3.3.12 \(\int \frac {x (a+b \log (c x^n))}{d+e x^2} \, dx\) [212]

Optimal. Leaf size=49 \[ \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e}+\frac {b n \text {Li}_2\left (-\frac {e x^2}{d}\right )}{4 e} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2375, 2438} \begin {gather*} \frac {b n \text {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e}+\frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2375

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c*x^n])^p/(e*r)), x] - Dist[b*f^m*n*(p/(e*r)), Int[Log[1 + e*(x^r/d)]*((
a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

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

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Mathematica [A]
time = 0.02, size = 94, normalized size = 1.92 \begin {gather*} \frac {\left (a+b \log \left (c x^n\right )\right ) \left (\log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+\log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )\right )+b n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )+b n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 299, normalized size = 6.10

method	result	size
risch	\(\frac {b \ln \left (e \,x^{2}+d \right ) \ln \left (x^{n}\right )}{2 e}+\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e}+\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e}-\frac {b n \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e}+\frac {b n \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e}+\frac {b n \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e}-\frac {i \ln \left (e \,x^{2}+d \right ) b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 e}+\frac {i \ln \left (e \,x^{2}+d \right ) b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e}+\frac {i \ln \left (e \,x^{2}+d \right ) b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e}-\frac {i \ln \left (e \,x^{2}+d \right ) b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 e}+\frac {\ln \left (e \,x^{2}+d \right ) b \ln \left (c \right )}{2 e}+\frac {a \ln \left (e \,x^{2}+d \right )}{2 e}\)	\(299\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x*(a+b*log(c*x^n))/(e*x^2+d),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [A]
time = 3.73, size = 141, normalized size = 2.88 \begin {gather*} \frac {a \log {\left (d + e x^{2} \right )}}{2 e} - \frac {b n \left (\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \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 )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}\right )}{2 e} + \frac {b \log {\left (c x^{n} \right )} \log {\left (d + e x^{2} \right )}}{2 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^n))/(e*x^2+d),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{e\,x^2+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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