2\,c\,d\,e^3+8\,B\,a\,c\,d^3\,e\,\sqrt {-a\,c}+8\,B\,a\,c^2\,d^3\,e\,x-8\,B\,a^2\,c\,d\,e^3\,x+8\,A\,c^2\,d^3\,e\,x\,\sqrt {-a\,c}-14\,A\,a\,c^2\,d^2\,e^2\,x+10\,B\,a\,c\,d^2\,e^2\,x\,\sqrt {-a\,c}\right )\,\left (c\,\left (\frac {B\,a\,d^2}{2}+\frac {A\,d^2\,\sqrt {-a\,c}}{2}-A\,a\,d\,e\right )-e^2\,\left (\frac {B\,a^2}{2}+\frac {A\,a\,\sqrt {-a\,c}}{2}\right )+B\,a\,d\,e\,\sqrt {-a\,c}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {\ln \left (d+e\,x\right )\,\left (c\,\left (B\,d^2-2\,A\,d\,e\right )-B\,a\,e^2\right )}{a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4}-\frac {\ln \left (3\,B\,a^3\,e^4+8\,B\,d^3\,e\,{\left (-a\,c\right )}^{3/2}+3\,B\,a\,c^2\,d^4+A\,c^3\,d^4\,x-A\,a^2\,e^4\,\sqrt {-a\,c}-A\,c^2\,d^4\,\sqrt {-a\,c}+A\,a^2\,c\,e^4\,x+8\,B\,a^2\,d\,e^3\,\sqrt {-a\,c}+3\,B\,a^2\,e^4\,x\,\sqrt {-a\,c}+3\,B\,c^2\,d^4\,x\,\sqrt {-a\,c}+10\,B\,d^2\,e^2\,x\,{\left (-a\,c\right )}^{3/2}-10\,B\,a^2\,c\,d^2\,e^2-8\,A\,a\,c^2\,d^3\,e+8\,A\,a^2\,c\,d\,e^3+8\,B\,a\,c^2\,d^3\,e\,x-8\,B\,a^2\,c\,d\,e^3\,x+14\,A\,a\,c\,d^2\,e^2\,\sqrt {-a\,c}-8\,A\,c^2\,d^3\,e\,x\,\sqrt {-a\,c}-14\,A\,a\,c^2\,d^2\,e^2\,x+8\,A\,a\,c\,d\,e^3\,x\,\sqrt {-a\,c}\right )\,\left (e^2\,\left (\frac {B\,a^2}{2}-\frac {A\,a\,\sqrt {-a\,c}}{2}\right )+c\,\left (\frac {A\,d^2\,\sqrt {-a\,c}}{2}-\frac {B\,a\,d^2}{2}+A\,a\,d\,e\right )+B\,a\,d\,e\,\sqrt {-a\,c}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {A\,e-B\,d}{\left (c\,d^2+a\,e^2\right )\,\left (d+e\,x\right )} \] Input:

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

Output:

(log(3*B*a^3*e^4 + 3*B*a*c^2*d^4 + A*c^3*d^4*x + A*a^2*e^4*(-a*c)^(1/2) + 
A*c^2*d^4*(-a*c)^(1/2) + 14*A*d^2*e^2*(-a*c)^(3/2) + A*a^2*c*e^4*x - 8*B*a 
^2*d*e^3*(-a*c)^(1/2) - 3*B*a^2*e^4*x*(-a*c)^(1/2) - 3*B*c^2*d^4*x*(-a*c)^ 
(1/2) - 10*B*a^2*c*d^2*e^2 + 8*A*d*e^3*x*(-a*c)^(3/2) - 8*A*a*c^2*d^3*e + 
8*A*a^2*c*d*e^3 + 8*B*a*c*d^3*e*(-a*c)^(1/2) + 8*B*a*c^2*d^3*e*x - 8*B*a^2 
*c*d*e^3*x + 8*A*c^2*d^3*e*x*(-a*c)^(1/2) - 14*A*a*c^2*d^2*e^2*x + 10*B*a* 
c*d^2*e^2*x*(-a*c)^(1/2))*(c*((B*a*d^2)/2 + (A*d^2*(-a*c)^(1/2))/2 - A*a*d 
*e) - e^2*((B*a^2)/2 + (A*a*(-a*c)^(1/2))/2) + B*a*d*e*(-a*c)^(1/2)))/(a^3 
*e^4 + a*c^2*d^4 + 2*a^2*c*d^2*e^2) - (log(d + e*x)*(c*(B*d^2 - 2*A*d*e) - 
 B*a*e^2))/(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2) - (log(3*B*a^3*e^4 + 8*B*d^ 
3*e*(-a*c)^(3/2) + 3*B*a*c^2*d^4 + A*c^3*d^4*x - A*a^2*e^4*(-a*c)^(1/2) - 
A*c^2*d^4*(-a*c)^(1/2) + A*a^2*c*e^4*x + 8*B*a^2*d*e^3*(-a*c)^(1/2) + 3*B* 
a^2*e^4*x*(-a*c)^(1/2) + 3*B*c^2*d^4*x*(-a*c)^(1/2) + 10*B*d^2*e^2*x*(-a*c 
)^(3/2) - 10*B*a^2*c*d^2*e^2 - 8*A*a*c^2*d^3*e + 8*A*a^2*c*d*e^3 + 8*B*a*c 
^2*d^3*e*x - 8*B*a^2*c*d*e^3*x + 14*A*a*c*d^2*e^2*(-a*c)^(1/2) - 8*A*c^2*d 
^3*e*x*(-a*c)^(1/2) - 14*A*a*c^2*d^2*e^2*x + 8*A*a*c*d*e^3*x*(-a*c)^(1/2)) 
*(e^2*((B*a^2)/2 - (A*a*(-a*c)^(1/2))/2) + c*((A*d^2*(-a*c)^(1/2))/2 - (B* 
a*d^2)/2 + A*a*d*e) + B*a*d*e*(-a*c)^(1/2)))/(a^3*e^4 + a*c^2*d^4 + 2*a^2* 
c*d^2*e^2) - (A*e - B*d)/((a*e^2 + c*d^2)*(d + e*x))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.51 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {-2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a \,d^{2} e^{2}-2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a d \,e^{3} x +4 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) b \,d^{3} e +4 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) b \,d^{2} e^{2} x +2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) c \,d^{4}+2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) c \,d^{3} e x -\mathrm {log}\left (c \,x^{2}+a \right ) a b \,d^{2} e^{2}-\mathrm {log}\left (c \,x^{2}+a \right ) a b d \,e^{3} x -2 \,\mathrm {log}\left (c \,x^{2}+a \right ) a c \,d^{3} e -2 \,\mathrm {log}\left (c \,x^{2}+a \right ) a c \,d^{2} e^{2} x +\mathrm {log}\left (c \,x^{2}+a \right ) b c \,d^{4}+\mathrm {log}\left (c \,x^{2}+a \right ) b c \,d^{3} e x +2 \,\mathrm {log}\left (e x +d \right ) a b \,d^{2} e^{2}+2 \,\mathrm {log}\left (e x +d \right ) a b d \,e^{3} x +4 \,\mathrm {log}\left (e x +d \right ) a c \,d^{3} e +4 \,\mathrm {log}\left (e x +d \right ) a c \,d^{2} e^{2} x -2 \,\mathrm {log}\left (e x +d \right ) b c \,d^{4}-2 \,\mathrm {log}\left (e x +d \right ) b c \,d^{3} e x +2 a^{2} e^{4} x -2 a b d \,e^{3} x +2 a c \,d^{2} e^{2} x -2 b c \,d^{3} e x}{2 d \left (a^{2} e^{5} x +2 a c \,d^{2} e^{3} x +c^{2} d^{4} e x +a^{2} d \,e^{4}+2 a c \,d^{3} e^{2}+c^{2} d^{5}\right )} \] Input:

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

Output:

( - 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*d**2*e**2 - 2*sqrt(c 
)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*d*e**3*x + 4*sqrt(c)*sqrt(a)*ata 
n((c*x)/(sqrt(c)*sqrt(a)))*b*d**3*e + 4*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c 
)*sqrt(a)))*b*d**2*e**2*x + 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)) 
)*c*d**4 + 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*c*d**3*e*x - lo 
g(a + c*x**2)*a*b*d**2*e**2 - log(a + c*x**2)*a*b*d*e**3*x - 2*log(a + c*x 
**2)*a*c*d**3*e - 2*log(a + c*x**2)*a*c*d**2*e**2*x + log(a + c*x**2)*b*c* 
d**4 + log(a + c*x**2)*b*c*d**3*e*x + 2*log(d + e*x)*a*b*d**2*e**2 + 2*log 
(d + e*x)*a*b*d*e**3*x + 4*log(d + e*x)*a*c*d**3*e + 4*log(d + e*x)*a*c*d* 
*2*e**2*x - 2*log(d + e*x)*b*c*d**4 - 2*log(d + e*x)*b*c*d**3*e*x + 2*a**2 
*e**4*x - 2*a*b*d*e**3*x + 2*a*c*d**2*e**2*x - 2*b*c*d**3*e*x)/(2*d*(a**2* 
d*e**4 + a**2*e**5*x + 2*a*c*d**3*e**2 + 2*a*c*d**2*e**3*x + c**2*d**5 + c 
**2*d**4*e*x))