Integrand size = 25, antiderivative size = 252 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)} \, dx=-\frac {1}{2 c d x^2}+\frac {b d+c e}{c^2 d^2 x}-\frac {\left (b^4 e+a^2 c (3 b d+2 c e)-a b^2 (b d+4 c e)\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\left (b^2 d^2+b c d e-c \left (a d^2-c e^2\right )\right ) \log (x)}{c^3 d^3}-\frac {e^4 \log (d+e x)}{d^3 \left (a d^2-e (b d-c e)\right )}+\frac {\left (a^2 c d+b^3 e-a b (b d+2 c e)\right ) \log \left (c+b x+a x^2\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )} \]
-1/2/c/d/x^2+(b*d+c*e)/c^2/d^2/x+(b^2*d^2+b*c*d*e-c*(a*d^2-c*e^2))*ln(x)/c ^3/d^3-e^4*ln(e*x+d)/d^3/(a*d^2-e*(b*d-c*e))+1/2*(a^2*c*d+b^3*e-a*b*(b*d+2 *c*e))*ln(a*x^2+b*x+c)/c^3/(a*d^2-e*(b*d-c*e))-(b^4*e+a^2*c*(3*b*d+2*c*e)- a*b^2*(b*d+4*c*e))*arctanh((2*a*x+b)/(-4*a*c+b^2)^(1/2))/c^3/(a*d^2-e*(b*d -c*e))/(-4*a*c+b^2)^(1/2)
Time = 0.14 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)} \, dx=-\frac {1}{2 c d x^2}+\frac {b d+c e}{c^2 d^2 x}-\frac {\left (b^4 e+a^2 c (3 b d+2 c e)-a b^2 (b d+4 c e)\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )}{c^3 \sqrt {-b^2+4 a c} \left (-a d^2+e (b d-c e)\right )}+\frac {\left (b^2 d^2+b c d e+c \left (-a d^2+c e^2\right )\right ) \log (x)}{c^3 d^3}-\frac {e^4 \log (d+e x)}{a d^5+d^3 e (-b d+c e)}+\frac {\left (a^2 c d+b^3 e-a b (b d+2 c e)\right ) \log (c+x (b+a x))}{2 c^3 \left (a d^2+e (-b d+c e)\right )} \]
-1/2*1/(c*d*x^2) + (b*d + c*e)/(c^2*d^2*x) - ((b^4*e + a^2*c*(3*b*d + 2*c* e) - a*b^2*(b*d + 4*c*e))*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]])/(c^3*Sqr t[-b^2 + 4*a*c]*(-(a*d^2) + e*(b*d - c*e))) + ((b^2*d^2 + b*c*d*e + c*(-(a *d^2) + c*e^2))*Log[x])/(c^3*d^3) - (e^4*Log[d + e*x])/(a*d^5 + d^3*e*(-(b *d) + c*e)) + ((a^2*c*d + b^3*e - a*b*(b*d + 2*c*e))*Log[c + x*(b + a*x)]) /(2*c^3*(a*d^2 + e*(-(b*d) + c*e)))
Time = 0.60 (sec) , antiderivative size = 252, 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.
\(\displaystyle \int \frac {1}{x^5 (d+e x) \left (a+\frac {b}{x}+\frac {c}{x^2}\right )} \, dx\) |
\(\Big \downarrow \) 1893 |
\(\displaystyle \int \frac {1}{x^3 (d+e x) \left (a x^2+b x+c\right )}dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {a x \left (a^2 c d-a b (b d+2 c e)+b^3 e\right )+a^2 c (2 b d+c e)-a b^2 (b d+3 c e)+b^4 e}{c^3 \left (a x^2+b x+c\right ) \left (a d^2-e (b d-c e)\right )}+\frac {-c \left (a d^2-c e^2\right )+b^2 d^2+b c d e}{c^3 d^3 x}+\frac {e^5}{d^3 (d+e x) \left (e (b d-c e)-a d^2\right )}+\frac {-b d-c e}{c^2 d^2 x^2}+\frac {1}{c d x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right ) \left (a^2 c (3 b d+2 c e)-a b^2 (b d+4 c e)+b^4 e\right )}{c^3 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\left (a^2 c d-a b (b d+2 c e)+b^3 e\right ) \log \left (a x^2+b x+c\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )}+\frac {\log (x) \left (-c \left (a d^2-c e^2\right )+b^2 d^2+b c d e\right )}{c^3 d^3}-\frac {e^4 \log (d+e x)}{d^3 \left (a d^2-e (b d-c e)\right )}+\frac {b d+c e}{c^2 d^2 x}-\frac {1}{2 c d x^2}\) |
-1/2*1/(c*d*x^2) + (b*d + c*e)/(c^2*d^2*x) - ((b^4*e + a^2*c*(3*b*d + 2*c* e) - a*b^2*(b*d + 4*c*e))*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(c^3*Sqr t[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))) + ((b^2*d^2 + b*c*d*e - c*(a*d^2 - c*e^2))*Log[x])/(c^3*d^3) - (e^4*Log[d + e*x])/(d^3*(a*d^2 - e*(b*d - c*e ))) + ((a^2*c*d + b^3*e - a*b*(b*d + 2*c*e))*Log[c + b*x + a*x^2])/(2*c^3* (a*d^2 - e*(b*d - c*e)))
3.1.69.3.1 Defintions of rubi rules used
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 = 0.74 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.10
method | result | size |
default | \(-\frac {1}{2 c d \,x^{2}}-\frac {-b d -e c}{x \,c^{2} d^{2}}+\frac {\left (-d^{2} a c +b^{2} d^{2}+b c d e +e^{2} c^{2}\right ) \ln \left (x \right )}{d^{3} c^{3}}+\frac {\frac {\left (a^{3} c d -a^{2} b^{2} d -2 a^{2} b c e +a \,b^{3} e \right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (2 a^{2} b c d +a^{2} c^{2} e -a \,b^{3} d -3 a \,b^{2} c e +b^{4} e -\frac {\left (a^{3} c d -a^{2} b^{2} d -2 a^{2} b c e +a \,b^{3} e \right ) b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right ) c^{3}}-\frac {e^{4} \ln \left (e x +d \right )}{d^{3} \left (a \,d^{2}-b d e +c \,e^{2}\right )}\) | \(277\) |
risch | \(\text {Expression too large to display}\) | \(1315\) |
-1/2/c/d/x^2-(-b*d-c*e)/x/c^2/d^2+1/d^3/c^3*(-a*c*d^2+b^2*d^2+b*c*d*e+c^2* e^2)*ln(x)+1/(a*d^2-b*d*e+c*e^2)/c^3*(1/2*(a^3*c*d-a^2*b^2*d-2*a^2*b*c*e+a *b^3*e)/a*ln(a*x^2+b*x+c)+2*(2*a^2*b*c*d+a^2*c^2*e-a*b^3*d-3*a*b^2*c*e+b^4 *e-1/2*(a^3*c*d-a^2*b^2*d-2*a^2*b*c*e+a*b^3*e)*b/a)/(4*a*c-b^2)^(1/2)*arct an((2*a*x+b)/(4*a*c-b^2)^(1/2)))-e^4/d^3/(a*d^2-b*d*e+c*e^2)*ln(e*x+d)
Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.32 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)} \, dx=-\frac {e^{5} \log \left ({\left | e x + d \right |}\right )}{a d^{5} e - b d^{4} e^{2} + c d^{3} e^{3}} - \frac {{\left (a b^{2} d - a^{2} c d - b^{3} e + 2 \, a b c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a c^{3} d^{2} - b c^{3} d e + c^{4} e^{2}\right )}} - \frac {{\left (a b^{3} d - 3 \, a^{2} b c d - b^{4} e + 4 \, a b^{2} c e - 2 \, a^{2} c^{2} e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a c^{3} d^{2} - b c^{3} d e + c^{4} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {{\left (b^{2} d^{2} - a c d^{2} + b c d e + c^{2} e^{2}\right )} \log \left ({\left | x \right |}\right )}{c^{3} d^{3}} - \frac {c^{2} d^{2} - 2 \, {\left (b c d^{2} + c^{2} d e\right )} x}{2 \, c^{3} d^{3} x^{2}} \]
-e^5*log(abs(e*x + d))/(a*d^5*e - b*d^4*e^2 + c*d^3*e^3) - 1/2*(a*b^2*d - a^2*c*d - b^3*e + 2*a*b*c*e)*log(a*x^2 + b*x + c)/(a*c^3*d^2 - b*c^3*d*e + c^4*e^2) - (a*b^3*d - 3*a^2*b*c*d - b^4*e + 4*a*b^2*c*e - 2*a^2*c^2*e)*ar ctan((2*a*x + b)/sqrt(-b^2 + 4*a*c))/((a*c^3*d^2 - b*c^3*d*e + c^4*e^2)*sq rt(-b^2 + 4*a*c)) + (b^2*d^2 - a*c*d^2 + b*c*d*e + c^2*e^2)*log(abs(x))/(c ^3*d^3) - 1/2*(c^2*d^2 - 2*(b*c*d^2 + c^2*d*e)*x)/(c^3*d^3*x^2)
Time = 28.67 (sec) , antiderivative size = 3530, normalized size of antiderivative = 14.01 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^5 (d+e x)} \, dx=\text {Too large to display} \]
(log((a^4*e^4*(b^2*d^2 + c^2*e^2 - a*c*d^2 + b*c*d*e))/(c^4*d^4) - (((((a* e*(a^2*b^3*d^5 - 4*a*c^4*e^5 + b^2*c^3*e^5 + b^5*d^3*e^2 - 3*a^3*c^2*d^4*e + b^3*c^2*d*e^4 + b^4*c*d^2*e^3 + 4*a^2*c^3*d^2*e^3 - 2*a^3*b*c*d^5 - 2*a *b^4*d^4*e - 4*a*b*c^3*d*e^4 - 6*a*b^3*c*d^3*e^2 + 7*a^2*b^2*c*d^4*e - 5*a *b^2*c^2*d^2*e^3 + 8*a^2*b*c^2*d^3*e^2))/(c^2*d^2) + (a*e*x*(2*a^3*b^2*d^5 - 3*a^4*c*d^5 + 2*b^3*c^2*e^5 + 2*b^5*d^2*e^3 - 2*a*b^4*d^3*e^2 - 2*a^2*b ^3*d^4*e + 8*a^2*c^3*d*e^4 - 8*a^3*c^2*d^3*e^2 - 8*a*b*c^3*e^5 + b^4*c*d*e ^4 + 4*a^3*b*c*d^4*e - 6*a*b^2*c^2*d*e^4 - 12*a*b^3*c*d^2*e^3 + 16*a^2*b*c ^2*d^2*e^3 + 10*a^2*b^2*c*d^3*e^2))/(c^2*d^2) - (a*e*(b^4*e*(b^2 - 4*a*c)^ (1/2) - b^5*e + 4*a^3*c^2*d + a*b^4*d + 6*a*b^3*c*e - a*b^3*d*(b^2 - 4*a*c )^(1/2) - 5*a^2*b^2*c*d - 8*a^2*b*c^2*e + 2*a^2*c^2*e*(b^2 - 4*a*c)^(1/2) + 3*a^2*b*c*d*(b^2 - 4*a*c)^(1/2) - 4*a*b^2*c*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^3*(4*a*c - b^2)*(a*d^2 + c*e^2 - b*d*e)))*(b ^4*e*(b^2 - 4*a*c)^(1/2) - b^5*e + 4*a^3*c^2*d + a*b^4*d + 6*a*b^3*c*e - a *b^3*d*(b^2 - 4*a*c)^(1/2) - 5*a^2*b^2*c*d - 8*a^2*b*c^2*e + 2*a^2*c^2*e*( b^2 - 4*a*c)^(1/2) + 3*a^2*b*c*d*(b^2 - 4*a*c)^(1/2) - 4*a*b^2*c*e*(b^2...