\(\int \frac {a+b \log (c x^n)}{(d+e x^2)^3} \, dx\) [238]

Optimal result

Integrand size = 20, antiderivative size = 210 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {b n x}{8 d^2 \left (d+e x^2\right )}-\frac {b n \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{8 d^2 \left (d+e x^2\right )}+\frac {3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 d^{5/2} \sqrt {e}}-\frac {3 i b n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 d^{5/2} \sqrt {e}}+\frac {3 i b n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 d^{5/2} \sqrt {e}} \]

[Out]

-1/8*b*n*x/d^2/(e*x^2+d)+1/4*x*(a+b*ln(c*x^n))/d/(e*x^2+d)^2+3/8*x*(a+b*ln(c*x^n))/d^2/(e*x^2+d)-1/2*b*n*arcta
n(x*e^(1/2)/d^(1/2))/d^(5/2)/e^(1/2)+3/8*arctan(x*e^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(5/2)/e^(1/2)-3/16*I*b*n*
polylog(2,-I*x*e^(1/2)/d^(1/2))/d^(5/2)/e^(1/2)+3/16*I*b*n*polylog(2,I*x*e^(1/2)/d^(1/2))/d^(5/2)/e^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2360, 211, 2361, 12, 4940, 2438, 205} \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 d^{5/2} \sqrt {e}}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{8 d^2 \left (d+e x^2\right )}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}-\frac {3 i b n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 d^{5/2} \sqrt {e}}+\frac {3 i b n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 d^{5/2} \sqrt {e}}-\frac {b n x}{8 d^2 \left (d+e x^2\right )} \]

[In]

Int[(a + b*Log[c*x^n])/(d + e*x^2)^3,x]

[Out]

-1/8*(b*n*x)/(d^2*(d + e*x^2)) - (b*n*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(5/2)*Sqrt[e]) + (x*(a + b*Log[c*x^n])
)/(4*d*(d + e*x^2)^2) + (3*x*(a + b*Log[c*x^n]))/(8*d^2*(d + e*x^2)) + (3*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b*L
og[c*x^n]))/(8*d^(5/2)*Sqrt[e]) - (((3*I)/16)*b*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]])/(d^(5/2)*Sqrt[e]) + ((
(3*I)/16)*b*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/(d^(5/2)*Sqrt[e])

Rule 12

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(q +
1)*((a + b*Log[c*x^n])/(2*d*(q + 1))), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*Log[
c*x^n]), x], x] + Dist[b*(n/(2*d*(q + 1))), Int[(d + e*x^2)^(q + 1), x], x]) /; FreeQ[{a, b, c, d, e, n}, x] &
& LtQ[q, -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 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 {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}+\frac {3 \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^2} \, dx}{4 d}-\frac {(b n) \int \frac {1}{\left (d+e x^2\right )^2} \, dx}{4 d} \\ & = -\frac {b n x}{8 d^2 \left (d+e x^2\right )}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{8 d^2 \left (d+e x^2\right )}+\frac {3 \int \frac {a+b \log \left (c x^n\right )}{d+e x^2} \, dx}{8 d^2}-\frac {(b n) \int \frac {1}{d+e x^2} \, dx}{8 d^2}-\frac {(3 b n) \int \frac {1}{d+e x^2} \, dx}{8 d^2} \\ & = -\frac {b n x}{8 d^2 \left (d+e x^2\right )}-\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{8 d^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 d^{5/2} \sqrt {e}}-\frac {(3 b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{8 d^2} \\ & = -\frac {b n x}{8 d^2 \left (d+e x^2\right )}-\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{8 d^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 d^{5/2} \sqrt {e}}-\frac {(3 b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{8 d^{5/2} \sqrt {e}} \\ & = -\frac {b n x}{8 d^2 \left (d+e x^2\right )}-\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{8 d^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 d^{5/2} \sqrt {e}}-\frac {(3 i b n) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{16 d^{5/2} \sqrt {e}}+\frac {(3 i b n) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{16 d^{5/2} \sqrt {e}} \\ & = -\frac {b n x}{8 d^2 \left (d+e x^2\right )}-\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )}{8 d^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 d^{5/2} \sqrt {e}}-\frac {3 i b n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 d^{5/2} \sqrt {e}}+\frac {3 i b n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 d^{5/2} \sqrt {e}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(544\) vs. \(2(210)=420\).

Time = 0.59 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.59 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=\frac {1}{16} \left (\frac {d \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2} \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )^2}+\frac {a+b \log \left (c x^n\right )}{(-d)^{3/2} \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )^2}+\frac {3 \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2} \sqrt {e}+d^2 e x}+\frac {3 \left (a+b \log \left (c x^n\right )\right )}{(-d)^{3/2} d \sqrt {e}+d^2 e x}+\frac {3 b n \left (\log (x)-\log \left (\sqrt {-d}-\sqrt {e} x\right )\right )}{(-d)^{5/2} \sqrt {e}}-\frac {3 b n \left (\log (x)-\log \left (\sqrt {-d}+\sqrt {e} x\right )\right )}{(-d)^{5/2} \sqrt {e}}-\frac {b n \left (d+\left (d-\sqrt {-d} \sqrt {e} x\right ) \log (x)+\left (-d+\sqrt {-d} \sqrt {e} x\right ) \log \left (\sqrt {-d}+\sqrt {e} x\right )\right )}{d^3 \left (\sqrt {-d} \sqrt {e}+e x\right )}-\frac {3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2} \sqrt {e}}-\frac {b n \left (d+\left (d+\sqrt {-d} \sqrt {e} x\right ) \log (x)-\left (d+\sqrt {-d} \sqrt {e} x\right ) \log \left ((-d)^{3/2}+d \sqrt {e} x\right )\right )}{(-d)^{7/2} \sqrt {e}+d^3 e x}+\frac {3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2} \sqrt {e}}+\frac {3 b n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2} \sqrt {e}}-\frac {3 b n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2} \sqrt {e}}\right ) \]

[In]

Integrate[(a + b*Log[c*x^n])/(d + e*x^2)^3,x]

[Out]

((d*(a + b*Log[c*x^n]))/((-d)^(5/2)*Sqrt[e]*(Sqrt[-d] - Sqrt[e]*x)^2) + (a + b*Log[c*x^n])/((-d)^(3/2)*Sqrt[e]
*(Sqrt[-d] + Sqrt[e]*x)^2) + (3*(a + b*Log[c*x^n]))/((-d)^(5/2)*Sqrt[e] + d^2*e*x) + (3*(a + b*Log[c*x^n]))/((
-d)^(3/2)*d*Sqrt[e] + d^2*e*x) + (3*b*n*(Log[x] - Log[Sqrt[-d] - Sqrt[e]*x]))/((-d)^(5/2)*Sqrt[e]) - (3*b*n*(L
og[x] - Log[Sqrt[-d] + Sqrt[e]*x]))/((-d)^(5/2)*Sqrt[e]) - (b*n*(d + (d - Sqrt[-d]*Sqrt[e]*x)*Log[x] + (-d + S
qrt[-d]*Sqrt[e]*x)*Log[Sqrt[-d] + Sqrt[e]*x]))/(d^3*(Sqrt[-d]*Sqrt[e] + e*x)) - (3*(a + b*Log[c*x^n])*Log[1 +
(Sqrt[e]*x)/Sqrt[-d]])/((-d)^(5/2)*Sqrt[e]) - (b*n*(d + (d + Sqrt[-d]*Sqrt[e]*x)*Log[x] - (d + Sqrt[-d]*Sqrt[e
]*x)*Log[(-d)^(3/2) + d*Sqrt[e]*x]))/((-d)^(7/2)*Sqrt[e] + d^3*e*x) + (3*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]
*x)/(-d)^(3/2)])/((-d)^(5/2)*Sqrt[e]) + (3*b*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]])/((-d)^(5/2)*Sqrt[e]) - (3*b*n
*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/((-d)^(5/2)*Sqrt[e]))/16

Maple [C] (warning: unable to verify)

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

Time = 0.63 (sec) , antiderivative size = 664, normalized size of antiderivative = 3.16

method	result	size
risch	\(\frac {3 b n \ln \left (x \right ) x}{8 d \left (e \,x^{2}+d \right )^{2}}+\frac {b x \ln \left (x^{n}\right )}{4 d \left (e \,x^{2}+d \right )^{2}}-\frac {3 b x n \ln \left (x \right )}{8 d^{2} \left (e \,x^{2}+d \right )}+\frac {3 b x \ln \left (x^{n}\right )}{8 d^{2} \left (e \,x^{2}+d \right )}-\frac {3 b \arctan \left (\frac {x e}{\sqrt {d e}}\right ) n \ln \left (x \right )}{8 d^{2} \sqrt {d e}}+\frac {3 b \arctan \left (\frac {x e}{\sqrt {d e}}\right ) \ln \left (x^{n}\right )}{8 d^{2} \sqrt {d e}}-\frac {b n x}{8 d^{2} \left (e \,x^{2}+d \right )}-\frac {b n \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{2 d^{2} \sqrt {d e}}+\frac {3 b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{4} e^{2}}{16 d^{2} \left (e \,x^{2}+d \right )^{2} \sqrt {-d e}}-\frac {3 b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{4} e^{2}}{16 d^{2} \left (e \,x^{2}+d \right )^{2} \sqrt {-d e}}+\frac {3 b n \ln \left (x \right ) x^{3} e}{8 d^{2} \left (e \,x^{2}+d \right )^{2}}+\frac {3 b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{2} e}{8 d \left (e \,x^{2}+d \right )^{2} \sqrt {-d e}}-\frac {3 b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{2} e}{8 d \left (e \,x^{2}+d \right )^{2} \sqrt {-d e}}+\frac {3 b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{16 \left (e \,x^{2}+d \right )^{2} \sqrt {-d e}}-\frac {3 b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{16 \left (e \,x^{2}+d \right )^{2} \sqrt {-d e}}+\frac {3 b n \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{16 \sqrt {-d e}\, d^{2}}-\frac {3 b n \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{16 \sqrt {-d e}\, d^{2}}+\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 {x}{4 d \left (e \,x^{2}+d \right )^{2}}+\frac {\frac {3 x}{8 d \left (e \,x^{2}+d \right )}+\frac {3 \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{8 d \sqrt {d e}}}{d}\right )\)	\(664\)

[In]

int((a+b*ln(c*x^n))/(e*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

3/8*b*n*ln(x)/d/(e*x^2+d)^2*x+1/4*b*x/d/(e*x^2+d)^2*ln(x^n)-3/8*b/d^2*x/(e*x^2+d)*n*ln(x)+3/8*b/d^2*x/(e*x^2+d
)*ln(x^n)-3/8*b/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*n*ln(x)+3/8*b/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*
ln(x^n)-1/8*b*n*x/d^2/(e*x^2+d)-1/2*b*n/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+3/16*b*n*ln(x)/d^2/(e*x^2+d)^2
/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^4*e^2-3/16*b*n*ln(x)/d^2/(e*x^2+d)^2/(-d*e)^(1/2)*ln((e*x
+(-d*e)^(1/2))/(-d*e)^(1/2))*x^4*e^2+3/8*b*n*ln(x)/d^2/(e*x^2+d)^2*x^3*e+3/8*b*n*ln(x)/d/(e*x^2+d)^2/(-d*e)^(1
/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2*e-3/8*b*n*ln(x)/d/(e*x^2+d)^2/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/
(-d*e)^(1/2))*x^2*e+3/16*b*n*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*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/(-d*e)^(1/2)/d^2*dilog((-e*x+(-d*e)^(1
/2))/(-d*e)^(1/2))-3/16*b*n/(-d*e)^(1/2)/d^2*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+(-1/2*I*b*Pi*csgn(I*c)*csg
n(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)*csgn(I*c*x^n)^2-1/2*I*b*Pi*
csgn(I*c*x^n)^3+b*ln(c)+a)*(1/4*x/d/(e*x^2+d)^2+3/4/d*(1/2*x/d/(e*x^2+d)+1/2/d/(d*e)^(1/2)*arctan(x*e/(d*e)^(1
/2))))

Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

Sympy [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{\left (d + e x^{2}\right )^{3}}\, dx \]

[In]

integrate((a+b*ln(c*x**n))/(e*x**2+d)**3,x)

[Out]

Integral((a + b*log(c*x**n))/(d + e*x**2)**3, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*log(c*x^n))/(e*x^2+d)^3,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]

\[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/(e*x^2 + d)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]

[In]

int((a + b*log(c*x^n))/(d + e*x^2)^3,x)

[Out]

int((a + b*log(c*x^n))/(d + e*x^2)^3, x)