3.1805 $$\int \frac{1}{a c+(b c+a d) x+b d x^2} \, dx$$

Optimal. Leaf size=36 $\frac{\log (a+b x)}{b c-a d}-\frac{\log (c+d x)}{b c-a d}$

[Out]

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

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Rubi [A]  time = 0.0122863, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.095, Rules used = {616, 31} $\frac{\log (a+b x)}{b c-a d}-\frac{\log (c+d x)}{b c-a d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

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

Rubi steps

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

Mathematica [A]  time = 0.0109429, size = 26, normalized size = 0.72 $\frac{\log (a+b x)-\log (c+d x)}{b c-a d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 37, normalized size = 1. \begin{align*}{\frac{\ln \left ( dx+c \right ) }{ad-bc}}-{\frac{\ln \left ( bx+a \right ) }{ad-bc}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00709, size = 49, normalized size = 1.36 \begin{align*} \frac{\log \left (b x + a\right )}{b c - a d} - \frac{\log \left (d x + c\right )}{b c - a d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.56742, size = 58, normalized size = 1.61 \begin{align*} \frac{\log \left (b x + a\right ) - \log \left (d x + c\right )}{b c - a d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.410633, size = 128, normalized size = 3.56 \begin{align*} \frac{\log{\left (x + \frac{- \frac{a^{2} d^{2}}{a d - b c} + \frac{2 a b c d}{a d - b c} + a d - \frac{b^{2} c^{2}}{a d - b c} + b c}{2 b d} \right )}}{a d - b c} - \frac{\log{\left (x + \frac{\frac{a^{2} d^{2}}{a d - b c} - \frac{2 a b c d}{a d - b c} + a d + \frac{b^{2} c^{2}}{a d - b c} + b c}{2 b d} \right )}}{a d - b c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError