\(\int \frac {1}{(d+e x) (a+b x+c x^2)} \, dx\) [2187]

Optimal result

Integrand size = 20, antiderivative size = 122 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {e \log (d+e x)}{c d^2-b d e+a e^2}-\frac {e \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {719, 31, 648, 632, 212, 642} \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac {e \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}+\frac {e \log (d+e x)}{a e^2-b d e+c d^2} \]

[In]

Int[1/((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]])/(Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2))) + (e*Log
[d + e*x])/(c*d^2 - b*d*e + a*e^2) - (e*Log[a + b*x + c*x^2])/(2*(c*d^2 - b*d*e + a*e^2))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

Rule 719

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {(-4 c d+2 b e) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} e (-2 \log (d+e x)+\log (a+x (b+c x)))}{2 \sqrt {-b^2+4 a c} \left (-c d^2+e (b d-a e)\right )} \]

[In]

Integrate[1/((d + e*x)*(a + b*x + c*x^2)),x]

[Out]

((-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)))

Maple [A] (verified)

Time = 29.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {-\frac {e \ln \left (c \,x^{2}+b x +a \right )}{2}+\frac {2 \left (c d -\frac {b e}{2}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{e^{2} a -b d e +c \,d^{2}}+\frac {e \ln \left (e x +d \right )}{e^{2} a -b d e +c \,d^{2}}\)	\(104\)
risch	\(\frac {e \ln \left (e x +d \right )}{e^{2} a -b d e +c \,d^{2}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{2} c \,e^{2}-a \,b^{2} e^{2}-4 a b c d e +4 a \,c^{2} d^{2}+b^{3} d e -b^{2} c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 a c e -b^{2} e \right ) \textit {\_Z} +c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (6 a c \,e^{2}-2 b^{2} e^{2}+2 b c d e -2 c^{2} d^{2}\right ) \textit {\_R} +3 c e \right ) x +\left (-a b \,e^{2}+8 a c d e -b^{2} d e -b c \,d^{2}\right ) \textit {\_R} +b e +c d \right )\right )\)	\(181\)

[In]

int(1/(e*x+d)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.73 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.50 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} e \log \left (e x + d\right ) + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c d - b e\right )} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right )}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}, -\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} e \log \left (e x + d\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (2 \, c d - b e\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right )}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}\right ] \]

[In]

integrate(1/(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) - 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)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*x+d)/(c*x**2+b*x+a),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(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 = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {e^{2} \log \left ({\left | e x + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} - \frac {e \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}} + \frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \]

[In]

integrate(1/(e*x+d)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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))

Mupad [B] (verification not implemented)

Time = 10.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {e\,\ln \left (\frac {{\left (d+e\,x\right )}^2}{c\,x^2+b\,x+a}\right )}{2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2}-\frac {\ln \left (\frac {b+2\,c\,x-\sqrt {b^2-4\,a\,c}}{b+2\,c\,x+\sqrt {b^2-4\,a\,c}}\right )\,\left (b\,e-2\,c\,d\right )}{\sqrt {b^2-4\,a\,c}\,\left (2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2\right )} \]

[In]

int(1/((d + e*x)*(a + b*x + c*x^2)),x)

[Out]

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