Integrand size = 19, antiderivative size = 65 \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=-\frac {(\text {b1} c-b \text {c1}) \text {arctanh}\left (\frac {b+c x}{\sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c}}+\frac {\text {c1} \log \left (a+2 b x+c x^2\right )}{2 c} \]
1/2*c1*ln(c*x^2+2*b*x+a)/c-(-b*c1+b1*c)*arctanh((c*x+b)/(-a*c+b^2)^(1/2))/ c/(-a*c+b^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=\frac {(\text {b1} c-b \text {c1}) \arctan \left (\frac {b+c x}{\sqrt {-b^2+a c}}\right )}{c \sqrt {-b^2+a c}}+\frac {\text {c1} \log \left (a+2 b x+c x^2\right )}{2 c} \]
((b1*c - b*c1)*ArcTan[(b + c*x)/Sqrt[-b^2 + a*c]])/(c*Sqrt[-b^2 + a*c]) + (c1*Log[a + 2*b*x + c*x^2])/(2*c)
Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1142, 27, 1083, 219, 1103}
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.
\(\displaystyle \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {(\text {b1} c-b \text {c1}) \int \frac {1}{c x^2+2 b x+a}dx}{c}+\frac {\text {c1} \int \frac {2 (b+c x)}{c x^2+2 b x+a}dx}{2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(\text {b1} c-b \text {c1}) \int \frac {1}{c x^2+2 b x+a}dx}{c}+\frac {\text {c1} \int \frac {b+c x}{c x^2+2 b x+a}dx}{c}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\text {c1} \int \frac {b+c x}{c x^2+2 b x+a}dx}{c}-\frac {2 (\text {b1} c-b \text {c1}) \int \frac {1}{4 \left (b^2-a c\right )-(2 b+2 c x)^2}d(2 b+2 c x)}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {c1} \int \frac {b+c x}{c x^2+2 b x+a}dx}{c}-\frac {(\text {b1} c-b \text {c1}) \text {arctanh}\left (\frac {2 b+2 c x}{2 \sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c}}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\text {c1} \log \left (a+2 b x+c x^2\right )}{2 c}-\frac {(\text {b1} c-b \text {c1}) \text {arctanh}\left (\frac {2 b+2 c x}{2 \sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c}}\) |
-(((b1*c - b*c1)*ArcTanh[(2*b + 2*c*x)/(2*Sqrt[b^2 - a*c])])/(c*Sqrt[b^2 - a*c])) + (c1*Log[a + 2*b*x + c*x^2])/(2*c)
3.2.94.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Time = 0.46 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97
method | result | size |
default | \(\frac {\operatorname {c1} \ln \left (c \,x^{2}+2 b x +a \right )}{2 c}+\frac {\left (\operatorname {b1} -\frac {\operatorname {c1} b}{c}\right ) \arctan \left (\frac {2 c x +2 b}{2 \sqrt {a c -b^{2}}}\right )}{\sqrt {a c -b^{2}}}\) | \(63\) |
risch | \(\frac {\ln \left (-a b c \operatorname {c1} +a \operatorname {b1} \,c^{2}+b^{3} \operatorname {c1} -b^{2} \operatorname {b1} c -\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, c x -\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, b \right ) a \operatorname {c1}}{2 a c -2 b^{2}}-\frac {\ln \left (-a b c \operatorname {c1} +a \operatorname {b1} \,c^{2}+b^{3} \operatorname {c1} -b^{2} \operatorname {b1} c -\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, c x -\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, b \right ) b^{2} \operatorname {c1}}{2 c \left (a c -b^{2}\right )}+\frac {\ln \left (-a b c \operatorname {c1} +a \operatorname {b1} \,c^{2}+b^{3} \operatorname {c1} -b^{2} \operatorname {b1} c -\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, c x -\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, b \right ) \sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}}{2 c \left (a c -b^{2}\right )}+\frac {\ln \left (-a b c \operatorname {c1} +a \operatorname {b1} \,c^{2}+b^{3} \operatorname {c1} -b^{2} \operatorname {b1} c +\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, c x +\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, b \right ) a \operatorname {c1}}{2 a c -2 b^{2}}-\frac {\ln \left (-a b c \operatorname {c1} +a \operatorname {b1} \,c^{2}+b^{3} \operatorname {c1} -b^{2} \operatorname {b1} c +\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, c x +\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, b \right ) b^{2} \operatorname {c1}}{2 c \left (a c -b^{2}\right )}-\frac {\ln \left (-a b c \operatorname {c1} +a \operatorname {b1} \,c^{2}+b^{3} \operatorname {c1} -b^{2} \operatorname {b1} c +\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, c x +\sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}\, b \right ) \sqrt {-\left (b \operatorname {c1} -\operatorname {b1} c \right )^{2} \left (a c -b^{2}\right )}}{2 c \left (a c -b^{2}\right )}\) | \(618\) |
1/2*c1*ln(c*x^2+2*b*x+a)/c+(b1-c1*b/c)/(a*c-b^2)^(1/2)*arctan(1/2*(2*c*x+2 *b)/(a*c-b^2)^(1/2))
Time = 0.25 (sec) , antiderivative size = 203, normalized size of antiderivative = 3.12 \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=\left [\frac {{\left (b^{2} - a c\right )} c_{1} \log \left (c x^{2} + 2 \, b x + a\right ) - \sqrt {b^{2} - a c} {\left (b_{1} c - b c_{1}\right )} \log \left (\frac {c^{2} x^{2} + 2 \, b c x + 2 \, b^{2} - a c + 2 \, \sqrt {b^{2} - a c} {\left (c x + b\right )}}{c x^{2} + 2 \, b x + a}\right )}{2 \, {\left (b^{2} c - a c^{2}\right )}}, \frac {{\left (b^{2} - a c\right )} c_{1} \log \left (c x^{2} + 2 \, b x + a\right ) - 2 \, \sqrt {-b^{2} + a c} {\left (b_{1} c - b c_{1}\right )} \arctan \left (-\frac {\sqrt {-b^{2} + a c} {\left (c x + b\right )}}{b^{2} - a c}\right )}{2 \, {\left (b^{2} c - a c^{2}\right )}}\right ] \]
[1/2*((b^2 - a*c)*c1*log(c*x^2 + 2*b*x + a) - sqrt(b^2 - a*c)*(b1*c - b*c1 )*log((c^2*x^2 + 2*b*c*x + 2*b^2 - a*c + 2*sqrt(b^2 - a*c)*(c*x + b))/(c*x ^2 + 2*b*x + a)))/(b^2*c - a*c^2), 1/2*((b^2 - a*c)*c1*log(c*x^2 + 2*b*x + a) - 2*sqrt(-b^2 + a*c)*(b1*c - b*c1)*arctan(-sqrt(-b^2 + a*c)*(c*x + b)/ (b^2 - a*c)))/(b^2*c - a*c^2)]
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (53) = 106\).
Time = 0.40 (sec) , antiderivative size = 246, normalized size of antiderivative = 3.78 \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=\left (\frac {c_{1}}{2 c} - \frac {\sqrt {- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 2 a c \left (\frac {c_{1}}{2 c} - \frac {\sqrt {- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) + a c_{1} + 2 b^{2} \left (\frac {c_{1}}{2 c} - \frac {\sqrt {- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) - b b_{1}}{b c_{1} - b_{1} c} \right )} + \left (\frac {c_{1}}{2 c} + \frac {\sqrt {- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 2 a c \left (\frac {c_{1}}{2 c} + \frac {\sqrt {- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) + a c_{1} + 2 b^{2} \left (\frac {c_{1}}{2 c} + \frac {\sqrt {- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) - b b_{1}}{b c_{1} - b_{1} c} \right )} \]
(c1/(2*c) - sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2)))*log(x + (- 2*a*c*(c1/(2*c) - sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2))) + a* c1 + 2*b**2*(c1/(2*c) - sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2)) ) - b*b1)/(b*c1 - b1*c)) + (c1/(2*c) + sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2* c*(a*c - b**2)))*log(x + (-2*a*c*(c1/(2*c) + sqrt(-a*c + b**2)*(b*c1 - b1* c)/(2*c*(a*c - b**2))) + a*c1 + 2*b**2*(c1/(2*c) + sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2))) - b*b1)/(b*c1 - b1*c))
Exception generated. \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=\text {Exception raised: ValueError} \]
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*b^2-4*a*c>0)', see `assume?` f or more de
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=\frac {c_{1} \log \left (c x^{2} + 2 \, b x + a\right )}{2 \, c} + \frac {{\left (b_{1} c - b c_{1}\right )} \arctan \left (\frac {c x + b}{\sqrt {-b^{2} + a c}}\right )}{\sqrt {-b^{2} + a c} c} \]
1/2*c1*log(c*x^2 + 2*b*x + a)/c + (b1*c - b*c1)*arctan((c*x + b)/sqrt(-b^2 + a*c))/(sqrt(-b^2 + a*c)*c)
Time = 0.31 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.38 \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=\frac {b_{1}\,\mathrm {atan}\left (\frac {b}{\sqrt {a\,c-b^2}}+\frac {c\,x}{\sqrt {a\,c-b^2}}\right )}{\sqrt {a\,c-b^2}}-\frac {2\,b^2\,c_{1}\,\ln \left (c\,x^2+2\,b\,x+a\right )}{4\,a\,c^2-4\,b^2\,c}+\frac {2\,a\,c\,c_{1}\,\ln \left (c\,x^2+2\,b\,x+a\right )}{4\,a\,c^2-4\,b^2\,c}-\frac {b\,c_{1}\,\mathrm {atan}\left (\frac {b}{\sqrt {a\,c-b^2}}+\frac {c\,x}{\sqrt {a\,c-b^2}}\right )}{c\,\sqrt {a\,c-b^2}} \]
(b1*atan(b/(a*c - b^2)^(1/2) + (c*x)/(a*c - b^2)^(1/2)))/(a*c - b^2)^(1/2) - (2*b^2*c1*log(a + 2*b*x + c*x^2))/(4*a*c^2 - 4*b^2*c) + (2*a*c*c1*log(a + 2*b*x + c*x^2))/(4*a*c^2 - 4*b^2*c) - (b*c1*atan(b/(a*c - b^2)^(1/2) + (c*x)/(a*c - b^2)^(1/2)))/(c*(a*c - b^2)^(1/2))
Time = 0.00 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.80 \[ \int \frac {\text {b1}+\text {c1} x}{a+2 b x+c x^2} \, dx=\frac {-2 \sqrt {a c -b^{2}}\, \mathit {atan} \left (\frac {c x +b}{\sqrt {a c -b^{2}}}\right ) b \mathit {c1} +2 \sqrt {a c -b^{2}}\, \mathit {atan} \left (\frac {c x +b}{\sqrt {a c -b^{2}}}\right ) \mathit {b1} c +\mathrm {log}\left (c \,x^{2}+2 b x +a \right ) a c \mathit {c1} -\mathrm {log}\left (c \,x^{2}+2 b x +a \right ) b^{2} \mathit {c1}}{2 c \left (a c -b^{2}\right )} \]