\(\int \frac {1}{(a+\frac {c}{x^2}+\frac {b}{x}) x^3 (d+e x)^2} \, dx\) [215]

Optimal result

Integrand size = 25, antiderivative size = 248 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx=\frac {e^2}{d \left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {\left (b^3 e^2-a b e (2 b d+3 c e)+a^2 d (b d+4 c e)\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac {\log (x)}{c d^2}-\frac {e^2 \left (3 a d^2-e (2 b d-c e)\right ) \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )^2}-\frac {\left (a^2 d^2+b^2 e^2-a e (2 b d+c e)\right ) \log \left (c+b x+a x^2\right )}{2 c \left (a d^2-e (b d-c e)\right )^2} \] Output:

e^2/d/(a*d^2-b*d*e+c*e^2)/(e*x+d)+(b^3*e^2-a*b*e*(2*b*d+3*c*e)+a^2*d*(b*d+ 
4*c*e))*arctanh((2*a*x+b)/(-4*a*c+b^2)^(1/2))/c/(-4*a*c+b^2)^(1/2)/(a*d^2- 
e*(b*d-c*e))^2+ln(x)/c/d^2-e^2*(3*a*d^2-e*(2*b*d-c*e))*ln(e*x+d)/d^2/(a*d^ 
2-e*(b*d-c*e))^2-1/2*(a^2*d^2+b^2*e^2-a*e*(2*b*d+c*e))*ln(a*x^2+b*x+c)/c/( 
a*d^2-e*(b*d-c*e))^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx=\frac {e^2}{d \left (a d^2+e (-b d+c e)\right ) (d+e x)}-\frac {\left (b^3 e^2-a b e (2 b d+3 c e)+a^2 d (b d+4 c e)\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )}{c \sqrt {-b^2+4 a c} \left (a d^2+e (-b d+c e)\right )^2}+\frac {\log (x)}{c d^2}-\frac {e^2 \left (3 a d^2+e (-2 b d+c e)\right ) \log (d+e x)}{\left (a d^3+d e (-b d+c e)\right )^2}+\frac {\left (-a^2 d^2-b^2 e^2+a e (2 b d+c e)\right ) \log (c+x (b+a x))}{2 c \left (a d^2+e (-b d+c e)\right )^2} \] Input:

Integrate[1/((a + c/x^2 + b/x)*x^3*(d + e*x)^2),x]
 

Output:

e^2/(d*(a*d^2 + e*(-(b*d) + c*e))*(d + e*x)) - ((b^3*e^2 - a*b*e*(2*b*d + 
3*c*e) + a^2*d*(b*d + 4*c*e))*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]])/(c*S 
qrt[-b^2 + 4*a*c]*(a*d^2 + e*(-(b*d) + c*e))^2) + Log[x]/(c*d^2) - (e^2*(3 
*a*d^2 + e*(-2*b*d + c*e))*Log[d + e*x])/(a*d^3 + d*e*(-(b*d) + c*e))^2 + 
((-(a^2*d^2) - b^2*e^2 + a*e*(2*b*d + c*e))*Log[c + x*(b + a*x)])/(2*c*(a* 
d^2 + e*(-(b*d) + c*e))^2)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1893, 1200, 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.

\(\displaystyle \int \frac {1}{x^3 (d+e x)^2 \left (a+\frac {b}{x}+\frac {c}{x^2}\right )} \, dx\)

\(\Big \downarrow \) 1893

\(\displaystyle \int \frac {1}{x (d+e x)^2 \left (a x^2+b x+c\right )}dx\)

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\text {arctanh}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right ) \left (a^2 d (b d+4 c e)-a b e (2 b d+3 c e)+b^3 e^2\right )}{c \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac {\left (a^2 d^2-a e (2 b d+c e)+b^2 e^2\right ) \log \left (a x^2+b x+c\right )}{2 c \left (a d^2-e (b d-c e)\right )^2}+\frac {e^2}{d (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac {e^2 \log (d+e x) \left (3 a d^2-e (2 b d-c e)\right )}{d^2 \left (a d^2-e (b d-c e)\right )^2}+\frac {\log (x)}{c d^2}\)

Input:

Int[1/((a + c/x^2 + b/x)*x^3*(d + e*x)^2),x]
 

Output:

e^2/(d*(a*d^2 - e*(b*d - c*e))*(d + e*x)) + ((b^3*e^2 - a*b*e*(2*b*d + 3*c 
*e) + a^2*d*(b*d + 4*c*e))*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt 
[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))^2) + Log[x]/(c*d^2) - (e^2*(3*a*d^2 
- e*(2*b*d - c*e))*Log[d + e*x])/(d^2*(a*d^2 - e*(b*d - c*e))^2) - ((a^2*d 
^2 + b^2*e^2 - a*e*(2*b*d + c*e))*Log[c + b*x + a*x^2])/(2*c*(a*d^2 - e*(b 
*d - c*e))^2)
 

Defintions of rubi rules used

rule 1200

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

rule 1893

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) 
 + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c 
 + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && EqQ[mn 
, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]
 

rule 2009

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {\frac {\left (-a^{3} d^{2}+2 a^{2} b d e +a^{2} c \,e^{2}-a \,b^{2} e^{2}\right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (-a^{2} b \,d^{2}-2 a^{2} c d e +2 a \,b^{2} d e +2 a b c \,e^{2}-b^{3} e^{2}-\frac {\left (-a^{3} d^{2}+2 a^{2} b d e +a^{2} c \,e^{2}-a \,b^{2} e^{2}\right ) b}{2 a}\right ) \arctan \left (\frac {2 x a +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right )^{2} c}+\frac {e^{2}}{d \left (a \,d^{2}-b d e +c \,e^{2}\right ) \left (e x +d \right )}-\frac {e^{2} \left (3 a \,d^{2}-2 b d e +c \,e^{2}\right ) \ln \left (e x +d \right )}{d^{2} \left (a \,d^{2}-b d e +c \,e^{2}\right )^{2}}+\frac {\ln \left (x \right )}{c \,d^{2}}\)	\(281\)
risch	\(\text {Expression too large to display}\)	\(1713\)

Input:

int(1/(a+c/x^2+b/x)/x^3/(e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

1/(a*d^2-b*d*e+c*e^2)^2/c*(1/2*(-a^3*d^2+2*a^2*b*d*e+a^2*c*e^2-a*b^2*e^2)/ 
a*ln(a*x^2+b*x+c)+2*(-a^2*b*d^2-2*a^2*c*d*e+2*a*b^2*d*e+2*a*b*c*e^2-b^3*e^ 
2-1/2*(-a^3*d^2+2*a^2*b*d*e+a^2*c*e^2-a*b^2*e^2)*b/a)/(4*a*c-b^2)^(1/2)*ar 
ctan((2*a*x+b)/(4*a*c-b^2)^(1/2)))+e^2/d/(a*d^2-b*d*e+c*e^2)/(e*x+d)-e^2*( 
3*a*d^2-2*b*d*e+c*e^2)/d^2/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)+ln(x)/c/d^2
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx=\text {Timed out} \] Input:

integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx=\text {Timed out} \] Input:

integrate(1/(a+c/x**2+b/x)/x**3/(e*x+d)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d)^2,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.62 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx=\frac {e^{5}}{{\left (a d^{3} e^{3} - b d^{2} e^{4} + c d e^{5}\right )} {\left (e x + d\right )}} - \frac {{\left (a^{2} d^{2} - 2 \, a b d e + b^{2} e^{2} - a c e^{2}\right )} \log \left (-a + \frac {2 \, a d}{e x + d} - \frac {a d^{2}}{{\left (e x + d\right )}^{2}} - \frac {b e}{e x + d} + \frac {b d e}{{\left (e x + d\right )}^{2}} - \frac {c e^{2}}{{\left (e x + d\right )}^{2}}\right )}{2 \, {\left (a^{2} c d^{4} - 2 \, a b c d^{3} e + b^{2} c d^{2} e^{2} + 2 \, a c^{2} d^{2} e^{2} - 2 \, b c^{2} d e^{3} + c^{3} e^{4}\right )}} + \frac {{\left (a^{2} b d^{2} e^{2} - 2 \, a b^{2} d e^{3} + 4 \, a^{2} c d e^{3} + b^{3} e^{4} - 3 \, a b c e^{4}\right )} \arctan \left (-\frac {2 \, a d - \frac {2 \, a d^{2}}{e x + d} - b e + \frac {2 \, b d e}{e x + d} - \frac {2 \, c e^{2}}{e x + d}}{\sqrt {-b^{2} + 4 \, a c} e}\right )}{{\left (a^{2} c d^{4} - 2 \, a b c d^{3} e + b^{2} c d^{2} e^{2} + 2 \, a c^{2} d^{2} e^{2} - 2 \, b c^{2} d e^{3} + c^{3} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c} e^{2}} + \frac {\log \left ({\left | -\frac {d}{e x + d} + 1 \right |}\right )}{c d^{2}} \] Input:

integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 61.38 (sec) , antiderivative size = 3510, normalized size of antiderivative = 14.15 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx=\text {Too large to display} \] Input:

int(1/(x^3*(d + e*x)^2*(a + b/x + c/x^2)),x)
 

Output:

(log((a^4*e^4)/(d*(a*d^2 + c*e^2 - b*d*e)^2) + (a^4*e^5*x)/(d^2*(a*d^2 + c 
*e^2 - b*d*e)^2) - (((a*e^3*(3*a^3*b*d^4 + b^3*c*e^4 - b^4*d*e^3 + 5*a*b^3 
*d^2*e^2 - 7*a^2*b^2*d^3*e + 8*a^2*c^2*d*e^3 - 3*a*b*c^2*e^4 + 9*a^3*c*d^3 
*e - a*b^2*c*d*e^3 - 8*a^2*b*c*d^2*e^2))/(d^2*(a*d^2 + c*e^2 - b*d*e)^2) + 
 (((a*e*(a^3*b*d^5 - 4*a*c^3*e^5 + b^2*c^2*e^5 - b^4*d^2*e^3 + 3*a*b^3*d^3 
*e^2 - 3*a^2*b^2*d^4*e - 8*a^2*c^2*d^2*e^3 + 4*a^3*c*d^4*e - b^3*c*d*e^4 + 
 4*a*b*c^2*d*e^4 + 6*a*b^2*c*d^2*e^3 - 9*a^2*b*c*d^3*e^2))/(a*d^3 - b*d^2* 
e + c*d*e^2) + (a*e*x*(3*a^4*d^5 + 2*b^3*c*e^5 - 4*b^4*d*e^4 + 9*a*b^3*d^2 
*e^3 + 4*a^2*c^2*d*e^4 + 19*a^3*c*d^3*e^2 - 3*a^2*b^2*d^3*e^2 - 8*a*b*c^2* 
e^5 - 5*a^3*b*d^4*e + 15*a*b^2*c*d*e^4 - 36*a^2*b*c*d^2*e^3))/(a*d^3 - b*d 
^2*e + c*d*e^2) - (a*e*(b^4*e^2 - 4*a^3*c*d^2 + b^3*e^2*(b^2 - 4*a*c)^(1/2 
) + a^2*b^2*d^2 + 4*a^2*c^2*e^2 - 2*a*b^3*d*e - 5*a*b^2*c*e^2 + a^2*b*d^2* 
(b^2 - 4*a*c)^(1/2) + 8*a^2*b*c*d*e - 3*a*b*c*e^2*(b^2 - 4*a*c)^(1/2) - 2* 
a*b^2*d*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c*d*e*(b^2 - 4*a*c)^(1/2))*(4*a^2*c^ 
2*d^3*e + b^2*c^2*d*e^3 + b^3*c*d^2*e^2 + 2*a^2*b^2*d^4*x + 2*b^2*c^2*e^4* 
x + 2*b^4*d^2*e^2*x + a^2*b*c*d^4 - 4*a*c^3*d*e^3 - 6*a^3*c*d^4*x - 8*a*c^ 
3*e^4*x - 2*a*b^2*c*d^3*e - 4*a*b^3*d^3*e*x - 2*b^3*c*d*e^3*x - 3*a*b*c^2* 
d^2*e^2 - 6*a^2*c^2*d^2*e^2*x + 8*a*b*c^2*d*e^3*x + 14*a^2*b*c*d^3*e*x - 6 
*a*b^2*c*d^2*e^2*x))/(2*c*(4*a*c - b^2)*(a*d^2 + c*e^2 - b*d*e)^2))*(b^4*e 
^2 - 4*a^3*c*d^2 + b^3*e^2*(b^2 - 4*a*c)^(1/2) + a^2*b^2*d^2 + 4*a^2*c^...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1629, normalized size of antiderivative = 6.57 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)^2} \, dx =\text {Too large to display} \] Input:

int(1/(a+c/x^2+b/x)/x^3/(e*x+d)^2,x)
 

Output:

( - 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*b*d**5 
- 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*b*d**4*e* 
x - 8*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*c*d**4* 
e - 8*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*c*d**3* 
e**2*x + 4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**2* 
d**4*e + 4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**2* 
d**3*e**2*x + 6*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a* 
b*c*d**3*e**2 + 6*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))* 
a*b*c*d**2*e**3*x - 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b** 
2))*b**3*d**3*e**2 - 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b* 
*2))*b**3*d**2*e**3*x - 4*log(a*x**2 + b*x + c)*a**3*c*d**5 - 4*log(a*x**2 
 + b*x + c)*a**3*c*d**4*e*x + log(a*x**2 + b*x + c)*a**2*b**2*d**5 + log(a 
*x**2 + b*x + c)*a**2*b**2*d**4*e*x + 8*log(a*x**2 + b*x + c)*a**2*b*c*d** 
4*e + 8*log(a*x**2 + b*x + c)*a**2*b*c*d**3*e**2*x + 4*log(a*x**2 + b*x + 
c)*a**2*c**2*d**3*e**2 + 4*log(a*x**2 + b*x + c)*a**2*c**2*d**2*e**3*x - 2 
*log(a*x**2 + b*x + c)*a*b**3*d**4*e - 2*log(a*x**2 + b*x + c)*a*b**3*d**3 
*e**2*x - 5*log(a*x**2 + b*x + c)*a*b**2*c*d**3*e**2 - 5*log(a*x**2 + b*x 
+ c)*a*b**2*c*d**2*e**3*x + log(a*x**2 + b*x + c)*b**4*d**3*e**2 + log(a*x 
**2 + b*x + c)*b**4*d**2*e**3*x - 24*log(d + e*x)*a**2*c**2*d**3*e**2 - 24 
*log(d + e*x)*a**2*c**2*d**2*e**3*x + 6*log(d + e*x)*a*b**2*c*d**3*e**2...