3.336 \(\int \frac {a+b \log (c x^n)}{(d+\frac {e}{x}) x^4} \, dx\)

Optimal. Leaf size=135 \[ \frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}-\frac {b d^2 n \text {Li}_2\left (-\frac {d x}{e}\right )}{e^3}+\frac {b d n}{e^2 x}-\frac {b n}{4 e x^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {263, 44, 2351, 2304, 2301, 2317, 2391} \[ -\frac {b d^2 n \text {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^3}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}+\frac {b d n}{e^2 x}-\frac {b n}{4 e x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/((d + e/x)*x^4),x]

[Out]

-(b*n)/(4*e*x^2) + (b*d*n)/(e^2*x) - (a + b*Log[c*x^n])/(2*e*x^2) + (d*(a + b*Log[c*x^n]))/(e^2*x) + (d^2*(a +
b*Log[c*x^n])^2)/(2*b*e^3*n) - (d^2*(a + b*Log[c*x^n])*Log[1 + (d*x)/e])/e^3 - (b*d^2*n*PolyLog[2, -((d*x)/e)
])/e^3

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 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2304

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 2317

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

Rule 2351

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 2391

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 )}{\left (d+\frac {e}{x}\right ) x^4} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{e x^3}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x^2}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 x}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^3 (e+d x)}\right ) \, dx\\ &=\frac {d^2 \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{e^3}-\frac {d^3 \int \frac {a+b \log \left (c x^n\right )}{e+d x} \, dx}{e^3}-\frac {d \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{e^2}+\frac {\int \frac {a+b \log \left (c x^n\right )}{x^3} \, dx}{e}\\ &=-\frac {b n}{4 e x^2}+\frac {b d n}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^3}+\frac {\left (b d^2 n\right ) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{e^3}\\ &=-\frac {b n}{4 e x^2}+\frac {b d n}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^3}-\frac {b d^2 n \text {Li}_2\left (-\frac {d x}{e}\right )}{e^3}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 124, normalized size = 0.92 \[ -\frac {4 d^2 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}-\frac {4 d e \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {2 e^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}+4 b d^2 n \text {Li}_2\left (-\frac {d x}{e}\right )-\frac {4 b d e n}{x}+\frac {b e^2 n}{x^2}}{4 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/((d + e/x)*x^4),x]

[Out]

-1/4*((b*e^2*n)/x^2 - (4*b*d*e*n)/x + (2*e^2*(a + b*Log[c*x^n]))/x^2 - (4*d*e*(a + b*Log[c*x^n]))/x - (2*d^2*(
a + b*Log[c*x^n])^2)/(b*n) + 4*d^2*(a + b*Log[c*x^n])*Log[1 + (d*x)/e] + 4*b*d^2*n*PolyLog[2, -((d*x)/e)])/e^3

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{d x^{4} + e x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/(d+e/x)/x^4,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/(d+e/x)/x^4,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.20, size = 689, normalized size = 5.10 \[ \frac {b \,d^{2} \ln \relax (c ) \ln \relax (x )}{e^{3}}+\frac {b d \ln \relax (c )}{e^{2} x}-\frac {b \,d^{2} \ln \relax (c ) \ln \left (d x +e \right )}{e^{3}}-\frac {b \,d^{2} n \ln \relax (x )^{2}}{2 e^{3}}-\frac {a \,d^{2} \ln \left (d x +e \right )}{e^{3}}+\frac {a \,d^{2} \ln \relax (x )}{e^{3}}+\frac {a d}{e^{2} x}+\frac {b \,d^{2} n \ln \left (-\frac {d x}{e}\right ) \ln \left (d x +e \right )}{e^{3}}-\frac {a}{2 e \,x^{2}}-\frac {b \ln \relax (c )}{2 e \,x^{2}}-\frac {b \ln \left (x^{n}\right )}{2 e \,x^{2}}-\frac {b \,d^{2} \ln \left (x^{n}\right ) \ln \left (d x +e \right )}{e^{3}}+\frac {b \,d^{2} \ln \relax (x ) \ln \left (x^{n}\right )}{e^{3}}+\frac {b d \ln \left (x^{n}\right )}{e^{2} x}-\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2 e^{3}}+\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (d x +e \right )}{2 e^{3}}-\frac {i \pi b d \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 e^{2} x}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e \,x^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e \,x^{2}}-\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 )}{2 e^{2} x}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 e \,x^{2}}-\frac {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 ) \ln \relax (x )}{2 e^{3}}+\frac {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 ) \ln \left (d x +e \right )}{2 e^{3}}+\frac {i \pi b d \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e^{2} x}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 e \,x^{2}}-\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (d x +e \right )}{2 e^{3}}+\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 e^{3}}-\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (d x +e \right )}{2 e^{3}}+\frac {i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e^{2} x}+\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 e^{3}}+\frac {b \,d^{2} n \dilog \left (-\frac {d x}{e}\right )}{e^{3}}-\frac {b n}{4 e \,x^{2}}+\frac {b d n}{e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*x^n)+a)/(d+e/x)/x^4,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {2 \, d^{2} \log \left (d x + e\right )}{e^{3}} - \frac {2 \, d^{2} \log \relax (x)}{e^{3}} - \frac {2 \, d x - e}{e^{2} x^{2}}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{d x^{4} + e x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/(d+e/x)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\left (d+\frac {e}{x}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 88.47, size = 246, normalized size = 1.82 \[ - \frac {a d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {a d^{2} \log {\relax (x )}}{e^{3}} + \frac {a d}{e^{2} x} - \frac {a}{2 e x^{2}} + \frac {b d^{3} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} \log {\relax (e )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (e )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (e )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (e )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} - \frac {b d^{2} n \log {\relax (x )}^{2}}{2 e^{3}} + \frac {b d^{2} \log {\relax (x )} \log {\left (c x^{n} \right )}}{e^{3}} + \frac {b d n}{e^{2} x} + \frac {b d \log {\left (c x^{n} \right )}}{e^{2} x} - \frac {b n}{4 e x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 e x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/(d+e/x)/x**4,x)

[Out]

-a*d**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/e**3 + a*d**2*log(x)/e**3 + a*d/(e**2*x) - a/(2*e*x
**2) + b*d**3*n*Piecewise((x/e, Eq(d, 0)), (Piecewise((log(e)*log(x) - polylog(2, d*x*exp_polar(I*pi)/e), Abs(
x) < 1), (-log(e)*log(1/x) - polylog(2, d*x*exp_polar(I*pi)/e), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0)
, ()), x)*log(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) - polylog(2, d*x*exp_polar(I*pi)/e), True))/d
, True))/e**3 - b*d**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x**n)/e**3 - b*d**2*n*log(x)**
2/(2*e**3) + b*d**2*log(x)*log(c*x**n)/e**3 + b*d*n/(e**2*x) + b*d*log(c*x**n)/(e**2*x) - b*n/(4*e*x**2) - b*l
og(c*x**n)/(2*e*x**2)

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