Integrand size = 25, antiderivative size = 291 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx=-\frac {1}{c d^2 x}-\frac {e^3}{d^2 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac {\left (2 a^3 c d^2-b^4 e^2+2 a b^2 e (b d+2 c e)-a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac {(b d+2 c e) \log (x)}{c^2 d^3}+\frac {e^3 \left (4 a d^2-e (3 b d-2 c e)\right ) \log (d+e x)}{d^3 \left (a d^2-e (b d-c e)\right )^2}+\frac {(a d-b e) \left (a b d-b^2 e+2 a c e\right ) \log \left (c+b x+a x^2\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )^2} \] Output:
-1/c/d^2/x-e^3/d^2/(a*d^2-e*(b*d-c*e))/(e*x+d)+(2*a^3*c*d^2-b^4*e^2+2*a*b^ 2*e*(b*d+2*c*e)-a^2*(b^2*d^2+6*b*c*d*e+2*c^2*e^2))*arctanh((2*a*x+b)/(-4*a *c+b^2)^(1/2))/c^2/(-4*a*c+b^2)^(1/2)/(a*d^2-e*(b*d-c*e))^2-(b*d+2*c*e)*ln (x)/c^2/d^3+e^3*(4*a*d^2-e*(3*b*d-2*c*e))*ln(e*x+d)/d^3/(a*d^2-e*(b*d-c*e) )^2+1/2*(a*d-b*e)*(a*b*d+2*a*c*e-b^2*e)*ln(a*x^2+b*x+c)/c^2/(a*d^2-e*(b*d- c*e))^2
Time = 0.34 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx=-\frac {1}{c d^2 x}-\frac {e^3}{d^2 \left (a d^2+e (-b d+c e)\right ) (d+e x)}+\frac {\left (-2 a^3 c d^2+b^4 e^2-2 a b^2 e (b d+2 c e)+a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )}{c^2 \sqrt {-b^2+4 a c} \left (a d^2+e (-b d+c e)\right )^2}-\frac {(b d+2 c e) \log (x)}{c^2 d^3}+\frac {e^3 \left (4 a d^2+e (-3 b d+2 c e)\right ) \log (d+e x)}{d^3 \left (a d^2+e (-b d+c e)\right )^2}+\frac {(a d-b e) \left (a b d-b^2 e+2 a c e\right ) \log (c+x (b+a x))}{2 c^2 \left (a d^2+e (-b d+c e)\right )^2} \] Input:
Integrate[1/((a + c/x^2 + b/x)*x^4*(d + e*x)^2),x]
Output:
-(1/(c*d^2*x)) - e^3/(d^2*(a*d^2 + e*(-(b*d) + c*e))*(d + e*x)) + ((-2*a^3 *c*d^2 + b^4*e^2 - 2*a*b^2*e*(b*d + 2*c*e) + a^2*(b^2*d^2 + 6*b*c*d*e + 2* c^2*e^2))*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]])/(c^2*Sqrt[-b^2 + 4*a*c]* (a*d^2 + e*(-(b*d) + c*e))^2) - ((b*d + 2*c*e)*Log[x])/(c^2*d^3) + (e^3*(4 *a*d^2 + e*(-3*b*d + 2*c*e))*Log[d + e*x])/(d^3*(a*d^2 + e*(-(b*d) + c*e)) ^2) + ((a*d - b*e)*(a*b*d - b^2*e + 2*a*c*e)*Log[c + x*(b + a*x)])/(2*c^2* (a*d^2 + e*(-(b*d) + c*e))^2)
Time = 0.72 (sec) , antiderivative size = 291, 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.
\(\displaystyle \int \frac {1}{x^4 (d+e x)^2 \left (a+\frac {b}{x}+\frac {c}{x^2}\right )} \, dx\) |
\(\Big \downarrow \) 1893 |
\(\displaystyle \int \frac {1}{x^2 (d+e x)^2 \left (a x^2+b x+c\right )}dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {-a^3 c d^2+a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )+a x (a d-b e) \left (a b d+2 a c e+b^2 (-e)\right )-a b^2 e (2 b d+3 c e)+b^4 e^2}{c^2 \left (a x^2+b x+c\right ) \left (a d^2-e (b d-c e)\right )^2}+\frac {e^4}{d^2 (d+e x)^2 \left (a d^2-e (b d-c e)\right )}+\frac {e^4 \left (4 a d^2-e (3 b d-2 c e)\right )}{d^3 (d+e x) \left (a d^2-e (b d-c e)\right )^2}+\frac {-b d-2 c e}{c^2 d^3 x}+\frac {1}{c d^2 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right ) \left (2 a^3 c d^2-a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )+2 a b^2 e (b d+2 c e)+b^4 \left (-e^2\right )\right )}{c^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac {(a d-b e) \left (a b d+2 a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )^2}-\frac {e^3}{d^2 (d+e x) \left (a d^2-e (b d-c e)\right )}+\frac {e^3 \log (d+e x) \left (4 a d^2-e (3 b d-2 c e)\right )}{d^3 \left (a d^2-e (b d-c e)\right )^2}-\frac {\log (x) (b d+2 c e)}{c^2 d^3}-\frac {1}{c d^2 x}\) |
Input:
Int[1/((a + c/x^2 + b/x)*x^4*(d + e*x)^2),x]
Output:
-(1/(c*d^2*x)) - e^3/(d^2*(a*d^2 - e*(b*d - c*e))*(d + e*x)) + ((2*a^3*c*d ^2 - b^4*e^2 + 2*a*b^2*e*(b*d + 2*c*e) - a^2*(b^2*d^2 + 6*b*c*d*e + 2*c^2* e^2))*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c]*(a*d^ 2 - e*(b*d - c*e))^2) - ((b*d + 2*c*e)*Log[x])/(c^2*d^3) + (e^3*(4*a*d^2 - e*(3*b*d - 2*c*e))*Log[d + e*x])/(d^3*(a*d^2 - e*(b*d - c*e))^2) + ((a*d - b*e)*(a*b*d - b^2*e + 2*a*c*e)*Log[c + b*x + a*x^2])/(2*c^2*(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 = 0.18 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.19
method | result | size |
default | \(\frac {\frac {\left (a^{3} b \,d^{2}+2 a^{3} c d e -2 a^{2} b^{2} d e -2 a^{2} b c \,e^{2}+a \,b^{3} e^{2}\right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (-a^{3} c \,d^{2}+a^{2} b^{2} d^{2}+4 a^{2} b c d e +a^{2} c^{2} e^{2}-2 a \,b^{3} d e -3 a \,b^{2} c \,e^{2}+b^{4} e^{2}-\frac {\left (a^{3} b \,d^{2}+2 a^{3} c d e -2 a^{2} b^{2} d e -2 a^{2} b c \,e^{2}+a \,b^{3} e^{2}\right ) b}{2 a}\right ) \arctan \left (\frac {2 x a +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right )^{2} c^{2}}-\frac {e^{3}}{d^{2} \left (a \,d^{2}-b d e +c \,e^{2}\right ) \left (e x +d \right )}+\frac {e^{3} \left (4 a \,d^{2}-3 b d e +2 c \,e^{2}\right ) \ln \left (e x +d \right )}{d^{3} \left (a \,d^{2}-b d e +c \,e^{2}\right )^{2}}-\frac {1}{c \,d^{2} x}+\frac {\left (-b d -2 c e \right ) \ln \left (x \right )}{c^{2} d^{3}}\) | \(346\) |
risch | \(\text {Expression too large to display}\) | \(2377\) |
Input:
int(1/(a+c/x^2+b/x)/x^4/(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
1/(a*d^2-b*d*e+c*e^2)^2/c^2*(1/2*(a^3*b*d^2+2*a^3*c*d*e-2*a^2*b^2*d*e-2*a^ 2*b*c*e^2+a*b^3*e^2)/a*ln(a*x^2+b*x+c)+2*(-a^3*c*d^2+a^2*b^2*d^2+4*a^2*b*c *d*e+a^2*c^2*e^2-2*a*b^3*d*e-3*a*b^2*c*e^2+b^4*e^2-1/2*(a^3*b*d^2+2*a^3*c* d*e-2*a^2*b^2*d*e-2*a^2*b*c*e^2+a*b^3*e^2)*b/a)/(4*a*c-b^2)^(1/2)*arctan(( 2*a*x+b)/(4*a*c-b^2)^(1/2)))-e^3/d^2/(a*d^2-b*d*e+c*e^2)/(e*x+d)+e^3*(4*a* d^2-3*b*d*e+2*c*e^2)/d^3/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)-1/c/d^2/x+(-b*d-2 *c*e)/c^2/d^3*ln(x)
Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx=\text {Timed out} \] Input:
integrate(1/(a+c/x^2+b/x)/x^4/(e*x+d)^2,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx=\text {Timed out} \] Input:
integrate(1/(a+c/x**2+b/x)/x**4/(e*x+d)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+c/x^2+b/x)/x^4/(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 = 493, normalized size of antiderivative = 1.69 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx=-\frac {e^{7}}{{\left (a d^{4} e^{4} - b d^{3} e^{5} + c d^{2} e^{6}\right )} {\left (e x + d\right )}} + \frac {{\left (a^{2} b d^{2} - 2 \, a b^{2} d e + 2 \, a^{2} c d e + b^{3} e^{2} - 2 \, a b c e^{2}\right )} \log \left (-a + \frac {2 \, a d}{e x + d} - \frac {a d^{2}}{{\left (e x + d\right )}^{2}} - \frac {b e}{e x + d} + \frac {b d e}{{\left (e x + d\right )}^{2}} - \frac {c e^{2}}{{\left (e x + d\right )}^{2}}\right )}{2 \, {\left (a^{2} c^{2} d^{4} - 2 \, a b c^{2} d^{3} e + b^{2} c^{2} d^{2} e^{2} + 2 \, a c^{3} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + c^{4} e^{4}\right )}} - \frac {{\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} c d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 6 \, a^{2} b c d e^{3} + b^{4} e^{4} - 4 \, a b^{2} c e^{4} + 2 \, a^{2} c^{2} e^{4}\right )} \arctan \left (-\frac {2 \, a d - \frac {2 \, a d^{2}}{e x + d} - b e + \frac {2 \, b d e}{e x + d} - \frac {2 \, c e^{2}}{e x + d}}{\sqrt {-b^{2} + 4 \, a c} e}\right )}{{\left (a^{2} c^{2} d^{4} - 2 \, a b c^{2} d^{3} e + b^{2} c^{2} d^{2} e^{2} + 2 \, a c^{3} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + c^{4} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c} e^{2}} + \frac {e}{c d^{3} {\left (\frac {d}{e x + d} - 1\right )}} - \frac {{\left (b d e + 2 \, c e^{2}\right )} \log \left ({\left | -\frac {d}{e x + d} + 1 \right |}\right )}{c^{2} d^{3} e} \] Input:
integrate(1/(a+c/x^2+b/x)/x^4/(e*x+d)^2,x, algorithm="giac")
Output:
-e^7/((a*d^4*e^4 - b*d^3*e^5 + c*d^2*e^6)*(e*x + d)) + 1/2*(a^2*b*d^2 - 2* a*b^2*d*e + 2*a^2*c*d*e + b^3*e^2 - 2*a*b*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^2*d^4 - 2*a*b*c^2*d^3*e + b^2*c^2*d^2*e^2 + 2*a*c^3*d^2*e^2 - 2* b*c^3*d*e^3 + c^4*e^4) - (a^2*b^2*d^2*e^2 - 2*a^3*c*d^2*e^2 - 2*a*b^3*d*e^ 3 + 6*a^2*b*c*d*e^3 + b^4*e^4 - 4*a*b^2*c*e^4 + 2*a^2*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))/(sq rt(-b^2 + 4*a*c)*e))/((a^2*c^2*d^4 - 2*a*b*c^2*d^3*e + b^2*c^2*d^2*e^2 + 2 *a*c^3*d^2*e^2 - 2*b*c^3*d*e^3 + c^4*e^4)*sqrt(-b^2 + 4*a*c)*e^2) + e/(c*d ^3*(d/(e*x + d) - 1)) - (b*d*e + 2*c*e^2)*log(abs(-d/(e*x + d) + 1))/(c^2* d^3*e)
Time = 69.10 (sec) , antiderivative size = 4948, normalized size of antiderivative = 17.00 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx=\text {Too large to display} \] Input:
int(1/(x^4*(d + e*x)^2*(a + b/x + c/x^2)),x)
Output:
(log(d + e*x)*(2*c*e^5 + 4*a*d^2*e^3 - 3*b*d*e^4))/(a^2*d^7 + b^2*d^5*e^2 + c^2*d^3*e^4 - 2*a*b*d^6*e + 2*a*c*d^5*e^2 - 2*b*c*d^4*e^3) - (1/(c*d) + (x*(2*c*e^3 + a*d^2*e - b*d*e^2))/(c*d^2*(a*d^2 + c*e^2 - b*d*e)))/(d*x + e*x^2) - (log((((a*e*(a^5*b*d^8 + 4*b^3*c^3*e^8 + b^6*d^3*e^5 - 2*a*b^5*d^ 4*e^4 - 2*a^4*b^2*d^7*e + 16*a^2*c^4*d*e^7 - 4*b^4*c^2*d*e^7 - b^5*c*d^2*e ^6 + a^2*b^4*d^5*e^3 + a^3*b^3*d^6*e^2 + 16*a^3*c^3*d^3*e^5 + a^4*c^2*d^5* e^3 - 12*a*b*c^4*e^8 + 2*a^5*c*d^7*e - 16*a^2*b^2*c^2*d^3*e^5 + 4*a*b^2*c^ 3*d*e^7 - 2*a^4*b*c*d^6*e^2 + 13*a*b^3*c^2*d^2*e^6 - 20*a^2*b*c^3*d^2*e^6 + a^2*b^3*c*d^4*e^4 + 8*a^3*b*c^2*d^4*e^4))/(c^2*d^4*(a*d^2 + c*e^2 - b*d* e)^2) - (((a*e*(a^4*c*d^6 + 8*a*c^4*e^6 - a^3*b^2*d^6 - 2*b^2*c^3*e^6 + b^ 5*d^3*e^3 - 3*a*b^4*d^4*e^2 + 3*a^2*b^3*d^5*e + b^3*c^2*d*e^5 + b^4*c*d^2* e^4 + 8*a^2*c^3*d^2*e^4 - 7*a^3*c^2*d^4*e^2 - 4*a*b*c^3*d*e^5 - 7*a^3*b*c* d^5*e - 7*a*b^3*c*d^3*e^3 - 6*a*b^2*c^2*d^2*e^4 + 12*a^2*b*c^2*d^3*e^3 + 1 2*a^2*b^2*c*d^4*e^2))/(c*d^2*(a*d^2 + c*e^2 - b*d*e)) + (a*e*(b^5*e^2 + b^ 4*e^2*(b^2 - 4*a*c)^(1/2) + a^2*b^3*d^2 + 8*a^2*b*c^2*e^2 + a^2*b^2*d^2*(b ^2 - 4*a*c)^(1/2) + 2*a^2*c^2*e^2*(b^2 - 4*a*c)^(1/2) - 2*a*b^4*d*e - 4*a^ 3*b*c*d^2 - 6*a*b^3*c*e^2 - 8*a^3*c^2*d*e - 2*a^3*c*d^2*(b^2 - 4*a*c)^(1/2 ) + 10*a^2*b^2*c*d*e - 4*a*b^2*c*e^2*(b^2 - 4*a*c)^(1/2) - 2*a*b^3*d*e*(b^ 2 - 4*a*c)^(1/2) + 6*a^2*b*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^...
Time = 0.16 (sec) , antiderivative size = 2345, normalized size of antiderivative = 8.06 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)^2} \, dx =\text {Too large to display} \] Input:
int(1/(a+c/x^2+b/x)/x^4/(e*x+d)^2,x)
Output:
( - 4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**3*c*d**6* x - 4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**3*c*d**5* e*x**2 + 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*b* *2*d**6*x + 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2 *b**2*d**5*e*x**2 + 12*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b* *2))*a**2*b*c*d**5*e*x + 12*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a**2*b*c*d**4*e**2*x**2 + 4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/ sqrt(4*a*c - b**2))*a**2*c**2*d**4*e**2*x + 4*sqrt(4*a*c - b**2)*atan((2*a *x + b)/sqrt(4*a*c - b**2))*a**2*c**2*d**3*e**3*x**2 - 4*sqrt(4*a*c - b**2 )*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**3*d**5*e*x - 4*sqrt(4*a*c - b* *2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**3*d**4*e**2*x**2 - 8*sqrt(4* a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**2*c*d**4*e**2*x - 8* sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b**2*c*d**3*e**3 *x**2 + 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b**4*d** 4*e**2*x + 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b**4* d**3*e**3*x**2 + 4*log(a*x**2 + b*x + c)*a**3*b*c*d**6*x + 4*log(a*x**2 + b*x + c)*a**3*b*c*d**5*e*x**2 + 8*log(a*x**2 + b*x + c)*a**3*c**2*d**5*e*x + 8*log(a*x**2 + b*x + c)*a**3*c**2*d**4*e**2*x**2 - log(a*x**2 + b*x + c )*a**2*b**3*d**6*x - log(a*x**2 + b*x + c)*a**2*b**3*d**5*e*x**2 - 10*log( a*x**2 + b*x + c)*a**2*b**2*c*d**5*e*x - 10*log(a*x**2 + b*x + c)*a**2*...