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

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

[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 = 0.14, 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 = {2340, 2345, 2391} \[ \frac {b n \text {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^2}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (2 a+2 b \log \left (c x^n\right )-b n\right )}{4 d^2}+\frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )} \]

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 2340

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 2345

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 2391

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]  time = 0.42, size = 279, normalized size = 3.40 \[ -\frac {\log \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{2 d^2}+\frac {a+b \log \left (c x^n\right )-b n \log (x)}{2 d^2+2 d e x^2}+\frac {\log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d^2}+\frac {b n \left (-2 \left (\text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )-2 \left (\text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )+\frac {\sqrt {e} x \log (x)}{-\sqrt {e} x+i \sqrt {d}}-\frac {\sqrt {e} x \log (x)}{\sqrt {e} x+i \sqrt {d}}+\log \left (-\sqrt {e} x+i \sqrt {d}\right )+\log \left (\sqrt {e} x+i \sqrt {d}\right )+2 \log ^2(x)\right )}{4 d^2} \]

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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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e^{2} x^{5} + 2 \, d e x^{3} + d^{2} x}, x\right ) \]

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)/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x}\,{d x} \]

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)/((e*x^2 + d)^2*x), x)

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maple [C]  time = 0.20, size = 644, normalized size = 7.85 \[ \frac {b n \ln \relax (x ) \ln \left (e \,x^{2}+d \right )}{2 d^{2}}-\frac {b n \ln \relax (x ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}-\frac {b n \ln \relax (x ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}+\frac {b \ln \relax (x ) \ln \left (x^{n}\right )}{d^{2}}-\frac {b n \ln \relax (x )^{2}}{2 d^{2}}-\frac {b n \ln \relax (x )}{2 d^{2}}+\frac {b \ln \relax (c )}{2 \left (e \,x^{2}+d \right ) d}-\frac {b \ln \relax (c ) \ln \left (e \,x^{2}+d \right )}{2 d^{2}}+\frac {a}{2 \left (e \,x^{2}+d \right ) d}-\frac {a \ln \left (e \,x^{2}+d \right )}{2 d^{2}}-\frac {b n \dilog \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}-\frac {b n \dilog \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}+\frac {b \ln \left (x^{n}\right )}{2 \left (e \,x^{2}+d \right ) d}-\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 d^{2}}+\frac {b \ln \relax (c ) \ln \relax (x )}{d^{2}}+\frac {i \pi b \,\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 {a \ln \relax (x )}{d^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )}{2 d^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 \left (e \,x^{2}+d \right ) d}-\frac {i \pi b \,\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 \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e \,x^{2}+d \right ) d}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e \,x^{2}+d \right ) d}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{2}}-\frac {i \pi b \,\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 \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 \left (e \,x^{2}+d \right ) d}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2 d^{2}}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e \,x^{2}+d \right )}{4 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {1}{d e x^{2} + d^{2}} - \frac {\log \left (e x^{2} + d\right )}{d^{2}} + \frac {2 \, \log \relax (x)}{d^{2}}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{e^{2} x^{5} + 2 \, d e x^{3} + d^{2} x}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (e\,x^2+d\right )}^2} \,d x \]

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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