3.7.17 \(\int \frac {\log (x)}{(a+b x)^2} \, dx\) [617]

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

[Out]

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

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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 = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2351, 31} \begin {gather*} \frac {x \log (x)}{a (a+b x)}-\frac {\log (a+b x)}{a b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[x]/(a + b*x)^2,x]

[Out]

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

Rule 31

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

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 27, normalized size = 0.93 \begin {gather*} \frac {\frac {x \log (x)}{a+b x}-\frac {\log (a+b x)}{b}}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[x]/(a + b*x)^2,x]

[Out]

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

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Mathics [A]
time = 1.92, size = 41, normalized size = 1.41 \begin {gather*} \frac {-a \text {Log}\left [x\right ]+\left (a+b x\right ) \left (\text {Log}\left [x\right ]-\text {Log}\left [\frac {a+b x}{b}\right ]\right )}{a b \left (a+b x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[Log[x]/(a + b*x)^2,x]')

[Out]

(-a Log[x] + (a + b x) (Log[x] - Log[(a + b x) / b])) / (a b (a + b x))

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Maple [A]
time = 0.08, size = 30, normalized size = 1.03

method	result	size
default	\(\frac {x \ln \left (x \right )}{a \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{a b}\)	\(30\)
norman	\(\frac {x \ln \left (x \right )}{a \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{a b}\)	\(30\)
risch	\(-\frac {\ln \left (x \right )}{b \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{a b}+\frac {\ln \left (-x \right )}{b a}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 38, normalized size = 1.31 \begin {gather*} -\frac {\frac {\log \left (b x + a\right )}{a} - \frac {\log \left (x\right )}{a}}{b} - \frac {\log \left (x\right )}{{\left (b x + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.32, size = 34, normalized size = 1.17 \begin {gather*} \frac {b x \log \left (x\right ) - {\left (b x + a\right )} \log \left (b x + a\right )}{a b^{2} x + a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.12, size = 24, normalized size = 0.83 \begin {gather*} - \frac {\log {\left (x \right )}}{a b + b^{2} x} + \frac {\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}}{a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 37, normalized size = 1.28 \begin {gather*} \frac {\ln \left |x\right |}{a b}-\frac {b \ln \left |x b+a\right |}{a b^{2}}-\frac {\ln x}{b \left (x b+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)/(b*x+a)^2,x)

[Out]

-log(x)/((b*x + a)*b) - log(abs(b*x + a))/(a*b) + log(abs(x))/(a*b)

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Mupad [B]
time = 0.39, size = 35, normalized size = 1.21 \begin {gather*} \frac {x^2\,\ln \left (x\right )}{a\,\left (b\,x^2+a\,x\right )}-\frac {\ln \left (a+b\,x\right )}{a\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)/(a + b*x)^2,x)

[Out]

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

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