Integrand size = 25, antiderivative size = 372 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx=-\frac {1}{2 c d^2 x^2}+\frac {b d+2 c e}{c^2 d^3 x}+\frac {e^4}{d^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac {\left (b^5 e^2-a^3 c d (3 b d+4 c e)-a b^3 e (2 b d+5 c e)+a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac {\left (b^2 d^2+2 b c d e-c \left (a d^2-3 c e^2\right )\right ) \log (x)}{c^3 d^4}-\frac {e^4 \left (5 a d^2-e (4 b d-3 c e)\right ) \log (d+e x)}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac {\left (a^3 c d^2-b^4 e^2+a b^2 e (2 b d+3 c e)-a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )\right ) \log \left (c+b x+a x^2\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )^2} \] Output:
-1/2/c/d^2/x^2+(b*d+2*c*e)/c^2/d^3/x+e^4/d^3/(a*d^2-e*(b*d-c*e))/(e*x+d)+( b^5*e^2-a^3*c*d*(3*b*d+4*c*e)-a*b^3*e*(2*b*d+5*c*e)+a^2*b*(b^2*d^2+8*b*c*d *e+5*c^2*e^2))*arctanh((2*a*x+b)/(-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(1/2 )/(a*d^2-e*(b*d-c*e))^2+(b^2*d^2+2*b*c*d*e-c*(a*d^2-3*c*e^2))*ln(x)/c^3/d^ 4-e^4*(5*a*d^2-e*(4*b*d-3*c*e))*ln(e*x+d)/d^4/(a*d^2-e*(b*d-c*e))^2+1/2*(a ^3*c*d^2-b^4*e^2+a*b^2*e*(2*b*d+3*c*e)-a^2*(b^2*d^2+4*b*c*d*e+c^2*e^2))*ln (a*x^2+b*x+c)/c^3/(a*d^2-e*(b*d-c*e))^2
Time = 0.41 (sec) , antiderivative size = 370, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx=-\frac {1}{2 c d^2 x^2}+\frac {b d+2 c e}{c^2 d^3 x}+\frac {e^4}{d^3 \left (a d^2+e (-b d+c e)\right ) (d+e x)}+\frac {\left (-b^5 e^2+a^3 c d (3 b d+4 c e)+a b^3 e (2 b d+5 c e)-a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )}{c^3 \sqrt {-b^2+4 a c} \left (a d^2+e (-b d+c e)\right )^2}+\frac {\left (b^2 d^2+2 b c d e+c \left (-a d^2+3 c e^2\right )\right ) \log (x)}{c^3 d^4}-\frac {e^4 \left (5 a d^2+e (-4 b d+3 c e)\right ) \log (d+e x)}{d^4 \left (a d^2+e (-b d+c e)\right )^2}-\frac {\left (-a^3 c d^2+b^4 e^2-a b^2 e (2 b d+3 c e)+a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )\right ) \log (c+x (b+a x))}{2 c^3 \left (a d^2+e (-b d+c e)\right )^2} \] Input:
Integrate[1/((a + c/x^2 + b/x)*x^5*(d + e*x)^2),x]
Output:
-1/2*1/(c*d^2*x^2) + (b*d + 2*c*e)/(c^2*d^3*x) + e^4/(d^3*(a*d^2 + e*(-(b* d) + c*e))*(d + e*x)) + ((-(b^5*e^2) + a^3*c*d*(3*b*d + 4*c*e) + a*b^3*e*( 2*b*d + 5*c*e) - a^2*b*(b^2*d^2 + 8*b*c*d*e + 5*c^2*e^2))*ArcTan[(b + 2*a* x)/Sqrt[-b^2 + 4*a*c]])/(c^3*Sqrt[-b^2 + 4*a*c]*(a*d^2 + e*(-(b*d) + c*e)) ^2) + ((b^2*d^2 + 2*b*c*d*e + c*(-(a*d^2) + 3*c*e^2))*Log[x])/(c^3*d^4) - (e^4*(5*a*d^2 + e*(-4*b*d + 3*c*e))*Log[d + e*x])/(d^4*(a*d^2 + e*(-(b*d) + c*e))^2) - ((-(a^3*c*d^2) + b^4*e^2 - a*b^2*e*(2*b*d + 3*c*e) + a^2*(b^2 *d^2 + 4*b*c*d*e + c^2*e^2))*Log[c + x*(b + a*x)])/(2*c^3*(a*d^2 + e*(-(b* d) + c*e))^2)
Time = 0.95 (sec) , antiderivative size = 372, 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.
\(\displaystyle \int \frac {1}{x^5 (d+e x)^2 \left (a+\frac {b}{x}+\frac {c}{x^2}\right )} \, dx\) |
\(\Big \downarrow \) 1893 |
\(\displaystyle \int \frac {1}{x^3 (d+e x)^2 \left (a x^2+b x+c\right )}dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {a x \left (a^3 c d^2-a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )+a b^2 e (2 b d+3 c e)+b^4 \left (-e^2\right )\right )-\left (a b d+a c e+b^2 (-e)\right ) \left (-2 a^2 c d+a b^2 d+3 a b c e+b^3 (-e)\right )}{c^3 \left (a x^2+b x+c\right ) \left (a d^2-e (b d-c e)\right )^2}+\frac {-c \left (a d^2-3 c e^2\right )+b^2 d^2+2 b c d e}{c^3 d^4 x}+\frac {e^5 \left (e (4 b d-3 c e)-5 a d^2\right )}{d^4 (d+e x) \left (a d^2-e (b d-c e)\right )^2}+\frac {e^5}{d^3 (d+e x)^2 \left (e (b d-c e)-a d^2\right )}+\frac {-b d-2 c e}{c^2 d^3 x^2}+\frac {1}{c d^2 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right ) \left (-a^3 c d (3 b d+4 c e)+a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )-a b^3 e (2 b d+5 c e)+b^5 e^2\right )}{c^3 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac {\left (a^3 c d^2-a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )+a b^2 e (2 b d+3 c e)+b^4 \left (-e^2\right )\right ) \log \left (a x^2+b x+c\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )^2}+\frac {\log (x) \left (-c \left (a d^2-3 c e^2\right )+b^2 d^2+2 b c d e\right )}{c^3 d^4}-\frac {e^4 \log (d+e x) \left (5 a d^2-e (4 b d-3 c e)\right )}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac {e^4}{d^3 (d+e x) \left (a d^2-e (b d-c e)\right )}+\frac {b d+2 c e}{c^2 d^3 x}-\frac {1}{2 c d^2 x^2}\) |
Input:
Int[1/((a + c/x^2 + b/x)*x^5*(d + e*x)^2),x]
Output:
-1/2*1/(c*d^2*x^2) + (b*d + 2*c*e)/(c^2*d^3*x) + e^4/(d^3*(a*d^2 - e*(b*d - c*e))*(d + e*x)) + ((b^5*e^2 - a^3*c*d*(3*b*d + 4*c*e) - a*b^3*e*(2*b*d + 5*c*e) + a^2*b*(b^2*d^2 + 8*b*c*d*e + 5*c^2*e^2))*ArcTanh[(b + 2*a*x)/Sq rt[b^2 - 4*a*c]])/(c^3*Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))^2) + ((b^ 2*d^2 + 2*b*c*d*e - c*(a*d^2 - 3*c*e^2))*Log[x])/(c^3*d^4) - (e^4*(5*a*d^2 - e*(4*b*d - 3*c*e))*Log[d + e*x])/(d^4*(a*d^2 - e*(b*d - c*e))^2) + ((a^ 3*c*d^2 - b^4*e^2 + a*b^2*e*(2*b*d + 3*c*e) - a^2*(b^2*d^2 + 4*b*c*d*e + c ^2*e^2))*Log[c + b*x + a*x^2])/(2*c^3*(a*d^2 - e*(b*d - c*e))^2)
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]
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]
Time = 0.40 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.22
method | result | size |
default | \(\frac {\frac {\left (c \,a^{4} d^{2}-a^{3} b^{2} d^{2}-4 a^{3} b c d e -a^{3} c^{2} e^{2}+2 a^{2} b^{3} d e +3 a^{2} b^{2} c \,e^{2}-a \,b^{4} e^{2}\right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (2 a^{3} b c \,d^{2}+2 a^{3} c^{2} d e -a^{2} b^{3} d^{2}-6 a^{2} b^{2} c d e -3 a^{2} c^{2} e^{2} b +2 a \,b^{4} d e +4 a c \,e^{2} b^{3}-e^{2} b^{5}-\frac {\left (c \,a^{4} d^{2}-a^{3} b^{2} d^{2}-4 a^{3} b c d e -a^{3} c^{2} e^{2}+2 a^{2} b^{3} d e +3 a^{2} b^{2} c \,e^{2}-a \,b^{4} e^{2}\right ) b}{2 a}\right ) \arctan \left (\frac {2 x a +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right )^{2} c^{3}}+\frac {e^{4}}{d^{3} \left (a \,d^{2}-b d e +c \,e^{2}\right ) \left (e x +d \right )}-\frac {e^{4} \left (5 a \,d^{2}-4 b d e +3 c \,e^{2}\right ) \ln \left (e x +d \right )}{d^{4} \left (a \,d^{2}-b d e +c \,e^{2}\right )^{2}}-\frac {1}{2 c \,d^{2} x^{2}}-\frac {-b d -2 c e}{x \,c^{2} d^{3}}+\frac {\left (-a c \,d^{2}+b^{2} d^{2}+2 b c d e +3 c^{2} e^{2}\right ) \ln \left (x \right )}{c^{3} d^{4}}\) | \(455\) |
risch | \(\text {Expression too large to display}\) | \(636436\) |
Input:
int(1/(a+c/x^2+b/x)/x^5/(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
1/(a*d^2-b*d*e+c*e^2)^2/c^3*(1/2*(a^4*c*d^2-a^3*b^2*d^2-4*a^3*b*c*d*e-a^3* c^2*e^2+2*a^2*b^3*d*e+3*a^2*b^2*c*e^2-a*b^4*e^2)/a*ln(a*x^2+b*x+c)+2*(2*a^ 3*b*c*d^2+2*a^3*c^2*d*e-a^2*b^3*d^2-6*a^2*b^2*c*d*e-3*a^2*c^2*e^2*b+2*a*b^ 4*d*e+4*a*c*e^2*b^3-e^2*b^5-1/2*(a^4*c*d^2-a^3*b^2*d^2-4*a^3*b*c*d*e-a^3*c ^2*e^2+2*a^2*b^3*d*e+3*a^2*b^2*c*e^2-a*b^4*e^2)*b/a)/(4*a*c-b^2)^(1/2)*arc tan((2*a*x+b)/(4*a*c-b^2)^(1/2)))+e^4/d^3/(a*d^2-b*d*e+c*e^2)/(e*x+d)-e^4* (5*a*d^2-4*b*d*e+3*c*e^2)/d^4/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)-1/2/c/d^2/x^ 2-(-b*d-2*c*e)/x/c^2/d^3+1/c^3/d^4*(-a*c*d^2+b^2*d^2+2*b*c*d*e+3*c^2*e^2)* ln(x)
Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx=\text {Timed out} \] Input:
integrate(1/(a+c/x^2+b/x)/x^5/(e*x+d)^2,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx=\text {Timed out} \] Input:
integrate(1/(a+c/x**2+b/x)/x**5/(e*x+d)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+c/x^2+b/x)/x^5/(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
Time = 0.12 (sec) , antiderivative size = 598, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx=\frac {e^{9}}{{\left (a d^{5} e^{5} - b d^{4} e^{6} + c d^{3} e^{7}\right )} {\left (e x + d\right )}} - \frac {{\left (a^{2} b^{2} d^{2} - a^{3} c d^{2} - 2 \, a b^{3} d e + 4 \, a^{2} b c d e + b^{4} e^{2} - 3 \, a b^{2} c e^{2} + a^{2} c^{2} e^{2}\right )} \log \left (-a + \frac {2 \, a d}{e x + d} - \frac {a d^{2}}{{\left (e x + d\right )}^{2}} - \frac {b e}{e x + d} + \frac {b d e}{{\left (e x + d\right )}^{2}} - \frac {c e^{2}}{{\left (e x + d\right )}^{2}}\right )}{2 \, {\left (a^{2} c^{3} d^{4} - 2 \, a b c^{3} d^{3} e + b^{2} c^{3} d^{2} e^{2} + 2 \, a c^{4} d^{2} e^{2} - 2 \, b c^{4} d e^{3} + c^{5} e^{4}\right )}} + \frac {{\left (a^{2} b^{3} d^{2} e^{2} - 3 \, a^{3} b c d^{2} e^{2} - 2 \, a b^{4} d e^{3} + 8 \, a^{2} b^{2} c d e^{3} - 4 \, a^{3} c^{2} d e^{3} + b^{5} e^{4} - 5 \, a b^{3} c e^{4} + 5 \, a^{2} b c^{2} e^{4}\right )} \arctan \left (-\frac {2 \, a d - \frac {2 \, a d^{2}}{e x + d} - b e + \frac {2 \, b d e}{e x + d} - \frac {2 \, c e^{2}}{e x + d}}{\sqrt {-b^{2} + 4 \, a c} e}\right )}{{\left (a^{2} c^{3} d^{4} - 2 \, a b c^{3} d^{3} e + b^{2} c^{3} d^{2} e^{2} + 2 \, a c^{4} d^{2} e^{2} - 2 \, b c^{4} d e^{3} + c^{5} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c} e^{2}} + \frac {{\left (b^{2} d^{2} e - a c d^{2} e + 2 \, b c d e^{2} + 3 \, c^{2} e^{3}\right )} \log \left ({\left | -\frac {d}{e x + d} + 1 \right |}\right )}{c^{3} d^{4} e} + \frac {2 \, b c d e + 5 \, c^{2} e^{2} - \frac {2 \, {\left (b c d^{2} e^{2} + 3 \, c^{2} d e^{3}\right )}}{{\left (e x + d\right )} e}}{2 \, c^{3} d^{4} {\left (\frac {d}{e x + d} - 1\right )}^{2}} \] Input:
integrate(1/(a+c/x^2+b/x)/x^5/(e*x+d)^2,x, algorithm="giac")
Output:
e^9/((a*d^5*e^5 - b*d^4*e^6 + c*d^3*e^7)*(e*x + d)) - 1/2*(a^2*b^2*d^2 - a ^3*c*d^2 - 2*a*b^3*d*e + 4*a^2*b*c*d*e + b^4*e^2 - 3*a*b^2*c*e^2 + a^2*c^2 *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^3*d^4 - 2*a*b*c^3*d^3*e + b^2*c^3 *d^2*e^2 + 2*a*c^4*d^2*e^2 - 2*b*c^4*d*e^3 + c^5*e^4) + (a^2*b^3*d^2*e^2 - 3*a^3*b*c*d^2*e^2 - 2*a*b^4*d*e^3 + 8*a^2*b^2*c*d*e^3 - 4*a^3*c^2*d*e^3 + b^5*e^4 - 5*a*b^3*c*e^4 + 5*a^2*b*c^2*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^3*d^4 - 2*a*b*c^3*d^3*e + b^2*c^3*d^2*e^2 + 2*a*c^4*d^2*e^2 - 2* b*c^4*d*e^3 + c^5*e^4)*sqrt(-b^2 + 4*a*c)*e^2) + (b^2*d^2*e - a*c*d^2*e + 2*b*c*d*e^2 + 3*c^2*e^3)*log(abs(-d/(e*x + d) + 1))/(c^3*d^4*e) + 1/2*(2*b *c*d*e + 5*c^2*e^2 - 2*(b*c*d^2*e^2 + 3*c^2*d*e^3)/((e*x + d)*e))/(c^3*d^4 *(d/(e*x + d) - 1)^2)
Time = 105.00 (sec) , antiderivative size = 7144, normalized size of antiderivative = 19.20 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx=\text {Too large to display} \] Input:
int(1/(x^5*(d + e*x)^2*(a + b/x + c/x^2)),x)
Output:
((x*(2*b*d + 3*c*e))/(2*c^2*d^2) - 1/(2*c*d) + (x^2*(3*c^2*e^4 - b^2*d^2*e ^2 + a*b*d^3*e - b*c*d*e^3 + 2*a*c*d^2*e^2))/(c^2*d^3*(a*d^2 + c*e^2 - b*d *e)))/(d*x^2 + e*x^3) - (log(d + e*x)*(3*c*e^6 + 5*a*d^2*e^4 - 4*b*d*e^5)) /(a^2*d^8 + b^2*d^6*e^2 + c^2*d^4*e^4 - 2*a*b*d^7*e + 2*a*c*d^6*e^2 - 2*b* c*d^5*e^3) + (log((((27*a^2*b*c^6*e^11 - 9*a*b^3*c^5*e^11 - a*b^8*d^5*e^6 - a^6*b^3*d^10*e - 36*a^3*c^6*d*e^10 + 2*a^2*b^7*d^6*e^5 - a^3*b^6*d^7*e^4 - a^4*b^5*d^8*e^3 + 2*a^5*b^4*d^9*e^2 - 36*a^4*c^5*d^3*e^8 + 4*a^5*c^4*d^ 5*e^6 + 3*a^6*c^3*d^7*e^4 + a^7*b*c*d^10*e - 39*a^2*b^3*c^4*d^2*e^9 - 15*a ^2*b^4*c^3*d^3*e^8 + 7*a^2*b^5*c^2*d^4*e^7 + 53*a^3*b^2*c^4*d^3*e^8 + 7*a^ 3*b^3*c^3*d^4*e^7 - 33*a^3*b^4*c^2*d^5*e^6 + 20*a^4*b^2*c^3*d^5*e^6 + 33*a ^4*b^3*c^2*d^6*e^5 - 9*a^5*b^2*c^2*d^7*e^4 + 6*a*b^4*c^4*d*e^10 - 2*a*b^7* c*d^4*e^7 + 5*a*b^5*c^3*d^2*e^9 + a*b^6*c^2*d^3*e^8 + 12*a^2*b^6*c*d^5*e^6 + 51*a^3*b*c^5*d^2*e^9 - 16*a^3*b^5*c*d^6*e^5 - 27*a^4*b*c^4*d^4*e^7 + 6* a^4*b^4*c*d^7*e^4 - 19*a^5*b*c^3*d^6*e^5 + 3*a^5*b^3*c*d^8*e^3 - a^6*b*c^2 *d^8*e^3 - 4*a^6*b^2*c*d^9*e^2)/(c^4*d^6*(a*d^2 + c*e^2 - b*d*e)^2) + (((a *e*(12*a*c^5*e^7 - a^3*b^3*d^7 - 3*b^2*c^4*e^7 + b^6*d^4*e^3 - 3*a*b^5*d^5 *e^2 + 3*a^2*b^4*d^6*e + 4*a^4*c^2*d^6*e + b^3*c^3*d*e^6 + b^5*c*d^3*e^4 + 8*a^2*c^4*d^2*e^5 - 8*a^3*c^3*d^4*e^3 + b^4*c^2*d^2*e^5 + 2*a^4*b*c*d^7 - 4*a*b*c^4*d*e^6 + 18*a^2*b^2*c^2*d^4*e^3 - 8*a*b^4*c*d^4*e^3 - 10*a^3*b^2 *c*d^6*e - 6*a*b^2*c^3*d^2*e^5 - 7*a*b^3*c^2*d^3*e^4 + 12*a^2*b*c^3*d^3...
Time = 0.16 (sec) , antiderivative size = 2986, normalized size of antiderivative = 8.03 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)^2} \, dx =\text {Too large to display} \] Input:
int(1/(a+c/x^2+b/x)/x^5/(e*x+d)^2,x)
Output:
(6*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**3*b*c*d**7*x **2 + 6*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**3*b*c*d **6*e*x**3 + 8*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a** 3*c**2*d**6*e*x**2 + 8*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b* *2))*a**3*c**2*d**5*e**2*x**3 - 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt (4*a*c - b**2))*a**2*b**3*d**7*x**2 - 2*sqrt(4*a*c - b**2)*atan((2*a*x + b )/sqrt(4*a*c - b**2))*a**2*b**3*d**6*e*x**3 - 16*sqrt(4*a*c - b**2)*atan(( 2*a*x + b)/sqrt(4*a*c - b**2))*a**2*b**2*c*d**6*e*x**2 - 16*sqrt(4*a*c - b **2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*b**2*c*d**5*e**2*x**3 - 10* sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*b*c**2*d**5*e **2*x**2 - 10*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2 *b*c**2*d**4*e**3*x**3 + 4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**4*d**6*e*x**2 + 4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4 *a*c - b**2))*a*b**4*d**5*e**2*x**3 + 10*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**3*c*d**5*e**2*x**2 + 10*sqrt(4*a*c - b**2)*ata n((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**3*c*d**4*e**3*x**3 - 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b**5*d**5*e**2*x**2 - 2*sqrt(4 *a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b**5*d**4*e**3*x**3 + 4* log(a*x**2 + b*x + c)*a**4*c**2*d**7*x**2 + 4*log(a*x**2 + b*x + c)*a**4*c **2*d**6*e*x**3 - 5*log(a*x**2 + b*x + c)*a**3*b**2*c*d**7*x**2 - 5*log...