3.4.43 \(\int \frac {a+b \log (c x)}{(d+\frac {e}{x}) x^3} \, dx\) [343]

Optimal. Leaf size=84 \[ -\frac {b}{e x}-\frac {a+b \log (c x)}{e x}-\frac {d (a+b \log (c x))^2}{2 b e^2}+\frac {d (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{e^2}+\frac {b d \text {Li}_2\left (-\frac {d x}{e}\right )}{e^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {269, 46, 2393, 2341, 2338, 2354, 2438} \begin {gather*} \frac {b d \text {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^2}-\frac {d (a+b \log (c x))^2}{2 b e^2}+\frac {d \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{e^2}-\frac {a+b \log (c x)}{e x}-\frac {b}{e x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 269

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 2338

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 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 2354

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

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Mathematica [A]
time = 0.06, size = 77, normalized size = 0.92 \begin {gather*} -\frac {\frac {2 b e}{x}+\frac {2 e (a+b \log (c x))}{x}+\frac {d (a+b \log (c x))^2}{b}-2 d (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )-2 b d \text {Li}_2\left (-\frac {d x}{e}\right )}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 148, normalized size = 1.76

method	result	size
risch	\(-\frac {a}{e x}-\frac {a d \ln \left (x \right )}{e^{2}}+\frac {a d \ln \left (d x +e \right )}{e^{2}}-\frac {b \ln \left (c x \right )^{2} d}{2 e^{2}}-\frac {b \ln \left (c x \right )}{e x}-\frac {b}{e x}+\frac {b d \dilog \left (\frac {c d x +c e}{e c}\right )}{e^{2}}+\frac {b d \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{e^{2}}\)	\(115\)
derivativedivides	\(c^{2} \left (-\frac {a}{e \,c^{2} x}-\frac {a d \ln \left (c x \right )}{e^{2} c^{2}}+\frac {a d \ln \left (c d x +c e \right )}{e^{2} c^{2}}-\frac {b \ln \left (c x \right )^{2} d}{2 e^{2} c^{2}}+\frac {b d \dilog \left (\frac {c d x +c e}{e c}\right )}{e^{2} c^{2}}+\frac {b d \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{e^{2} c^{2}}-\frac {b \ln \left (c x \right )}{e \,c^{2} x}-\frac {b}{e \,c^{2} x}\right )\)	\(148\)
default	\(c^{2} \left (-\frac {a}{e \,c^{2} x}-\frac {a d \ln \left (c x \right )}{e^{2} c^{2}}+\frac {a d \ln \left (c d x +c e \right )}{e^{2} c^{2}}-\frac {b \ln \left (c x \right )^{2} d}{2 e^{2} c^{2}}+\frac {b d \dilog \left (\frac {c d x +c e}{e c}\right )}{e^{2} c^{2}}+\frac {b d \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{e^{2} c^{2}}-\frac {b \ln \left (c x \right )}{e \,c^{2} x}-\frac {b}{e \,c^{2} x}\right )\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 92, normalized size = 1.10 \begin {gather*} {\left (\log \left (d x e^{\left (-1\right )} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-d x e^{\left (-1\right )}\right )\right )} b d e^{\left (-2\right )} + {\left (b d \log \left (c\right ) + a d\right )} e^{\left (-2\right )} \log \left (d x + e\right ) - \frac {{\left (b d x \log \left (x\right )^{2} + 2 \, {\left (b {\left (\log \left (c\right ) + 1\right )} + a\right )} e + 2 \, {\left ({\left (b d \log \left (c\right ) + a d\right )} x + b e\right )} \log \left (x\right )\right )} e^{\left (-2\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(d*x*e^(-1) + 1)*log(x) + dilog(-d*x*e^(-1)))*b*d*e^(-2) + (b*d*log(c) + a*d)*e^(-2)*log(d*x + e) - 1/2*(b
*d*x*log(x)^2 + 2*(b*(log(c) + 1) + a)*e + 2*((b*d*log(c) + a*d)*x + b*e)*log(x))*e^(-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))/(d+e/x)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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))/(d+e/x)/x^3,x, algorithm="giac")

[Out]

integrate((b*log(c*x) + a)/((d + e/x)*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\right )}{x^3\,\left (d+\frac {e}{x}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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