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*ln(c*x^n))/d/(e*x+d)-b*n*ln(e*x+d)/d/e

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 31

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

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.47, size = 51, normalized size = 1.31 \[ \frac {b e n x \log \relax (x) - b d \log \relax (c) - a d - {\left (b e n x + b d n\right )} \log \left (e x + d\right )}{d e^{2} x + d^{2} e} \]

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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giac [A]  time = 0.29, size = 58, normalized size = 1.49 \[ -\frac {b n x e \log \left (x e + d\right ) - b n x e \log \relax (x) + b d n \log \left (x e + d\right ) + b d \log \relax (c) + a d}{d x e^{2} + d^{2} e} \]

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)

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maple [C]  time = 0.20, size = 173, normalized size = 4.44 \[ -\frac {b \ln \left (x^{n}\right )}{\left (e x +d \right ) e}-\frac {-i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b d \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b d \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-2 b e n x \ln \left (-x \right )+2 b e n x \ln \left (e x +d \right )-2 b d n \ln \left (-x \right )+2 b d n \ln \left (e x +d \right )+2 b d \ln \relax (c )+2 a d}{2 \left (e x +d \right ) d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.54, size = 63, normalized size = 1.62 \[ -b n {\left (\frac {\log \left (e x + d\right )}{d e} - \frac {\log \relax (x)}{d e}\right )} - \frac {b \log \left (c x^{n}\right )}{e^{2} x + d e} - \frac {a}{e^{2} x + d e} \]

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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mupad [B]  time = 4.56, size = 54, normalized size = 1.38 \[ -\frac {a}{x\,e^2+d\,e}-\frac {b\,\ln \left (c\,x^n\right )}{e\,\left (d+e\,x\right )}-\frac {2\,b\,n\,\mathrm {atanh}\left (\frac {2\,e\,x}{d}+1\right )}{d\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))/(d + e*x)^2,x)

[Out]

- a/(d*e + e^2*x) - (b*log(c*x^n))/(e*(d + e*x)) - (2*b*n*atanh((2*e*x)/d + 1))/(d*e)

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sympy [A]  time = 2.35, size = 187, normalized size = 4.79 \[ \begin {cases} \tilde {\infty } \left (- \frac {a}{x} - \frac {b n \log {\relax (x )}}{x} - \frac {b n}{x} - \frac {b \log {\relax (c )}}{x}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {a x + b n x \log {\relax (x )} - b n x + b x \log {\relax (c )}}{d^{2}} & \text {for}\: e = 0 \\\frac {- \frac {a}{x} - \frac {b n \log {\relax (x )}}{x} - \frac {b n}{x} - \frac {b \log {\relax (c )}}{x}}{e^{2}} & \text {for}\: d = 0 \\- \frac {a d}{d^{2} e + d e^{2} x} - \frac {b d n \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac {b e n x \log {\relax (x )}}{d^{2} e + d e^{2} x} - \frac {b e n x \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac {b e x \log {\relax (c )}}{d^{2} e + d e^{2} x} & \text {otherwise} \end {cases} \]

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