Integrand size = 21, antiderivative size = 49 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e} \] Output:
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
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.92 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right ) \left (\log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+\log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )\right )+b n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+b n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{2 e} \] Input:
Integrate[(x*(a + b*Log[c*x^n]))/(d + e*x^2),x]
Output:
((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*S qrt[e]*x)/(-d)^(3/2)])/(2*e)
Time = 0.25 (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 = {2775, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx\) |
\(\Big \downarrow \) 2775 |
\(\displaystyle \frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b n \int \frac {\log \left (\frac {e x^2}{d}+1\right )}{x}dx}{2 e}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e}\) |
Input:
Int[(x*(a + b*Log[c*x^n]))/(d + e*x^2),x]
Output:
((a + b*Log[c*x^n])*Log[1 + (e*x^2)/d])/(2*e) + (b*n*PolyLog[2, -((e*x^2)/ d)])/(4*e)
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Simp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c* x^n])^p/(e*r)), x] - Simp[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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.42 (sec) , antiderivative size = 244, normalized size of antiderivative = 4.98
method | result | size |
risch | \(\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 e}-\frac {b n \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e}+\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {b n \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}+\frac {\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \ln \left (e \,x^{2}+d \right )}{2 e}\) | \(244\) |
Input:
int(x*(a+b*ln(c*x^n))/(e*x^2+d),x,method=_RETURNVERBOSE)
Output:
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*d ilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*(1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c *x^n)^2-1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*I*Pi*b*csgn(I*c *x^n)^3+1/2*I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+b*ln(c)+a)/e*ln(e*x^2+d)
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{e x^{2} + d} \,d x } \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x^2+d),x, algorithm="fricas")
Output:
integral((b*x*log(c*x^n) + a*x)/(e*x^2 + d), x)
Time = 3.51 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.88 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {a \log {\left (d + e x^{2} \right )}}{2 e} - \frac {b n \left (\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}\right )}{2 e} + \frac {b \log {\left (c x^{n} \right )} \log {\left (d + e x^{2} \right )}}{2 e} \] Input:
integrate(x*(a+b*ln(c*x**n))/(e*x**2+d),x)
Output:
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)
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{e x^{2} + d} \,d x } \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x^2+d),x, algorithm="maxima")
Output:
b*integrate((x*log(c) + x*log(x^n))/(e*x^2 + d), x) + 1/2*a*log(e*x^2 + d) /e
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{e x^{2} + d} \,d x } \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x^2+d),x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)*x/(e*x^2 + d), x)
Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{e\,x^2+d} \,d x \] Input:
int((x*(a + b*log(c*x^n)))/(d + e*x^2),x)
Output:
int((x*(a + b*log(c*x^n)))/(d + e*x^2), x)
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {-2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{3}+d x}d x \right ) b d n +\mathrm {log}\left (e \,x^{2}+d \right ) a n +\mathrm {log}\left (x^{n} c \right )^{2} b}{2 e n} \] Input:
int(x*(a+b*log(c*x^n))/(e*x^2+d),x)
Output:
( - 2*int(log(x**n*c)/(d*x + e*x**3),x)*b*d*n + log(d + e*x**2)*a*n + log( x**n*c)**2*b)/(2*e*n)