\(\int \frac {A+B x}{(d+e x)^3 (a+c x^2)} \, dx\) [86]

Optimal result

Integrand size = 22, antiderivative size = 251 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx=\frac {B d-A e}{2 \left (c d^2+a e^2\right ) (d+e x)^2}+\frac {B c d^2-2 A c d e-a B e^2}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac {\sqrt {c} \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (c d^2+a e^2\right )^3}-\frac {c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx=\frac {\frac {\left (c d^2+a e^2\right ) \left (B \left (-a e^2 (d+2 e x)+c d^2 (3 d+2 e x)\right )-A e \left (a e^2+c d (5 d+4 e x)\right )\right )}{(d+e x)^2}+\frac {2 \sqrt {c} \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a}}-2 c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log (d+e x)+c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 251, 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.

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 1.36 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {A e -B d}{2 \left (a \,e^{2}+c \,d^{2}\right ) \left (e x +d \right )^{2}}-\frac {2 A c d e +B a \,e^{2}-B c \,d^{2}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {c \left (A a \,e^{3}-3 A c \,d^{2} e -3 B a d \,e^{2}+B c \,d^{3}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {c \left (\frac {\left (-A a c \,e^{3}+3 A \,c^{2} d^{2} e +3 B a c d \,e^{2}-B \,c^{2} d^{3}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (3 A a c d \,e^{2}-A \,c^{2} d^{3}+B \,e^{3} a^{2}-3 B a c \,d^{2} e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}\)	\(243\)
risch	\(\text {Expression too large to display}\)	\(11304\)

Input:

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

Output:

-1/2*(A*e-B*d)/(a*e^2+c*d^2)/(e*x+d)^2-(2*A*c*d*e+B*a*e^2-B*c*d^2)/(a*e^2+ 
c*d^2)^2/(e*x+d)-c*(A*a*e^3-3*A*c*d^2*e-3*B*a*d*e^2+B*c*d^3)*ln(e*x+d)/(a* 
e^2+c*d^2)^3-c/(a*e^2+c*d^2)^3*(1/2*(-A*a*c*e^3+3*A*c^2*d^2*e+3*B*a*c*d*e^ 
2-B*c^2*d^3)/c*ln(c*x^2+a)+(3*A*a*c*d*e^2-A*c^2*d^3+B*a^2*e^3-3*B*a*c*d^2* 
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. 663 vs. \(2 (239) = 478\).

Time = 19.06 (sec) , antiderivative size = 1350, normalized size of antiderivative = 5.38 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.67 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx=\frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A a c e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A a c e^{3}\right )} \log \left (e x + d\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac {{\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e - 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} \sqrt {a c}} + \frac {3 \, B c d^{3} - 5 \, A c d^{2} e - B a d e^{2} - A a e^{3} + 2 \, {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x}{2 \, {\left (c^{2} d^{6} + 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4} + {\left (c^{2} d^{4} e^{2} + 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e + 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.59 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx=\frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A a c e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac {{\left (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} - 3 \, B a c d e^{3} + A a c e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{c^{3} d^{6} e + 3 \, a c^{2} d^{4} e^{3} + 3 \, a^{2} c d^{2} e^{5} + a^{3} e^{7}} + \frac {{\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e - 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} \sqrt {a c}} + \frac {3 \, B c^{2} d^{5} - 5 \, A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} - B a^{2} d e^{4} - A a^{2} e^{5} + 2 \, {\left (B c^{2} d^{4} e - 2 \, A c^{2} d^{3} e^{2} - 2 \, A a c d e^{4} - B a^{2} e^{5}\right )} x}{2 \, {\left (c d^{2} + a e^{2}\right )}^{3} {\left (e x + d\right )}^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 7.21 (sec) , antiderivative size = 1680, normalized size of antiderivative = 6.69 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

(log(B^2*a^7*e^10*(-a*c)^(1/2) - A^2*c^5*d^10*(-a*c)^(3/2) - 9*A^2*a^5*e^1 
0*(-a*c)^(3/2) + 9*B^2*c^3*d^10*(-a*c)^(5/2) + A^2*a*c^7*d^10*x + B^2*a^7* 
c*e^10*x + 6*A^2*a*d^4*e^6*(-a*c)^(7/2) + 6*B^2*a*d^6*e^4*(-a*c)^(7/2) - 1 
06*A^2*c*d^6*e^4*(-a*c)^(7/2) + 27*B^2*c*d^8*e^2*(-a*c)^(7/2) + 9*A^2*a^6* 
c^2*e^10*x + 9*B^2*a^2*c^6*d^10*x - 27*A^2*a^3*d^2*e^8*(-a*c)^(5/2) + 106* 
B^2*a^3*d^4*e^6*(-a*c)^(5/2) - 77*B^2*a^5*d^2*e^8*(-a*c)^(3/2) + 77*A^2*c^ 
3*d^8*e^2*(-a*c)^(5/2) - 224*A*B*a*d^5*e^5*(-a*c)^(7/2) + 48*A*B*a^5*d*e^9 
*(-a*c)^(3/2) - 64*A*B*c*d^7*e^3*(-a*c)^(7/2) - 48*A*B*c^3*d^9*e*(-a*c)^(5 
/2) + 77*A^2*a^2*c^6*d^8*e^2*x + 106*A^2*a^3*c^5*d^6*e^4*x - 6*A^2*a^4*c^4 
*d^4*e^6*x - 27*A^2*a^5*c^3*d^2*e^8*x - 27*B^2*a^3*c^5*d^8*e^2*x - 6*B^2*a 
^4*c^4*d^6*e^4*x + 106*B^2*a^5*c^3*d^4*e^6*x + 77*B^2*a^6*c^2*d^2*e^8*x + 
64*A*B*a^3*d^3*e^7*(-a*c)^(5/2) - 48*A*B*a^2*c^6*d^9*e*x - 48*A*B*a^6*c^2* 
d*e^9*x + 64*A*B*a^3*c^5*d^7*e^3*x + 224*A*B*a^4*c^4*d^5*e^5*x + 64*A*B*a^ 
5*c^3*d^3*e^7*x)*(a^2*e^3*((A*c)/2 - (B*(-a*c)^(1/2))/2) - e^2*((3*B*a^2*c 
*d)/2 + (3*A*a*c*d*(-a*c)^(1/2))/2) - a*e*((3*A*c^2*d^2)/2 - (3*B*c*d^2*(- 
a*c)^(1/2))/2) + (B*a*c^2*d^3)/2 + (A*c^2*d^3*(-a*c)^(1/2))/2))/(a^4*e^6 + 
 a*c^3*d^6 + 3*a^3*c*d^2*e^4 + 3*a^2*c^2*d^4*e^2) - (log(d + e*x)*(c^2*(B* 
d^3 - 3*A*d^2*e) + a*c*(A*e^3 - 3*B*d*e^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) - (log(9*A^2*a^5*e^10*(-a*c)^(3/2) + A^2*c^5*d^ 
10*(-a*c)^(3/2) - B^2*a^7*e^10*(-a*c)^(1/2) - 9*B^2*c^3*d^10*(-a*c)^(5/...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1036, normalized size of antiderivative = 4.13 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*b*d**3*e**3 - 4*sqrt 
(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*b*d**2*e**4*x - 2*sqrt(c)*sqrt 
(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*b*d*e**5*x**2 - 6*sqrt(c)*sqrt(a)*atan 
((c*x)/(sqrt(c)*sqrt(a)))*a*c*d**4*e**2 - 12*sqrt(c)*sqrt(a)*atan((c*x)/(s 
qrt(c)*sqrt(a)))*a*c*d**3*e**3*x - 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*s 
qrt(a)))*a*c*d**2*e**4*x**2 + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a 
)))*b*c*d**5*e + 12*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*b*c*d**4 
*e**2*x + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*b*c*d**3*e**3*x* 
*2 + 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*c**2*d**6 + 4*sqrt(c) 
*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*c**2*d**5*e*x + 2*sqrt(c)*sqrt(a)*a 
tan((c*x)/(sqrt(c)*sqrt(a)))*c**2*d**4*e**2*x**2 + log(a + c*x**2)*a**2*c* 
d**3*e**3 + 2*log(a + c*x**2)*a**2*c*d**2*e**4*x + log(a + c*x**2)*a**2*c* 
d*e**5*x**2 - 3*log(a + c*x**2)*a*b*c*d**4*e**2 - 6*log(a + c*x**2)*a*b*c* 
d**3*e**3*x - 3*log(a + c*x**2)*a*b*c*d**2*e**4*x**2 - 3*log(a + c*x**2)*a 
*c**2*d**5*e - 6*log(a + c*x**2)*a*c**2*d**4*e**2*x - 3*log(a + c*x**2)*a* 
c**2*d**3*e**3*x**2 + log(a + c*x**2)*b*c**2*d**6 + 2*log(a + c*x**2)*b*c* 
*2*d**5*e*x + log(a + c*x**2)*b*c**2*d**4*e**2*x**2 - 2*log(d + e*x)*a**2* 
c*d**3*e**3 - 4*log(d + e*x)*a**2*c*d**2*e**4*x - 2*log(d + e*x)*a**2*c*d* 
e**5*x**2 + 6*log(d + e*x)*a*b*c*d**4*e**2 + 12*log(d + e*x)*a*b*c*d**3*e* 
*3*x + 6*log(d + e*x)*a*b*c*d**2*e**4*x**2 + 6*log(d + e*x)*a*c**2*d**5...