\(\int \frac {A+B x}{(d+e x)^2 (a+c x^2)^3} \, dx\) [102]

Optimal result

Integrand size = 22, antiderivative size = 461 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^3} \, dx=-\frac {e \left (2 a B d e \left (c d^2-11 a e^2\right )-3 A \left (c^2 d^4+4 a c d^2 e^2-5 a^2 e^4\right )\right )}{8 a^2 \left (c d^2+a e^2\right )^3 (d+e x)}-\frac {a \left (B c d^2-2 A c d e-a B e^2\right )-c \left (A c d^2+2 a B d e-a A e^2\right ) x}{4 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )^2}-\frac {a e \left (3 A c d^2+10 a B d e-7 a A e^2\right )+\left (2 a B e \left (c d^2-2 a e^2\right )-3 A c d \left (c d^2+3 a e^2\right )\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 (d+e x) \left (a+c x^2\right )}-\frac {\sqrt {c} \left (2 a B d e \left (c^2 d^4+10 a c d^2 e^2-15 a^2 e^4\right )-3 A \left (c^3 d^6+5 a c^2 d^4 e^2+15 a^2 c d^2 e^4-5 a^3 e^6\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} \left (c d^2+a e^2\right )^4}-\frac {e^4 \left (5 B c d^2-6 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^4}+\frac {e^4 \left (5 B c d^2-6 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^4} \] Output:

-1/8*e*(2*a*B*d*e*(-11*a*e^2+c*d^2)-3*A*(-5*a^2*e^4+4*a*c*d^2*e^2+c^2*d^4) 
)/a^2/(a*e^2+c*d^2)^3/(e*x+d)-1/4*(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)^2-1/8*(a*e*(-7*A*a*e 
^2+3*A*c*d^2+10*B*a*d*e)+(2*a*B*e*(-2*a*e^2+c*d^2)-3*A*c*d*(3*a*e^2+c*d^2) 
)*x)/a^2/(a*e^2+c*d^2)^2/(e*x+d)/(c*x^2+a)-1/8*c^(1/2)*(2*a*B*d*e*(-15*a^2 
*e^4+10*a*c*d^2*e^2+c^2*d^4)-3*A*(-5*a^3*e^6+15*a^2*c*d^2*e^4+5*a*c^2*d^4* 
e^2+c^3*d^6))*arctan(c^(1/2)*x/a^(1/2))/a^(5/2)/(a*e^2+c*d^2)^4-e^4*(-6*A* 
c*d*e-B*a*e^2+5*B*c*d^2)*ln(e*x+d)/(a*e^2+c*d^2)^4+1/2*e^4*(-6*A*c*d*e-B*a 
*e^2+5*B*c*d^2)*ln(c*x^2+a)/(a*e^2+c*d^2)^4
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 378, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^3} \, dx=\frac {-\frac {8 e^4 (-B d+A e) \left (c d^2+a e^2\right )}{d+e x}+\frac {\left (c d^2+a e^2\right ) \left (4 a^3 B e^4+3 A c^3 d^4 x-2 a c^2 d^2 e (B d-6 A e) x+a^2 c e^2 (-2 B d (6 d-7 e x)+A e (16 d-7 e x))\right )}{a^2 \left (a+c x^2\right )}+\frac {2 \left (c d^2+a e^2\right )^2 \left (a^2 B e^2+A c^2 d^2 x-a c (B d (d-2 e x)+A e (-2 d+e x))\right )}{a \left (a+c x^2\right )^2}+\frac {\sqrt {c} \left (2 a B d e \left (-c^2 d^4-10 a c d^2 e^2+15 a^2 e^4\right )+3 A \left (c^3 d^6+5 a c^2 d^4 e^2+15 a^2 c d^2 e^4-5 a^3 e^6\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{5/2}}+8 e^4 \left (-5 B c d^2+6 A c d e+a B e^2\right ) \log (d+e x)-4 e^4 \left (-5 B c d^2+6 A c d e+a B e^2\right ) \log \left (a+c x^2\right )}{8 \left (c d^2+a e^2\right )^4} \] Input:

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

Output:

((-8*e^4*(-(B*d) + A*e)*(c*d^2 + a*e^2))/(d + e*x) + ((c*d^2 + a*e^2)*(4*a 
^3*B*e^4 + 3*A*c^3*d^4*x - 2*a*c^2*d^2*e*(B*d - 6*A*e)*x + a^2*c*e^2*(-2*B 
*d*(6*d - 7*e*x) + A*e*(16*d - 7*e*x))))/(a^2*(a + c*x^2)) + (2*(c*d^2 + a 
*e^2)^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)^2) + (Sqrt[c]*(2*a*B*d*e*(-(c^2*d^4) - 10*a*c*d^2*e^2 + 
 15*a^2*e^4) + 3*A*(c^3*d^6 + 5*a*c^2*d^4*e^2 + 15*a^2*c*d^2*e^4 - 5*a^3*e 
^6))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(5/2) + 8*e^4*(-5*B*c*d^2 + 6*A*c*d*e 
+ a*B*e^2)*Log[d + e*x] - 4*e^4*(-5*B*c*d^2 + 6*A*c*d*e + a*B*e^2)*Log[a + 
 c*x^2])/(8*(c*d^2 + a*e^2)^4)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 480, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {686, 25, 27, 686, 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.

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

\(\Big \downarrow \) 686

\(\displaystyle -\frac {\int -\frac {c \left (3 A c d^2-2 a B e d+5 a A e^2+4 e (A c d+a B e) x\right )}{(d+e x)^2 \left (c x^2+a\right )^2}dx}{4 a c \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 (d+e x) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {c \left (3 A c d^2-2 a B e d+5 a A e^2+4 e (A c d+a B e) x\right )}{(d+e x)^2 \left (c x^2+a\right )^2}dx}{4 a c \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 (d+e x) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 A c d^2-2 a B e d+5 a A e^2+4 e (A c d+a B e) x}{(d+e x)^2 \left (c x^2+a\right )^2}dx}{4 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 (d+e x) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 686

\(\displaystyle \frac {-\frac {\int \frac {c \left (2 a B d e \left (c d^2+7 a e^2\right )-3 A \left (c^2 d^4+2 a c e^2 d^2+5 a^2 e^4\right )+2 e \left (2 a B e \left (c d^2-2 a e^2\right )-3 A c d \left (c d^2+3 a e^2\right )\right ) x\right )}{(d+e x)^2 \left (c x^2+a\right )}dx}{2 a c \left (a e^2+c d^2\right )}-\frac {x \left (2 a B e \left (c d^2-2 a e^2\right )-3 A c d \left (3 a e^2+c d^2\right )\right )+a e \left (-5 a A e^2+6 a B d e+A c d^2\right )}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 (d+e x) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {2 a B d e \left (c d^2+7 a e^2\right )-3 A \left (c^2 d^4+2 a c e^2 d^2+5 a^2 e^4\right )+2 e \left (2 a B e \left (c d^2-2 a e^2\right )-3 A c d \left (c d^2+3 a e^2\right )\right ) x}{(d+e x)^2 \left (c x^2+a\right )}dx}{2 a \left (a e^2+c d^2\right )}-\frac {x \left (2 a B e \left (c d^2-2 a e^2\right )-3 A c d \left (3 a e^2+c d^2\right )\right )+a e \left (-5 a A e^2+6 a B d e+A c d^2\right )}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 (d+e x) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {-\frac {\int \left (-\frac {8 a^2 \left (-5 B c d^2+6 A c e d+a B e^2\right ) e^5}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac {\left (3 A \left (c^2 d^4+4 a c e^2 d^2-5 a^2 e^4\right )-2 a B d e \left (c d^2-11 a e^2\right )\right ) e^2}{\left (c d^2+a e^2\right ) (d+e x)^2}+\frac {c \left (-8 a^2 \left (5 B c d^2-6 A c e d-a B e^2\right ) x e^4+2 a B d \left (c^2 d^4+10 a c e^2 d^2-15 a^2 e^4\right ) e-3 A \left (c^3 d^6+5 a c^2 e^2 d^4+15 a^2 c e^4 d^2-5 a^3 e^6\right )\right )}{\left (c d^2+a e^2\right )^2 \left (c x^2+a\right )}\right )dx}{2 a \left (a e^2+c d^2\right )}-\frac {x \left (2 a B e \left (c d^2-2 a e^2\right )-3 A c d \left (3 a e^2+c d^2\right )\right )+a e \left (-5 a A e^2+6 a B d e+A c d^2\right )}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 (d+e x) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\frac {e \left (2 a B d e \left (c d^2-11 a e^2\right )-3 A \left (-5 a^2 e^4+4 a c d^2 e^2+c^2 d^4\right )\right )}{(d+e x) \left (a e^2+c d^2\right )}-\frac {4 a^2 e^4 \log \left (a+c x^2\right ) \left (-a B e^2-6 A c d e+5 B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac {8 a^2 e^4 \log (d+e x) \left (-a B e^2-6 A c d e+5 B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (2 a B d e \left (-15 a^2 e^4+10 a c d^2 e^2+c^2 d^4\right )-3 A \left (-5 a^3 e^6+15 a^2 c d^2 e^4+5 a c^2 d^4 e^2+c^3 d^6\right )\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2}}{2 a \left (a e^2+c d^2\right )}-\frac {x \left (2 a B e \left (c d^2-2 a e^2\right )-3 A c d \left (3 a e^2+c d^2\right )\right )+a e \left (-5 a A e^2+6 a B d e+A c d^2\right )}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 (d+e x) \left (a e^2+c d^2\right )}\)

Input:

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

Output:

-1/4*(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)^2) + (-1/2*(a*e*(A*c*d^2 + 6*a*B*d*e - 5*a*A*e^2) + (2*a*B*e*(c*d^ 
2 - 2*a*e^2) - 3*A*c*d*(c*d^2 + 3*a*e^2))*x)/(a*(c*d^2 + a*e^2)*(d + e*x)* 
(a + c*x^2)) - ((e*(2*a*B*d*e*(c*d^2 - 11*a*e^2) - 3*A*(c^2*d^4 + 4*a*c*d^ 
2*e^2 - 5*a^2*e^4)))/((c*d^2 + a*e^2)*(d + e*x)) + (Sqrt[c]*(2*a*B*d*e*(c^ 
2*d^4 + 10*a*c*d^2*e^2 - 15*a^2*e^4) - 3*A*(c^3*d^6 + 5*a*c^2*d^4*e^2 + 15 
*a^2*c*d^2*e^4 - 5*a^3*e^6))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*(c*d^2 
+ a*e^2)^2) + (8*a^2*e^4*(5*B*c*d^2 - 6*A*c*d*e - a*B*e^2)*Log[d + e*x])/( 
c*d^2 + a*e^2)^2 - (4*a^2*e^4*(5*B*c*d^2 - 6*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)))/(4*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 = 567, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {e^{4} \left (6 A c d e +B a \,e^{2}-5 B c \,d^{2}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{4}}-\frac {\left (A e -B d \right ) e^{4}}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \left (e x +d \right )}-\frac {c \left (\frac {\frac {c \left (7 A \,a^{3} e^{6}-5 A \,a^{2} c \,d^{2} e^{4}-15 A a \,c^{2} d^{4} e^{2}-3 A \,c^{3} d^{6}-14 B \,a^{3} d \,e^{5}-12 B \,a^{2} c \,d^{3} e^{3}+2 B a \,c^{2} d^{5} e \right ) x^{3}}{8 a^{2}}+\left (-2 A a c d \,e^{5}-2 A \,c^{2} d^{3} e^{3}-\frac {1}{2} B \,a^{2} e^{6}+B a c \,d^{2} e^{4}+\frac {3}{2} B \,c^{2} d^{4} e^{2}\right ) x^{2}+\frac {\left (9 A \,a^{3} e^{6}-3 A \,a^{2} c \,d^{2} e^{4}-17 A a \,c^{2} d^{4} e^{2}-5 A \,c^{3} d^{6}-18 B \,a^{3} d \,e^{5}-20 B \,a^{2} c \,d^{3} e^{3}-2 B a \,c^{2} d^{5} e \right ) x}{8 a}-\frac {10 A \,a^{2} c d \,e^{5}+12 A a \,c^{2} d^{3} e^{3}+2 A \,c^{3} d^{5} e +3 B \,a^{3} e^{6}-3 B \,a^{2} c \,d^{2} e^{4}-7 B a \,c^{2} d^{4} e^{2}-B \,c^{3} d^{6}}{4 c}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\frac {\left (48 A \,a^{2} c d \,e^{5}+8 B \,a^{3} e^{6}-40 B \,a^{2} c \,d^{2} e^{4}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (15 A \,a^{3} e^{6}-45 A \,a^{2} c \,d^{2} e^{4}-15 A a \,c^{2} d^{4} e^{2}-3 A \,c^{3} d^{6}-30 B \,a^{3} d \,e^{5}+20 B \,a^{2} c \,d^{3} e^{3}+2 B a \,c^{2} d^{5} e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{8 a^{2}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{4}}\)	\(567\)
risch	\(\text {Expression too large to display}\)	\(1797\)

Input:

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

Output:

e^4*(6*A*c*d*e+B*a*e^2-5*B*c*d^2)/(a*e^2+c*d^2)^4*ln(e*x+d)-(A*e-B*d)*e^4/ 
(a*e^2+c*d^2)^3/(e*x+d)-c/(a*e^2+c*d^2)^4*((1/8*c*(7*A*a^3*e^6-5*A*a^2*c*d 
^2*e^4-15*A*a*c^2*d^4*e^2-3*A*c^3*d^6-14*B*a^3*d*e^5-12*B*a^2*c*d^3*e^3+2* 
B*a*c^2*d^5*e)/a^2*x^3+(-2*A*a*c*d*e^5-2*A*c^2*d^3*e^3-1/2*B*a^2*e^6+B*a*c 
*d^2*e^4+3/2*B*c^2*d^4*e^2)*x^2+1/8*(9*A*a^3*e^6-3*A*a^2*c*d^2*e^4-17*A*a* 
c^2*d^4*e^2-5*A*c^3*d^6-18*B*a^3*d*e^5-20*B*a^2*c*d^3*e^3-2*B*a*c^2*d^5*e) 
/a*x-1/4*(10*A*a^2*c*d*e^5+12*A*a*c^2*d^3*e^3+2*A*c^3*d^5*e+3*B*a^3*e^6-3* 
B*a^2*c*d^2*e^4-7*B*a*c^2*d^4*e^2-B*c^3*d^6)/c)/(c*x^2+a)^2+1/8/a^2*(1/2*( 
48*A*a^2*c*d*e^5+8*B*a^3*e^6-40*B*a^2*c*d^2*e^4)/c*ln(c*x^2+a)+(15*A*a^3*e 
^6-45*A*a^2*c*d^2*e^4-15*A*a*c^2*d^4*e^2-3*A*c^3*d^6-30*B*a^3*d*e^5+20*B*a 
^2*c*d^3*e^3+2*B*a*c^2*d^5*e)/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))))
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 997 vs. \(2 (444) = 888\).

Time = 0.15 (sec) , antiderivative size = 997, normalized size of antiderivative = 2.16 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*(5*B*c*d^2*e^4 - 6*A*c*d*e^5 - B*a*e^6)*log(c*x^2 + a)/(c^4*d^8 + 4*a* 
c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 + 4*a^3*c*d^2*e^6 + a^4*e^8) - (5*B*c*d^2* 
e^4 - 6*A*c*d*e^5 - B*a*e^6)*log(e*x + d)/(c^4*d^8 + 4*a*c^3*d^6*e^2 + 6*a 
^2*c^2*d^4*e^4 + 4*a^3*c*d^2*e^6 + a^4*e^8) + 1/8*(3*A*c^4*d^6 - 2*B*a*c^3 
*d^5*e + 15*A*a*c^3*d^4*e^2 - 20*B*a^2*c^2*d^3*e^3 + 45*A*a^2*c^2*d^2*e^4 
+ 30*B*a^3*c*d*e^5 - 15*A*a^3*c*e^6)*arctan(c*x/sqrt(a*c))/((a^2*c^4*d^8 + 
 4*a^3*c^3*d^6*e^2 + 6*a^4*c^2*d^4*e^4 + 4*a^5*c*d^2*e^6 + a^6*e^8)*sqrt(a 
*c)) - 1/8*(2*B*a^2*c^2*d^5 - 4*A*a^2*c^2*d^4*e + 12*B*a^3*c*d^3*e^2 - 20* 
A*a^3*c*d^2*e^3 - 14*B*a^4*d*e^4 + 8*A*a^4*e^5 - (3*A*c^4*d^4*e - 2*B*a*c^ 
3*d^3*e^2 + 12*A*a*c^3*d^2*e^3 + 22*B*a^2*c^2*d*e^4 - 15*A*a^2*c^2*e^5)*x^ 
4 - (3*A*c^4*d^5 - 2*B*a*c^3*d^4*e + 12*A*a*c^3*d^3*e^2 + 2*B*a^2*c^2*d^2* 
e^3 + 9*A*a^2*c^2*d*e^4 + 4*B*a^3*c*e^5)*x^3 - (5*A*a*c^3*d^4*e - 10*B*a^2 
*c^2*d^3*e^2 + 28*A*a^2*c^2*d^2*e^3 + 38*B*a^3*c*d*e^4 - 25*A*a^3*c*e^5)*x 
^2 - (5*A*a*c^3*d^5 + 16*A*a^2*c^2*d^3*e^2 + 6*B*a^3*c*d^2*e^3 + 11*A*a^3* 
c*d*e^4 + 6*B*a^4*e^5)*x)/(a^4*c^3*d^7 + 3*a^5*c^2*d^5*e^2 + 3*a^6*c*d^3*e 
^4 + a^7*d*e^6 + (a^2*c^5*d^6*e + 3*a^3*c^4*d^4*e^3 + 3*a^4*c^3*d^2*e^5 + 
a^5*c^2*e^7)*x^5 + (a^2*c^5*d^7 + 3*a^3*c^4*d^5*e^2 + 3*a^4*c^3*d^3*e^4 + 
a^5*c^2*d*e^6)*x^4 + 2*(a^3*c^4*d^6*e + 3*a^4*c^3*d^4*e^3 + 3*a^5*c^2*d^2* 
e^5 + a^6*c*e^7)*x^3 + 2*(a^3*c^4*d^7 + 3*a^4*c^3*d^5*e^2 + 3*a^5*c^2*d^3* 
e^4 + a^6*c*d*e^6)*x^2 + (a^4*c^3*d^6*e + 3*a^5*c^2*d^4*e^3 + 3*a^6*c*d...
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 884, normalized size of antiderivative = 1.92 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*(5*B*c*d^2*e^4 - 6*A*c*d*e^5 - B*a*e^6)*log(c - 2*c*d/(e*x + d) + c*d^ 
2/(e*x + d)^2 + a*e^2/(e*x + d)^2)/(c^4*d^8 + 4*a*c^3*d^6*e^2 + 6*a^2*c^2* 
d^4*e^4 + 4*a^3*c*d^2*e^6 + a^4*e^8) + (B*d*e^10/(e*x + d) - A*e^11/(e*x + 
 d))/(c^3*d^6*e^6 + 3*a*c^2*d^4*e^8 + 3*a^2*c*d^2*e^10 + a^3*e^12) + 1/8*( 
3*A*c^4*d^6*e^2 - 2*B*a*c^3*d^5*e^3 + 15*A*a*c^3*d^4*e^4 - 20*B*a^2*c^2*d^ 
3*e^5 + 45*A*a^2*c^2*d^2*e^6 + 30*B*a^3*c*d*e^7 - 15*A*a^3*c*e^8)*arctan(( 
c*d - c*d^2/(e*x + d) - a*e^2/(e*x + d))/(sqrt(a*c)*e))/((a^2*c^4*d^8 + 4* 
a^3*c^3*d^6*e^2 + 6*a^4*c^2*d^4*e^4 + 4*a^5*c*d^2*e^6 + a^6*e^8)*sqrt(a*c) 
*e^2) + 1/8*(3*A*c^5*d^5*e - 2*B*a*c^4*d^4*e^2 + 14*A*a*c^4*d^3*e^3 + 32*B 
*a^2*c^3*d^2*e^4 - 29*A*a^2*c^3*d*e^5 - 6*B*a^3*c^2*e^6 - (9*A*c^5*d^6*e^2 
 - 6*B*a*c^4*d^5*e^3 + 41*A*a*c^4*d^4*e^4 + 116*B*a^2*c^3*d^3*e^5 - 121*A* 
a^2*c^3*d^2*e^6 - 38*B*a^3*c^2*d*e^7 + 7*A*a^3*c^2*e^8)/((e*x + d)*e) + (9 
*A*c^5*d^7*e^3 - 6*B*a*c^4*d^6*e^4 + 45*A*a*c^4*d^5*e^5 + 140*B*a^2*c^3*d^ 
4*e^6 - 145*A*a^2*c^3*d^3*e^7 - 22*B*a^3*c^2*d^2*e^8 - 21*A*a^3*c^2*d*e^9 
- 8*B*a^4*c*e^10)/((e*x + d)^2*e^2) - (3*A*c^5*d^8*e^4 - 2*B*a*c^4*d^7*e^5 
 + 18*A*a*c^4*d^6*e^6 + 58*B*a^2*c^3*d^5*e^7 - 60*A*a^2*c^3*d^4*e^8 + 26*B 
*a^3*c^2*d^3*e^9 - 66*A*a^3*c^2*d^2*e^10 - 34*B*a^4*c*d*e^11 + 9*A*a^4*c*e 
^12)/((e*x + d)^3*e^3))/((c*d^2 + a*e^2)^4*a^2*(c - 2*c*d/(e*x + d) + c*d^ 
2/(e*x + d)^2 + a*e^2/(e*x + d)^2)^2)
 

Mupad [B] (verification not implemented)

Time = 8.70 (sec) , antiderivative size = 3015, normalized size of antiderivative = 6.54 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((x*(5*A*c^2*d^3 + 6*B*a^2*e^3 + 11*A*a*c*d*e^2))/(8*a*(a^2*e^4 + c^2*d^4 
+ 2*a*c*d^2*e^2)) - (4*A*a^2*e^5 + B*c^2*d^5 - 7*B*a^2*d*e^4 - 2*A*c^2*d^4 
*e - 10*A*a*c*d^2*e^3 + 6*B*a*c*d^3*e^2)/(4*(a*e^2 + c*d^2)*(a^2*e^4 + c^2 
*d^4 + 2*a*c*d^2*e^2)) + (x^3*(3*A*c^3*d^3 + 4*B*a^2*c*e^3 + 9*A*a*c^2*d*e 
^2 - 2*B*a*c^2*d^2*e))/(8*a^2*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)) + (x^4* 
(3*A*c^4*d^4*e - 15*A*a^2*c^2*e^5 + 12*A*a*c^3*d^2*e^3 - 2*B*a*c^3*d^3*e^2 
 + 22*B*a^2*c^2*d*e^4))/(8*a^2*(a^3*e^6 + c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^ 
2*c*d^2*e^4)) + (x^2*(5*A*c^3*d^4*e - 25*A*a^2*c*e^5 + 28*A*a*c^2*d^2*e^3 
- 10*B*a*c^2*d^3*e^2 + 38*B*a^2*c*d*e^4))/(8*a*(a*e^2 + c*d^2)*(a^2*e^4 + 
c^2*d^4 + 2*a*c*d^2*e^2)))/(a^2*d + c^2*d*x^4 + c^2*e*x^5 + a^2*e*x + 2*a* 
c*d*x^2 + 2*a*c*e*x^3) - (log(d + e*x)*(c*(5*B*d^2*e^4 - 6*A*d*e^5) - B*a* 
e^6))/(a^4*e^8 + c^4*d^8 + 4*a*c^3*d^6*e^2 + 4*a^3*c*d^2*e^6 + 6*a^2*c^2*d 
^4*e^4) + (log(576*B^2*a^14*e^16*(-a^5*c)^(1/2) - 9*A^2*c^8*d^16*(-a^5*c)^ 
(3/2) - 225*A^2*a^8*e^16*(-a^5*c)^(3/2) + 19836*A^2*a^2*d^2*e^14*(-a^5*c)^ 
(5/2) + 4056*B^2*a^2*d^4*e^12*(-a^5*c)^(5/2) + 3708*B^2*a^8*d^2*e^14*(-a^5 
*c)^(3/2) + 23796*A^2*c^2*d^6*e^10*(-a^5*c)^(5/2) + 13840*B^2*c^2*d^8*e^8* 
(-a^5*c)^(5/2) + 576*B^2*a^16*c*e^16*x + 9*A^2*a^7*c^10*d^16*x + 225*A^2*a 
^15*c^2*e^16*x - 19236*A*B*a^2*d^3*e^13*(-a^5*c)^(5/2) - 33540*A*B*c^2*d^7 
*e^9*(-a^5*c)^(5/2) + 40572*A^2*a*c*d^4*e^12*(-a^5*c)^(5/2) + 21820*B^2*a* 
c*d^6*e^10*(-a^5*c)^(5/2) + 108*A^2*a^8*c^9*d^14*e^2*x + 684*A^2*a^9*c^...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 3096, normalized size of antiderivative = 6.72 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**5*d**2*e**6 - 15*s 
qrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**5*d*e**7*x + 30*sqrt(c)*sq 
rt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**4*b*d**3*e**5 + 30*sqrt(c)*sqrt(a)* 
atan((c*x)/(sqrt(c)*sqrt(a)))*a**4*b*d**2*e**6*x + 45*sqrt(c)*sqrt(a)*atan 
((c*x)/(sqrt(c)*sqrt(a)))*a**4*c*d**4*e**4 + 45*sqrt(c)*sqrt(a)*atan((c*x) 
/(sqrt(c)*sqrt(a)))*a**4*c*d**3*e**5*x - 30*sqrt(c)*sqrt(a)*atan((c*x)/(sq 
rt(c)*sqrt(a)))*a**4*c*d**2*e**6*x**2 - 30*sqrt(c)*sqrt(a)*atan((c*x)/(sqr 
t(c)*sqrt(a)))*a**4*c*d*e**7*x**3 - 20*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c) 
*sqrt(a)))*a**3*b*c*d**5*e**3 - 20*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqr 
t(a)))*a**3*b*c*d**4*e**4*x + 60*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt( 
a)))*a**3*b*c*d**3*e**5*x**2 + 60*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt 
(a)))*a**3*b*c*d**2*e**6*x**3 + 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqr 
t(a)))*a**3*c**2*d**6*e**2 + 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a 
)))*a**3*c**2*d**5*e**3*x + 90*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a) 
))*a**3*c**2*d**4*e**4*x**2 + 90*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt( 
a)))*a**3*c**2*d**3*e**5*x**3 - 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqr 
t(a)))*a**3*c**2*d**2*e**6*x**4 - 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*s 
qrt(a)))*a**3*c**2*d*e**7*x**5 - 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqr 
t(a)))*a**2*b*c**2*d**7*e - 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)) 
)*a**2*b*c**2*d**6*e**2*x - 40*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt...