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

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

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Rubi [A]  time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {705, 31, 634, 618, 206, 628} \begin {gather*} -\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} \end {gather*}

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 31

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

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 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]

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*}

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Mathematica [A]  time = 0.08, size = 105, normalized size = 0.86 \begin {gather*} \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 )} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((d + e*x)*(a + b*x + c*x^2)),x]

[Out]

IntegrateAlgebraic[1/((d + e*x)*(a + b*x + c*x^2)), x]

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fricas [A]  time = 0.68, size = 305, normalized size = 2.50 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.16, size = 126, normalized size = 1.03 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.06, size = 168, normalized size = 1.38 \begin {gather*} -\frac {b e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {2 c d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {e \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}-\frac {e \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(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 = 0.98, size = 122, normalized size = 1.00 \begin {gather*} \frac {e\,\ln \left (\frac {{\left (d+e\,x\right )}^2}{c\,x^2+b\,x+a}\right )}{2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2}-\frac {\ln \left (\frac {b+2\,c\,x-\sqrt {b^2-4\,a\,c}}{b+2\,c\,x+\sqrt {b^2-4\,a\,c}}\right )\,\left (b\,e-2\,c\,d\right )}{\sqrt {b^2-4\,a\,c}\,\left (2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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