3.42 \(\int \frac{a+b \log (c x^n)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=39 \[ \frac{x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac{b n \log (d+e x)}{d e} \]

[Out]

(x*(a + b*Log[c*x^n]))/(d*(d + e*x)) - (b*n*Log[d + e*x])/(d*e)

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Rubi [A]  time = 0.0187661, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2314, 31} \[ \frac{x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac{b n \log (d+e x)}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(d + e*x)^2,x]

[Out]

(x*(a + b*Log[c*x^n]))/(d*(d + e*x)) - (b*n*Log[d + e*x])/(d*e)

Rule 2314

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]

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

Mathematica [A]  time = 0.0287581, size = 41, normalized size = 1.05 \[ \frac{\frac{b n (\log (x)-\log (d+e x))}{d}-\frac{a+b \log \left (c x^n\right )}{d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(d + e*x)^2,x]

[Out]

(-((a + b*Log[c*x^n])/(d + e*x)) + (b*n*(Log[x] - Log[d + e*x]))/d)/e

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Maple [C]  time = 0.1, size = 173, normalized size = 4.4 \begin{align*} -{\frac{b\ln \left ({x}^{n} \right ) }{ \left ( ex+d \right ) e}}-{\frac{i\pi \,bd{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-i\pi \,bd{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -i\pi \,bd \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+i\pi \,bd \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -2\,\ln \left ( -x \right ) benx+2\,\ln \left ( ex+d \right ) benx-2\,\ln \left ( -x \right ) bdn+2\,\ln \left ( ex+d \right ) bdn+2\,\ln \left ( c \right ) bd+2\,ad}{ \left ( 2\,ex+2\,d \right ) ed}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/(e*x+d)^2,x)

[Out]

-b/e/(e*x+d)*ln(x^n)-1/2*(I*Pi*b*d*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*d*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*
Pi*b*d*csgn(I*c*x^n)^3+I*Pi*b*d*csgn(I*c*x^n)^2*csgn(I*c)-2*ln(-x)*b*e*n*x+2*ln(e*x+d)*b*e*n*x-2*ln(-x)*b*d*n+
2*ln(e*x+d)*b*d*n+2*ln(c)*b*d+2*a*d)/(e*x+d)/e/d

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Maxima [A]  time = 1.14125, size = 85, normalized size = 2.18 \begin{align*} -b n{\left (\frac{\log \left (e x + d\right )}{d e} - \frac{\log \left (x\right )}{d e}\right )} - \frac{b \log \left (c x^{n}\right )}{e^{2} x + d e} - \frac{a}{e^{2} x + d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*n*(log(e*x + d)/(d*e) - log(x)/(d*e)) - b*log(c*x^n)/(e^2*x + d*e) - a/(e^2*x + d*e)

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Fricas [A]  time = 1.04868, size = 119, normalized size = 3.05 \begin{align*} \frac{b e n x \log \left (x\right ) - b d \log \left (c\right ) - a d -{\left (b e n x + b d n\right )} \log \left (e x + d\right )}{d e^{2} x + d^{2} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*e*n*x*log(x) - b*d*log(c) - a*d - (b*e*n*x + b*d*n)*log(e*x + d))/(d*e^2*x + d^2*e)

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Sympy [A]  time = 4.00992, size = 189, normalized size = 4.85 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{a}{x} - \frac{b n \log{\left (x \right )}}{x} - \frac{b n}{x} - \frac{b \log{\left (c \right )}}{x}\right ) & \text{for}\: d = 0 \wedge e = 0 \\\frac{- \frac{a}{x} - \frac{b n \log{\left (x \right )}}{x} - \frac{b n}{x} - \frac{b \log{\left (c \right )}}{x}}{e^{2}} & \text{for}\: d = 0 \\\frac{a x + b n x \log{\left (x \right )} - b n x + b x \log{\left (c \right )}}{d^{2}} & \text{for}\: e = 0 \\\frac{a e x}{d^{2} e + d e^{2} x} - \frac{b d n \log{\left (\frac{d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac{b e n x \log{\left (x \right )}}{d^{2} e + d e^{2} x} - \frac{b e n x \log{\left (\frac{d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac{b e x \log{\left (c \right )}}{d^{2} e + d e^{2} x} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/(e*x+d)**2,x)

[Out]

Piecewise((zoo*(-a/x - b*n*log(x)/x - b*n/x - b*log(c)/x), Eq(d, 0) & Eq(e, 0)), ((-a/x - b*n*log(x)/x - b*n/x
- b*log(c)/x)/e**2, Eq(d, 0)), ((a*x + b*n*x*log(x) - b*n*x + b*x*log(c))/d**2, Eq(e, 0)), (a*e*x/(d**2*e + d
*e**2*x) - b*d*n*log(d/e + x)/(d**2*e + d*e**2*x) + b*e*n*x*log(x)/(d**2*e + d*e**2*x) - b*e*n*x*log(d/e + x)/
(d**2*e + d*e**2*x) + b*e*x*log(c)/(d**2*e + d*e**2*x), True))

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Giac [A]  time = 1.29645, size = 78, normalized size = 2. \begin{align*} -\frac{b n x e \log \left (x e + d\right ) - b n x e \log \left (x\right ) + b d n \log \left (x e + d\right ) + b d \log \left (c\right ) + a d}{d x e^{2} + d^{2} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/(e*x+d)^2,x, algorithm="giac")

[Out]

-(b*n*x*e*log(x*e + d) - b*n*x*e*log(x) + b*d*n*log(x*e + d) + b*d*log(c) + a*d)/(d*x*e^2 + d^2*e)