3.2367 \(\int \frac{A+B x}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=64 \[ \frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c}}+\frac{B \log \left (a+b x+c x^2\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.0389597, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {634, 618, 206, 628} \[ \frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c}}+\frac{B \log \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(a + b*x + c*x^2),x]

[Out]

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

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{A+B x}{a+b x+c x^2} \, dx &=\frac{B \int \frac{b+2 c x}{a+b x+c x^2} \, dx}{2 c}+\frac{(-b B+2 A c) \int \frac{1}{a+b x+c x^2} \, dx}{2 c}\\ &=\frac{B \log \left (a+b x+c x^2\right )}{2 c}-\frac{(-b B+2 A c) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c}\\ &=\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c}}+\frac{B \log \left (a+b x+c x^2\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.0605093, size = 66, normalized size = 1.03 \[ \frac{B \log (a+x (b+c x))-\frac{2 (b B-2 A c) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(a + b*x + c*x^2),x]

[Out]

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

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Maple [A]  time = 0.002, size = 93, normalized size = 1.5 \begin{align*}{\frac{B\ln \left ( c{x}^{2}+bx+a \right ) }{2\,c}}+2\,{\frac{A}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{bB}{c}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+b*x+a),x)

[Out]

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

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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((B*x+A)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37421, size = 471, normalized size = 7.36 \begin{align*} \left [-\frac{{\left (B b - 2 \, A c\right )} \sqrt{b^{2} - 4 \, a c} \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 ) -{\left (B b^{2} - 4 \, B a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac{2 \,{\left (B b - 2 \, A c\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) +{\left (B b^{2} - 4 \, B a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((B*b - 2*A*c)*sqrt(b^2 - 4*a*c)*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*b^2 - 4*B*a*c)*log(c*x^2 + b*x + a))/(b^2*c - 4*a*c^2), 1/2*(2*(B*b - 2*A*c)*sqrt(-b^
2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + (B*b^2 - 4*B*a*c)*log(c*x^2 + b*x + a))/(b^
2*c - 4*a*c^2)]

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Sympy [B]  time = 0.649333, size = 280, normalized size = 4.38 \begin{align*} \left (\frac{B}{2 c} - \frac{\left (- 2 A c + B b\right ) \sqrt{- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )}\right ) \log{\left (x + \frac{- A b + 2 B a - 4 a c \left (\frac{B}{2 c} - \frac{\left (- 2 A c + B b\right ) \sqrt{- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )}\right ) + b^{2} \left (\frac{B}{2 c} - \frac{\left (- 2 A c + B b\right ) \sqrt{- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )}\right )}{- 2 A c + B b} \right )} + \left (\frac{B}{2 c} + \frac{\left (- 2 A c + B b\right ) \sqrt{- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )}\right ) \log{\left (x + \frac{- A b + 2 B a - 4 a c \left (\frac{B}{2 c} + \frac{\left (- 2 A c + B b\right ) \sqrt{- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )}\right ) + b^{2} \left (\frac{B}{2 c} + \frac{\left (- 2 A c + B b\right ) \sqrt{- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )}\right )}{- 2 A c + B b} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x**2+b*x+a),x)

[Out]

(B/(2*c) - (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2)))*log(x + (-A*b + 2*B*a - 4*a*c*(B/(2*c) - (
-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2))) + b**2*(B/(2*c) - (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(
2*c*(4*a*c - b**2))))/(-2*A*c + B*b)) + (B/(2*c) + (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2)))*lo
g(x + (-A*b + 2*B*a - 4*a*c*(B/(2*c) + (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2))) + b**2*(B/(2*c
) + (-2*A*c + B*b)*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2))))/(-2*A*c + B*b))

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Giac [A]  time = 1.10795, size = 85, normalized size = 1.33 \begin{align*} \frac{B \log \left (c x^{2} + b x + a\right )}{2 \, c} - \frac{{\left (B b - 2 \, A c\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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