3.1.45 \(\int \frac {a+b \log (c x^n)}{x^3 (d+e x)^2} \, dx\) [45]

Optimal. Leaf size=154 \[ -\frac {b n}{4 d^2 x^2}+\frac {2 b e n}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{2 d^2 x^2}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {3 e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {b e^2 n \log (d+e x)}{d^4}+\frac {3 b e^2 n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^4} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {3 b e^2 n \text {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4}-\frac {e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {3 e^2 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{2 d^2 x^2}+\frac {b e^2 n \log (d+e x)}{d^4}+\frac {2 b e n}{d^3 x}-\frac {b n}{4 d^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.16, size = 165, normalized size = 1.07 \begin {gather*} -\frac {\frac {b d^2 n}{x^2}-\frac {8 b d e n}{x}+\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {8 d e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {4 d e^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-\frac {6 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+4 b e^2 n (\log (x)-\log (d+e x))+12 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+12 b e^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{4 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((b*d^2*n)/x^2 - (8*b*d*e*n)/x + (2*d^2*(a + b*Log[c*x^n]))/x^2 - (8*d*e*(a + b*Log[c*x^n]))/x - (4*d*e^2
*(a + b*Log[c*x^n]))/(d + e*x) - (6*e^2*(a + b*Log[c*x^n])^2)/(b*n) + 4*b*e^2*n*(Log[x] - Log[d + e*x]) + 12*e
^2*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 12*b*e^2*n*PolyLog[2, -((e*x)/d)])/d^4

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 910, normalized size = 5.91

method	result	size
risch	\(\text {Expression too large to display}\)	\(910\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x^3/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*e^2/d^3/(e*x+d)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^3*e/x-3*a/d^4*e^2*ln(e
*x+d)+a*e^2/d^3/(e*x+d)+3*a/d^4*e^2*ln(x)+2*a/d^3*e/x+3*b*n/d^4*e^2*ln(e*x+d)*ln(-e*x/d)+3/2*I*b*Pi*csgn(I*c*x
^n)^3/d^4*e^2*ln(e*x+d)-1/2*I*b*Pi*csgn(I*c*x^n)^3*e^2/d^3/(e*x+d)-3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^4*
e^2*ln(e*x+d)+3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^4*e^2*ln(x)-3*b*ln(c)/d^4*e^2*ln(e*x+d)+3*b*ln(c)/d^4*e
^2*ln(x)+b*ln(c)*e^2/d^3/(e*x+d)+2*b*ln(c)/d^3*e/x-b*n/d^4*e^2*ln(x)-3/2*b*n/d^4*e^2*ln(x)^2+3*b*n/d^4*e^2*dil
og(-e*x/d)+1/4*I*b*Pi*csgn(I*c*x^n)^3/d^2/x^2-3/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^4*e^2*ln(e*x+d)-1/4*I*b*P
i*csgn(I*c)*csgn(I*c*x^n)^2/d^2/x^2-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^2/x^2+I*b*Pi*csgn(I*x^n)*csgn(I*c
*x^n)^2/d^3*e/x+3/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^4*e^2*ln(x)-1/2*a/d^2/x^2-3/2*I*b*Pi*csgn(I*c*x^n)^3/d^
4*e^2*ln(x)-I*b*Pi*csgn(I*c*x^n)^3/d^3*e/x+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^2/d^3/(e*x+d)+1/4*I*b*Pi*c
sgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d^2/x^2-3*b*ln(x^n)/d^4*e^2*ln(e*x+d)-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn
(I*c*x^n)*e^2/d^3/(e*x+d)-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d^3*e/x-3/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*
csgn(I*c*x^n)/d^4*e^2*ln(x)+3/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d^4*e^2*ln(e*x+d)-1/2*b*ln(c)/d^2/x
^2-1/2*b*ln(x^n)/d^2/x^2+b*e^2*n*ln(e*x+d)/d^4+b*ln(x^n)*e^2/d^3/(e*x+d)+3*b*ln(x^n)/d^4*e^2*ln(x)+2*b*ln(x^n)
/d^3*e/x-1/4*b*n/d^2/x^2+2*b*e*n/d^3/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^3/(e*x+d)^2,x, algorithm="maxima")

[Out]

1/2*a*((6*x^2*e^2 + 3*d*x*e - d^2)/(d^3*x^3*e + d^4*x^2) - 6*e^2*log(x*e + d)/d^4 + 6*e^2*log(x)/d^4) + b*inte
grate((log(c) + log(x^n))/(x^5*e^2 + 2*d*x^4*e + d^2*x^3), 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^3/(e*x+d)^2,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)/(x^5*e^2 + 2*d*x^4*e + d^2*x^3), x)

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Sympy [A]
time = 59.49, size = 376, normalized size = 2.44 \begin {gather*} - \frac {a}{2 d^{2} x^{2}} - \frac {a e^{3} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {2 a e}{d^{3} x} - \frac {3 a e^{3} \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 {3 a e^{2} \log {\left (x \right )}}{d^{4}} - \frac {b n}{4 d^{2} x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 d^{2} x^{2}} + \frac {b e^{3} 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^{3}} - \frac {b e^{3} \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^{3}} + \frac {2 b e n}{d^{3} x} + \frac {2 b e \log {\left (c x^{n} \right )}}{d^{3} x} + \frac {3 b e^{3} 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^{4}} - \frac {3 b e^{3} \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 {3 b e^{2} n \log {\left (x \right )}^{2}}{2 d^{4}} + \frac {3 b e^{2} \log {\left (x \right )} \log {\left (c x^{n} \right )}}{d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**3/(e*x+d)**2,x)

[Out]

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

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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^3/(e*x+d)^2,x, algorithm="giac")

[Out]

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

[Out]

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

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