3.4.39 \(\int \frac {x (a+b \log (c x))}{d+\frac {e}{x}} \, dx\) [339]

Optimal. Leaf size=98 \[ -\frac {a e x}{d^2}+\frac {b e x}{d^2}-\frac {b x^2}{4 d}-\frac {b e x \log (c x)}{d^2}+\frac {x^2 (a+b \log (c x))}{2 d}+\frac {e^2 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (-\frac {d x}{e}\right )}{d^3} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {269, 45, 2393, 2332, 2341, 2354, 2438} \begin {gather*} \frac {b e^2 \text {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^3}+\frac {e^2 \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{d^3}+\frac {x^2 (a+b \log (c x))}{2 d}-\frac {a e x}{d^2}-\frac {b e x \log (c x)}{d^2}+\frac {b e x}{d^2}-\frac {b x^2}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*Log[c*x]))/(d + e/x),x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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 {x (a+b \log (c x))}{d+\frac {e}{x}} \, dx &=\int \left (-\frac {e (a+b \log (c x))}{d^2}+\frac {x (a+b \log (c x))}{d}+\frac {e^2 (a+b \log (c x))}{d^2 (e+d x)}\right ) \, dx\\ &=\frac {\int x (a+b \log (c x)) \, dx}{d}-\frac {e \int (a+b \log (c x)) \, dx}{d^2}+\frac {e^2 \int \frac {a+b \log (c x)}{e+d x} \, dx}{d^2}\\ &=-\frac {a e x}{d^2}-\frac {b x^2}{4 d}+\frac {x^2 (a+b \log (c x))}{2 d}+\frac {e^2 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d^3}-\frac {(b e) \int \log (c x) \, dx}{d^2}-\frac {\left (b e^2\right ) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{d^3}\\ &=-\frac {a e x}{d^2}+\frac {b e x}{d^2}-\frac {b x^2}{4 d}-\frac {b e x \log (c x)}{d^2}+\frac {x^2 (a+b \log (c x))}{2 d}+\frac {e^2 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (-\frac {d x}{e}\right )}{d^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 99, normalized size = 1.01 \begin {gather*} -\frac {a e x}{d^2}+\frac {b e x}{d^2}-\frac {b x^2}{4 d}-\frac {b e x \log (c x)}{d^2}+\frac {x^2 (a+b \log (c x))}{2 d}+\frac {e^2 (a+b \log (c x)) \log \left (\frac {e+d x}{e}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (-\frac {d x}{e}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 160, normalized size = 1.63

method	result	size
risch	\(\frac {a \,x^{2}}{2 d}-\frac {a e x}{d^{2}}+\frac {a \,e^{2} \ln \left (d x +e \right )}{d^{3}}+\frac {b \,x^{2} \ln \left (c x \right )}{2 d}-\frac {b \,x^{2}}{4 d}-\frac {b e x \ln \left (c x \right )}{d^{2}}+\frac {b e x}{d^{2}}+\frac {b \,e^{2} \dilog \left (\frac {c d x +c e}{e c}\right )}{d^{3}}+\frac {b \,e^{2} \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d^{3}}\)	\(126\)
derivativedivides	\(\frac {-\frac {a \,c^{2} e x}{d^{2}}+\frac {a \,c^{2} x^{2}}{2 d}+\frac {a \,c^{2} e^{2} \ln \left (c d x +c e \right )}{d^{3}}+\frac {b \,c^{2} x^{2} \ln \left (c x \right )}{2 d}-\frac {b \,c^{2} x^{2}}{4 d}-\frac {b e \,c^{2} x \ln \left (c x \right )}{d^{2}}+\frac {b \,c^{2} e x}{d^{2}}+\frac {b \,c^{2} e^{2} \dilog \left (\frac {c d x +c e}{e c}\right )}{d^{3}}+\frac {b \,c^{2} e^{2} \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d^{3}}}{c^{2}}\)	\(160\)
default	\(\frac {-\frac {a \,c^{2} e x}{d^{2}}+\frac {a \,c^{2} x^{2}}{2 d}+\frac {a \,c^{2} e^{2} \ln \left (c d x +c e \right )}{d^{3}}+\frac {b \,c^{2} x^{2} \ln \left (c x \right )}{2 d}-\frac {b \,c^{2} x^{2}}{4 d}-\frac {b e \,c^{2} x \ln \left (c x \right )}{d^{2}}+\frac {b \,c^{2} e x}{d^{2}}+\frac {b \,c^{2} e^{2} \dilog \left (\frac {c d x +c e}{e c}\right )}{d^{3}}+\frac {b \,c^{2} e^{2} \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d^{3}}}{c^{2}}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 102, normalized size = 1.04 \begin {gather*} \frac {{\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 e^{2}}{d^{3}} + \frac {{\left (b \log \left (c\right ) + a\right )} e^{2} \log \left (d x + e\right )}{d^{3}} + \frac {{\left ({\left (2 \, d \log \left (c\right ) - d\right )} b + 2 \, a d\right )} x^{2} - 4 \, {\left (b {\left (\log \left (c\right ) - 1\right )} + a\right )} x e + 2 \, {\left (b d x^{2} - 2 \, b x e\right )} \log \left (x\right )}{4 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Sympy [A]
time = 87.50, size = 207, normalized size = 2.11 \begin {gather*} \frac {a x^{2}}{2 d} + \frac {a e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a e x}{d^{2}} + \frac {b x^{2} \log {\left (c x \right )}}{2 d} - \frac {b x^{2}}{4 d} - \frac {b e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \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 {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {b e^{2} \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 \right )}}{d^{2}} - \frac {b e x \log {\left (c x \right )}}{d^{2}} + \frac {b e x}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**2/(2*d) + a*e**2*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/d**2 - a*e*x/d**2 + b*x**2*log(c*x)/(
2*d) - b*x**2/(4*d) - b*e**2*Piecewise((x/e, Eq(d, 0)), (Piecewise((-polylog(2, d*x*exp_polar(I*pi)/e), (Abs(x
) < 1) & (1/Abs(x) < 1)), (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))/d**2 + b*e**2*Piece
wise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x)/d**2 - b*e*x*log(c*x)/d**2 + b*e*x/d**2

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*log(c*x)))/(d + e/x),x)

[Out]

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

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