Integrand size = 23, antiderivative size = 121 \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=-\frac {b n}{16 d x^4}+\frac {b e n}{4 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \log \left (1+\frac {d}{e x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^3} \] Output:
-1/16*b*n/d/x^4+1/4*b*e*n/d^2/x^2-1/4*(a+b*ln(c*x^n))/d/x^4+1/2*e*(a+b*ln( c*x^n))/d^2/x^2-1/2*e^2*ln(1+d/e/x^2)*(a+b*ln(c*x^n))/d^3+1/4*b*e^2*n*poly log(2,-d/e/x^2)/d^3
Time = 0.22 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.62 \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=-\frac {\frac {b d^2 n}{x^4}-\frac {4 b d e n}{x^2}+\frac {4 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^4}-\frac {8 d e \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {8 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+8 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+8 b e^2 n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 b e^2 n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{16 d^3} \] Input:
Integrate[(a + b*Log[c*x^n])/(x^5*(d + e*x^2)),x]
Output:
-1/16*((b*d^2*n)/x^4 - (4*b*d*e*n)/x^2 + (4*d^2*(a + b*Log[c*x^n]))/x^4 - (8*d*e*(a + b*Log[c*x^n]))/x^2 - (8*e^2*(a + b*Log[c*x^n])^2)/(b*n) + 8*e^ 2*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]] + 8*e^2*(a + b*Log[c*x^ n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)] + 8*b*e^2*n*PolyLog[2, (Sqrt[e]*x)/S qrt[-d]] + 8*b*e^2*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/d^3
Time = 0.67 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2780, 2741, 2780, 2741, 2779, 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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx\) |
\(\Big \downarrow \) 2780 |
\(\displaystyle \frac {\int \frac {a+b \log \left (c x^n\right )}{x^5}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (e x^2+d\right )}dx}{d}\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{4 x^4}-\frac {b n}{16 x^4}}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (e x^2+d\right )}dx}{d}\) |
\(\Big \downarrow \) 2780 |
\(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{4 x^4}-\frac {b n}{16 x^4}}{d}-\frac {e \left (\frac {\int \frac {a+b \log \left (c x^n\right )}{x^3}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x \left (e x^2+d\right )}dx}{d}\right )}{d}\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{4 x^4}-\frac {b n}{16 x^4}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x \left (e x^2+d\right )}dx}{d}\right )}{d}\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{4 x^4}-\frac {b n}{16 x^4}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \left (\frac {b n \int \frac {\log \left (\frac {d}{e x^2}+1\right )}{x}dx}{2 d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d}\right )}{d}\right )}{d}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{4 x^4}-\frac {b n}{16 x^4}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \left (\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d}\right )}{d}\right )}{d}\) |
Input:
Int[(a + b*Log[c*x^n])/(x^5*(d + e*x^2)),x]
Output:
(-1/16*(b*n)/x^4 - (a + b*Log[c*x^n])/(4*x^4))/d - (e*((-1/4*(b*n)/x^2 - ( a + b*Log[c*x^n])/(2*x^2))/d - (e*(-1/2*(Log[1 + d/(e*x^2)]*(a + b*Log[c*x ^n]))/d + (b*n*PolyLog[2, -(d/(e*x^2))])/(4*d)))/d))/d
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]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)* (x_)^(r_.)), x_Symbol] :> Simp[1/d Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Simp[e/d Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /; Fre eQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]
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 = 1.00 (sec) , antiderivative size = 369, normalized size of antiderivative = 3.05
method | result | size |
risch | \(-\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {b \ln \left (x^{n}\right )}{4 d \,x^{4}}+\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {b \ln \left (x^{n}\right ) e}{2 d^{2} x^{2}}-\frac {b n}{16 d \,x^{4}}+\frac {b e n}{4 d^{2} x^{2}}-\frac {b n \,e^{2} \ln \left (x \right )^{2}}{2 d^{3}}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}-\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}-\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {e^{2} \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {1}{4 d \,x^{4}}+\frac {e^{2} \ln \left (x \right )}{d^{3}}+\frac {e}{2 d^{2} x^{2}}\right )\) | \(369\) |
Input:
int((a+b*ln(c*x^n))/x^5/(e*x^2+d),x,method=_RETURNVERBOSE)
Output:
-1/2*b*ln(x^n)*e^2/d^3*ln(e*x^2+d)-1/4*b*ln(x^n)/d/x^4+b*ln(x^n)*e^2/d^3*l n(x)+1/2*b*ln(x^n)*e/d^2/x^2-1/16*b*n/d/x^4+1/4*b*e*n/d^2/x^2-1/2*b*n*e^2/ d^3*ln(x)^2+1/2*b*n*e^2/d^3*ln(x)*ln(e*x^2+d)-1/2*b*n*e^2/d^3*ln(x)*ln((-e *x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*e^2/d^3*ln(x)*ln((e*x+(-d*e)^(1/2)) /(-d*e)^(1/2))-1/2*b*n*e^2/d^3*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2 *b*n*e^2/d^3*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(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)*(-1/2*e^ 2/d^3*ln(e*x^2+d)-1/4/d/x^4+e^2/d^3*ln(x)+1/2*e/d^2/x^2)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{5}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^5/(e*x^2+d),x, algorithm="fricas")
Output:
integral((b*log(c*x^n) + a)/(e*x^7 + d*x^5), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*x**n))/x**5/(e*x**2+d),x)
Output:
Timed out
\[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{5}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^5/(e*x^2+d),x, algorithm="maxima")
Output:
-1/4*a*(2*e^2*log(e*x^2 + d)/d^3 - 4*e^2*log(x)/d^3 - (2*e*x^2 - d)/(d^2*x ^4)) + b*integrate((log(c) + log(x^n))/(e*x^7 + d*x^5), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{5}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^5/(e*x^2+d),x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)/((e*x^2 + d)*x^5), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^5\,\left (e\,x^2+d\right )} \,d x \] Input:
int((a + b*log(c*x^n))/(x^5*(d + e*x^2)),x)
Output:
int((a + b*log(c*x^n))/(x^5*(d + e*x^2)), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\frac {4 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{7}+d \,x^{5}}d x \right ) b \,d^{3} x^{4}-2 \,\mathrm {log}\left (e \,x^{2}+d \right ) a \,e^{2} x^{4}+4 \,\mathrm {log}\left (x \right ) a \,e^{2} x^{4}-a \,d^{2}+2 a d e \,x^{2}}{4 d^{3} x^{4}} \] Input:
int((a+b*log(c*x^n))/x^5/(e*x^2+d),x)
Output:
(4*int(log(x**n*c)/(d*x**5 + e*x**7),x)*b*d**3*x**4 - 2*log(d + e*x**2)*a* e**2*x**4 + 4*log(x)*a*e**2*x**4 - a*d**2 + 2*a*d*e*x**2)/(4*d**3*x**4)