### 3.2187 $$\int \frac{1}{(d+e x) (a+b x+c x^2)} \, dx$$

Optimal. Leaf size=122 $-\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac{e \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}+\frac{e \log (d+e x)}{a e^2-b d e+c d^2}$

[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))

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Rubi [A]  time = 0.0942328, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {705, 31, 634, 618, 206, 628} $-\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac{e \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}+\frac{e \log (d+e x)}{a e^2-b d e+c d^2}$

Antiderivative was successfully veriﬁed.

[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 705

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]

Rule 31

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

Rule 634

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 618

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x) \left (a+b x+c x^2\right )} \, dx &=\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) \operatorname{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]  time = 0.0806766, size = 105, normalized size = 0.86 $\frac{e \sqrt{4 a c-b^2} (\log (a+x (b+c x))-2 \log (d+e x))+(2 b e-4 c d) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{2 \sqrt{4 a c-b^2} \left (e (b d-a e)-c d^2\right )}$

Antiderivative was successfully veriﬁed.

[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)))

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Maple [A]  time = 0.155, size = 168, normalized size = 1.4 \begin{align*} -{\frac{e\ln \left ( c{x}^{2}+bx+a \right ) }{2\,a{e}^{2}-2\,bde+2\,c{d}^{2}}}-{\frac{be}{a{e}^{2}-bde+c{d}^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+2\,{\frac{cd}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{e\ln \left ( ex+d \right ) }{a{e}^{2}-bde+c{d}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.85503, size = 697, normalized size = 5.71 \begin{align*} \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 ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.11311, size = 170, normalized size = 1.39 \begin{align*} -\frac{e \log \left (c x^{2} + b x + a\right )}{2 \,{\left (c d^{2} - b d e + a e^{2}\right )}} + \frac{e^{2} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} + \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e*log(c*x^2 + b*x + a)/(c*d^2 - b*d*e + a*e^2) + e^2*log(abs(x*e + d))/(c*d^2*e - b*d*e^2 + a*e^3) + (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))