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

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

[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 = 0.05, 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 = {2337, 2391} \[ \frac {b n \text {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e}+\frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e} \]

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 2337

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 2391

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 = 0.03, size = 94, normalized size = 1.92 \[ \frac {\left (\log \left (\frac {\sqrt {e} x}{\sqrt {-d}}+1\right )+\log \left (\frac {d \sqrt {e} x}{(-d)^{3/2}}+1\right )\right ) \left (a+b \log \left (c x^n\right )\right )+b n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )+b n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{2 e} \]

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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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \log \left (c x^{n}\right ) + a x}{e x^{2} + d}, x\right ) \]

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)/(e*x^2 + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{e x^{2} + d}\,{d x} \]

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/(e*x^2 + d), x)

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maple [C]  time = 0.18, size = 299, normalized size = 6.10 \[ -\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{4 e}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e \,x^{2}+d \right )}{4 e}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e \,x^{2}+d \right )}{4 e}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e \,x^{2}+d \right )}{4 e}+\frac {b n \ln \relax (x ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \ln \relax (x ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}-\frac {b n \ln \relax (x ) \ln \left (e \,x^{2}+d \right )}{2 e}+\frac {b n \dilog \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \dilog \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b \ln \relax (c ) \ln \left (e \,x^{2}+d \right )}{2 e}+\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 e}+\frac {a \ln \left (e \,x^{2}+d \right )}{2 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b/e*ln(e*x^2+d)*ln(x^n)-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))+1/4*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*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/4*I/e*ln(e*x^2+d)*b*Pi*csgn(I*c*x^n)^3+1/4*I/e*ln(e*x^2+d)*b*Pi*
csgn(I*c*x^n)^2*csgn(I*c)+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 \[ b \int \frac {x \log \relax (c) + x \log \left (x^{n}\right )}{e x^{2} + d}\,{d x} + \frac {a \log \left (e x^{2} + d\right )}{2 \, e} \]

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]

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

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

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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sympy [A]  time = 7.83, size = 119, normalized size = 2.43 \[ \frac {a \log {\left (d + e x^{2} \right )}}{2 e} - \frac {b n \left (\begin {cases} \log {\relax (d )} \log {\relax (x )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}\right )}{2 e} + \frac {b \log {\left (c x^{n} \right )} \log {\left (d + e x^{2} \right )}}{2 e} \]

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((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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