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

Optimal result

Integrand size = 21, antiderivative size = 49 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2375, 2438} \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.92 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right ) \left (\log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+\log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )\right )+b n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+b n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{2 e} \]

[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)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.40 (sec) , antiderivative size = 244, normalized size of antiderivative = 4.98

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 e}-\frac {b n \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e}+\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \ln \left (e \,x^{2}+d \right )}{2 e}\)	\(244\)

[In]

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

[Out]

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

Fricas [F]

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

[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)/(e*x^2 + d), x)

Sympy [A] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.88 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {a \log {\left (d + e x^{2} \right )}}{2 e} - \frac {b n \left (\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}\right )}{2 e} + \frac {b \log {\left (c x^{n} \right )} \log {\left (d + e x^{2} \right )}}{2 e} \]

[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)

Maxima [F]

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

[In]

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

[Out]

b*integrate((x*log(c) + x*log(x^n))/(e*x^2 + d), x) + 1/2*a*log(e*x^2 + d)/e

Giac [F]

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

[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/(e*x^2 + d), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{e\,x^2+d} \,d x \]

[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)