3.194 \(\int \frac{\text{b1}+\text{c1} x}{a+2 b x+c x^2} \, dx\)

Optimal. Leaf size=65 \[ \frac{\text{c1} \log \left (a+2 b x+c x^2\right )}{2 c}-\frac{(\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{c \sqrt{b^2-a c}} \]

[Out]

-(((b1*c - b*c1)*ArcTanh[(b + c*x)/Sqrt[b^2 - a*c]])/(c*Sqrt[b^2 - a*c])) + (c1*Log[a + 2*b*x + c*x^2])/(2*c)

________________________________________________________________________________________

Rubi [A]  time = 0.0415227, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {634, 618, 206, 628} \[ \frac{\text{c1} \log \left (a+2 b x+c x^2\right )}{2 c}-\frac{(\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{c \sqrt{b^2-a c}} \]

Antiderivative was successfully verified.

[In]

Int[(b1 + c1*x)/(a + 2*b*x + c*x^2),x]

[Out]

-(((b1*c - b*c1)*ArcTanh[(b + c*x)/Sqrt[b^2 - a*c]])/(c*Sqrt[b^2 - a*c])) + (c1*Log[a + 2*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{\text{b1}+\text{c1} x}{a+2 b x+c x^2} \, dx &=\frac{\text{c1} \int \frac{2 b+2 c x}{a+2 b x+c x^2} \, dx}{2 c}+\frac{(2 \text{b1} c-2 b \text{c1}) \int \frac{1}{a+2 b x+c x^2} \, dx}{2 c}\\ &=\frac{\text{c1} \log \left (a+2 b x+c x^2\right )}{2 c}-\frac{(2 \text{b1} c-2 b \text{c1}) \operatorname{Subst}\left (\int \frac{1}{4 \left (b^2-a c\right )-x^2} \, dx,x,2 b+2 c x\right )}{c}\\ &=-\frac{(\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{c \sqrt{b^2-a c}}+\frac{\text{c1} \log \left (a+2 b x+c x^2\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.0431969, size = 66, normalized size = 1.02 \[ \frac{(\text{b1} c-b \text{c1}) \tan ^{-1}\left (\frac{b+c x}{\sqrt{a c-b^2}}\right )}{c \sqrt{a c-b^2}}+\frac{\text{c1} \log \left (a+2 b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(b1 + c1*x)/(a + 2*b*x + c*x^2),x]

[Out]

((b1*c - b*c1)*ArcTan[(b + c*x)/Sqrt[-b^2 + a*c]])/(c*Sqrt[-b^2 + a*c]) + (c1*Log[a + 2*b*x + c*x^2])/(2*c)

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 95, normalized size = 1.5 \begin{align*}{\frac{{\it c1}\,\ln \left ( c{x}^{2}+2\,bx+a \right ) }{2\,c}}+{{\it b1}\arctan \left ({\frac{2\,cx+2\,b}{2}{\frac{1}{\sqrt{ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{ac-{b}^{2}}}}}-{\frac{b{\it c1}}{c}\arctan \left ({\frac{2\,cx+2\,b}{2}{\frac{1}{\sqrt{ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c1*x+b1)/(c*x^2+2*b*x+a),x)

[Out]

1/2*c1*ln(c*x^2+2*b*x+a)/c+1/(a*c-b^2)^(1/2)*arctan(1/2*(2*c*x+2*b)/(a*c-b^2)^(1/2))*b1-1/(a*c-b^2)^(1/2)*arct
an(1/2*(2*c*x+2*b)/(a*c-b^2)^(1/2))*c1*b/c

________________________________________________________________________________________

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((c1*x+b1)/(c*x^2+2*b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.75041, size = 446, normalized size = 6.86 \begin{align*} \left [\frac{{\left (b^{2} - a c\right )} c_{1} \log \left (c x^{2} + 2 \, b x + a\right ) - \sqrt{b^{2} - a c}{\left (b_{1} c - b c_{1}\right )} \log \left (\frac{c^{2} x^{2} + 2 \, b c x + 2 \, b^{2} - a c + 2 \, \sqrt{b^{2} - a c}{\left (c x + b\right )}}{c x^{2} + 2 \, b x + a}\right )}{2 \,{\left (b^{2} c - a c^{2}\right )}}, \frac{{\left (b^{2} - a c\right )} c_{1} \log \left (c x^{2} + 2 \, b x + a\right ) - 2 \, \sqrt{-b^{2} + a c}{\left (b_{1} c - b c_{1}\right )} \arctan \left (-\frac{\sqrt{-b^{2} + a c}{\left (c x + b\right )}}{b^{2} - a c}\right )}{2 \,{\left (b^{2} c - a c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x^2+2*b*x+a),x, algorithm="fricas")

[Out]

[1/2*((b^2 - a*c)*c1*log(c*x^2 + 2*b*x + a) - sqrt(b^2 - a*c)*(b1*c - b*c1)*log((c^2*x^2 + 2*b*c*x + 2*b^2 - a
*c + 2*sqrt(b^2 - a*c)*(c*x + b))/(c*x^2 + 2*b*x + a)))/(b^2*c - a*c^2), 1/2*((b^2 - a*c)*c1*log(c*x^2 + 2*b*x
+ a) - 2*sqrt(-b^2 + a*c)*(b1*c - b*c1)*arctan(-sqrt(-b^2 + a*c)*(c*x + b)/(b^2 - a*c)))/(b^2*c - a*c^2)]

________________________________________________________________________________________

Sympy [B]  time = 0.609779, size = 246, normalized size = 3.78 \begin{align*} \left (\frac{c_{1}}{2 c} - \frac{\sqrt{- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) \log{\left (x + \frac{- 2 a c \left (\frac{c_{1}}{2 c} - \frac{\sqrt{- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) + a c_{1} + 2 b^{2} \left (\frac{c_{1}}{2 c} - \frac{\sqrt{- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) - b b_{1}}{b c_{1} - b_{1} c} \right )} + \left (\frac{c_{1}}{2 c} + \frac{\sqrt{- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) \log{\left (x + \frac{- 2 a c \left (\frac{c_{1}}{2 c} + \frac{\sqrt{- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) + a c_{1} + 2 b^{2} \left (\frac{c_{1}}{2 c} + \frac{\sqrt{- a c + b^{2}} \left (b c_{1} - b_{1} c\right )}{2 c \left (a c - b^{2}\right )}\right ) - b b_{1}}{b c_{1} - b_{1} c} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x**2+2*b*x+a),x)

[Out]

(c1/(2*c) - sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2)))*log(x + (-2*a*c*(c1/(2*c) - sqrt(-a*c + b**2)*
(b*c1 - b1*c)/(2*c*(a*c - b**2))) + a*c1 + 2*b**2*(c1/(2*c) - sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2
))) - b*b1)/(b*c1 - b1*c)) + (c1/(2*c) + sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2)))*log(x + (-2*a*c*(
c1/(2*c) + sqrt(-a*c + b**2)*(b*c1 - b1*c)/(2*c*(a*c - b**2))) + a*c1 + 2*b**2*(c1/(2*c) + sqrt(-a*c + b**2)*(
b*c1 - b1*c)/(2*c*(a*c - b**2))) - b*b1)/(b*c1 - b1*c))

________________________________________________________________________________________

Giac [A]  time = 1.06189, size = 81, normalized size = 1.25 \begin{align*} \frac{c_{1} \log \left (c x^{2} + 2 \, b x + a\right )}{2 \, c} + \frac{{\left (b_{1} c - b c_{1}\right )} \arctan \left (\frac{c x + b}{\sqrt{-b^{2} + a c}}\right )}{\sqrt{-b^{2} + a c} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x^2+2*b*x+a),x, algorithm="giac")

[Out]

1/2*c1*log(c*x^2 + 2*b*x + a)/c + (b1*c - b*c1)*arctan((c*x + b)/sqrt(-b^2 + a*c))/(sqrt(-b^2 + a*c)*c)