\(\int \frac {a+b \log (c x^n)}{x^4 (d+e x^2)^2} \, dx\) [230]

Optimal result

Integrand size = 23, antiderivative size = 224 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=-\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 i b e^{3/2} n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {5 i b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}} \]

[Out]

-5/18*b*n/d^2/x^3+5/2*b*e*n/d^3/x+1/2*(a+b*ln(c*x^n))/d/x^3/(e*x^2+d)+1/6*(-5*a+b*n-5*b*ln(c*x^n))/d^2/x^3+1/2
*e*(5*a-b*n+5*b*ln(c*x^n))/d^3/x+1/2*e^(3/2)*arctan(x*e^(1/2)/d^(1/2))*(5*a-b*n+5*b*ln(c*x^n))/d^(7/2)-5/4*I*b
*e^(3/2)*n*polylog(2,-I*x*e^(1/2)/d^(1/2))/d^(7/2)+5/4*I*b*e^(3/2)*n*polylog(2,I*x*e^(1/2)/d^(1/2))/d^(7/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2385, 2380, 2341, 211, 2361, 12, 4940, 2438} \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a+5 b \log \left (c x^n\right )-b n\right )}{2 d^{7/2}}+\frac {e \left (5 a+5 b \log \left (c x^n\right )-b n\right )}{2 d^3 x}-\frac {5 a+5 b \log \left (c x^n\right )-b n}{6 d^2 x^3}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 i b e^{3/2} n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {5 i b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {5 b e n}{2 d^3 x}-\frac {5 b n}{18 d^2 x^3} \]

[In]

Int[(a + b*Log[c*x^n])/(x^4*(d + e*x^2)^2),x]

[Out]

(-5*b*n)/(18*d^2*x^3) + (5*b*e*n)/(2*d^3*x) + (a + b*Log[c*x^n])/(2*d*x^3*(d + e*x^2)) - (5*a - b*n + 5*b*Log[
c*x^n])/(6*d^2*x^3) + (e*(5*a - b*n + 5*b*Log[c*x^n]))/(2*d^3*x) + (e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(5*a -
b*n + 5*b*Log[c*x^n]))/(2*d^(7/2)) - (((5*I)/4)*b*e^(3/2)*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]])/d^(7/2) + (
((5*I)/4)*b*e^(3/2)*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/d^(7/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2341

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
]

Rule 2361

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] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 2380

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)*(x_)^(r_.)), x_Symbol] :> Dist[1/d,
Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Dist[e/d, Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /;
FreeQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]

Rule 2385

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] + Dist[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]

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]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[I*(b/2), Int[Log[1 - I*c*x
]/x, x], x] - Dist[I*(b/2), Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {\int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx}{2 d} \\ & = \frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {\int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{x^4} \, dx}{2 d^2}+\frac {e \int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )} \, dx}{2 d^2} \\ & = -\frac {5 b n}{18 d^2 x^3}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{x^2} \, dx}{2 d^3}-\frac {e^2 \int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{d+e x^2} \, dx}{2 d^3} \\ & = -\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {\left (5 b e^2 n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{2 d^3} \\ & = -\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {\left (5 b e^{3/2} n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 d^{7/2}} \\ & = -\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {\left (5 i b e^{3/2} n\right ) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{4 d^{7/2}}+\frac {\left (5 i b e^{3/2} n\right ) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{4 d^{7/2}} \\ & = -\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 i b e^{3/2} n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {5 i b e^{3/2} n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.61 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=\frac {1}{36} \left (-\frac {4 b n}{d^2 x^3}+\frac {72 b e n}{d^3 x}-\frac {12 \left (a+b \log \left (c x^n\right )\right )}{d^2 x^3}+\frac {72 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right )}{d^3 \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right )}{d^3 \left (\sqrt {-d}+\sqrt {e} x\right )}-\frac {9 b e^{3/2} n \left (\log (x)-\log \left (\sqrt {-d}-\sqrt {e} x\right )\right )}{(-d)^{7/2}}+\frac {9 b e^{3/2} n \left (\log (x)-\log \left (\sqrt {-d}+\sqrt {e} x\right )\right )}{(-d)^{7/2}}+\frac {45 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{7/2}}-\frac {45 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{7/2}}-\frac {45 b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{7/2}}+\frac {45 b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{7/2}}\right ) \]

[In]

Integrate[(a + b*Log[c*x^n])/(x^4*(d + e*x^2)^2),x]

[Out]

((-4*b*n)/(d^2*x^3) + (72*b*e*n)/(d^3*x) - (12*(a + b*Log[c*x^n]))/(d^2*x^3) + (72*e*(a + b*Log[c*x^n]))/(d^3*
x) - (9*e^(3/2)*(a + b*Log[c*x^n]))/(d^3*(Sqrt[-d] - Sqrt[e]*x)) + (9*e^(3/2)*(a + b*Log[c*x^n]))/(d^3*(Sqrt[-
d] + Sqrt[e]*x)) - (9*b*e^(3/2)*n*(Log[x] - Log[Sqrt[-d] - Sqrt[e]*x]))/(-d)^(7/2) + (9*b*e^(3/2)*n*(Log[x] -
Log[Sqrt[-d] + Sqrt[e]*x]))/(-d)^(7/2) + (45*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(7
/2) - (45*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(7/2) - (45*b*e^(3/2)*n*PolyLog[2
, (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(7/2) + (45*b*e^(3/2)*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(7/2))/36

Maple [C] (warning: unable to verify)

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

Time = 0.86 (sec) , antiderivative size = 622, normalized size of antiderivative = 2.78

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) e^{2} x}{2 d^{3} \left (e \,x^{2}+d \right )}-\frac {5 b \,e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right ) n \ln \left (x \right )}{2 d^{3} \sqrt {d e}}+\frac {5 b \,e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right ) \ln \left (x^{n}\right )}{2 d^{3} \sqrt {d e}}-\frac {b \ln \left (x^{n}\right )}{3 d^{2} x^{3}}+\frac {2 b \ln \left (x^{n}\right ) e}{d^{3} x}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{d^{3} \sqrt {-d e}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{d^{3} \sqrt {-d e}}+\frac {5 b n \,e^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 d^{3} \sqrt {-d e}}-\frac {5 b n \,e^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 d^{3} \sqrt {-d e}}-\frac {b n}{9 d^{2} x^{3}}-\frac {b n \,e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{2 d^{3} \sqrt {d e}}+\frac {b n \,e^{3} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{2}}{4 d^{3} \left (e \,x^{2}+d \right ) \sqrt {-d e}}-\frac {b n \,e^{3} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{2}}{4 d^{3} \left (e \,x^{2}+d \right ) \sqrt {-d e}}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 d^{2} \left (e \,x^{2}+d \right ) \sqrt {-d e}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 d^{2} \left (e \,x^{2}+d \right ) \sqrt {-d e}}+\frac {2 b e n}{d^{3} x}+\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 ) \left (\frac {e^{2} \left (\frac {x}{2 e \,x^{2}+2 d}+\frac {5 \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{2 \sqrt {d e}}\right )}{d^{3}}-\frac {1}{3 d^{2} x^{3}}+\frac {2 e}{d^{3} x}\right )\)	\(622\)

[In]

int((a+b*ln(c*x^n))/x^4/(e*x^2+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*b*ln(x^n)*e^2/d^3*x/(e*x^2+d)-5/2*b*e^2/d^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*n*ln(x)+5/2*b*e^2/d^3/(d*e
)^(1/2)*arctan(x*e/(d*e)^(1/2))*ln(x^n)-1/3*b*ln(x^n)/d^2/x^3+2*b*ln(x^n)/d^3*e/x+b*n*e^2/d^3*ln(x)/(-d*e)^(1/
2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-b*n*e^2/d^3*ln(x)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+5/4
*b*n*e^2/d^3/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-5/4*b*n*e^2/d^3/(-d*e)^(1/2)*dilog((e*x+(-d*
e)^(1/2))/(-d*e)^(1/2))-1/9*b*n/d^2/x^3-1/2*b*n*e^2/d^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+1/4*b*n*e^3/d^3*ln
(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2-1/4*b*n*e^3/d^3*ln(x)/(e*x^2+d)/(-d*e)^(1/
2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2+1/4*b*n*e^2/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^2/d^2*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+2*b*e*n/d^3/x+
(-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)*c
sgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*ln(c)+a)*(e^2/d^3*(1/2*x/(e*x^2+d)+5/2/(d*e)^(1/2)*arctan(x*e/(d*e
)^(1/2)))-1/3/d^2/x^3+2/d^3*e/x)

Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)/(e^2*x^8 + 2*d*e*x^6 + d^2*x^4), x)

Sympy [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x^{4} \left (d + e x^{2}\right )^{2}}\, dx \]

[In]

integrate((a+b*ln(c*x**n))/x**4/(e*x**2+d)**2,x)

[Out]

Integral((a + b*log(c*x**n))/(x**4*(d + e*x**2)**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d)^2,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((e*x^2 + d)^2*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,{\left (e\,x^2+d\right )}^2} \,d x \]

[In]

int((a + b*log(c*x^n))/(x^4*(d + e*x^2)^2),x)

[Out]

int((a + b*log(c*x^n))/(x^4*(d + e*x^2)^2), x)