\(\int \frac {a+b \log (c x^n)}{x^4 (d+e x^2)} \, dx\) [220]

Optimal result

Integrand size = 23, antiderivative size = 165 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=-\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {i b e^{3/2} n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {i b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}} \]

[Out]

-1/9*b*n/d/x^3+b*e*n/d^2/x+1/3*(-a-b*ln(c*x^n))/d/x^3+e*(a+b*ln(c*x^n))/d^2/x+e^(3/2)*arctan(x*e^(1/2)/d^(1/2)
)*(a+b*ln(c*x^n))/d^(5/2)-1/2*I*b*e^(3/2)*n*polylog(2,-I*x*e^(1/2)/d^(1/2))/d^(5/2)+1/2*I*b*e^(3/2)*n*polylog(
2,I*x*e^(1/2)/d^(1/2))/d^(5/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2380, 2341, 211, 2361, 12, 4940, 2438} \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}-\frac {i b e^{3/2} n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {i b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {b e n}{d^2 x}-\frac {b n}{9 d x^3} \]

[In]

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

[Out]

-1/9*(b*n)/(d*x^3) + (b*e*n)/(d^2*x) - (a + b*Log[c*x^n])/(3*d*x^3) + (e*(a + b*Log[c*x^n]))/(d^2*x) + (e^(3/2
)*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/d^(5/2) - ((I/2)*b*e^(3/2)*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqr
t[d]])/d^(5/2) + ((I/2)*b*e^(3/2)*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 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 {\int \frac {a+b \log \left (c x^n\right )}{x^4} \, dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )} \, dx}{d} \\ & = -\frac {b n}{9 d x^3}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {a+b \log \left (c x^n\right )}{d+e x^2} \, dx}{d^2} \\ & = -\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {\left (b e^2 n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{d^2} \\ & = -\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {\left (b e^{3/2} n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{d^{5/2}} \\ & = -\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {\left (i b e^{3/2} n\right ) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 d^{5/2}}+\frac {\left (i b e^{3/2} n\right ) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 d^{5/2}} \\ & = -\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {i b e^{3/2} n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {i b e^{3/2} n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.28 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\frac {1}{18} \left (-\frac {2 b n}{d x^3}+\frac {18 b e n}{d^2 x}-\frac {6 \left (a+b \log \left (c x^n\right )\right )}{d x^3}+\frac {18 e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}+\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}+\frac {9 b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}-\frac {9 b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}\right ) \]

[In]

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

[Out]

((-2*b*n)/(d*x^3) + (18*b*e*n)/(d^2*x) - (6*(a + b*Log[c*x^n]))/(d*x^3) + (18*e*(a + b*Log[c*x^n]))/(d^2*x) -
(9*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) + (9*e^(3/2)*(a + b*Log[c*x^n])*Log[1
+ (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(5/2) + (9*b*e^(3/2)*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) - (9*b*e
^(3/2)*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(5/2))/18

Maple [C] (warning: unable to verify)

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

Time = 0.56 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.24

method	result	size
risch	\(-\frac {b \,e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right ) n \ln \left (x \right )}{d^{2} \sqrt {d e}}+\frac {b \,e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right ) \ln \left (x^{n}\right )}{d^{2} \sqrt {d e}}-\frac {b \ln \left (x^{n}\right )}{3 d \,x^{3}}+\frac {b \ln \left (x^{n}\right ) e}{d^{2} x}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}+\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}-\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}-\frac {b n}{9 d \,x^{3}}+\frac {b e n}{d^{2} x}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{d^{2} \sqrt {d e}}-\frac {1}{3 d \,x^{3}}+\frac {e}{d^{2} x}\right )\)	\(369\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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