Integrand size = 23, antiderivative size = 218 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=-\frac {15 b n}{8 d^3 x}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}+\frac {5 a-b n+5 b \log \left (c x^n\right )}{8 d^2 x \left (d+e x^2\right )}-\frac {15 a-8 b n+15 b \log \left (c x^n\right )}{8 d^3 x}-\frac {\sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 a-8 b n+15 b \log \left (c x^n\right )\right )}{8 d^{7/2}}-\frac {15 b \sqrt {e} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{16 (-d)^{7/2}}+\frac {15 b \sqrt {e} n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{16 (-d)^{7/2}} \] Output:
-15/8*b*n/d^3/x+1/4*(a+b*ln(c*x^n))/d/x/(e*x^2+d)^2+1/8*(5*a-b*n+5*b*ln(c* x^n))/d^2/x/(e*x^2+d)-1/8*(15*a-8*b*n+15*b*ln(c*x^n))/d^3/x-1/8*e^(1/2)*ar ctan(e^(1/2)*x/d^(1/2))*(15*a-8*b*n+15*b*ln(c*x^n))/d^(7/2)-15/16*b*e^(1/2 )*n*polylog(2,-e^(1/2)*x/(-d)^(1/2))/(-d)^(7/2)+15/16*b*e^(1/2)*n*polylog( 2,e^(1/2)*x/(-d)^(1/2))/(-d)^(7/2)
Leaf count is larger than twice the leaf count of optimal. \(552\) vs. \(2(218)=436\).
Time = 1.85 (sec) , antiderivative size = 552, normalized size of antiderivative = 2.53 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=\frac {1}{16} \left (-\frac {16 b n}{d^3 x}-\frac {16 \left (a+b \log \left (c x^n\right )\right )}{d^3 x}+\frac {d \sqrt {e} \left (a+b \log \left (c x^n\right )\right )}{(-d)^{7/2} \left (\sqrt {-d}-\sqrt {e} x\right )^2}+\frac {7 \sqrt {e} \left (a+b \log \left (c x^n\right )\right )}{d^3 \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\sqrt {e} \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2} \left (\sqrt {-d}+\sqrt {e} x\right )^2}-\frac {7 \sqrt {e} \left (a+b \log \left (c x^n\right )\right )}{d^3 \left (\sqrt {-d}+\sqrt {e} x\right )}+\frac {7 b \sqrt {e} n \left (\log (x)-\log \left (\sqrt {-d}-\sqrt {e} x\right )\right )}{(-d)^{7/2}}-\frac {7 b \sqrt {e} n \left (\log (x)-\log \left (\sqrt {-d}+\sqrt {e} x\right )\right )}{(-d)^{7/2}}+\frac {b d \sqrt {e} n \left (\frac {1}{\sqrt {-d} \left (\sqrt {-d}+\sqrt {e} x\right )}-\frac {\log (x)}{d}+\frac {\log \left (\sqrt {-d}+\sqrt {e} x\right )}{d}\right )}{(-d)^{7/2}}-\frac {15 \sqrt {e} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{7/2}}+\frac {b \sqrt {e} n \left (\frac {1}{\sqrt {-d} \left (\sqrt {-d}-\sqrt {e} x\right )}-\frac {\log (x)}{d}+\frac {\log \left ((-d)^{3/2}+d \sqrt {e} x\right )}{d}\right )}{(-d)^{5/2}}+\frac {15 \sqrt {e} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{7/2}}+\frac {15 b \sqrt {e} n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{7/2}}-\frac {15 b \sqrt {e} n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{7/2}}\right ) \] Input:
Integrate[(a + b*Log[c*x^n])/(x^2*(d + e*x^2)^3),x]
Output:
((-16*b*n)/(d^3*x) - (16*(a + b*Log[c*x^n]))/(d^3*x) + (d*Sqrt[e]*(a + b*L og[c*x^n]))/((-d)^(7/2)*(Sqrt[-d] - Sqrt[e]*x)^2) + (7*Sqrt[e]*(a + b*Log[ c*x^n]))/(d^3*(Sqrt[-d] - Sqrt[e]*x)) + (Sqrt[e]*(a + b*Log[c*x^n]))/((-d) ^(5/2)*(Sqrt[-d] + Sqrt[e]*x)^2) - (7*Sqrt[e]*(a + b*Log[c*x^n]))/(d^3*(Sq rt[-d] + Sqrt[e]*x)) + (7*b*Sqrt[e]*n*(Log[x] - Log[Sqrt[-d] - Sqrt[e]*x]) )/(-d)^(7/2) - (7*b*Sqrt[e]*n*(Log[x] - Log[Sqrt[-d] + Sqrt[e]*x]))/(-d)^( 7/2) + (b*d*Sqrt[e]*n*(1/(Sqrt[-d]*(Sqrt[-d] + Sqrt[e]*x)) - Log[x]/d + Lo g[Sqrt[-d] + Sqrt[e]*x]/d))/(-d)^(7/2) - (15*Sqrt[e]*(a + b*Log[c*x^n])*Lo g[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(7/2) + (b*Sqrt[e]*n*(1/(Sqrt[-d]*(Sqrt[ -d] - Sqrt[e]*x)) - Log[x]/d + Log[(-d)^(3/2) + d*Sqrt[e]*x]/d))/(-d)^(5/2 ) + (15*Sqrt[e]*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d) ^(7/2) + (15*b*Sqrt[e]*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(7/2) - (1 5*b*Sqrt[e]*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(7/2))/16
Result contains complex when optimal does not.
Time = 0.90 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2785, 25, 2785, 25, 2780, 2741, 2761, 27, 5355, 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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2785 |
\(\displaystyle \frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}-\frac {\int -\frac {5 a-b n+5 b \log \left (c x^n\right )}{x^2 \left (e x^2+d\right )^2}dx}{4 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {5 a-b n+5 b \log \left (c x^n\right )}{x^2 \left (e x^2+d\right )^2}dx}{4 d}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 2785 |
\(\displaystyle \frac {\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}-\frac {\int -\frac {15 a-8 b n+15 b \log \left (c x^n\right )}{x^2 \left (e x^2+d\right )}dx}{2 d}}{4 d}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {15 a-8 b n+15 b \log \left (c x^n\right )}{x^2 \left (e x^2+d\right )}dx}{2 d}+\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}}{4 d}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 2780 |
\(\displaystyle \frac {\frac {\frac {\int \frac {15 a-8 b n+15 b \log \left (c x^n\right )}{x^2}dx}{d}-\frac {e \int \frac {15 a-8 b n+15 b \log \left (c x^n\right )}{e x^2+d}dx}{d}}{2 d}+\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}}{4 d}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {\frac {\frac {-\frac {15 a+15 b \log \left (c x^n\right )-8 b n}{x}-\frac {15 b n}{x}}{d}-\frac {e \int \frac {15 a-8 b n+15 b \log \left (c x^n\right )}{e x^2+d}dx}{d}}{2 d}+\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}}{4 d}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 2761 |
\(\displaystyle \frac {\frac {\frac {-\frac {15 a+15 b \log \left (c x^n\right )-8 b n}{x}-\frac {15 b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 a+15 b \log \left (c x^n\right )-8 b n\right )}{\sqrt {d} \sqrt {e}}-15 b n \int \frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x}dx\right )}{d}}{2 d}+\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}}{4 d}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {-\frac {15 a+15 b \log \left (c x^n\right )-8 b n}{x}-\frac {15 b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 a+15 b \log \left (c x^n\right )-8 b n\right )}{\sqrt {d} \sqrt {e}}-\frac {15 b n \int \frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x}dx}{\sqrt {d} \sqrt {e}}\right )}{d}}{2 d}+\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}}{4 d}+\frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 5355 |
\(\displaystyle \frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}+\frac {\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}+\frac {\frac {-\frac {15 a+15 b \log \left (c x^n\right )-8 b n}{x}-\frac {15 b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 a+15 b \log \left (c x^n\right )-8 b n\right )}{\sqrt {d} \sqrt {e}}-\frac {15 b n \left (\frac {1}{2} i \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x}dx-\frac {1}{2} i \int \frac {\log \left (\frac {i \sqrt {e} x}{\sqrt {d}}+1\right )}{x}dx\right )}{\sqrt {d} \sqrt {e}}\right )}{d}}{2 d}}{4 d}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {a+b \log \left (c x^n\right )}{4 d x \left (d+e x^2\right )^2}+\frac {\frac {5 a+5 b \log \left (c x^n\right )-b n}{2 d x \left (d+e x^2\right )}+\frac {\frac {-\frac {15 a+15 b \log \left (c x^n\right )-8 b n}{x}-\frac {15 b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 a+15 b \log \left (c x^n\right )-8 b n\right )}{\sqrt {d} \sqrt {e}}-\frac {15 b n \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )}{\sqrt {d} \sqrt {e}}\right )}{d}}{2 d}}{4 d}\) |
Input:
Int[(a + b*Log[c*x^n])/(x^2*(d + e*x^2)^3),x]
Output:
(a + b*Log[c*x^n])/(4*d*x*(d + e*x^2)^2) + ((5*a - b*n + 5*b*Log[c*x^n])/( 2*d*x*(d + e*x^2)) + (((-15*b*n)/x - (15*a - 8*b*n + 15*b*Log[c*x^n])/x)/d - (e*((ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(15*a - 8*b*n + 15*b*Log[c*x^n]))/(Sqr t[d]*Sqrt[e]) - (15*b*n*((I/2)*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]] - (I/2 )*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]]))/(Sqrt[d]*Sqrt[e])))/d)/(2*d))/(4*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2), x]}, Simp[u*(a + b*Log[c*x^n]), x] - Si mp[b*n Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]
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[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^2)^(q_.), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(d + e*x^2)^(q + 1)*((a + b*Log[c*x^n])/(2*d*f*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(f*x)^m*(d + e*x^2)^(q + 1)*(a*(m + 2*q + 3) + b*n + b*(m + 2*q + 3)*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && ILtQ[m, 0]
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]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[I*(b/2) Int[Log[1 - I*c*x]/x, x], x] - Simp[I*(b/2) Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.22 (sec) , antiderivative size = 964, normalized size of antiderivative = 4.42
Input:
int((a+b*ln(c*x^n))/x^2/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
Output:
(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/d^3*e*((7/8*e*x^3+9/8*d*x)/(e*x^2+d)^2+15/8/(d*e)^(1/2)* arctan(x*e/(d*e)^(1/2)))-1/d^3/x)+b*n*e/d^3/(d*e)^(1/2)*arctan(x*e/(d*e)^( 1/2))-15/16*b*n*e/d^3/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2)) +15/16*b*n*e/d^3/(-d*e)^(1/2)*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-7/8*b /d^3*e^2/(e*x^2+d)^2*x^3*ln(x^n)-15/8*b*e/d^3/(d*e)^(1/2)*arctan(x*e/(d*e) ^(1/2))*ln(x^n)-1/2*b*n/d^3*e*ln(x)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(- d*e)^(1/2))+1/2*b*n/d^3*e*ln(x)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^ (1/2))+1/4*b*n*e/d^2*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(- d*e)^(1/2))-1/4*b*n*e/d^2*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/ 2))/(-d*e)^(1/2))+3/16*b*n*e/d*ln(x)/(e*x^2+d)^2/(-d*e)^(1/2)*ln((e*x+(-d* e)^(1/2))/(-d*e)^(1/2))-3/16*b*n*e/d*ln(x)/(e*x^2+d)^2/(-d*e)^(1/2)*ln((-e *x+(-d*e)^(1/2))/(-d*e)^(1/2))+15/8*b*e/d^3/(d*e)^(1/2)*arctan(x*e/(d*e)^( 1/2))*n*ln(x)+1/2*b*n*e^2/d^3*ln(x)/(e*x^2+d)^2*x^3+1/2*b*n*e/d^2*ln(x)/(e *x^2+d)^2*x+1/8*b*n/d^3*e*x/(e*x^2+d)-1/2*b*n/d^3*ln(x)*e*x/(e*x^2+d)-9/8* b*ln(x^n)/d^2/(e*x^2+d)^2*e*x-3/16*b*n*e^3/d^3*ln(x)/(e*x^2+d)^2/(-d*e)^(1 /2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^4+3/16*b*n*e^3/d^3*ln(x)/(e*x^2 +d)^2/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^4-3/8*b*n*e^2/d^2 *ln(x)/(e*x^2+d)^2/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^...
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{3} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^2/(e*x^2+d)^3,x, algorithm="fricas")
Output:
integral((b*log(c*x^n) + a)/(e^3*x^8 + 3*d*e^2*x^6 + 3*d^2*e*x^4 + d^3*x^2 ), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*x**n))/x**2/(e*x**2+d)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*log(c*x^n))/x^2/(e*x^2+d)^3,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{3} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^2/(e*x^2+d)^3,x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)/((e*x^2 + d)^3*x^2), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,{\left (e\,x^2+d\right )}^3} \,d x \] Input:
int((a + b*log(c*x^n))/(x^2*(d + e*x^2)^3),x)
Output:
int((a + b*log(c*x^n))/(x^2*(d + e*x^2)^3), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^3} \, dx=\frac {-15 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,d^{2} x -30 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d e \,x^{3}-15 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,e^{2} x^{5}+8 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{3} x^{8}+3 d \,e^{2} x^{6}+3 d^{2} e \,x^{4}+d^{3} x^{2}}d x \right ) b \,d^{6} x +16 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{3} x^{8}+3 d \,e^{2} x^{6}+3 d^{2} e \,x^{4}+d^{3} x^{2}}d x \right ) b \,d^{5} e \,x^{3}+8 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{3} x^{8}+3 d \,e^{2} x^{6}+3 d^{2} e \,x^{4}+d^{3} x^{2}}d x \right ) b \,d^{4} e^{2} x^{5}-8 a \,d^{3}-25 a \,d^{2} e \,x^{2}-15 a d \,e^{2} x^{4}}{8 d^{4} x \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:
int((a+b*log(c*x^n))/x^2/(e*x^2+d)^3,x)
Output:
( - 15*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d**2*x - 30*sqrt(e) *sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d*e*x**3 - 15*sqrt(e)*sqrt(d)*ata n((e*x)/(sqrt(e)*sqrt(d)))*a*e**2*x**5 + 8*int(log(x**n*c)/(d**3*x**2 + 3* d**2*e*x**4 + 3*d*e**2*x**6 + e**3*x**8),x)*b*d**6*x + 16*int(log(x**n*c)/ (d**3*x**2 + 3*d**2*e*x**4 + 3*d*e**2*x**6 + e**3*x**8),x)*b*d**5*e*x**3 + 8*int(log(x**n*c)/(d**3*x**2 + 3*d**2*e*x**4 + 3*d*e**2*x**6 + e**3*x**8) ,x)*b*d**4*e**2*x**5 - 8*a*d**3 - 25*a*d**2*e*x**2 - 15*a*d*e**2*x**4)/(8* d**4*x*(d**2 + 2*d*e*x**2 + e**2*x**4))