3.1.44 \(\int \frac {a+b \log (c x^n)}{x^2 (d+e x)^2} \, dx\) [44]

Optimal. Leaf size=114 \[ -\frac {b n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{d^2 x}+\frac {e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}+\frac {2 e \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {b e n \log (d+e x)}{d^3}-\frac {2 b e n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^3} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {46, 2393, 2341, 2351, 31, 2379, 2438} \begin {gather*} -\frac {2 b e n \text {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^3}+\frac {e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}+\frac {2 e \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {a+b \log \left (c x^n\right )}{d^2 x}-\frac {b e n \log (d+e x)}{d^3}-\frac {b n}{d^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*n)/(d^2*x)) - (a + b*Log[c*x^n])/(d^2*x) + (e^2*x*(a + b*Log[c*x^n]))/(d^3*(d + e*x)) + (2*e*Log[1 + d/(e
*x)]*(a + b*Log[c*x^n]))/d^3 - (b*e*n*Log[d + e*x])/d^3 - (2*b*e*n*PolyLog[2, -(d/(e*x))])/d^3

Rule 31

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

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]

Rule 2379

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

Rule 2393

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^2} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{d^2 x^2}-\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^2 (d+e x)^2}+\frac {2 e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{d^2}-\frac {(2 e) \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{d^3}+\frac {\left (2 e^2\right ) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{d^3}+\frac {e^2 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{d^2}\\ &=-\frac {b n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{d^2 x}+\frac {e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{b d^3 n}+\frac {2 e \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^3}-\frac {(2 b e n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d^3}-\frac {\left (b e^2 n\right ) \int \frac {1}{d+e x} \, dx}{d^3}\\ &=-\frac {b n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{d^2 x}+\frac {e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{b d^3 n}-\frac {b e n \log (d+e x)}{d^3}+\frac {2 e \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^3}+\frac {2 b e n \text {Li}_2\left (-\frac {e x}{d}\right )}{d^3}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 120, normalized size = 1.05 \begin {gather*} -\frac {\frac {b d n}{x}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {d e \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{b n}-b e n (\log (x)-\log (d+e x))-2 e \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 b e n \text {Li}_2\left (-\frac {e x}{d}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*d*n)/x + (d*(a + b*Log[c*x^n]))/x + (d*e*(a + b*Log[c*x^n]))/(d + e*x) + (e*(a + b*Log[c*x^n])^2)/(b*n)
- b*e*n*(Log[x] - Log[d + e*x]) - 2*e*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] - 2*b*e*n*PolyLog[2, -((e*x)/d)])/d^
3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 703, normalized size = 6.17

method	result	size
risch	\(-\frac {2 b n e \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{3}}-\frac {2 b n e \dilog \left (-\frac {e x}{d}\right )}{d^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2} x}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2} x}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e \ln \left (x \right )}{d^{3}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e}{2 d^{2} \left (e x +d \right )}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e \ln \left (e x +d \right )}{d^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e}{2 d^{2} \left (e x +d \right )}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (x \right )}{d^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 d^{2} x}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (e x +d \right )}{d^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (x \right )}{d^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e}{2 d^{2} \left (e x +d \right )}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 d^{2} x}-\frac {a}{d^{2} x}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (e x +d \right )}{d^{3}}+\frac {2 b \ln \left (x^{n}\right ) e \ln \left (e x +d \right )}{d^{3}}-\frac {2 b \ln \left (x^{n}\right ) e \ln \left (x \right )}{d^{3}}-\frac {b \ln \left (x^{n}\right ) e}{d^{2} \left (e x +d \right )}+\frac {2 a e \ln \left (e x +d \right )}{d^{3}}-\frac {2 a e \ln \left (x \right )}{d^{3}}-\frac {a e}{d^{2} \left (e x +d \right )}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) e \ln \left (e x +d \right )}{d^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) e}{2 d^{2} \left (e x +d \right )}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) e \ln \left (x \right )}{d^{3}}+\frac {2 b \ln \left (c \right ) e \ln \left (e x +d \right )}{d^{3}}-\frac {2 b \ln \left (c \right ) e \ln \left (x \right )}{d^{3}}-\frac {b \ln \left (c \right ) e}{d^{2} \left (e x +d \right )}+\frac {b n e \ln \left (x \right )^{2}}{d^{3}}+\frac {b n e \ln \left (x \right )}{d^{3}}-\frac {b \ln \left (x^{n}\right )}{d^{2} x}-\frac {b e n \ln \left (e x +d \right )}{d^{3}}-\frac {b \ln \left (c \right )}{d^{2} x}-\frac {b n}{d^{2} x}\)	\(703\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 42.01, size = 318, normalized size = 2.79 \begin {gather*} \frac {a e^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a}{d^{2} x} + \frac {2 a e^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {2 a e \log {\left (x \right )}}{d^{3}} - \frac {b e^{2} n \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\left (x \right )}}{d e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {b e^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} - \frac {b n}{d^{2} x} - \frac {b \log {\left (c x^{n} \right )}}{d^{2} x} - \frac {2 b e^{2} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {2 b e^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} + \frac {b e n \log {\left (x \right )}^{2}}{d^{3}} - \frac {2 b e \log {\left (x \right )} \log {\left (c x^{n} \right )}}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/d**2 - a/(d**2*x) + 2*a*e**2*Piecewise((x/d, E
q(e, 0)), (log(d + e*x)/e, True))/d**3 - 2*a*e*log(x)/d**3 - b*e**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(
d*e) + log(d/e + x)/(d*e), True))/d**2 + b*e**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c
*x**n)/d**2 - b*n/(d**2*x) - b*log(c*x**n)/(d**2*x) - 2*b*e**2*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((-polyl
og(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) - polylog(2, e*x*exp_polar(I*pi)/
d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (1, 1)),
((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*pi)/d),
True))/e, True))/d**3 + 2*b*e**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/d**3 + b*e*n*l
og(x)**2/d**3 - 2*b*e*log(x)*log(c*x**n)/d**3

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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