3.220 \(\int \frac {a+b \log (c x^n)}{x^4 (d+e x^2)} \, dx\)

Optimal. Leaf size=165 \[ \frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}-\frac {i b e^{3/2} n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {i b e^{3/2} n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {b e n}{d^2 x}-\frac {b n}{9 d x^3} \]

[Out]

-1/9*b*n/d/x^3+b*e*n/d^2/x+1/3*(-a-b*ln(c*x^n))/d/x^3+e*(a+b*ln(c*x^n))/d^2/x+e^(3/2)*arctan(x*e^(1/2)/d^(1/2)
)*(a+b*ln(c*x^n))/d^(5/2)-1/2*I*b*e^(3/2)*n*polylog(2,-I*x*e^(1/2)/d^(1/2))/d^(5/2)+1/2*I*b*e^(3/2)*n*polylog(
2,I*x*e^(1/2)/d^(1/2))/d^(5/2)

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Rubi [A]  time = 0.20, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {325, 205, 2351, 2304, 2324, 12, 4848, 2391} \[ -\frac {i b e^{3/2} n \text {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {i b e^{3/2} n \text {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {b e n}{d^2 x}-\frac {b n}{9 d x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n)/(9*d*x^3) + (b*e*n)/(d^2*x) - (a + b*Log[c*x^n])/(3*d*x^3) + (e*(a + b*Log[c*x^n]))/(d^2*x) + (e^(3/2)*
ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/d^(5/2) - ((I/2)*b*e^(3/2)*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[
d]])/d^(5/2) + ((I/2)*b*e^(3/2)*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/d^(5/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2304

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 2324

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{d x^4}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x^2}+\frac {e^2 \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^4} \, dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {a+b \log \left (c x^n\right )}{d+e x^2} \, dx}{d^2}\\ &=-\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {\left (b e^2 n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{d^2}\\ &=-\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {\left (b e^{3/2} n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{d^{5/2}}\\ &=-\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {\left (i b e^{3/2} n\right ) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 d^{5/2}}+\frac {\left (i b e^{3/2} n\right ) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 d^{5/2}}\\ &=-\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {i b e^{3/2} n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {i b e^{3/2} n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 211, normalized size = 1.28 \[ \frac {1}{18} \left (\frac {18 e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {9 e^{3/2} \log \left (\frac {\sqrt {e} x}{\sqrt {-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2}}+\frac {9 e^{3/2} \log \left (\frac {d \sqrt {e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2}}-\frac {6 \left (a+b \log \left (c x^n\right )\right )}{d x^3}+\frac {18 b e n}{d^2 x}+\frac {9 b e^{3/2} n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}-\frac {9 b e^{3/2} n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}-\frac {2 b n}{d x^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*n)/(d*x^3) + (18*b*e*n)/(d^2*x) - (6*(a + b*Log[c*x^n]))/(d*x^3) + (18*e*(a + b*Log[c*x^n]))/(d^2*x) -
(9*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) + (9*e^(3/2)*(a + b*Log[c*x^n])*Log[1
+ (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(5/2) + (9*b*e^(3/2)*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) - (9*b*e
^(3/2)*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(5/2))/18

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e x^{6} + d x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(c*x^n) + a)/(e*x^6 + d*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((e*x^2 + d)*x^4), x)

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maple [C]  time = 0.30, size = 706, normalized size = 4.28 \[ \frac {a \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}\, d^{2}}+\frac {b e \ln \left (x^{n}\right )}{d^{2} x}+\frac {b e \ln \relax (c )}{d^{2} x}+\frac {b \,e^{2} n \dilog \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 \sqrt {-d e}\, d^{2}}-\frac {b \,e^{2} n \dilog \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 \sqrt {-d e}\, d^{2}}+\frac {b \,e^{2} n \ln \relax (x ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 \sqrt {-d e}\, d^{2}}-\frac {b \,e^{2} n \ln \relax (x ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 \sqrt {-d e}\, d^{2}}+\frac {i \pi b e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2} x}+\frac {i \pi b e \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2} x}-\frac {b \ln \left (x^{n}\right )}{3 d \,x^{3}}+\frac {i \pi b \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \sqrt {d e}\, d^{2}}+\frac {b \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \ln \left (x^{n}\right )}{\sqrt {d e}\, d^{2}}-\frac {b \ln \relax (c )}{3 d \,x^{3}}+\frac {i \pi b \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \sqrt {d e}\, d^{2}}-\frac {b \,e^{2} n \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \ln \relax (x )}{\sqrt {d e}\, d^{2}}+\frac {a e}{d^{2} x}-\frac {a}{3 d \,x^{3}}+\frac {b \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \ln \relax (c )}{\sqrt {d e}\, 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 )}{6 d \,x^{3}}-\frac {i \pi b \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 \sqrt {d e}\, d^{2}}-\frac {i \pi b e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 d^{2} x}-\frac {i \pi b \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 \sqrt {d e}\, d^{2}}-\frac {b n}{9 d \,x^{3}}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{6 d \,x^{3}}-\frac {i \pi b e \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 d^{2} x}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6 d \,x^{3}}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6 d \,x^{3}}+\frac {b e n}{d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e^2/d^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)+a*e^2/d^2/(d*e)^(1/2)*arcta
n(1/(d*e)^(1/2)*e*x)+b/d^2*e/x*ln(x^n)+b/d^2*e/x*ln(c)+1/2*b*n*e^2/d^2/(-d*e)^(1/2)*ln(x)*ln((-e*x+(-d*e)^(1/2
))/(-d*e)^(1/2))-1/2*b*n*e^2/d^2/(-d*e)^(1/2)*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)^2*e^2/d^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)-1/2*I*b*Pi*csgn(I*c*x^n)^3*e^2/d^2/(d*e)^(1/2)*
arctan(1/(d*e)^(1/2)*e*x)+1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d/x^3-1/3*b*ln(x^n)/d/x^3+1/6*I*b*Pi*
csgn(I*c*x^n)^3/d/x^3+b*e^2/d^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*ln(x^n)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*
x^n)*csgn(I*c)*e^2/d^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e/d^2/x-1/
3*b/d/x^3*ln(c)+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e/d^2/x-b*e^2/d^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*n
*ln(x)+1/2*b*n*e^2/d^2/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*e^2/d^2/(-d*e)^(1/2)*dilog
((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+a/d^2*e/x-1/3*a/d/x^3+b*ln(c)*e^2/d^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)-
1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e/d^2/x-1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d/x^3-1/9*b/d*n/
x^3-1/6*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d/x^3-1/2*I*b*Pi*csgn(I*c*x^n)^3*e/d^2/x+b/d^2*e*n/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a {\left (\frac {3 \, e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} d^{2}} + \frac {3 \, e x^{2} - d}{d^{2} x^{3}}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{e x^{6} + d x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d),x, algorithm="maxima")

[Out]

1/3*a*(3*e^2*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^2) + (3*e*x^2 - d)/(d^2*x^3)) + b*integrate((log(c) + log(x^n)
)/(e*x^6 + d*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\left (e\,x^2+d\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))/(x^4*(d + e*x^2)),x)

[Out]

int((a + b*log(c*x^n))/(x^4*(d + e*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \log {\left (c x^{n} \right )}}{x^{4} \left (d + e x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**4/(e*x**2+d),x)

[Out]

Integral((a + b*log(c*x**n))/(x**4*(d + e*x**2)), x)

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