3.15.77 \(\int \frac {A+B x}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=32 \[ \frac {B \log (a+b x)}{b^2}-\frac {A b-a B}{b^2 (a+b x)} \]

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 43} \begin {gather*} \frac {B \log (a+b x)}{b^2}-\frac {A b-a B}{b^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b - a*B)/(b^2*(a + b*x))) + (B*Log[a + b*x])/b^2

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.97 \begin {gather*} \frac {a B-A b}{b^2 (a+b x)}+\frac {B \log (a+b x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b) + a*B)/(b^2*(a + b*x)) + (B*Log[a + b*x])/b^2

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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 = 0.40, size = 37, normalized size = 1.16 \begin {gather*} \frac {B a - A b + {\left (B b x + B a\right )} \log \left (b x + a\right )}{b^{3} x + a b^{2}} \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]

(B*a - A*b + (B*b*x + B*a)*log(b*x + a))/(b^3*x + a*b^2)

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giac [A]  time = 0.15, size = 32, normalized size = 1.00 \begin {gather*} \frac {B \log \left ({\left | b x + a \right |}\right )}{b^{2}} + \frac {B a - A b}{{\left (b x + a\right )} b^{2}} \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]

B*log(abs(b*x + a))/b^2 + (B*a - A*b)/((b*x + a)*b^2)

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maple [A]  time = 0.05, size = 39, normalized size = 1.22 \begin {gather*} -\frac {A}{\left (b x +a \right ) b}+\frac {B a}{\left (b x +a \right ) b^{2}}+\frac {B \ln \left (b x +a \right )}{b^{2}} \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]

B/b^2*ln(b*x+a)-1/b/(b*x+a)*A+1/b^2/(b*x+a)*B*a

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maxima [A]  time = 0.52, size = 34, normalized size = 1.06 \begin {gather*} \frac {B a - A b}{b^{3} x + a b^{2}} + \frac {B \log \left (b x + a\right )}{b^{2}} \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]

(B*a - A*b)/(b^3*x + a*b^2) + B*log(b*x + a)/b^2

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mupad [B]  time = 2.05, size = 32, normalized size = 1.00 \begin {gather*} \frac {B\,\ln \left (a+b\,x\right )}{b^2}-\frac {A\,b-B\,a}{b^2\,\left (a+b\,x\right )} \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]

(B*log(a + b*x))/b^2 - (A*b - B*a)/(b^2*(a + b*x))

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sympy [A]  time = 0.18, size = 27, normalized size = 0.84 \begin {gather*} \frac {B \log {\left (a + b x \right )}}{b^{2}} + \frac {- A b + B a}{a b^{2} + b^{3} x} \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]

B*log(a + b*x)/b**2 + (-A*b + B*a)/(a*b**2 + b**3*x)

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