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

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

[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)

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Rubi [A]  time = 0.197425, antiderivative size = 165, normalized size of antiderivative = 1., 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 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 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 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 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 12

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

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]

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^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*}

Mathematica [A]  time = 0.182793, size = 211, normalized size = 1.28 \[ \frac{1}{18} \left (\frac{9 b e^{3/2} n \text{PolyLog}\left (2,\frac{\sqrt{e} x}{\sqrt{-d}}\right )}{(-d)^{5/2}}-\frac{9 b e^{3/2} n \text{PolyLog}\left (2,\frac{d \sqrt{e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}+\frac{18 e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac{9 e^{3/2} \log \left (\frac{\sqrt{e} x}{\sqrt{-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2}}+\frac{9 e^{3/2} \log \left (\frac{d \sqrt{e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2}}-\frac{6 \left (a+b \log \left (c x^n\right )\right )}{d x^3}+\frac{18 b e n}{d^2 x}-\frac{2 b n}{d x^3}\right ) \]

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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Maple [C]  time = 0.208, size = 706, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left (c x^{n}\right ) + a}{e x^{6} + d x^{4}}, x\right ) \end{align*}

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

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., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{4}}\,{d x} \end{align*}

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)