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

Optimal. Leaf size=150 \[ \frac {e^3 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}-\frac {b e^3 n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^4}-\frac {b e^2 n}{d^3 x}+\frac {b e n}{4 d^2 x^2}-\frac {b n}{9 d x^3} \]

[Out]

-1/9*b*n/d/x^3+1/4*b*e*n/d^2/x^2-b*e^2*n/d^3/x+1/3*(-a-b*ln(c*x^n))/d/x^3+1/2*e*(a+b*ln(c*x^n))/d^2/x^2-e^2*(a
+b*ln(c*x^n))/d^3/x+e^3*ln(1+d/e/x)*(a+b*ln(c*x^n))/d^4-b*e^3*n*polylog(2,-d/e/x)/d^4

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Rubi [A]  time = 0.21, antiderivative size = 173, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {44, 2351, 2304, 2301, 2317, 2391} \[ \frac {b e^3 n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^4}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac {e^3 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}-\frac {b e^2 n}{d^3 x}+\frac {b e n}{4 d^2 x^2}-\frac {b n}{9 d x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n)/(9*d*x^3) + (b*e*n)/(4*d^2*x^2) - (b*e^2*n)/(d^3*x) - (a + b*Log[c*x^n])/(3*d*x^3) + (e*(a + b*Log[c*x^
n]))/(2*d^2*x^2) - (e^2*(a + b*Log[c*x^n]))/(d^3*x) - (e^3*(a + b*Log[c*x^n])^2)/(2*b*d^4*n) + (e^3*(a + b*Log
[c*x^n])*Log[1 + (e*x)/d])/d^4 + (b*e^3*n*PolyLog[2, -((e*x)/d)])/d^4

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 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 )}{x^4 (d+e x)} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{d x^4}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x^3}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x^2}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^4 x}+\frac {e^4 \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \log \left (c x^n\right )}{x^4} \, dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^3} \, dx}{d^2}+\frac {e^2 \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{d^3}-\frac {e^3 \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{d^4}+\frac {e^4 \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{d^4}\\ &=-\frac {b n}{9 d x^3}+\frac {b e n}{4 d^2 x^2}-\frac {b e^2 n}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac {e^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^4}-\frac {\left (b e^3 n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d^4}\\ &=-\frac {b n}{9 d x^3}+\frac {b e n}{4 d^2 x^2}-\frac {b e^2 n}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac {e^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^4}+\frac {b e^3 n \text {Li}_2\left (-\frac {e x}{d}\right )}{d^4}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 159, normalized size = 1.06 \[ \frac {-\frac {12 d^3 \left (a+b \log \left (c x^n\right )\right )}{x^3}+\frac {18 d^2 e \left (a+b \log \left (c x^n\right )\right )}{x^2}+36 e^3 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {36 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {18 e^3 \left (a+b \log \left (c x^n\right )\right )^2}{b n}-\frac {4 b d^3 n}{x^3}+\frac {9 b d^2 e n}{x^2}+36 b e^3 n \text {Li}_2\left (-\frac {e x}{d}\right )-\frac {36 b d e^2 n}{x}}{36 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b*d^3*n)/x^3 + (9*b*d^2*e*n)/x^2 - (36*b*d*e^2*n)/x - (12*d^3*(a + b*Log[c*x^n]))/x^3 + (18*d^2*e*(a + b*
Log[c*x^n]))/x^2 - (36*d*e^2*(a + b*Log[c*x^n]))/x - (18*e^3*(a + b*Log[c*x^n])^2)/(b*n) + 36*e^3*(a + b*Log[c
*x^n])*Log[1 + (e*x)/d] + 36*b*e^3*n*PolyLog[2, -((e*x)/d)])/(36*d^4)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e x^{5} + d x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(c*x^n) + a)/(e*x^5 + d*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.20, size = 868, normalized size = 5.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*ln(x^n)*e/d^2/x^2-b*ln(x^n)*e^3/d^4*ln(x)+b*ln(x^n)*e^3/d^4*ln(e*x+d)-b*ln(x^n)*e^2/d^3/x+1/2*b*n*e^3/d^
4*ln(x)^2-b*n*e^3/d^4*dilog(-1/d*e*x)-b*ln(c)*e^2/d^3/x+1/2*b*ln(c)*e/d^2/x^2-b*ln(c)*e^3/d^4*ln(x)+b*ln(c)*e^
3/d^4*ln(e*x+d)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^3/d^4*ln(e*x+d)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*
csgn(I*c)*e^3/d^4*ln(x)-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e/d^2/x^2+1/6*I*b*Pi*csgn(I*x^n)*csgn(I
*c*x^n)*csgn(I*c)/d/x^3-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^3/d^4*ln(x)-1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I
*c)*e^3/d^4*ln(x)-a*e^3/d^4*ln(x)+a*e^3/d^4*ln(e*x+d)-a*e^2/d^3/x+1/2*a*e/d^2/x^2-1/3*b*ln(c)/d/x^3+1/4*I*b*Pi
*csgn(I*x^n)*csgn(I*c*x^n)^2*e/d^2/x^2+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e^3/d^4*ln(e*x+d)-1/2*I*b*Pi*csgn(
I*c*x^n)^2*csgn(I*c)*e^2/d^3/x-1/3*b*ln(x^n)/d/x^3-b*n*e^3/d^4*ln(e*x+d)*ln(-1/d*e*x)+1/6*I*b*Pi*csgn(I*c*x^n)
^3/d/x^3-1/3*a/d/x^3-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^2/d^3/x+1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e/d
^2/x^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e^2/d^3/x-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)
*e^3/d^4*ln(e*x+d)-1/9*b*n/d/x^3-1/6*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d/x^3-1/4*I*b*Pi*csgn(I*c*x^n)^3*e/d^2/x
^2+1/2*I*b*Pi*csgn(I*c*x^n)^3*e^2/d^3/x-1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d/x^3+1/2*I*b*Pi*csgn(I*c*x^n)^
3*e^3/d^4*ln(x)-1/2*I*b*Pi*csgn(I*c*x^n)^3*e^3/d^4*ln(e*x+d)+1/4*b*e*n/d^2/x^2-b*e^2*n/d^3/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, a {\left (\frac {6 \, e^{3} \log \left (e x + d\right )}{d^{4}} - \frac {6 \, e^{3} \log \relax (x)}{d^{4}} - \frac {6 \, e^{2} x^{2} - 3 \, d e x + 2 \, d^{2}}{d^{3} x^{3}}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{e x^{5} + d x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*(6*e^3*log(e*x + d)/d^4 - 6*e^3*log(x)/d^4 - (6*e^2*x^2 - 3*d*e*x + 2*d^2)/(d^3*x^3)) + b*integrate((log
(c) + log(x^n))/(e*x^5 + d*x^4), x)

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

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 = 91.86, size = 296, normalized size = 1.97 \[ - \frac {a}{3 d x^{3}} + \frac {a e}{2 d^{2} x^{2}} - \frac {a e^{2}}{d^{3} x} + \frac {a e^{4} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{4}} - \frac {a e^{3} \log {\relax (x )}}{d^{4}} - \frac {b n}{9 d x^{3}} - \frac {b \log {\left (c x^{n} \right )}}{3 d x^{3}} + \frac {b e n}{4 d^{2} x^{2}} + \frac {b e \log {\left (c x^{n} \right )}}{2 d^{2} x^{2}} - \frac {b e^{2} n}{d^{3} x} - \frac {b e^{2} \log {\left (c x^{n} \right )}}{d^{3} x} - \frac {b e^{4} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} \log {\relax (d )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{d^{4}} + \frac {b e^{4} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{4}} + \frac {b e^{3} n \log {\relax (x )}^{2}}{2 d^{4}} - \frac {b e^{3} \log {\relax (x )} \log {\left (c x^{n} \right )}}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(3*d*x**3) + a*e/(2*d**2*x**2) - a*e**2/(d**3*x) + a*e**4*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True)
)/d**4 - a*e**3*log(x)/d**4 - b*n/(9*d*x**3) - b*log(c*x**n)/(3*d*x**3) + b*e*n/(4*d**2*x**2) + b*e*log(c*x**n
)/(2*d**2*x**2) - b*e**2*n/(d**3*x) - b*e**2*log(c*x**n)/(d**3*x) - b*e**4*n*Piecewise((x/d, Eq(e, 0)), (Piece
wise((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_p
olar(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**4 + b*e**4*Piecewise((x/d, Eq(e, 0)),
(log(d + e*x)/e, True))*log(c*x**n)/d**4 + b*e**3*n*log(x)**2/(2*d**4) - b*e**3*log(x)*log(c*x**n)/d**4

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