3.49 \(\int \frac {a+b \log (c x^n)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=76 \[ -\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}+\frac {b n \log (x)}{2 d^2 e}-\frac {b n \log (d+e x)}{2 d^2 e}+\frac {b n}{2 d e (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2319, 44} \[ -\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}+\frac {b n \log (x)}{2 d^2 e}-\frac {b n \log (d+e x)}{2 d^2 e}+\frac {b n}{2 d e (d+e x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(d + e*x)^3,x]

[Out]

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

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])

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx &=-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}+\frac {(b n) \int \frac {1}{x (d+e x)^2} \, dx}{2 e}\\ &=-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}+\frac {(b n) \int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx}{2 e}\\ &=\frac {b n}{2 d e (d+e x)}+\frac {b n \log (x)}{2 d^2 e}-\frac {a+b \log \left (c x^n\right )}{2 e (d+e x)^2}-\frac {b n \log (d+e x)}{2 d^2 e}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 53, normalized size = 0.70 \[ \frac {\frac {b n \left (\frac {d}{d+e x}-\log (d+e x)+\log (x)\right )}{d^2}-\frac {a+b \log \left (c x^n\right )}{(d+e x)^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(d + e*x)^3,x]

[Out]

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

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fricas [A]  time = 0.59, size = 107, normalized size = 1.41 \[ \frac {b d e n x + b d^{2} n - b d^{2} \log \relax (c) - a d^{2} - {\left (b e^{2} n x^{2} + 2 \, b d e n x + b d^{2} n\right )} \log \left (e x + d\right ) + {\left (b e^{2} n x^{2} + 2 \, b d e n x\right )} \log \relax (x)}{2 \, {\left (d^{2} e^{3} x^{2} + 2 \, d^{3} e^{2} x + d^{4} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*(b*d*e*n*x + b*d^2*n - b*d^2*log(c) - a*d^2 - (b*e^2*n*x^2 + 2*b*d*e*n*x + b*d^2*n)*log(e*x + d) + (b*e^2*
n*x^2 + 2*b*d*e*n*x)*log(x))/(d^2*e^3*x^2 + 2*d^3*e^2*x + d^4*e)

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giac [A]  time = 0.30, size = 120, normalized size = 1.58 \[ -\frac {b n x^{2} e^{2} \log \left (x e + d\right ) + 2 \, b d n x e \log \left (x e + d\right ) - b n x^{2} e^{2} \log \relax (x) - 2 \, b d n x e \log \relax (x) - b d n x e + b d^{2} n \log \left (x e + d\right ) - b d^{2} n + b d^{2} \log \relax (c) + a d^{2}}{2 \, {\left (d^{2} x^{2} e^{3} + 2 \, d^{3} x e^{2} + d^{4} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/(e*x+d)^3,x, algorithm="giac")

[Out]

-1/2*(b*n*x^2*e^2*log(x*e + d) + 2*b*d*n*x*e*log(x*e + d) - b*n*x^2*e^2*log(x) - 2*b*d*n*x*e*log(x) - b*d*n*x*
e + b*d^2*n*log(x*e + d) - b*d^2*n + b*d^2*log(c) + a*d^2)/(d^2*x^2*e^3 + 2*d^3*x*e^2 + d^4*e)

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maple [C]  time = 0.20, size = 235, normalized size = 3.09 \[ -\frac {b \ln \left (x^{n}\right )}{2 \left (e x +d \right )^{2} e}-\frac {-2 b \,e^{2} n \,x^{2} \ln \left (-x \right )+2 b \,e^{2} n \,x^{2} \ln \left (e x +d \right )-i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-4 b d e n x \ln \left (-x \right )+4 b d e n x \ln \left (e x +d \right )-2 b \,d^{2} n \ln \left (-x \right )+2 b \,d^{2} n \ln \left (e x +d \right )-2 b d e n x -2 b \,d^{2} n +2 b \,d^{2} \ln \relax (c )+2 a \,d^{2}}{4 \left (e x +d \right )^{2} d^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*x^n)+a)/(e*x+d)^3,x)

[Out]

-1/2*b/e/(e*x+d)^2*ln(x^n)-1/4*(I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I
*c*x^n)-I*Pi*b*d^2*csgn(I*c*x^n)^3+I*Pi*b*d^2*csgn(I*c)*csgn(I*c*x^n)^2-2*b*e^2*n*x^2*ln(-x)+2*b*e^2*n*x^2*ln(
e*x+d)-4*b*d*e*n*x*ln(-x)+4*b*d*e*n*x*ln(e*x+d)-2*b*d^2*n*ln(-x)+2*b*d^2*n*ln(e*x+d)-2*b*d*e*n*x+2*b*d^2*ln(c)
-2*b*d^2*n+2*a*d^2)/e/d^2/(e*x+d)^2

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maxima [A]  time = 0.56, size = 99, normalized size = 1.30 \[ \frac {1}{2} \, b n {\left (\frac {1}{d e^{2} x + d^{2} e} - \frac {\log \left (e x + d\right )}{d^{2} e} + \frac {\log \relax (x)}{d^{2} e}\right )} - \frac {b \log \left (c x^{n}\right )}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {a}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/(e*x+d)^3,x, algorithm="maxima")

[Out]

1/2*b*n*(1/(d*e^2*x + d^2*e) - log(e*x + d)/(d^2*e) + log(x)/(d^2*e)) - 1/2*b*log(c*x^n)/(e^3*x^2 + 2*d*e^2*x
+ d^2*e) - 1/2*a/(e^3*x^2 + 2*d*e^2*x + d^2*e)

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mupad [B]  time = 4.05, size = 91, normalized size = 1.20 \[ \frac {b\,n-a+\frac {b\,e\,n\,x}{d}}{2\,d^2\,e+4\,d\,e^2\,x+2\,e^3\,x^2}-\frac {b\,\ln \left (c\,x^n\right )}{2\,e\,\left (d^2+2\,d\,e\,x+e^2\,x^2\right )}-\frac {b\,n\,\mathrm {atanh}\left (\frac {2\,e\,x}{d}+1\right )}{d^2\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))/(d + e*x)^3,x)

[Out]

(b*n - a + (b*e*n*x)/d)/(2*d^2*e + 2*e^3*x^2 + 4*d*e^2*x) - (b*log(c*x^n))/(2*e*(d^2 + e^2*x^2 + 2*d*e*x)) - (
b*n*atanh((2*e*x)/d + 1))/(d^2*e)

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sympy [A]  time = 6.63, size = 515, normalized size = 6.78 \[ \begin {cases} \tilde {\infty } \left (- \frac {a}{2 x^{2}} - \frac {b n \log {\relax (x )}}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\relax (c )}}{2 x^{2}}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {- \frac {a}{2 x^{2}} - \frac {b n \log {\relax (x )}}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\relax (c )}}{2 x^{2}}}{e^{3}} & \text {for}\: d = 0 \\\frac {a x + b n x \log {\relax (x )} - b n x + b x \log {\relax (c )}}{d^{3}} & \text {for}\: e = 0 \\- \frac {a d^{2}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} - \frac {b d^{2} n \log {\left (\frac {d}{e} + x \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {b d^{2} n}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {2 b d e n x \log {\relax (x )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} - \frac {2 b d e n x \log {\left (\frac {d}{e} + x \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {b d e n x}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {2 b d e x \log {\relax (c )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {b e^{2} n x^{2} \log {\relax (x )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} - \frac {b e^{2} n x^{2} \log {\left (\frac {d}{e} + x \right )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} + \frac {b e^{2} x^{2} \log {\relax (c )}}{2 d^{4} e + 4 d^{3} e^{2} x + 2 d^{2} e^{3} x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/(e*x+d)**3,x)

[Out]

Piecewise((zoo*(-a/(2*x**2) - b*n*log(x)/(2*x**2) - b*n/(4*x**2) - b*log(c)/(2*x**2)), Eq(d, 0) & Eq(e, 0)), (
(-a/(2*x**2) - b*n*log(x)/(2*x**2) - b*n/(4*x**2) - b*log(c)/(2*x**2))/e**3, Eq(d, 0)), ((a*x + b*n*x*log(x) -
b*n*x + b*x*log(c))/d**3, Eq(e, 0)), (-a*d**2/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) - b*d**2*n*log(d/
e + x)/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) + b*d**2*n/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2)
+ 2*b*d*e*n*x*log(x)/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) - 2*b*d*e*n*x*log(d/e + x)/(2*d**4*e + 4*d*
*3*e**2*x + 2*d**2*e**3*x**2) + b*d*e*n*x/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) + 2*b*d*e*x*log(c)/(2*
d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) + b*e**2*n*x**2*log(x)/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2
) - b*e**2*n*x**2*log(d/e + x)/(2*d**4*e + 4*d**3*e**2*x + 2*d**2*e**3*x**2) + b*e**2*x**2*log(c)/(2*d**4*e +
4*d**3*e**2*x + 2*d**2*e**3*x**2), True))

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