3.3.20 \(\int \frac {a+b \log (c x^n)}{x^4 (d+e x^2)} \, dx\) [220]

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

[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 = 0.16, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2380, 2341, 211, 2361, 12, 4940, 2438} \begin {gather*} -\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} \text {ArcTan}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/9*(b*n)/(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)/Sqr
t[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 211

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

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 2361

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 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]

Rule 4940

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 = 0.13, size = 211, normalized size = 1.28 \begin {gather*} \frac {1}{18} \left (-\frac {2 b n}{d x^3}+\frac {18 b e n}{d^2 x}-\frac {6 \left (a+b \log \left (c x^n\right )\right )}{d x^3}+\frac {18 e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}+\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}+\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}}\right ) \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 706, normalized size = 4.28

method	result	size
risch	\(-\frac {b \ln \left (c \right )}{3 d \,x^{3}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e}{2 d^{2} x}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e}{2 d^{2} x}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e}{2 d^{2} x}+\frac {b \,e^{2} \arctan \left (\frac {x e}{\sqrt {e d}}\right ) \ln \left (x^{n}\right )}{d^{2} \sqrt {e d}}-\frac {a}{3 d \,x^{3}}+\frac {a e}{d^{2} x}-\frac {b \,e^{2} \arctan \left (\frac {x e}{\sqrt {e d}}\right ) n \ln \left (x \right )}{d^{2} \sqrt {e d}}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2} \sqrt {-e d}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2} \sqrt {-e d}}+\frac {b \ln \left (x^{n}\right ) e}{d^{2} x}+\frac {b \ln \left (c \right ) e^{2} \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{d^{2} \sqrt {e d}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{6 d \,x^{3}}+\frac {a \,e^{2} \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{d^{2} \sqrt {e d}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6 d \,x^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6 d \,x^{3}}-\frac {b \ln \left (x^{n}\right )}{3 d \,x^{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 ) e}{2 d^{2} x}+\frac {b \ln \left (c \right ) e}{d^{2} x}+\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 )}{6 d \,x^{3}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e^{2} \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{2 d^{2} \sqrt {e d}}+\frac {b n \,e^{2} \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2} \sqrt {-e d}}-\frac {b n \,e^{2} \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 d^{2} \sqrt {-e d}}+\frac {b e n}{d^{2} x}-\frac {b n}{9 d \,x^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{2 d^{2} \sqrt {e d}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{2 d^{2} \sqrt {e d}}-\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} \arctan \left (\frac {x e}{\sqrt {e d}}\right )}{2 d^{2} \sqrt {e d}}\)	\(706\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^4/(e*x^2+d),x, algorithm="maxima")

[Out]

1/3*a*(3*arctan(x*e^(1/2)/sqrt(d))*e^(3/2)/d^(5/2) + (3*x^2*e - d)/(d^2*x^3)) + b*integrate((log(c) + log(x^n)
)/(x^6*e + d*x^4), 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^4/(e*x^2+d),x, algorithm="fricas")

[Out]

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

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

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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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^4/(e*x^2+d),x, algorithm="giac")

[Out]

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