Integrand size = 20, antiderivative size = 122 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {e \log (d+e x)}{c d^2-b d e+a e^2}-\frac {e \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )} \] Output:
-(-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)+e*ln(e*x+d)/(a*e^2-b*d*e+c*d^2)-e*ln(c*x^2+b*x+a)/(2*a*e^ 2-2*b*d*e+2*c*d^2)
Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {(-4 c d+2 b e) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} e (-2 \log (d+e x)+\log (a+x (b+c x)))}{2 \sqrt {-b^2+4 a c} \left (-c d^2+e (b d-a e)\right )} \] Input:
Integrate[1/((d + e*x)*(a + b*x + c*x^2)),x]
Output:
((-4*c*d + 2*b*e)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a *c]*e*(-2*Log[d + e*x] + Log[a + x*(b + c*x)]))/(2*Sqrt[-b^2 + 4*a*c]*(-(c *d^2) + e*(b*d - a*e)))
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1144, 1142, 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.
\(\displaystyle \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx\) |
\(\Big \downarrow \) 1144 |
\(\displaystyle \frac {\int \frac {c d-b e-c e x}{c x^2+b x+a}dx}{a e^2-b d e+c d^2}+\frac {e \log (d+e x)}{a e^2-b d e+c d^2}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {\frac {1}{2} (2 c d-b e) \int \frac {1}{c x^2+b x+a}dx-\frac {1}{2} e \int \frac {b+2 c x}{c x^2+b x+a}dx}{a e^2-b d e+c d^2}+\frac {e \log (d+e x)}{a e^2-b d e+c d^2}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {-(2 c d-b e) \int \frac {1}{b^2-(b+2 c x)^2-4 a c}d(b+2 c x)-\frac {1}{2} e \int \frac {b+2 c x}{c x^2+b x+a}dx}{a e^2-b d e+c d^2}+\frac {e \log (d+e x)}{a e^2-b d e+c d^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {-\frac {1}{2} e \int \frac {b+2 c x}{c x^2+b x+a}dx-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a e^2-b d e+c d^2}+\frac {e \log (d+e x)}{a e^2-b d e+c d^2}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {1}{2} e \log \left (a+b x+c x^2\right )}{a e^2-b d e+c d^2}+\frac {e \log (d+e x)}{a e^2-b d e+c d^2}\) |
Input:
Int[1/((d + e*x)*(a + b*x + c*x^2)),x]
Output:
(e*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2) + (-(((2*c*d - b*e)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4*a*c]) - (e*Log[a + b*x + c*x^2])/2 )/(c*d^2 - b*d*e + a*e^2)
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]
Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[e*(Log[RemoveContent[d + e*x, x]]/(c*d^2 - b*d*e + a*e^2)), x] + S imp[1/(c*d^2 - b*d*e + a*e^2) Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Time = 1.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85
method | result | size |
default | \(\frac {e \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}+\frac {-\frac {e \ln \left (c \,x^{2}+b x +a \right )}{2}+\frac {2 \left (c d -\frac {b e}{2}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a \,e^{2}-b d e +c \,d^{2}}\) | \(104\) |
risch | \(\text {Expression too large to display}\) | \(2835\) |
Input:
int(1/(e*x+d)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
e*ln(e*x+d)/(a*e^2-b*d*e+c*d^2)+1/(a*e^2-b*d*e+c*d^2)*(-1/2*e*ln(c*x^2+b*x +a)+2*(c*d-1/2*b*e)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))
Time = 0.25 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.50 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} e \log \left (e x + d\right ) + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c d - b e\right )} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right )}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}, -\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} e \log \left (e x + d\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (2 \, c d - b e\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right )}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}\right ] \] Input:
integrate(1/(e*x+d)/(c*x^2+b*x+a),x, algorithm="fricas")
Output:
[-1/2*((b^2 - 4*a*c)*e*log(c*x^2 + b*x + a) - 2*(b^2 - 4*a*c)*e*log(e*x + d) + sqrt(b^2 - 4*a*c)*(2*c*d - b*e)*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)))/((b^2*c - 4*a*c^2)* d^2 - (b^3 - 4*a*b*c)*d*e + (a*b^2 - 4*a^2*c)*e^2), -1/2*((b^2 - 4*a*c)*e* log(c*x^2 + b*x + a) - 2*(b^2 - 4*a*c)*e*log(e*x + d) + 2*sqrt(-b^2 + 4*a* c)*(2*c*d - b*e)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)))/(( b^2*c - 4*a*c^2)*d^2 - (b^3 - 4*a*b*c)*d*e + (a*b^2 - 4*a^2*c)*e^2)]
Timed out. \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x+d)/(c*x**2+b*x+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(e*x+d)/(c*x^2+b*x+a),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 = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {e^{2} \log \left ({\left | e x + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} - \frac {e \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}} + \frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \] Input:
integrate(1/(e*x+d)/(c*x^2+b*x+a),x, algorithm="giac")
Output:
e^2*log(abs(e*x + d))/(c*d^2*e - b*d*e^2 + a*e^3) - 1/2*e*log(c*x^2 + b*x + a)/(c*d^2 - b*d*e + a*e^2) + (2*c*d - b*e)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c*d^2 - b*d*e + a*e^2)*sqrt(-b^2 + 4*a*c))
Time = 6.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {e\,\ln \left (\frac {{\left (d+e\,x\right )}^2}{c\,x^2+b\,x+a}\right )}{2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2}-\frac {\ln \left (\frac {b+2\,c\,x-\sqrt {b^2-4\,a\,c}}{b+2\,c\,x+\sqrt {b^2-4\,a\,c}}\right )\,\left (b\,e-2\,c\,d\right )}{\sqrt {b^2-4\,a\,c}\,\left (2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2\right )} \] Input:
int(1/((d + e*x)*(a + b*x + c*x^2)),x)
Output:
(e*log((d + e*x)^2/(a + b*x + c*x^2)))/(2*a*e^2 + 2*c*d^2 - 2*b*d*e) - (lo g((b + 2*c*x - (b^2 - 4*a*c)^(1/2))/(b + 2*c*x + (b^2 - 4*a*c)^(1/2)))*(b* e - 2*c*d))/((b^2 - 4*a*c)^(1/2)*(2*a*e^2 + 2*c*d^2 - 2*b*d*e))
Time = 0.21 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {-2 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b e +4 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) c d -4 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a c e +\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{2} e +8 \,\mathrm {log}\left (e x +d \right ) a c e -2 \,\mathrm {log}\left (e x +d \right ) b^{2} e}{8 a^{2} c \,e^{2}-2 a \,b^{2} e^{2}-8 a b c d e +8 a \,c^{2} d^{2}+2 b^{3} d e -2 b^{2} c \,d^{2}} \] Input:
int(1/(e*x+d)/(c*x^2+b*x+a),x)
Output:
( - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*e + 4*sqrt (4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c*d - 4*log(a + b*x + c*x**2)*a*c*e + log(a + b*x + c*x**2)*b**2*e + 8*log(d + e*x)*a*c*e - 2*lo g(d + e*x)*b**2*e)/(2*(4*a**2*c*e**2 - a*b**2*e**2 - 4*a*b*c*d*e + 4*a*c** 2*d**2 + b**3*d*e - b**2*c*d**2))