Integrand size = 23, antiderivative size = 176 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\frac {x}{a e}+\frac {\left (b^3 d-3 a b c d-b^2 c e+2 a c^2 e\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {\left (b^2 d-a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )} \]
x/a/e-d^3*ln(e*x+d)/e^2/(a*d^2-e*(b*d-c*e))+1/2*(-a*c*d+b^2*d-b*c*e)*ln(a* x^2+b*x+c)/a^2/(a*d^2-e*(b*d-c*e))+(-3*a*b*c*d+2*a*c^2*e+b^3*d-b^2*c*e)*ar ctanh((2*a*x+b)/(-4*a*c+b^2)^(1/2))/a^2/(a*d^2-e*(b*d-c*e))/(-4*a*c+b^2)^( 1/2)
Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.01 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\frac {x}{a e}+\frac {\left (b^3 d-3 a b c d-b^2 c e+2 a c^2 e\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )}{a^2 \sqrt {-b^2+4 a c} \left (-a d^2+b d e-c e^2\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-b d e+c e^2\right )}+\frac {\left (b^2 d-a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-b d e+c e^2\right )} \]
x/(a*e) + ((b^3*d - 3*a*b*c*d - b^2*c*e + 2*a*c^2*e)*ArcTan[(b + 2*a*x)/Sq rt[-b^2 + 4*a*c]])/(a^2*Sqrt[-b^2 + 4*a*c]*(-(a*d^2) + b*d*e - c*e^2)) - ( d^3*Log[d + e*x])/(e^2*(a*d^2 - b*d*e + c*e^2)) + ((b^2*d - a*c*d - b*c*e) *Log[c + b*x + a*x^2])/(2*a^2*(a*d^2 - b*d*e + c*e^2))
Time = 0.43 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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.
| \(\displaystyle \int \frac {x}{(d+e x) \left (a+\frac {b}{x}+\frac {c}{x^2}\right )} \, dx\) |
| \(\Big \downarrow \) 1893 |
| \(\displaystyle \int \frac {x^3}{(d+e x) \left (a x^2+b x+c\right )}dx\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \int \left (\frac {x \left (-a c d+b^2 d-b c e\right )+c (b d-c e)}{a \left (a x^2+b x+c\right ) \left (a d^2-e (b d-c e)\right )}+\frac {d^3}{e (d+e x) \left (e (b d-c e)-a d^2\right )}+\frac {1}{a e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\text {arctanh}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right ) \left (-3 a b c d+2 a c^2 e+b^3 d-b^2 c e\right )}{a^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {x}{a e}\) |
x/(a*e) + ((b^3*d - 3*a*b*c*d - b^2*c*e + 2*a*c^2*e)*ArcTanh[(b + 2*a*x)/S qrt[b^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))) - (d^3* Log[d + e*x])/(e^2*(a*d^2 - e*(b*d - c*e))) + ((b^2*d - a*c*d - b*c*e)*Log [c + b*x + a*x^2])/(2*a^2*(a*d^2 - e*(b*d - c*e)))
3.1.63.3.1 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 = 0.82 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.93
| method | result | size |
| default | \(\frac {x}{a e}+\frac {\frac {\left (-a c d +b^{2} d -e b c \right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (b c d -e \,c^{2}-\frac {\left (-a c d +b^{2} d -e b c \right ) b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right ) a}-\frac {d^{3} \ln \left (e x +d \right )}{e^{2} \left (a \,d^{2}-b d e +c \,e^{2}\right )}\) | \(164\) |
| risch | \(\text {Expression too large to display}\) | \(15838\) |
x/a/e+1/(a*d^2-b*d*e+c*e^2)/a*(1/2*(-a*c*d+b^2*d-b*c*e)/a*ln(a*x^2+b*x+c)+ 2*(b*c*d-e*c^2-1/2*(-a*c*d+b^2*d-b*c*e)*b/a)/(4*a*c-b^2)^(1/2)*arctan((2*a *x+b)/(4*a*c-b^2)^(1/2)))-1/e^2*d^3/(a*d^2-b*d*e+c*e^2)*ln(e*x+d)
Time = 3.16 (sec) , antiderivative size = 596, normalized size of antiderivative = 3.39 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\left [-\frac {2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) - {\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} - {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, a^{2} x^{2} + 2 \, a b x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, a x + b\right )}}{a x^{2} + b x + c}\right ) - 2 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )} x - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} - {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )}}, -\frac {2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) - 2 \, {\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} - {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right ) - 2 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )} x - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} - {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )}}\right ] \]
[-1/2*(2*(a^2*b^2 - 4*a^3*c)*d^3*log(e*x + d) - ((b^3 - 3*a*b*c)*d*e^2 - ( b^2*c - 2*a*c^2)*e^3)*sqrt(b^2 - 4*a*c)*log((2*a^2*x^2 + 2*a*b*x + b^2 - 2 *a*c + sqrt(b^2 - 4*a*c)*(2*a*x + b))/(a*x^2 + b*x + c)) - 2*((a^2*b^2 - 4 *a^3*c)*d^2*e - (a*b^3 - 4*a^2*b*c)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3)*x - ((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d*e^2 - (b^3*c - 4*a*b*c^2)*e^3)*log(a*x^2 + b*x + c))/((a^3*b^2 - 4*a^4*c)*d^2*e^2 - (a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^2*b^2*c - 4*a^3*c^2)*e^4), -1/2*(2*(a^2*b^2 - 4*a^3*c)*d^3*log(e*x + d) - 2*((b^3 - 3*a*b*c)*d*e^2 - (b^2*c - 2*a*c^2)*e^3)*sqrt(-b^2 + 4*a*c)*ar ctan(-sqrt(-b^2 + 4*a*c)*(2*a*x + b)/(b^2 - 4*a*c)) - 2*((a^2*b^2 - 4*a^3* c)*d^2*e - (a*b^3 - 4*a^2*b*c)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3)*x - ((b^ 4 - 5*a*b^2*c + 4*a^2*c^2)*d*e^2 - (b^3*c - 4*a*b*c^2)*e^3)*log(a*x^2 + b* x + c))/((a^3*b^2 - 4*a^4*c)*d^2*e^2 - (a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^2* b^2*c - 4*a^3*c^2)*e^4)]
Timed out. \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.33 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.05 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=-\frac {d^{3} \log \left ({\left | e x + d \right |}\right )}{a d^{2} e^{2} - b d e^{3} + c e^{4}} + \frac {{\left (b^{2} d - a c d - b c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )}} - \frac {{\left (b^{3} d - 3 \, a b c d - b^{2} c e + 2 \, a c^{2} e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {x}{a e} \]
-d^3*log(abs(e*x + d))/(a*d^2*e^2 - b*d*e^3 + c*e^4) + 1/2*(b^2*d - a*c*d - b*c*e)*log(a*x^2 + b*x + c)/(a^3*d^2 - a^2*b*d*e + a^2*c*e^2) - (b^3*d - 3*a*b*c*d - b^2*c*e + 2*a*c^2*e)*arctan((2*a*x + b)/sqrt(-b^2 + 4*a*c))/( (a^3*d^2 - a^2*b*d*e + a^2*c*e^2)*sqrt(-b^2 + 4*a*c)) + x/(a*e)
Time = 10.18 (sec) , antiderivative size = 1367, normalized size of antiderivative = 7.77 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\frac {x}{a\,e}-\frac {\ln \left (c^3\,e^5\,\sqrt {b^2-4\,a\,c}-b\,c^3\,e^5-4\,a^3\,c\,d^5+a^2\,b^2\,d^5+b^4\,d^3\,e^2+3\,b^2\,c^2\,d\,e^4-3\,b^3\,c\,d^2\,e^3-b^3\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^3\,e^2-6\,a\,c^3\,d\,e^4-2\,a\,c^3\,e^5\,x-a^2\,b\,d^5\,\sqrt {b^2-4\,a\,c}-2\,a^3\,d^5\,x\,\sqrt {b^2-4\,a\,c}-8\,a^3\,c\,d^4\,e\,x+4\,a^2\,c\,d^4\,e\,\sqrt {b^2-4\,a\,c}-3\,b\,c^2\,d\,e^4\,\sqrt {b^2-4\,a\,c}+9\,a\,b\,c^2\,d^2\,e^3-5\,a\,b^2\,c\,d^3\,e^2+2\,a^2\,b^2\,d^4\,e\,x-3\,a\,c^2\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}+3\,b^2\,c\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^2\,e^3\,x-2\,a\,b^2\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}+3\,a^2\,c\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,b\,c^2\,d\,e^4\,x+a\,b\,c\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}+2\,a^2\,b\,d^4\,e\,x\,\sqrt {b^2-4\,a\,c}-3\,a\,c^2\,d\,e^4\,x\,\sqrt {b^2-4\,a\,c}-3\,a\,b^2\,c\,d^2\,e^3\,x+a^2\,b\,c\,d^3\,e^2\,x+3\,a\,b\,c\,d^2\,e^3\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^4\,d-b^3\,d\,\sqrt {b^2-4\,a\,c}+4\,a^2\,c^2\,d-b^3\,c\,e-5\,a\,b^2\,c\,d+4\,a\,b\,c^2\,e-2\,a\,c^2\,e\,\sqrt {b^2-4\,a\,c}+b^2\,c\,e\,\sqrt {b^2-4\,a\,c}+3\,a\,b\,c\,d\,\sqrt {b^2-4\,a\,c}\right )}{2\,\left (4\,a^4\,c\,d^2-a^3\,b^2\,d^2-4\,a^3\,b\,c\,d\,e+4\,a^3\,c^2\,e^2+a^2\,b^3\,d\,e-a^2\,b^2\,c\,e^2\right )}-\frac {\ln \left (a^2\,b^2\,d^5-b\,c^3\,e^5-c^3\,e^5\,\sqrt {b^2-4\,a\,c}-4\,a^3\,c\,d^5+b^4\,d^3\,e^2+3\,b^2\,c^2\,d\,e^4-3\,b^3\,c\,d^2\,e^3+b^3\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^3\,e^2-6\,a\,c^3\,d\,e^4-2\,a\,c^3\,e^5\,x+a^2\,b\,d^5\,\sqrt {b^2-4\,a\,c}+2\,a^3\,d^5\,x\,\sqrt {b^2-4\,a\,c}-8\,a^3\,c\,d^4\,e\,x-4\,a^2\,c\,d^4\,e\,\sqrt {b^2-4\,a\,c}+3\,b\,c^2\,d\,e^4\,\sqrt {b^2-4\,a\,c}+9\,a\,b\,c^2\,d^2\,e^3-5\,a\,b^2\,c\,d^3\,e^2+2\,a^2\,b^2\,d^4\,e\,x+3\,a\,c^2\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}-3\,b^2\,c\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^2\,e^3\,x+2\,a\,b^2\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}-3\,a^2\,c\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,b\,c^2\,d\,e^4\,x-a\,b\,c\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}-2\,a^2\,b\,d^4\,e\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,c^2\,d\,e^4\,x\,\sqrt {b^2-4\,a\,c}-3\,a\,b^2\,c\,d^2\,e^3\,x+a^2\,b\,c\,d^3\,e^2\,x-3\,a\,b\,c\,d^2\,e^3\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^4\,d+b^3\,d\,\sqrt {b^2-4\,a\,c}+4\,a^2\,c^2\,d-b^3\,c\,e-5\,a\,b^2\,c\,d+4\,a\,b\,c^2\,e+2\,a\,c^2\,e\,\sqrt {b^2-4\,a\,c}-b^2\,c\,e\,\sqrt {b^2-4\,a\,c}-3\,a\,b\,c\,d\,\sqrt {b^2-4\,a\,c}\right )}{2\,\left (4\,a^4\,c\,d^2-a^3\,b^2\,d^2-4\,a^3\,b\,c\,d\,e+4\,a^3\,c^2\,e^2+a^2\,b^3\,d\,e-a^2\,b^2\,c\,e^2\right )}-\frac {d^3\,\ln \left (d+e\,x\right )}{a\,d^2\,e^2-b\,d\,e^3+c\,e^4} \]
x/(a*e) - (log(c^3*e^5*(b^2 - 4*a*c)^(1/2) - b*c^3*e^5 - 4*a^3*c*d^5 + a^2 *b^2*d^5 + b^4*d^3*e^2 + 3*b^2*c^2*d*e^4 - 3*b^3*c*d^2*e^3 - b^3*d^3*e^2*( b^2 - 4*a*c)^(1/2) + 6*a^2*c^2*d^3*e^2 - 6*a*c^3*d*e^4 - 2*a*c^3*e^5*x - a ^2*b*d^5*(b^2 - 4*a*c)^(1/2) - 2*a^3*d^5*x*(b^2 - 4*a*c)^(1/2) - 8*a^3*c*d ^4*e*x + 4*a^2*c*d^4*e*(b^2 - 4*a*c)^(1/2) - 3*b*c^2*d*e^4*(b^2 - 4*a*c)^( 1/2) + 9*a*b*c^2*d^2*e^3 - 5*a*b^2*c*d^3*e^2 + 2*a^2*b^2*d^4*e*x - 3*a*c^2 *d^2*e^3*(b^2 - 4*a*c)^(1/2) + 3*b^2*c*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 6*a^2 *c^2*d^2*e^3*x - 2*a*b^2*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) + 3*a^2*c*d^3*e^2*x *(b^2 - 4*a*c)^(1/2) + 3*a*b*c^2*d*e^4*x + a*b*c*d^3*e^2*(b^2 - 4*a*c)^(1/ 2) + 2*a^2*b*d^4*e*x*(b^2 - 4*a*c)^(1/2) - 3*a*c^2*d*e^4*x*(b^2 - 4*a*c)^( 1/2) - 3*a*b^2*c*d^2*e^3*x + a^2*b*c*d^3*e^2*x + 3*a*b*c*d^2*e^3*x*(b^2 - 4*a*c)^(1/2))*(b^4*d - b^3*d*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*d - b^3*c*e - 5*a*b^2*c*d + 4*a*b*c^2*e - 2*a*c^2*e*(b^2 - 4*a*c)^(1/2) + b^2*c*e*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*d*(b^2 - 4*a*c)^(1/2)))/(2*(4*a^4*c*d^2 - a^3*b^2 *d^2 + 4*a^3*c^2*e^2 - a^2*b^2*c*e^2 + a^2*b^3*d*e - 4*a^3*b*c*d*e)) - (lo g(a^2*b^2*d^5 - b*c^3*e^5 - c^3*e^5*(b^2 - 4*a*c)^(1/2) - 4*a^3*c*d^5 + b^ 4*d^3*e^2 + 3*b^2*c^2*d*e^4 - 3*b^3*c*d^2*e^3 + b^3*d^3*e^2*(b^2 - 4*a*c)^ (1/2) + 6*a^2*c^2*d^3*e^2 - 6*a*c^3*d*e^4 - 2*a*c^3*e^5*x + a^2*b*d^5*(b^2 - 4*a*c)^(1/2) + 2*a^3*d^5*x*(b^2 - 4*a*c)^(1/2) - 8*a^3*c*d^4*e*x - 4*a^ 2*c*d^4*e*(b^2 - 4*a*c)^(1/2) + 3*b*c^2*d*e^4*(b^2 - 4*a*c)^(1/2) + 9*a...