3.3.24 \(\int \frac {a+b \log (c x^n)}{x (d+e x^2)^2} \, dx\) [224]

Optimal. Leaf size=82 \[ \frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )}-\frac {\log \left (1+\frac {d}{e x^2}\right ) \left (2 a-b n+2 b \log \left (c x^n\right )\right )}{4 d^2}+\frac {b n \text {Li}_2\left (-\frac {d}{e x^2}\right )}{4 d^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2385, 2379, 2438} \begin {gather*} \frac {b n \text {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^2}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (2 a+2 b \log \left (c x^n\right )-b n\right )}{4 d^2}+\frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2385

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(-
(f*x)^(m + 1))*(d + e*x^2)^(q + 1)*((a + b*Log[c*x^n])/(2*d*f*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(f*x)^
m*(d + e*x^2)^(q + 1)*(a*(m + 2*q + 3) + b*n + b*(m + 2*q + 3)*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && ILtQ[q, -1] && ILtQ[m, 0]

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.27, size = 279, normalized size = 3.40 \begin {gather*} \frac {a-b n \log (x)+b \log \left (c x^n\right )}{2 d^2+2 d e x^2}+\frac {\log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{d^2}-\frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+e x^2\right )}{2 d^2}+\frac {b n \left (\frac {\sqrt {e} x \log (x)}{i \sqrt {d}-\sqrt {e} x}-\frac {\sqrt {e} x \log (x)}{i \sqrt {d}+\sqrt {e} x}+2 \log ^2(x)+\log \left (i \sqrt {d}-\sqrt {e} x\right )+\log \left (i \sqrt {d}+\sqrt {e} x\right )-2 \left (\log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )-2 \left (\log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )\right )}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a - b*n*Log[x] + b*Log[c*x^n])/(2*d^2 + 2*d*e*x^2) + (Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]))/d^2 - ((a - b*n
*Log[x] + b*Log[c*x^n])*Log[d + e*x^2])/(2*d^2) + (b*n*((Sqrt[e]*x*Log[x])/(I*Sqrt[d] - Sqrt[e]*x) - (Sqrt[e]*
x*Log[x])/(I*Sqrt[d] + Sqrt[e]*x) + 2*Log[x]^2 + Log[I*Sqrt[d] - Sqrt[e]*x] + Log[I*Sqrt[d] + Sqrt[e]*x] - 2*(
Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]]) - 2*(Log[x]*Log[1 - (I*Sqrt[e]*x
)/Sqrt[d]] + PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])))/(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 = 644, normalized size = 7.85

method	result	size
risch	\(-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}+\frac {b n \ln \left (x \right ) \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 \left (e \,x^{2}+d \right )}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e \,x^{2}+d \right )}{4 d^{2}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (x \right )}{2 d^{2}}-\frac {b n \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}-\frac {b n \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}-\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (x \right )}{2 d^{2}}-\frac {b n \ln \left (x \right )^{2}}{2 d^{2}}-\frac {b n \ln \left (x \right )}{2 d^{2}}+\frac {a \ln \left (x \right )}{d^{2}}+\frac {b \ln \left (c \right ) \ln \left (x \right )}{d^{2}}-\frac {a \ln \left (e \,x^{2}+d \right )}{2 d^{2}}+\frac {a}{2 d \left (e \,x^{2}+d \right )}+\frac {b \ln \left (c \right )}{2 d \left (e \,x^{2}+d \right )}-\frac {b \ln \left (c \right ) \ln \left (e \,x^{2}+d \right )}{2 d^{2}}+\frac {b \ln \left (x^{n}\right ) \ln \left (x \right )}{d^{2}}+\frac {b \ln \left (x^{n}\right )}{2 d \left (e \,x^{2}+d \right )}-\frac {b \ln \left (x^{n}\right ) \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 ) \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 )}{4 d \left (e \,x^{2}+d \right )}+\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 ) \ln \left (e \,x^{2}+d \right )}{4 d^{2}}+\frac {b n \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} \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}}{4 d \left (e \,x^{2}+d \right )}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \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} \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}}{4 d \left (e \,x^{2}+d \right )}\)	\(644\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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