Integrand size = 26, antiderivative size = 210 \[ \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx=\frac {2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac {\sqrt {b^2-4 a c} e (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (c d^2-b d e+a e^2\right )^2}-\frac {\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}+\frac {\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^2} \]
(-b*e+2*c*d)/(a*e^2-b*d*e+c*d^2)/(e*x+d)-(2*c^2*d^2+b^2*e^2-2*c*e*(a*e+b*d ))*ln(e*x+d)/(a*e^2-b*d*e+c*d^2)^2+1/2*(2*c^2*d^2+b^2*e^2-2*c*e*(a*e+b*d)) *ln(c*x^2+b*x+a)/(a*e^2-b*d*e+c*d^2)^2+e*(-b*e+2*c*d)*arctanh((2*c*x+b)/(- 4*a*c+b^2)^(1/2))*(-4*a*c+b^2)^(1/2)/(a*e^2-b*d*e+c*d^2)^2
Time = 0.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.84 \[ \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx=\frac {\frac {2 (2 c d-b e) \left (c d^2+e (-b d+a e)\right )}{d+e x}-2 \sqrt {-b^2+4 a c} e (-2 c d+b e) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )+\left (-4 c^2 d^2-2 b^2 e^2+4 c e (b d+a e)\right ) \log (d+e x)+\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (a+x (b+c x))}{2 \left (c d^2+e (-b d+a e)\right )^2} \]
((2*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e)))/(d + e*x) - 2*Sqrt[-b^2 + 4* a*c]*e*(-2*c*d + b*e)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]] + (-4*c^2*d^2 - 2*b^2*e^2 + 4*c*e*(b*d + a*e))*Log[d + e*x] + (2*c^2*d^2 + b^2*e^2 - 2* c*e*(b*d + a*e))*Log[a + x*(b + c*x)])/(2*(c*d^2 + e*(-(b*d) + a*e))^2)
Time = 0.49 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {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.
| \(\displaystyle \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \int \left (\frac {e \left (2 c e (a e+b d)-b^2 e^2-2 c^2 d^2\right )}{(d+e x) \left (a e^2-b d e+c d^2\right )^2}+\frac {c x \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )+b c \left (c d^2-3 a e^2\right )+4 a c^2 d e+b^3 e^2-2 b^2 c d e}{\left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )^2}+\frac {e (b e-2 c d)}{(d+e x)^2 \left (a e^2-b d e+c d^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e \sqrt {b^2-4 a c} (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac {\log (d+e x) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac {2 c d-b e}{(d+e x) \left (a e^2-b d e+c d^2\right )}\) |
(2*c*d - b*e)/((c*d^2 - b*d*e + a*e^2)*(d + e*x)) + (Sqrt[b^2 - 4*a*c]*e*( 2*c*d - b*e)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c*d^2 - b*d*e + a*e^ 2)^2 - ((2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*Log[d + e*x])/(c*d^2 - b *d*e + a*e^2)^2 + ((2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*Log[a + b*x + c*x^2])/(2*(c*d^2 - b*d*e + a*e^2)^2)
3.16.30.3.1 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]
Time = 0.54 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.31
| method | result | size |
| default | \(\frac {\left (2 a c \,e^{2}-b^{2} e^{2}+2 b c d e -2 c^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (e^{2} a -b d e +c \,d^{2}\right )^{2}}-\frac {b e -2 c d}{\left (e^{2} a -b d e +c \,d^{2}\right ) \left (e x +d \right )}+\frac {\frac {\left (-2 a \,c^{2} e^{2}+b^{2} c \,e^{2}-2 b \,c^{2} d e +2 c^{3} d^{2}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-3 c \,e^{2} b a +4 a \,c^{2} d e +b^{3} e^{2}-2 b^{2} c d e +b \,d^{2} c^{2}-\frac {\left (-2 a \,c^{2} e^{2}+b^{2} c \,e^{2}-2 b \,c^{2} d e +2 c^{3} d^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (e^{2} a -b d e +c \,d^{2}\right )^{2}}\) | \(276\) |
| risch | \(\text {Expression too large to display}\) | \(11474\) |
(2*a*c*e^2-b^2*e^2+2*b*c*d*e-2*c^2*d^2)/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)-(b *e-2*c*d)/(a*e^2-b*d*e+c*d^2)/(e*x+d)+1/(a*e^2-b*d*e+c*d^2)^2*(1/2*(-2*a*c ^2*e^2+b^2*c*e^2-2*b*c^2*d*e+2*c^3*d^2)/c*ln(c*x^2+b*x+a)+2*(-3*c*e^2*b*a+ 4*a*c^2*d*e+b^3*e^2-2*b^2*c*d*e+b*d^2*c^2-1/2*(-2*a*c^2*e^2+b^2*c*e^2-2*b* c^2*d*e+2*c^3*d^2)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/ 2)))
Time = 1.33 (sec) , antiderivative size = 745, normalized size of antiderivative = 3.55 \[ \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx=\left [\frac {4 \, c^{2} d^{3} - 6 \, b c d^{2} e - 2 \, a b e^{3} + 2 \, {\left (b^{2} + 2 \, a c\right )} d e^{2} - {\left (2 \, c d^{2} e - b d e^{2} + {\left (2 \, c d e^{2} - b e^{3}\right )} x\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}, \frac {4 \, c^{2} d^{3} - 6 \, b c d^{2} e - 2 \, a b e^{3} + 2 \, {\left (b^{2} + 2 \, a c\right )} d e^{2} + 2 \, {\left (2 \, c d^{2} e - b d e^{2} + {\left (2 \, c d e^{2} - b e^{3}\right )} x\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}\right ] \]
[1/2*(4*c^2*d^3 - 6*b*c*d^2*e - 2*a*b*e^3 + 2*(b^2 + 2*a*c)*d*e^2 - (2*c*d ^2*e - b*d*e^2 + (2*c*d*e^2 - b*e^3)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + (2*c^2*d^3 - 2*b*c*d^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^2*d^2*e - 2*b*c*d *e^2 + (b^2 - 2*a*c)*e^3)*x)*log(c*x^2 + b*x + a) - 2*(2*c^2*d^3 - 2*b*c*d ^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^2*d^2*e - 2*b*c*d*e^2 + (b^2 - 2*a*c)*e^ 3)*x)*log(e*x + d))/(c^2*d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4 + ( b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4 + a^2*e^5 + (b^2 + 2*a*c)*d^2*e^3)*x), 1/2*(4*c^2*d^3 - 6*b*c*d^2*e - 2*a*b*e^3 + 2* (b^2 + 2*a*c)*d*e^2 + 2*(2*c*d^2*e - b*d*e^2 + (2*c*d*e^2 - b*e^3)*x)*sqrt (-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + (2* c^2*d^3 - 2*b*c*d^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^2*d^2*e - 2*b*c*d*e^2 + (b^2 - 2*a*c)*e^3)*x)*log(c*x^2 + b*x + a) - 2*(2*c^2*d^3 - 2*b*c*d^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^2*d^2*e - 2*b*c*d*e^2 + (b^2 - 2*a*c)*e^3)*x)* log(e*x + d))/(c^2*d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4 + (b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4 + a^2*e^5 + (b^2 + 2*a*c)*d^2*e^3)*x)]
Timed out. \[ \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \]
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 = 0.27 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.75 \[ \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx=\frac {{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} \log \left (c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {b e}{e x + d} - \frac {b d e}{{\left (e x + d\right )}^{2}} + \frac {a e^{2}}{{\left (e x + d\right )}^{2}}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} + \frac {\frac {2 \, c d e^{2}}{e x + d} - \frac {b e^{3}}{e x + d}}{c d^{2} e^{2} - b d e^{3} + a e^{4}} - \frac {{\left (2 \, b^{2} c d e^{3} - 8 \, a c^{2} d e^{3} - b^{3} e^{4} + 4 \, a b c e^{4}\right )} \arctan \left (\frac {2 \, c d - \frac {2 \, c d^{2}}{e x + d} - b e + \frac {2 \, b d e}{e x + d} - \frac {2 \, a e^{2}}{e x + d}}{\sqrt {-b^{2} + 4 \, a c} e}\right )}{{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c} e^{2}} \]
1/2*(2*c^2*d^2 - 2*b*c*d*e + b^2*e^2 - 2*a*c*e^2)*log(c - 2*c*d/(e*x + d) + c*d^2/(e*x + d)^2 + b*e/(e*x + d) - b*d*e/(e*x + d)^2 + a*e^2/(e*x + d)^ 2)/(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^ 2*e^4) + (2*c*d*e^2/(e*x + d) - b*e^3/(e*x + d))/(c*d^2*e^2 - b*d*e^3 + a* e^4) - (2*b^2*c*d*e^3 - 8*a*c^2*d*e^3 - b^3*e^4 + 4*a*b*c*e^4)*arctan((2*c *d - 2*c*d^2/(e*x + d) - b*e + 2*b*d*e/(e*x + d) - 2*a*e^2/(e*x + d))/(sqr t(-b^2 + 4*a*c)*e))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*sqrt(-b^2 + 4*a*c)*e^2)
Time = 13.09 (sec) , antiderivative size = 1637, normalized size of antiderivative = 7.80 \[ \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \]
(log(d + e*x)*(e^2*(2*a*c - b^2) - 2*c^2*d^2 + 2*b*c*d*e))/(a^2*e^4 + c^2* d^4 + b^2*d^2*e^2 - 2*a*b*d*e^3 - 2*b*c*d^3*e + 2*a*c*d^2*e^2) - (log(3*b^ 2*c^3*d^4 - 12*a*c^4*d^4 - 2*b^5*e^4*x - 12*a^3*c^2*e^4 - 2*a*b^4*e^4 + 2* b^4*e^4*x*(b^2 - 4*a*c)^(1/2) + 6*c^4*d^4*x*(b^2 - 4*a*c)^(1/2) + 11*a^2*b ^2*c*e^4 - 2*b^3*c^2*d^3*e + b^4*c*d^2*e^2 + 40*a^2*c^3*d^2*e^2 + 2*a*b^3* e^4*(b^2 - 4*a*c)^(1/2) + 3*b*c^3*d^4*(b^2 - 4*a*c)^(1/2) + 8*a*b*c^3*d^3* e + 6*a*b^3*c*d*e^3 + 12*a*b^3*c*e^4*x - 32*a*c^4*d^3*e*x + 8*b^4*c*d*e^3* x - 5*a^2*b*c*e^4*(b^2 - 4*a*c)^(1/2) - 16*a*c^3*d^3*e*(b^2 - 4*a*c)^(1/2) - 24*a^2*b*c^2*d*e^3 - 16*a^2*b*c^2*e^4*x + 32*a^2*c^3*d*e^3*x + 8*b^2*c^ 3*d^3*e*x + 16*a^2*c^2*d*e^3*(b^2 - 4*a*c)^(1/2) - 2*b^2*c^2*d^3*e*(b^2 - 4*a*c)^(1/2) + b^3*c*d^2*e^2*(b^2 - 4*a*c)^(1/2) + 6*a^2*c^2*e^4*x*(b^2 - 4*a*c)^(1/2) - 14*a*b^2*c^2*d^2*e^2 - 12*b^3*c^2*d^2*e^2*x + 14*a*b*c^2*d^ 2*e^2*(b^2 - 4*a*c)^(1/2) - 20*a*c^3*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) + 14*b^ 2*c^2*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) - 10*a*b^2*c*d*e^3*(b^2 - 4*a*c)^(1/2) - 8*a*b^2*c*e^4*x*(b^2 - 4*a*c)^(1/2) - 12*b*c^3*d^3*e*x*(b^2 - 4*a*c)^(1 /2) - 8*b^3*c*d*e^3*x*(b^2 - 4*a*c)^(1/2) + 48*a*b*c^3*d^2*e^2*x - 40*a*b^ 2*c^2*d*e^3*x + 20*a*b*c^2*d*e^3*x*(b^2 - 4*a*c)^(1/2))*(e^2*(a*c + (b*(b^ 2 - 4*a*c)^(1/2))/2 - b^2/2) + e*(b*c*d - c*d*(b^2 - 4*a*c)^(1/2)) - c^2*d ^2))/(a^2*e^4 + c^2*d^4 + b^2*d^2*e^2 - 2*a*b*d*e^3 - 2*b*c*d^3*e + 2*a*c* d^2*e^2) + (log(2*a*b^4*e^4 + 12*a*c^4*d^4 + 2*b^5*e^4*x + 12*a^3*c^2*e...