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

Optimal. Leaf size=121 \[ -\frac {b n}{16 d x^4}+\frac {b e n}{4 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \log \left (1+\frac {d}{e x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {b e^2 n \text {Li}_2\left (-\frac {d}{e x^2}\right )}{4 d^3} \]

[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

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Rubi [A]
time = 0.14, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2380, 2341, 2379, 2438} \begin {gather*} \frac {b e^2 n \text {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^3}-\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 n}{4 d^2 x^2}-\frac {b n}{16 d x^4} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.13, size = 196, normalized size = 1.62 \begin {gather*} -\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 \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 b e^2 n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{16 d^3} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 805, normalized size = 6.65

method	result	size
risch	\(-\frac {a \,e^{2} \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 {-e d}}{\sqrt {-e d}}\right )}{2 d^{3}}-\frac {b n \,e^{2} \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{3}}-\frac {b \ln \left (x^{n}\right )}{4 d \,x^{4}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e^{2} \ln \left (x \right )}{2 d^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \ln \left (x \right )}{2 d^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \ln \left (x \right )}{2 d^{3}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e^{2} \ln \left (e \,x^{2}+d \right )}{4 d^{3}}-\frac {a}{4 d \,x^{4}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{3}}+\frac {a \,e^{2} \ln \left (x \right )}{d^{3}}+\frac {b \ln \left (c \right ) e}{2 d^{2} x^{2}}+\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{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} \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{3}}+\frac {a e}{2 d^{2} x^{2}}-\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^{2} \ln \left (x \right )}{2 d^{3}}+\frac {b \ln \left (c \right ) e^{2} \ln \left (x \right )}{d^{3}}-\frac {b n \,e^{2} \ln \left (x \right )^{2}}{2 d^{3}}+\frac {b \ln \left (x^{n}\right ) e}{2 d^{2} x^{2}}-\frac {b \ln \left (c \right ) e^{2} \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {b \ln \left (c \right )}{4 d \,x^{4}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \ln \left (e \,x^{2}+d \right )}{4 d^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e}{4 d^{2} x^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e}{4 d^{2} x^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \ln \left (e \,x^{2}+d \right )}{4 d^{3}}+\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 )}{8 d \,x^{4}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8 d \,x^{4}}-\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (e \,x^{2}+d \right )}{2 d^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8 d \,x^{4}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8 d \,x^{4}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e}{4 d^{2} x^{2}}+\frac {b e n}{4 d^{2} x^{2}}-\frac {b n}{16 d \,x^{4}}-\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}{4 d^{2} x^{2}}+\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^{2} \ln \left (e \,x^{2}+d \right )}{4 d^{3}}\)	\(805\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^5/(e*x^2+d),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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^5/(e*x^2+d),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((x^2*e + d)*x^5), 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^5\,\left (e\,x^2+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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