3.2.75 \(\int \frac {1}{x (a+b x)^2} \, dx\)

Optimal. Leaf size=29 \[ -\frac {\log (a+b x)}{a^2}+\frac {\log (x)}{a^2}+\frac {1}{a (a+b x)} \]

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {44} \begin {gather*} -\frac {\log (a+b x)}{a^2}+\frac {\log (x)}{a^2}+\frac {1}{a (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x)^2),x]

[Out]

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.83 \begin {gather*} \frac {\frac {a}{a+b x}-\log (a+b x)+\log (x)}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x)^2),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x (a+b x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x*(a + b*x)^2),x]

[Out]

IntegrateAlgebraic[1/(x*(a + b*x)^2), x]

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fricas [A]  time = 1.28, size = 39, normalized size = 1.34 \begin {gather*} -\frac {{\left (b x + a\right )} \log \left (b x + a\right ) - {\left (b x + a\right )} \log \relax (x) - a}{a^{2} b x + a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.00, size = 38, normalized size = 1.31 \begin {gather*} b {\left (\frac {\log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{2} b} + \frac {1}{{\left (b x + a\right )} a b}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(log(abs(-a/(b*x + a) + 1))/(a^2*b) + 1/((b*x + a)*a*b))

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maple [A]  time = 0.01, size = 30, normalized size = 1.03 \begin {gather*} \frac {1}{\left (b x +a \right ) a}+\frac {\ln \relax (x )}{a^{2}}-\frac {\ln \left (b x +a \right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.30, size = 28, normalized size = 0.97 \begin {gather*} \frac {1}{a b x + a^{2}} - \frac {\log \left (b x + a\right )}{a^{2}} + \frac {\log \relax (x)}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.12, size = 26, normalized size = 0.90 \begin {gather*} \frac {1}{a^2+b\,x\,a}-\frac {\ln \left (\frac {a+b\,x}{x}\right )}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x)^2),x)

[Out]

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

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sympy [A]  time = 0.22, size = 22, normalized size = 0.76 \begin {gather*} \frac {1}{a^{2} + a b x} + \frac {\log {\relax (x )} - \log {\left (\frac {a}{b} + x \right )}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a**2 + a*b*x) + (log(x) - log(a/b + x))/a**2

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