\(\int \frac {a+b \log (c x^n)}{x^5 (d+e x^2)} \, dx\) [215]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=-\frac {b n}{16 d x^4}+\frac {b e n}{4 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \log \left (1+\frac {d}{e x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2380, 2341, 2379, 2438} \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=-\frac {e^2 \log \left (\frac {d}{e x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^3}+\frac {b e n}{4 d^2 x^2}-\frac {b n}{16 d x^4} \]

[In]

Int[(a + b*Log[c*x^n])/(x^5*(d + e*x^2)),x]

[Out]

-1/16*(b*n)/(d*x^4) + (b*e*n)/(4*d^2*x^2) - (a + b*Log[c*x^n])/(4*d*x^4) + (e*(a + b*Log[c*x^n]))/(2*d^2*x^2)
- (e^2*Log[1 + d/(e*x^2)]*(a + b*Log[c*x^n]))/(2*d^3) + (b*e^2*n*PolyLog[2, -(d/(e*x^2))])/(4*d^3)

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 2379

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {a+b \log \left (c x^n\right )}{x^5} \, dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )} \, dx}{d} \\ & = -\frac {b n}{16 d x^4}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^3} \, dx}{d^2}+\frac {e^2 \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx}{d^2} \\ & = -\frac {b n}{16 d x^4}+\frac {b e n}{4 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \log \left (1+\frac {d}{e x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {\left (b e^2 n\right ) \int \frac {\log \left (1+\frac {d}{e x^2}\right )}{x} \, dx}{2 d^3} \\ & = -\frac {b n}{16 d x^4}+\frac {b e n}{4 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \log \left (1+\frac {d}{e x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {b e^2 n \text {Li}_2\left (-\frac {d}{e x^2}\right )}{4 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.62 \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=-\frac {\frac {b d^2 n}{x^4}-\frac {4 b d e n}{x^2}+\frac {4 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^4}-\frac {8 d e \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {8 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+8 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+8 b e^2 n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 b e^2 n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{16 d^3} \]

[In]

Integrate[(a + b*Log[c*x^n])/(x^5*(d + e*x^2)),x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.64 (sec) , antiderivative size = 369, normalized size of antiderivative = 3.05

method	result	size
risch	\(-\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {b \ln \left (x^{n}\right )}{4 d \,x^{4}}+\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {b \ln \left (x^{n}\right ) e}{2 d^{2} x^{2}}+\frac {b e n}{4 d^{2} x^{2}}-\frac {b n}{16 d \,x^{4}}-\frac {b n \,e^{2} \ln \left (x \right )^{2}}{2 d^{3}}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}-\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}-\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {e^{2} \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {1}{4 d \,x^{4}}+\frac {e^{2} \ln \left (x \right )}{d^{3}}+\frac {e}{2 d^{2} x^{2}}\right )\)	\(369\)

[In]

int((a+b*ln(c*x^n))/x^5/(e*x^2+d),x,method=_RETURNVERBOSE)

[Out]

-1/2*b*ln(x^n)*e^2/d^3*ln(e*x^2+d)-1/4*b*ln(x^n)/d/x^4+b*ln(x^n)*e^2/d^3*ln(x)+1/2*b*ln(x^n)*e/d^2/x^2+1/4*b*e
*n/d^2/x^2-1/16*b*n/d/x^4-1/2*b*n*e^2/d^3*ln(x)^2+1/2*b*n*e^2/d^3*ln(x)*ln(e*x^2+d)-1/2*b*n*e^2/d^3*ln(x)*ln((
-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*e^2/d^3*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*e^2/d^3*dil
og((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*e^2/d^3*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+(-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)*(-1/2*e^2/d^3*ln(e*x^2+d)-1/4/d/x^4+e^2/d^3*ln(x)+1/2*e/d^2/x^2)

Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{5}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/x^5/(e*x^2+d),x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)/(e*x^7 + d*x^5), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate((a+b*ln(c*x**n))/x**5/(e*x**2+d),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{5}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/x^5/(e*x^2+d),x, algorithm="maxima")

[Out]

-1/4*a*(2*e^2*log(e*x^2 + d)/d^3 - 4*e^2*log(x)/d^3 - (2*e*x^2 - d)/(d^2*x^4)) + b*integrate((log(c) + log(x^n
))/(e*x^7 + d*x^5), x)

Giac [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{5}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/x^5/(e*x^2+d),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((e*x^2 + d)*x^5), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^5 \left (d+e x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^5\,\left (e\,x^2+d\right )} \,d x \]

[In]

int((a + b*log(c*x^n))/(x^5*(d + e*x^2)),x)

[Out]

int((a + b*log(c*x^n))/(x^5*(d + e*x^2)), x)