3.9.34 \(\int \frac {A+B x}{A^2+2 A B x+B^2 x^2} \, dx\)

Optimal. Leaf size=10 \[ \frac {\log (A+B x)}{B} \]

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Rubi [A]  time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 31} \begin {gather*} \frac {\log (A+B x)}{B} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(A^2 + 2*A*B*x + B^2*x^2),x]

[Out]

Log[A + B*x]/B

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 {A+B x}{A^2+2 A B x+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 = 10, 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 + 2*A*B*x + 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+2 A B x+B^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)/(A^2 + 2*A*B*x + B^2*x^2),x]

[Out]

IntegrateAlgebraic[(A + B*x)/(A^2 + 2*A*B*x + B^2*x^2), x]

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fricas [A]  time = 0.47, size = 10, normalized size = 1.00 \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+2*A*B*x+A^2),x, algorithm="fricas")

[Out]

log(B*x + A)/B

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giac [B]  time = 0.15, size = 22, normalized size = 2.20 \begin {gather*} \frac {\log \left (A^{2} + {\left (B x^{2} + 2 \, A x\right )} B\right )}{2 \, B} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(B^2*x^2+2*A*B*x+A^2),x, algorithm="giac")

[Out]

1/2*log(A^2 + (B*x^2 + 2*A*x)*B)/B

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maple [A]  time = 0.08, size = 11, normalized size = 1.10 \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+2*A*B*x+A^2),x)

[Out]

1/B*ln(B*x+A)

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maxima [B]  time = 0.71, size = 22, normalized size = 2.20 \begin {gather*} \frac {\log \left (B^{2} x^{2} + 2 \, A B x + A^{2}\right )}{2 \, B} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(B^2*x^2+2*A*B*x+A^2),x, algorithm="maxima")

[Out]

1/2*log(B^2*x^2 + 2*A*B*x + A^2)/B

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mupad [B]  time = 0.02, size = 10, normalized size = 1.00 \begin {gather*} \frac {\ln \left (A+B\,x\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 + 2*A*B*x),x)

[Out]

log(A + B*x)/B

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sympy [A]  time = 0.07, size = 7, normalized size = 0.70 \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+2*A*B*x+A**2),x)

[Out]

log(A + B*x)/B

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