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

Optimal. Leaf size=71 \[ \frac {(b d-2 a e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {d \log \left (a+b x+c x^2\right )}{2 a}+\frac {d \log (x)}{a} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {800, 634, 618, 206, 628} \begin {gather*} \frac {(b d-2 a e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {d \log \left (a+b x+c x^2\right )}{2 a}+\frac {d \log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 71, normalized size = 1.00 \begin {gather*} -\frac {\frac {2 (b d-2 a e) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+d (\log (a+x (b+c x))-2 \log (x))}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d+e x}{x \left (a+b x+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 228, normalized size = 3.21 \begin {gather*} \left [-\frac {{\left (b^{2} - 4 \, a c\right )} d \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} d \log \relax (x) + \sqrt {b^{2} - 4 \, a c} {\left (b d - 2 \, a 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 (a b^{2} - 4 \, a^{2} c\right )}}, -\frac {{\left (b^{2} - 4 \, a c\right )} d \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} d \log \relax (x) - 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (b d - 2 \, a 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 (a b^{2} - 4 \, a^{2} c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 72, normalized size = 1.01 \begin {gather*} -\frac {d \log \left (c x^{2} + b x + a\right )}{2 \, a} + \frac {d \log \left ({\left | x \right |}\right )}{a} - \frac {{\left (b d - 2 \, a e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 100, normalized size = 1.41 \begin {gather*} -\frac {b d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}+\frac {2 e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+\frac {d \ln \relax (x )}{a}-\frac {d \ln \left (c \,x^{2}+b x +a \right )}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/x/(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?

________________________________________________________________________________________

mupad [B]  time = 1.95, size = 375, normalized size = 5.28 \begin {gather*} \ln \left (a^2\,e\,\sqrt {b^2-4\,a\,c}-2\,a\,b^2\,d+a^2\,b\,e+6\,a^2\,c\,d-2\,b^3\,d\,x-2\,a\,b\,d\,\sqrt {b^2-4\,a\,c}+a\,b^2\,e\,x-2\,a^2\,c\,e\,x-2\,b^2\,d\,x\,\sqrt {b^2-4\,a\,c}+7\,a\,b\,c\,d\,x+a\,b\,e\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,c\,d\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {2\,a\,e\,\sqrt {b^2-4\,a\,c}-b\,d\,\sqrt {b^2-4\,a\,c}}{2\,a\,b^2-8\,a^2\,c}-\frac {d}{2\,a}\right )-\ln \left (a^2\,e\,\sqrt {b^2-4\,a\,c}+2\,a\,b^2\,d-a^2\,b\,e-6\,a^2\,c\,d+2\,b^3\,d\,x-2\,a\,b\,d\,\sqrt {b^2-4\,a\,c}-a\,b^2\,e\,x+2\,a^2\,c\,e\,x-2\,b^2\,d\,x\,\sqrt {b^2-4\,a\,c}-7\,a\,b\,c\,d\,x+a\,b\,e\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,c\,d\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {2\,a\,e\,\sqrt {b^2-4\,a\,c}-b\,d\,\sqrt {b^2-4\,a\,c}}{2\,a\,b^2-8\,a^2\,c}+\frac {d}{2\,a}\right )+\frac {d\,\ln \relax (x)}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a^2*e*(b^2 - 4*a*c)^(1/2) - 2*a*b^2*d + a^2*b*e + 6*a^2*c*d - 2*b^3*d*x - 2*a*b*d*(b^2 - 4*a*c)^(1/2) + a*
b^2*e*x - 2*a^2*c*e*x - 2*b^2*d*x*(b^2 - 4*a*c)^(1/2) + 7*a*b*c*d*x + a*b*e*x*(b^2 - 4*a*c)^(1/2) + 3*a*c*d*x*
(b^2 - 4*a*c)^(1/2))*((2*a*e*(b^2 - 4*a*c)^(1/2) - b*d*(b^2 - 4*a*c)^(1/2))/(2*a*b^2 - 8*a^2*c) - d/(2*a)) - l
og(a^2*e*(b^2 - 4*a*c)^(1/2) + 2*a*b^2*d - a^2*b*e - 6*a^2*c*d + 2*b^3*d*x - 2*a*b*d*(b^2 - 4*a*c)^(1/2) - a*b
^2*e*x + 2*a^2*c*e*x - 2*b^2*d*x*(b^2 - 4*a*c)^(1/2) - 7*a*b*c*d*x + a*b*e*x*(b^2 - 4*a*c)^(1/2) + 3*a*c*d*x*(
b^2 - 4*a*c)^(1/2))*((2*a*e*(b^2 - 4*a*c)^(1/2) - b*d*(b^2 - 4*a*c)^(1/2))/(2*a*b^2 - 8*a^2*c) + d/(2*a)) + (d
*log(x))/a

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________