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

Optimal. Leaf size=66 \[ \frac {e \log \left (a+b x+c x^2\right )}{2 c}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, 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} \[ \frac {e \log \left (a+b x+c x^2\right )}{2 c}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[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 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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {d+e x}{a+b x+c x^2} \, dx &=\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) \operatorname {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*}

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Mathematica [A]  time = 0.06, size = 66, normalized size = 1.00 \[ \frac {e \log (a+x (b+c x))-\frac {2 (b e-2 c d) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(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)

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fricas [A]  time = 1.12, size = 204, normalized size = 3.09 \[ \left [\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c d - b e\right )} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{2} + b x + a\right ) - 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (2 \, c d - b e\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.17, size = 65, normalized size = 0.98 \[ \frac {e \log \left (c x^{2} + b x + a\right )}{2 \, c} + \frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 93, normalized size = 1.41 \[ -\frac {b e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {2 d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+\frac {e \ln \left (c \,x^{2}+b x +a \right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 0.12, size = 162, normalized size = 2.45 \[ \frac {2\,d\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{\sqrt {4\,a\,c-b^2}}-\frac {b^2\,e\,\ln \left (c\,x^2+b\,x+a\right )}{2\,\left (4\,a\,c^2-b^2\,c\right )}+\frac {2\,a\,c\,e\,\ln \left (c\,x^2+b\,x+a\right )}{4\,a\,c^2-b^2\,c}-\frac {b\,e\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{c\,\sqrt {4\,a\,c-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [B]  time = 0.74, size = 280, normalized size = 4.24 \[ \left (\frac {e}{2 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 4 a c \left (\frac {e}{2 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) + 2 a e + b^{2} \left (\frac {e}{2 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) - b d}{b e - 2 c d} \right )} + \left (\frac {e}{2 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 4 a c \left (\frac {e}{2 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) + 2 a e + b^{2} \left (\frac {e}{2 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c \left (4 a c - b^{2}\right )}\right ) - b d}{b e - 2 c d} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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