3.6.93 \(\int \frac {a+b x}{a^2-b^2 x^2} \, dx\)

Optimal. Leaf size=12 \[ -\frac {\log (a-b x)}{b} \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {627, 31} \begin {gather*} -\frac {\log (a-b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/(a^2 - b^2*x^2),x]

[Out]

-(Log[a - b*x]/b)

Rule 31

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.00 \begin {gather*} -\frac {\log (a-b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/(a^2 - b^2*x^2),x]

[Out]

-(Log[a - b*x]/b)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x}{a^2-b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)/(a^2 - b^2*x^2),x]

[Out]

IntegrateAlgebraic[(a + b*x)/(a^2 - b^2*x^2), x]

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fricas [A]  time = 0.40, size = 13, normalized size = 1.08 \begin {gather*} -\frac {\log \left (b x - a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(b*x - a)/b

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giac [A]  time = 0.18, size = 14, normalized size = 1.17 \begin {gather*} -\frac {\log \left ({\left | b x - a \right |}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(b*x - a))/b

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maple [A]  time = 0.04, size = 14, normalized size = 1.17 \begin {gather*} -\frac {\ln \left (b x -a \right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(-b^2*x^2+a^2),x)

[Out]

-1/b*ln(b*x-a)

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maxima [A]  time = 1.32, size = 13, normalized size = 1.08 \begin {gather*} -\frac {\log \left (b x - a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(b*x - a)/b

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mupad [B]  time = 0.03, size = 13, normalized size = 1.08 \begin {gather*} -\frac {\ln \left (b\,x-a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)/(a^2 - b^2*x^2),x)

[Out]

-log(b*x - a)/b

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sympy [A]  time = 0.08, size = 8, normalized size = 0.67 \begin {gather*} - \frac {\log {\left (- a + b x \right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(-b**2*x**2+a**2),x)

[Out]

-log(-a + b*x)/b

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