\(\int \frac {A+B x}{(d+e x)^2 (a+c x^2)^2} \, dx\) [94]

Optimal result

Integrand size = 22, antiderivative size = 291 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {e^2 (B d-A e)}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a \left (B c d^2-2 A c d e-a B e^2\right )-c \left (A c d^2+2 a B d e-a A e^2\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac {\sqrt {c} \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )^3}-\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3} \] Output:

e^2*(-A*e+B*d)/(a*e^2+c*d^2)^2/(e*x+d)-1/2*(a*(-2*A*c*d*e-B*a*e^2+B*c*d^2) 
-c*(-A*a*e^2+A*c*d^2+2*B*a*d*e)*x)/a/(a*e^2+c*d^2)^2/(c*x^2+a)-1/2*c^(1/2) 
*(2*a*B*d*e*(-3*a*e^2+c*d^2)-A*(-3*a^2*e^4+6*a*c*d^2*e^2+c^2*d^4))*arctan( 
c^(1/2)*x/a^(1/2))/a^(3/2)/(a*e^2+c*d^2)^3-e^2*(-4*A*c*d*e-B*a*e^2+3*B*c*d 
^2)*ln(e*x+d)/(a*e^2+c*d^2)^3+1/2*e^2*(-4*A*c*d*e-B*a*e^2+3*B*c*d^2)*ln(c* 
x^2+a)/(a*e^2+c*d^2)^3
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {-\frac {2 e^2 (-B d+A e) \left (c d^2+a e^2\right )}{d+e x}+\frac {\left (c d^2+a e^2\right ) \left (a^2 B e^2+A c^2 d^2 x-a c (B d (d-2 e x)+A e (-2 d+e x))\right )}{a \left (a+c x^2\right )}+\frac {\sqrt {c} \left (2 a B d e \left (-c d^2+3 a e^2\right )+A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{3/2}}+2 e^2 \left (-3 B c d^2+4 A c d e+a B e^2\right ) \log (d+e x)-e^2 \left (-3 B c d^2+4 A c d e+a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3} \] Input:

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

Output:

((-2*e^2*(-(B*d) + A*e)*(c*d^2 + a*e^2))/(d + e*x) + ((c*d^2 + a*e^2)*(a^2 
*B*e^2 + A*c^2*d^2*x - a*c*(B*d*(d - 2*e*x) + A*e*(-2*d + e*x))))/(a*(a + 
c*x^2)) + (Sqrt[c]*(2*a*B*d*e*(-(c*d^2) + 3*a*e^2) + A*(c^2*d^4 + 6*a*c*d^ 
2*e^2 - 3*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(3/2) + 2*e^2*(-3*B*c*d 
^2 + 4*A*c*d*e + a*B*e^2)*Log[d + e*x] - e^2*(-3*B*c*d^2 + 4*A*c*d*e + a*B 
*e^2)*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {686, 25, 27, 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.

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

\(\Big \downarrow \) 686

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2}+\frac {a e^2 \log \left (a+c x^2\right ) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac {e \left (-3 a A e^2+4 a B d e+A c d^2\right )}{(d+e x) \left (a e^2+c d^2\right )}-\frac {2 a e^2 \log (d+e x) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{\left (a e^2+c d^2\right )^2}}{2 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}\)

Input:

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

Output:

-1/2*(a*(B*d - A*e) - (A*c*d + a*B*e)*x)/(a*(c*d^2 + a*e^2)*(d + e*x)*(a + 
 c*x^2)) + ((e*(A*c*d^2 + 4*a*B*d*e - 3*a*A*e^2))/((c*d^2 + a*e^2)*(d + e* 
x)) - (Sqrt[c]*(2*a*B*d*e*(c*d^2 - 3*a*e^2) - A*(c^2*d^4 + 6*a*c*d^2*e^2 - 
 3*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*(c*d^2 + a*e^2)^2) - (2 
*a*e^2*(3*B*c*d^2 - 4*A*c*d*e - a*B*e^2)*Log[d + e*x])/(c*d^2 + a*e^2)^2 + 
 (a*e^2*(3*B*c*d^2 - 4*A*c*d*e - a*B*e^2)*Log[a + c*x^2])/(c*d^2 + a*e^2)^ 
2)/(2*a*(c*d^2 + a*e^2))
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 686

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + 
a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 
1/(2*a*c*(p + 1)*(c*d^2 + a*e^2))   Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim 
p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f 
+ a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ 
[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 2009

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

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {e^{2} \left (4 A c d e +B a \,e^{2}-3 B c \,d^{2}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {\left (A e -B d \right ) e^{2}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {c \left (\frac {\frac {\left (A \,a^{2} e^{4}-A \,c^{2} d^{4}-2 B \,a^{2} d \,e^{3}-2 B a c \,d^{3} e \right ) x}{2 a}-\frac {2 A a c d \,e^{3}+2 A \,c^{2} d^{3} e +B \,e^{4} a^{2}-B \,c^{2} d^{4}}{2 c}}{c \,x^{2}+a}+\frac {\frac {\left (8 A a c d \,e^{3}+2 B \,e^{4} a^{2}-6 B a c \,d^{2} e^{2}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (3 A \,a^{2} e^{4}-6 A a c \,d^{2} e^{2}-A \,c^{2} d^{4}-6 B \,a^{2} d \,e^{3}+2 B a c \,d^{3} e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 a}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}\)	\(304\)
risch	\(\text {Expression too large to display}\)	\(11446\)

Input:

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

Output:

e^2*(4*A*c*d*e+B*a*e^2-3*B*c*d^2)/(a*e^2+c*d^2)^3*ln(e*x+d)-(A*e-B*d)*e^2/ 
(a*e^2+c*d^2)^2/(e*x+d)-c/(a*e^2+c*d^2)^3*((1/2*(A*a^2*e^4-A*c^2*d^4-2*B*a 
^2*d*e^3-2*B*a*c*d^3*e)/a*x-1/2*(2*A*a*c*d*e^3+2*A*c^2*d^3*e+B*a^2*e^4-B*c 
^2*d^4)/c)/(c*x^2+a)+1/2/a*(1/2*(8*A*a*c*d*e^3+2*B*a^2*e^4-6*B*a*c*d^2*e^2 
)/c*ln(c*x^2+a)+(3*A*a^2*e^4-6*A*a*c*d^2*e^2-A*c^2*d^4-6*B*a^2*d*e^3+2*B*a 
*c*d^3*e)/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 958 vs. \(2 (277) = 554\).

Time = 35.48 (sec) , antiderivative size = 1940, normalized size of antiderivative = 6.67 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(2*B*a*c^2*d^5 - 4*A*a*c^2*d^4*e - 4*B*a^2*c*d^3*e^2 - 6*B*a^3*d*e^4 
 + 4*A*a^3*e^5 - 2*(A*c^3*d^4*e + 4*B*a*c^2*d^3*e^2 - 2*A*a*c^2*d^2*e^3 + 
4*B*a^2*c*d*e^4 - 3*A*a^2*c*e^5)*x^2 + (A*a*c^2*d^5 - 2*B*a^2*c*d^4*e + 6* 
A*a^2*c*d^3*e^2 + 6*B*a^3*d^2*e^3 - 3*A*a^3*d*e^4 + (A*c^3*d^4*e - 2*B*a*c 
^2*d^3*e^2 + 6*A*a*c^2*d^2*e^3 + 6*B*a^2*c*d*e^4 - 3*A*a^2*c*e^5)*x^3 + (A 
*c^3*d^5 - 2*B*a*c^2*d^4*e + 6*A*a*c^2*d^3*e^2 + 6*B*a^2*c*d^2*e^3 - 3*A*a 
^2*c*d*e^4)*x^2 + (A*a*c^2*d^4*e - 2*B*a^2*c*d^3*e^2 + 6*A*a^2*c*d^2*e^3 + 
 6*B*a^3*d*e^4 - 3*A*a^3*e^5)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a) 
- a)/(c*x^2 + a)) - 2*(A*c^3*d^5 + B*a*c^2*d^4*e + 2*A*a*c^2*d^3*e^2 + 2*B 
*a^2*c*d^2*e^3 + A*a^2*c*d*e^4 + B*a^3*e^5)*x - 2*(3*B*a^2*c*d^3*e^2 - 4*A 
*a^2*c*d^2*e^3 - B*a^3*d*e^4 + (3*B*a*c^2*d^2*e^3 - 4*A*a*c^2*d*e^4 - B*a^ 
2*c*e^5)*x^3 + (3*B*a*c^2*d^3*e^2 - 4*A*a*c^2*d^2*e^3 - B*a^2*c*d*e^4)*x^2 
 + (3*B*a^2*c*d^2*e^3 - 4*A*a^2*c*d*e^4 - B*a^3*e^5)*x)*log(c*x^2 + a) + 4 
*(3*B*a^2*c*d^3*e^2 - 4*A*a^2*c*d^2*e^3 - B*a^3*d*e^4 + (3*B*a*c^2*d^2*e^3 
 - 4*A*a*c^2*d*e^4 - B*a^2*c*e^5)*x^3 + (3*B*a*c^2*d^3*e^2 - 4*A*a*c^2*d^2 
*e^3 - B*a^2*c*d*e^4)*x^2 + (3*B*a^2*c*d^2*e^3 - 4*A*a^2*c*d*e^4 - B*a^3*e 
^5)*x)*log(e*x + d))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3*a^4*c*d^3*e^4 + 
a^5*d*e^6 + (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3 + 3*a^3*c^2*d^2*e^5 + a^4*c*e 
^7)*x^3 + (a*c^4*d^7 + 3*a^2*c^3*d^5*e^2 + 3*a^3*c^2*d^3*e^4 + a^4*c*d*e^6 
)*x^2 + (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^2*e^5 + a^5*e^7)...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 511, normalized size of antiderivative = 1.76 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {{\left (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac {{\left (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\right )} \log \left (e x + d\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac {{\left (A c^{3} d^{4} - 2 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} + 6 \, B a^{2} c d e^{3} - 3 \, A a^{2} c e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c}} - \frac {B a c d^{3} - 2 \, A a c d^{2} e - 3 \, B a^{2} d e^{2} + 2 \, A a^{2} e^{3} - {\left (A c^{2} d^{2} e + 4 \, B a c d e^{2} - 3 \, A a c e^{3}\right )} x^{2} - {\left (A c^{2} d^{3} + B a c d^{2} e + A a c d e^{2} + B a^{2} e^{3}\right )} x}{2 \, {\left (a^{2} c^{2} d^{5} + 2 \, a^{3} c d^{3} e^{2} + a^{4} d e^{4} + {\left (a c^{3} d^{4} e + 2 \, a^{2} c^{2} d^{2} e^{3} + a^{3} c e^{5}\right )} x^{3} + {\left (a c^{3} d^{5} + 2 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x^{2} + {\left (a^{2} c^{2} d^{4} e + 2 \, a^{3} c d^{2} e^{3} + a^{4} e^{5}\right )} x\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 518, normalized size of antiderivative = 1.78 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {{\left (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\right )} \log \left (c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {a e^{2}}{{\left (e x + d\right )}^{2}}\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} + \frac {\frac {B d e^{6}}{e x + d} - \frac {A e^{7}}{e x + d}}{c^{2} d^{4} e^{4} + 2 \, a c d^{2} e^{6} + a^{2} e^{8}} + \frac {{\left (A c^{3} d^{4} e^{2} - 2 \, B a c^{2} d^{3} e^{3} + 6 \, A a c^{2} d^{2} e^{4} + 6 \, B a^{2} c d e^{5} - 3 \, A a^{2} c e^{6}\right )} \arctan \left (\frac {c d - \frac {c d^{2}}{e x + d} - \frac {a e^{2}}{e x + d}}{\sqrt {a c} e}\right )}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c} e^{2}} + \frac {\frac {A c^{3} d^{3} e + 3 \, B a c^{2} d^{2} e^{2} - 3 \, A a c^{2} d e^{3} - B a^{2} c e^{4}}{c d^{2} + a e^{2}} - \frac {A c^{3} d^{4} e^{2} + 4 \, B a c^{2} d^{3} e^{3} - 6 \, A a c^{2} d^{2} e^{4} - 4 \, B a^{2} c d e^{5} + A a^{2} c e^{6}}{{\left (c d^{2} + a e^{2}\right )} {\left (e x + d\right )} e}}{2 \, {\left (c d^{2} + a e^{2}\right )}^{2} a {\left (c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {a e^{2}}{{\left (e x + d\right )}^{2}}\right )}} \] Input:

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

Output:

1/2*(3*B*c*d^2*e^2 - 4*A*c*d*e^3 - B*a*e^4)*log(c - 2*c*d/(e*x + d) + c*d^ 
2/(e*x + d)^2 + a*e^2/(e*x + d)^2)/(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^ 
2*e^4 + a^3*e^6) + (B*d*e^6/(e*x + d) - A*e^7/(e*x + d))/(c^2*d^4*e^4 + 2* 
a*c*d^2*e^6 + a^2*e^8) + 1/2*(A*c^3*d^4*e^2 - 2*B*a*c^2*d^3*e^3 + 6*A*a*c^ 
2*d^2*e^4 + 6*B*a^2*c*d*e^5 - 3*A*a^2*c*e^6)*arctan((c*d - c*d^2/(e*x + d) 
 - a*e^2/(e*x + d))/(sqrt(a*c)*e))/((a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3 
*c*d^2*e^4 + a^4*e^6)*sqrt(a*c)*e^2) + 1/2*((A*c^3*d^3*e + 3*B*a*c^2*d^2*e 
^2 - 3*A*a*c^2*d*e^3 - B*a^2*c*e^4)/(c*d^2 + a*e^2) - (A*c^3*d^4*e^2 + 4*B 
*a*c^2*d^3*e^3 - 6*A*a*c^2*d^2*e^4 - 4*B*a^2*c*d*e^5 + A*a^2*c*e^6)/((c*d^ 
2 + a*e^2)*(e*x + d)*e))/((c*d^2 + a*e^2)^2*a*(c - 2*c*d/(e*x + d) + c*d^2 
/(e*x + d)^2 + a*e^2/(e*x + d)^2))
 

Mupad [B] (verification not implemented)

Time = 7.79 (sec) , antiderivative size = 2029, normalized size of antiderivative = 6.97 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((x*(A*c*d + B*a*e))/(2*a*(a*e^2 + c*d^2)) - (2*A*a*e^3 + B*c*d^3 - 3*B*a* 
d*e^2 - 2*A*c*d^2*e)/(2*(a*e^2 + c*d^2)^2) + (x^2*(A*c^2*d^2*e - 3*A*a*c*e 
^3 + 4*B*a*c*d*e^2))/(2*a*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)))/(a*d + a*e 
*x + c*d*x^2 + c*e*x^3) + (log(9*A^2*a^6*e^12*(-a^3*c)^(3/2) + A^2*c^6*d^1 
2*(-a^3*c)^(3/2) - 36*B^2*a^10*e^12*(-a^3*c)^(1/2) - 558*A^2*a^2*d^2*e^10* 
(-a^3*c)^(5/2) + 24*B^2*a^2*d^4*e^8*(-a^3*c)^(5/2) - 108*B^2*a^6*d^2*e^10* 
(-a^3*c)^(3/2) - 612*A^2*c^2*d^6*e^6*(-a^3*c)^(5/2) - 308*B^2*c^2*d^8*e^4* 
(-a^3*c)^(5/2) + 36*B^2*a^11*c*e^12*x + A^2*a^4*c^8*d^12*x + 9*A^2*a^10*c^ 
2*e^12*x + 276*A*B*a^2*d^3*e^9*(-a^3*c)^(5/2) + 808*A*B*c^2*d^7*e^5*(-a^3* 
c)^(5/2) - 1119*A^2*a*c*d^4*e^8*(-a^3*c)^(5/2) - 424*B^2*a*c*d^6*e^6*(-a^3 
*c)^(5/2) + 14*A^2*a^5*c^7*d^10*e^2*x + 55*A^2*a^6*c^6*d^8*e^4*x + 612*A^2 
*a^7*c^5*d^6*e^6*x + 1119*A^2*a^8*c^4*d^4*e^8*x + 558*A^2*a^9*c^3*d^2*e^10 
*x + 4*B^2*a^6*c^6*d^10*e^2*x + 308*B^2*a^7*c^5*d^8*e^4*x + 424*B^2*a^8*c^ 
4*d^6*e^6*x - 24*B^2*a^9*c^3*d^4*e^8*x - 108*B^2*a^10*c^2*d^2*e^10*x + 14* 
A^2*a*c^5*d^10*e^2*(-a^3*c)^(3/2) + 252*A*B*a^6*d*e^11*(-a^3*c)^(3/2) + 55 
*A^2*a^2*c^4*d^8*e^4*(-a^3*c)^(3/2) + 4*B^2*a^2*c^4*d^10*e^2*(-a^3*c)^(3/2 
) - 4*A*B*a^5*c^7*d^11*e*x + 252*A*B*a^10*c^2*d*e^11*x + 1320*A*B*a*c*d^5* 
e^7*(-a^3*c)^(5/2) - 4*A*B*a*c^5*d^11*e*(-a^3*c)^(3/2) - 20*A*B*a^6*c^6*d^ 
9*e^3*x - 808*A*B*a^7*c^5*d^7*e^5*x - 1320*A*B*a^8*c^4*d^5*e^7*x - 276*A*B 
*a^9*c^3*d^3*e^9*x - 20*A*B*a^2*c^4*d^9*e^3*(-a^3*c)^(3/2))*(c*(a^3*((3...
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1504, normalized size of antiderivative = 5.17 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 3*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**3*d**2*e**4 - 3*sqr 
t(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**3*d*e**5*x + 6*sqrt(c)*sqrt( 
a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**2*b*d**3*e**3 + 6*sqrt(c)*sqrt(a)*atan 
((c*x)/(sqrt(c)*sqrt(a)))*a**2*b*d**2*e**4*x + 6*sqrt(c)*sqrt(a)*atan((c*x 
)/(sqrt(c)*sqrt(a)))*a**2*c*d**4*e**2 + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt 
(c)*sqrt(a)))*a**2*c*d**3*e**3*x - 3*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*s 
qrt(a)))*a**2*c*d**2*e**4*x**2 - 3*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqr 
t(a)))*a**2*c*d*e**5*x**3 - 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)) 
)*a*b*c*d**5*e - 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*b*c*d** 
4*e**2*x + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*b*c*d**3*e**3 
*x**2 + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*b*c*d**2*e**4*x* 
*3 + sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*c**2*d**6 + sqrt(c)*s 
qrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*c**2*d**5*e*x + 6*sqrt(c)*sqrt(a)*a 
tan((c*x)/(sqrt(c)*sqrt(a)))*a*c**2*d**4*e**2*x**2 + 6*sqrt(c)*sqrt(a)*ata 
n((c*x)/(sqrt(c)*sqrt(a)))*a*c**2*d**3*e**3*x**3 - 2*sqrt(c)*sqrt(a)*atan( 
(c*x)/(sqrt(c)*sqrt(a)))*b*c**2*d**5*e*x**2 - 2*sqrt(c)*sqrt(a)*atan((c*x) 
/(sqrt(c)*sqrt(a)))*b*c**2*d**4*e**2*x**3 + sqrt(c)*sqrt(a)*atan((c*x)/(sq 
rt(c)*sqrt(a)))*c**3*d**6*x**2 + sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt( 
a)))*c**3*d**5*e*x**3 - log(a + c*x**2)*a**3*b*d**2*e**4 - log(a + c*x**2) 
*a**3*b*d*e**5*x - 4*log(a + c*x**2)*a**3*c*d**3*e**3 - 4*log(a + c*x**...