3.3.29 \(\int \frac {a+b \log (c x^n)}{x^2 (d+e x^2)^2} \, dx\) [229]

Optimal. Leaf size=183 \[ -\frac {3 b n}{2 d^2 x}+\frac {a+b \log \left (c x^n\right )}{2 d x \left (d+e x^2\right )}-\frac {3 a-b n+3 b \log \left (c x^n\right )}{2 d^2 x}-\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 a-b n+3 b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {3 i b \sqrt {e} n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{5/2}}-\frac {3 i b \sqrt {e} n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2385, 2380, 2341, 211, 2361, 12, 4940, 2438} \begin {gather*} \frac {3 i b \sqrt {e} n \text {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{5/2}}-\frac {3 i b \sqrt {e} n \text {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{5/2}}-\frac {\sqrt {e} \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 a+3 b \log \left (c x^n\right )-b n\right )}{2 d^{5/2}}-\frac {3 a+3 b \log \left (c x^n\right )-b n}{2 d^2 x}+\frac {a+b \log \left (c x^n\right )}{2 d x \left (d+e x^2\right )}-\frac {3 b n}{2 d^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.49, size = 328, normalized size = 1.79 \begin {gather*} \frac {1}{4} \left (-\frac {4 b n}{d^2 x}-\frac {4 \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {\sqrt {e} \left (a+b \log \left (c x^n\right )\right )}{d^2 \left (\sqrt {-d}-\sqrt {e} x\right )}-\frac {\sqrt {e} \left (a+b \log \left (c x^n\right )\right )}{d^2 \left (\sqrt {-d}+\sqrt {e} x\right )}+\frac {b \sqrt {e} n \left (-\log (x)+\log \left (\sqrt {-d}-\sqrt {e} x\right )\right )}{(-d)^{5/2}}+\frac {b \sqrt {e} n \left (\log (x)-\log \left (\sqrt {-d}+\sqrt {e} x\right )\right )}{(-d)^{5/2}}+\frac {3 \sqrt {e} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}-\frac {3 \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)^{5/2}}-\frac {3 b \sqrt {e} n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}+\frac {3 b \sqrt {e} n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b*n)/(d^2*x) - (4*(a + b*Log[c*x^n]))/(d^2*x) + (Sqrt[e]*(a + b*Log[c*x^n]))/(d^2*(Sqrt[-d] - Sqrt[e]*x))
- (Sqrt[e]*(a + b*Log[c*x^n]))/(d^2*(Sqrt[-d] + Sqrt[e]*x)) + (b*Sqrt[e]*n*(-Log[x] + Log[Sqrt[-d] - Sqrt[e]*
x]))/(-d)^(5/2) + (b*Sqrt[e]*n*(Log[x] - Log[Sqrt[-d] + Sqrt[e]*x]))/(-d)^(5/2) + (3*Sqrt[e]*(a + b*Log[c*x^n]
)*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) - (3*Sqrt[e]*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])
/(-d)^(5/2) - (3*b*Sqrt[e]*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) + (3*b*Sqrt[e]*n*PolyLog[2, (d*Sqrt[
e]*x)/(-d)^(3/2)])/(-d)^(5/2))/4

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 933, normalized size = 5.10

method	result	size
risch	\(-\frac {b e x \ln \left (x^{n}\right )}{2 d^{2} \left (e \,x^{2}+d \right )}-\frac {3 b e \arctan \left (\frac {x e}{\sqrt {e d}}\right ) \ln \left (x^{n}\right )}{2 d^{2} \sqrt {e d}}-\frac {a}{d^{2} x}-\frac {b \ln \left (x^{n}\right )}{d^{2} x}-\frac {3 a e \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{2 d^{2} \sqrt {e d}}-\frac {a e x}{2 d^{2} \left (e \,x^{2}+d \right )}-\frac {3 b n e \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{4 d^{2} \sqrt {-e d}}+\frac {3 b n e \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{4 d^{2} \sqrt {-e d}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2} x}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2} x}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right ) x^{2}}{4 d^{2} \left (e \,x^{2}+d \right ) \sqrt {-e d}}+\frac {3 b e \arctan \left (\frac {x e}{\sqrt {e d}}\right ) n \ln \left (x \right )}{2 d^{2} \sqrt {e d}}-\frac {b n e \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2} \sqrt {-e d}}+\frac {b n e \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2} \sqrt {-e d}}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right ) x^{2}}{4 d^{2} \left (e \,x^{2}+d \right ) \sqrt {-e d}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 d^{2} x}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 d^{2} x}+\frac {3 i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{4 d^{2} \sqrt {e d}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e x}{4 d^{2} \left (e \,x^{2}+d \right )}-\frac {b \ln \left (c \right )}{d^{2} x}-\frac {b \ln \left (c \right ) e x}{2 d^{2} \left (e \,x^{2}+d \right )}+\frac {b n e \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{2 d^{2} \sqrt {e d}}-\frac {b n e \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{4 d \left (e \,x^{2}+d \right ) \sqrt {-e d}}+\frac {b n e \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{4 d \left (e \,x^{2}+d \right ) \sqrt {-e d}}-\frac {3 b \ln \left (c \right ) e \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{2 d^{2} \sqrt {e d}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) e x}{4 d^{2} \left (e \,x^{2}+d \right )}-\frac {b n}{d^{2} x}-\frac {3 i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{4 d^{2} \sqrt {e d}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e x}{4 d^{2} \left (e \,x^{2}+d \right )}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e x}{4 d^{2} \left (e \,x^{2}+d \right )}-\frac {3 i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{4 d^{2} \sqrt {e d}}+\frac {3 i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) e \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{4 d^{2} \sqrt {e d}}\)	\(933\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*((3*x^2*e + 2*d)/(d^2*x^3*e + d^3*x) + 3*arctan(x*e^(1/2)/sqrt(d))*e^(1/2)/d^(5/2)) + b*integrate((log(
c) + log(x^n))/(x^6*e^2 + 2*d*x^4*e + d^2*x^2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,{\left (e\,x^2+d\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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