\(\int \frac {d+e x}{a+b x+c x^2} \, dx\) [2185]

Optimal result

Integrand size = 18, antiderivative size = 66 \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}+\frac {e \log \left (a+b x+c x^2\right )}{2 c} \]

[Out]

1/2*e*ln(c*x^2+b*x+a)/c-(-b*e+2*c*d)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/c/(-4*a*c+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {648, 632, 212, 642} \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=\frac {e \log \left (a+b x+c x^2\right )}{2 c}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}} \]

[In]

Int[(d + e*x)/(a + b*x + c*x^2),x]

[Out]

-(((2*c*d - b*e)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c])) + (e*Log[a + b*x + c*x^2])/(2*
c)

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = \frac {e \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c}+\frac {(2 c d-b e) \int \frac {1}{a+b x+c x^2} \, dx}{2 c} \\ & = \frac {e \log \left (a+b x+c x^2\right )}{2 c}-\frac {(2 c d-b e) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c} \\ & = -\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}+\frac {e \log \left (a+b x+c x^2\right )}{2 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=\frac {-\frac {2 (-2 c d+b e) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+e \log (a+x (b+c x))}{2 c} \]

[In]

Integrate[(d + e*x)/(a + b*x + c*x^2),x]

[Out]

((-2*(-2*c*d + b*e)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + e*Log[a + x*(b + c*x)])/(2*c)

Maple [A] (verified)

Time = 25.43 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {e \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (d -\frac {b e}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}\)	\(62\)
risch	\(\frac {2 \ln \left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d -2 \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, c x -\sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) a e}{4 a c -b^{2}}-\frac {\ln \left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d -2 \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, c x -\sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) b^{2} e}{2 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d -2 \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, c x -\sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}}{2 c \left (4 a c -b^{2}\right )}+\frac {2 \ln \left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d +2 \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, c x +\sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) a e}{4 a c -b^{2}}-\frac {\ln \left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d +2 \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, c x +\sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) b^{2} e}{2 c \left (4 a c -b^{2}\right )}-\frac {\ln \left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d +2 \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, c x +\sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) \sqrt {-\left (b e -2 c d \right )^{2} \left (4 a c -b^{2}\right )}}{2 c \left (4 a c -b^{2}\right )}\)	\(647\)

[In]

int((e*x+d)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/2*e*ln(c*x^2+b*x+a)/c+2*(d-1/2*b*e/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 204, normalized size of antiderivative = 3.09 \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=\left [\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c d - b e\right )} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (2 \, c d - b e\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \]

[In]

integrate((e*x+d)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

[1/2*((b^2 - 4*a*c)*e*log(c*x^2 + b*x + a) - 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), 1/2*((b^2 - 4*a*c)*e*log(c*x^2 +
b*x + a) - 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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (58) = 116\).

Time = 0.41 (sec) , antiderivative size = 280, normalized size of antiderivative = 4.24 \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=\left (\frac {e}{2 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 4 a c \left (\frac {e}{2 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) + 2 a e + b^{2} \left (\frac {e}{2 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) - b d}{b e - 2 c d} \right )} + \left (\frac {e}{2 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 4 a c \left (\frac {e}{2 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) + 2 a e + b^{2} \left (\frac {e}{2 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) - b d}{b e - 2 c d} \right )} \]

[In]

integrate((e*x+d)/(c*x**2+b*x+a),x)

[Out]

(e/(2*c) - sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(2*c*(4*a*c - b**2)))*log(x + (-4*a*c*(e/(2*c) - sqrt(-4*a*c + b*
*2)*(b*e - 2*c*d)/(2*c*(4*a*c - b**2))) + 2*a*e + b**2*(e/(2*c) - sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(2*c*(4*a*
c - b**2))) - b*d)/(b*e - 2*c*d)) + (e/(2*c) + sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(2*c*(4*a*c - b**2)))*log(x +
(-4*a*c*(e/(2*c) + sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(2*c*(4*a*c - b**2))) + 2*a*e + b**2*(e/(2*c) + sqrt(-4*
a*c + b**2)*(b*e - 2*c*d)/(2*c*(4*a*c - b**2))) - b*d)/(b*e - 2*c*d))

Maxima [F(-2)]

Exception generated. \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x+d)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=\frac {e \log \left (c x^{2} + b x + a\right )}{2 \, c} + \frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c} \]

[In]

integrate((e*x+d)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

1/2*e*log(c*x^2 + b*x + a)/c + (2*c*d - b*e)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.45 \[ \int \frac {d+e x}{a+b x+c x^2} \, dx=\frac {2\,d\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{\sqrt {4\,a\,c-b^2}}-\frac {b^2\,e\,\ln \left (c\,x^2+b\,x+a\right )}{2\,\left (4\,a\,c^2-b^2\,c\right )}+\frac {2\,a\,c\,e\,\ln \left (c\,x^2+b\,x+a\right )}{4\,a\,c^2-b^2\,c}-\frac {b\,e\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{c\,\sqrt {4\,a\,c-b^2}} \]

[In]

int((d + e*x)/(a + b*x + c*x^2),x)

[Out]

(2*d*atan(b/(4*a*c - b^2)^(1/2) + (2*c*x)/(4*a*c - b^2)^(1/2)))/(4*a*c - b^2)^(1/2) - (b^2*e*log(a + b*x + c*x
^2))/(2*(4*a*c^2 - b^2*c)) + (2*a*c*e*log(a + b*x + c*x^2))/(4*a*c^2 - b^2*c) - (b*e*atan(b/(4*a*c - b^2)^(1/2
) + (2*c*x)/(4*a*c - b^2)^(1/2)))/(c*(4*a*c - b^2)^(1/2))