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

Optimal. Leaf size=65 \[ -\frac {\log (x) (2 A b-a B)}{a^3}+\frac {(2 A b-a B) \log (a+b x)}{a^3}-\frac {A b-a B}{a^2 (a+b x)}-\frac {A}{a^2 x} \]

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Rubi [A]  time = 0.05, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {27, 77} \begin {gather*} -\frac {A b-a B}{a^2 (a+b x)}-\frac {\log (x) (2 A b-a B)}{a^3}+\frac {(2 A b-a B) \log (a+b x)}{a^3}-\frac {A}{a^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(a^2*x)) - (A*b - a*B)/(a^2*(a + b*x)) - ((2*A*b - a*B)*Log[x])/a^3 + ((2*A*b - a*B)*Log[a + b*x])/a^3

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac {A+B x}{x^2 (a+b x)^2} \, dx\\ &=\int \left (\frac {A}{a^2 x^2}+\frac {-2 A b+a B}{a^3 x}-\frac {b (-A b+a B)}{a^2 (a+b x)^2}-\frac {b (-2 A b+a B)}{a^3 (a+b x)}\right ) \, dx\\ &=-\frac {A}{a^2 x}-\frac {A b-a B}{a^2 (a+b x)}-\frac {(2 A b-a B) \log (x)}{a^3}+\frac {(2 A b-a B) \log (a+b x)}{a^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.86 \begin {gather*} \frac {\frac {a (a B-A b)}{a+b x}+\log (x) (a B-2 A b)+(2 A b-a B) \log (a+b x)-\frac {a A}{x}}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*A)/x) + (a*(-(A*b) + a*B))/(a + b*x) + (-2*A*b + a*B)*Log[x] + (2*A*b - a*B)*Log[a + b*x])/a^3

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 107, normalized size = 1.65 \begin {gather*} -\frac {A a^{2} - {\left (B a^{2} - 2 \, A a b\right )} x + {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{2} + {\left (B a^{2} - 2 \, A a b\right )} x\right )} \log \left (b x + a\right ) - {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{2} + {\left (B a^{2} - 2 \, A a b\right )} x\right )} \log \relax (x)}{a^{3} b x^{2} + a^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(A*a^2 - (B*a^2 - 2*A*a*b)*x + ((B*a*b - 2*A*b^2)*x^2 + (B*a^2 - 2*A*a*b)*x)*log(b*x + a) - ((B*a*b - 2*A*b^2
)*x^2 + (B*a^2 - 2*A*a*b)*x)*log(x))/(a^3*b*x^2 + a^4*x)

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giac [A]  time = 0.15, size = 71, normalized size = 1.09 \begin {gather*} \frac {{\left (B a - 2 \, A b\right )} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {B a x - 2 \, A b x - A a}{{\left (b x^{2} + a x\right )} a^{2}} - \frac {{\left (B a b - 2 \, A b^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 78, normalized size = 1.20 \begin {gather*} -\frac {A b}{\left (b x +a \right ) a^{2}}-\frac {2 A b \ln \relax (x )}{a^{3}}+\frac {2 A b \ln \left (b x +a \right )}{a^{3}}+\frac {B}{\left (b x +a \right ) a}+\frac {B \ln \relax (x )}{a^{2}}-\frac {B \ln \left (b x +a \right )}{a^{2}}-\frac {A}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a^3*ln(b*x+a)*A*b-1/a^2*ln(b*x+a)*B-1/a^2/(b*x+a)*A*b+1/a/(b*x+a)*B-A/a^2/x-2/a^3*ln(x)*A*b+B/a^2*ln(x)

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maxima [A]  time = 0.53, size = 67, normalized size = 1.03 \begin {gather*} -\frac {A a - {\left (B a - 2 \, A b\right )} x}{a^{2} b x^{2} + a^{3} x} - \frac {{\left (B a - 2 \, A b\right )} \log \left (b x + a\right )}{a^{3}} + \frac {{\left (B a - 2 \, A b\right )} \log \relax (x)}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.12, size = 58, normalized size = 0.89 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )\,\left (2\,A\,b-B\,a\right )}{a^3}-\frac {\frac {A}{a}+\frac {x\,\left (2\,A\,b-B\,a\right )}{a^2}}{b\,x^2+a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh((2*b*x)/a + 1)*(2*A*b - B*a))/a^3 - (A/a + (x*(2*A*b - B*a))/a^2)/(a*x + b*x^2)

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sympy [B]  time = 0.49, size = 128, normalized size = 1.97 \begin {gather*} \frac {- A a + x \left (- 2 A b + B a\right )}{a^{3} x + a^{2} b x^{2}} + \frac {\left (- 2 A b + B a\right ) \log {\left (x + \frac {- 2 A a b + B a^{2} - a \left (- 2 A b + B a\right )}{- 4 A b^{2} + 2 B a b} \right )}}{a^{3}} - \frac {\left (- 2 A b + B a\right ) \log {\left (x + \frac {- 2 A a b + B a^{2} + a \left (- 2 A b + B a\right )}{- 4 A b^{2} + 2 B a b} \right )}}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**2/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

(-A*a + x*(-2*A*b + B*a))/(a**3*x + a**2*b*x**2) + (-2*A*b + B*a)*log(x + (-2*A*a*b + B*a**2 - a*(-2*A*b + B*a
))/(-4*A*b**2 + 2*B*a*b))/a**3 - (-2*A*b + B*a)*log(x + (-2*A*a*b + B*a**2 + a*(-2*A*b + B*a))/(-4*A*b**2 + 2*
B*a*b))/a**3

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