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

Optimal. Leaf size=48 \[ \frac {\log \left (a+b x+c x^2\right )}{d \left (b^2-4 a c\right )}-\frac {2 \log (b+2 c x)}{d \left (b^2-4 a c\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {681, 31, 628} \begin {gather*} \frac {\log \left (a+b x+c x^2\right )}{d \left (b^2-4 a c\right )}-\frac {2 \log (b+2 c x)}{d \left (b^2-4 a c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 681

Int[1/(((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[(-4*b*c)/(d*(b^2 - 4*a*c)),
Int[1/(b + 2*c*x), x], x] + Dist[b^2/(d^2*(b^2 - 4*a*c)), Int[(d + e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a
, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.71 \begin {gather*} \frac {\log (a+x (b+c x))-2 \log (b+2 c x)}{d \left (b^2-4 a c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 35, normalized size = 0.73 \begin {gather*} \frac {\log \left (c x^{2} + b x + a\right ) - 2 \, \log \left (2 \, c x + b\right )}{{\left (b^{2} - 4 \, a c\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 57, normalized size = 1.19 \begin {gather*} -\frac {2 \, c^{2} \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{2} c^{2} d - 4 \, a c^{3} d} + \frac {\log \left (c x^{2} + b x + a\right )}{b^{2} d - 4 \, a c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

-2*c^2*log(abs(2*c*x + b))/(b^2*c^2*d - 4*a*c^3*d) + log(c*x^2 + b*x + a)/(b^2*d - 4*a*c*d)

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maple [A]  time = 0.05, size = 54, normalized size = 1.12 \begin {gather*} \frac {2 \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) d}-\frac {\ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right ) d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 48, normalized size = 1.00 \begin {gather*} \frac {\log \left (c x^{2} + b x + a\right )}{{\left (b^{2} - 4 \, a c\right )} d} - \frac {2 \, \log \left (2 \, c x + b\right )}{{\left (b^{2} - 4 \, a c\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.53, size = 47, normalized size = 0.98 \begin {gather*} \frac {2\,c\,d\,\ln \left (\frac {{\left (b+2\,c\,x\right )}^2}{c\,x^2+b\,x+a}\right )}{8\,a\,c^2\,d^2-2\,b^2\,c\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c*d*log((b + 2*c*x)^2/(a + b*x + c*x^2)))/(8*a*c^2*d^2 - 2*b^2*c*d^2)

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sympy [A]  time = 0.75, size = 42, normalized size = 0.88 \begin {gather*} \frac {2 \log {\left (\frac {b}{2 c} + x \right )}}{d \left (4 a c - b^{2}\right )} - \frac {\log {\left (\frac {a}{c} + \frac {b x}{c} + x^{2} \right )}}{d \left (4 a c - b^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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