3.3.14 \(\int \frac {a+b \log (c x^n)}{x^3 (d+e x^2)} \, dx\) [214]

Optimal. Leaf size=83 \[ -\frac {b n}{4 d x^2}-\frac {a+b \log \left (c x^n\right )}{2 d x^2}+\frac {e \log \left (1+\frac {d}{e x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^2}-\frac {b e n \text {Li}_2\left (-\frac {d}{e x^2}\right )}{4 d^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, 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 n \text {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^2}+\frac {e \log \left (\frac {d}{e x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^2}-\frac {a+b \log \left (c x^n\right )}{2 d x^2}-\frac {b n}{4 d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.08, size = 157, normalized size = 1.89 \begin {gather*} \frac {-\frac {b d n}{x^2}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {2 e \left (a+b \log \left (c x^n\right )\right )^2}{b n}+2 e \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+2 e \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+2 b e n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )+2 b e n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 611, normalized size = 7.36

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) e \ln \left (e \,x^{2}+d \right )}{2 d^{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 \ln \left (x \right )}{2 d^{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 \ln \left (e \,x^{2}+d \right )}{4 d^{2}}-\frac {b \ln \left (c \right )}{2 d \,x^{2}}+\frac {b n e \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}-\frac {b n e \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 d^{2}}+\frac {b n e \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}+\frac {b n e \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}+\frac {a e \ln \left (e \,x^{2}+d \right )}{2 d^{2}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 d \,x^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 d \,x^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 d \,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 )}{4 d \,x^{2}}-\frac {b \ln \left (x^{n}\right )}{2 d \,x^{2}}-\frac {a}{2 d \,x^{2}}-\frac {a e \ln \left (x \right )}{d^{2}}+\frac {b n e \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (e \,x^{2}+d \right )}{4 d^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (x \right )}{2 d^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (e \,x^{2}+d \right )}{4 d^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e \ln \left (x \right )}{2 d^{2}}+\frac {b \ln \left (c \right ) e \ln \left (e \,x^{2}+d \right )}{2 d^{2}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e \ln \left (e \,x^{2}+d \right )}{4 d^{2}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e \ln \left (x \right )}{2 d^{2}}-\frac {b \ln \left (x^{n}\right ) e \ln \left (x \right )}{d^{2}}-\frac {b \ln \left (c \right ) e \ln \left (x \right )}{d^{2}}+\frac {b n e \ln \left (x \right )^{2}}{2 d^{2}}-\frac {b n}{4 d \,x^{2}}\)	\(611\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*ln(x^n)*e/d^2*ln(e*x^2+d)+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*e/d^2*ln(e*x^2+d)-1/2*I*b*Pi*csgn(I*x^n)*
csgn(I*c*x^n)^2*e/d^2*ln(x)-1/2*b*ln(c)/d/x^2-1/2*b*n*e/d^2*ln(x)*ln(e*x^2+d)+1/2*b*n*e/d^2*ln(x)*ln((-e*x+(-e
*d)^(1/2))/(-e*d)^(1/2))+1/2*b*n*e/d^2*ln(x)*ln((e*x+(-e*d)^(1/2))/(-e*d)^(1/2))+1/2*a*e/d^2*ln(e*x^2+d)-1/4*I
*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d/x^2+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e/d^2*ln(e*x^2+d)-1/2*I*b*Pi*cs
gn(I*c)*csgn(I*c*x^n)^2*e/d^2*ln(x)-1/2*b*ln(x^n)/d/x^2-1/2*a/d/x^2-a*e/d^2*ln(x)-1/4*I*b*Pi*csgn(I*c*x^n)^3*e
/d^2*ln(e*x^2+d)+1/2*I*b*Pi*csgn(I*c*x^n)^3*e/d^2*ln(x)+1/4*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d/x^2-1
/4*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d/x^2+1/2*b*ln(c)*e/d^2*ln(e*x^2+d)+1/2*b*n*e/d^2*dilog((-e*x+(-e*d)^(1/2)
)/(-e*d)^(1/2))+1/2*b*n*e/d^2*dilog((e*x+(-e*d)^(1/2))/(-e*d)^(1/2))+1/4*I*b*Pi*csgn(I*c*x^n)^3/d/x^2-b*ln(x^n
)*e/d^2*ln(x)+1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*e/d^2*ln(x)-1/4*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn
(I*c*x^n)*e/d^2*ln(e*x^2+d)-b*ln(c)*e/d^2*ln(x)+1/2*b*n*e/d^2*ln(x)^2-1/4*b*n/d/x^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^3/(e*x^2+d),x, algorithm="maxima")

[Out]

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

[Out]

integral((b*log(c*x^n) + a)/(x^5*e + d*x^3), 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**3/(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^3/(e*x^2+d),x, algorithm="giac")

[Out]

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

[Out]

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

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