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

Optimal. Leaf size=183 \[ \frac{3 i b \sqrt{e} n \text{PolyLog}\left (2,-\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{4 d^{5/2}}-\frac{3 i b \sqrt{e} n \text{PolyLog}\left (2,\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{4 d^{5/2}}-\frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (3 a+3 b \log \left (c x^n\right )-b n\right )}{2 d^{5/2}}-\frac{3 a+3 b \log \left (c x^n\right )-b n}{2 d^2 x}+\frac{a+b \log \left (c x^n\right )}{2 d x \left (d+e x^2\right )}-\frac{3 b n}{2 d^2 x} \]

[Out]

(-3*b*n)/(2*d^2*x) + (a + b*Log[c*x^n])/(2*d*x*(d + e*x^2)) - (3*a - b*n + 3*b*Log[c*x^n])/(2*d^2*x) - (Sqrt[e
]*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(3*a - b*n + 3*b*Log[c*x^n]))/(2*d^(5/2)) + (((3*I)/4)*b*Sqrt[e]*n*PolyLog[2, ((
-I)*Sqrt[e]*x)/Sqrt[d]])/d^(5/2) - (((3*I)/4)*b*Sqrt[e]*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/d^(5/2)

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Rubi [A]  time = 0.283844, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {2340, 325, 205, 2351, 2304, 2324, 12, 4848, 2391} \[ \frac{3 i b \sqrt{e} n \text{PolyLog}\left (2,-\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{4 d^{5/2}}-\frac{3 i b \sqrt{e} n \text{PolyLog}\left (2,\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{4 d^{5/2}}-\frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (3 a+3 b \log \left (c x^n\right )-b n\right )}{2 d^{5/2}}-\frac{3 a+3 b \log \left (c x^n\right )-b n}{2 d^2 x}+\frac{a+b \log \left (c x^n\right )}{2 d x \left (d+e x^2\right )}-\frac{3 b n}{2 d^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*n)/(2*d^2*x) + (a + b*Log[c*x^n])/(2*d*x*(d + e*x^2)) - (3*a - b*n + 3*b*Log[c*x^n])/(2*d^2*x) - (Sqrt[e
]*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(3*a - b*n + 3*b*Log[c*x^n]))/(2*d^(5/2)) + (((3*I)/4)*b*Sqrt[e]*n*PolyLog[2, ((
-I)*Sqrt[e]*x)/Sqrt[d]])/d^(5/2) - (((3*I)/4)*b*Sqrt[e]*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/d^(5/2)

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

Mathematica [A]  time = 0.78284, size = 328, normalized size = 1.79 \[ \frac{1}{4} \left (-\frac{3 b \sqrt{e} n \text{PolyLog}\left (2,\frac{\sqrt{e} x}{\sqrt{-d}}\right )}{(-d)^{5/2}}+\frac{3 b \sqrt{e} n \text{PolyLog}\left (2,\frac{d \sqrt{e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}+\frac{\sqrt{e} \left (a+b \log \left (c x^n\right )\right )}{d^2 \left (\sqrt{-d}-\sqrt{e} x\right )}-\frac{\sqrt{e} \left (a+b \log \left (c x^n\right )\right )}{d^2 \left (\sqrt{-d}+\sqrt{e} x\right )}-\frac{4 \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac{3 \sqrt{e} \log \left (\frac{\sqrt{e} x}{\sqrt{-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2}}-\frac{3 \sqrt{e} \log \left (\frac{d \sqrt{e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{(-d)^{5/2}}-\frac{4 b n}{d^2 x}+\frac{b \sqrt{e} n \left (\log \left (\sqrt{-d}-\sqrt{e} x\right )-\log (x)\right )}{(-d)^{5/2}}+\frac{b \sqrt{e} n \left (\log (x)-\log \left (\sqrt{-d}+\sqrt{e} x\right )\right )}{(-d)^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b*n)/(d^2*x) - (4*(a + b*Log[c*x^n]))/(d^2*x) + (Sqrt[e]*(a + b*Log[c*x^n]))/(d^2*(Sqrt[-d] - Sqrt[e]*x))
- (Sqrt[e]*(a + b*Log[c*x^n]))/(d^2*(Sqrt[-d] + Sqrt[e]*x)) + (b*Sqrt[e]*n*(-Log[x] + Log[Sqrt[-d] - Sqrt[e]*
x]))/(-d)^(5/2) + (b*Sqrt[e]*n*(Log[x] - Log[Sqrt[-d] + Sqrt[e]*x]))/(-d)^(5/2) + (3*Sqrt[e]*(a + b*Log[c*x^n]
)*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) - (3*Sqrt[e]*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])
/(-d)^(5/2) - (3*b*Sqrt[e]*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) + (3*b*Sqrt[e]*n*PolyLog[2, (d*Sqrt[
e]*x)/(-d)^(3/2)])/(-d)^(5/2))/4

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Maple [C]  time = 0.35, size = 933, normalized size = 5.1 \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^2/(e*x^2+d)^2,x)

[Out]

1/2*I*b*Pi*csgn(I*c*x^n)^3/d^2/x+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e/d^2*x/(e*x^2+d)-3/4*I*b*Pi*c
sgn(I*x^n)*csgn(I*c*x^n)^2*e/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+1/2*b*n*e/d^2/(d*e)^(1/2)*arctan(x*e/(d*e
)^(1/2))-1/2*b*ln(c)*e/d^2*x/(e*x^2+d)-3/2*b*ln(c)*e/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+1/4*I*b*Pi*csgn(I
*c*x^n)^3*e/d^2*x/(e*x^2+d)+3/4*I*b*Pi*csgn(I*c*x^n)^3*e/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))-1/2*a*e/d^2*x
/(e*x^2+d)-3/2*a*e/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+3/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e/d^
2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))-1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^2/x-1/4*b*n*e/d*ln(x)/(e*x^2+d)/(
-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/4*b*n*e/d*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2)
)/(-d*e)^(1/2))-3/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e/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/d^2/x-1/4*b*n*e^2/d^2*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2)
)*x^2+1/4*b*n*e^2/d^2*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2-a/d^2/x-3/4*b*n*e/d
^2/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+3/4*b*n*e/d^2/(-d*e)^(1/2)*dilog((e*x+(-d*e)^(1/2))/(-
d*e)^(1/2))+3/2*b*e/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*n*ln(x)-1/2*b*n*e/d^2/(-d*e)^(1/2)*ln(x)*ln((-e*x+
(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n*e/d^2/(-d*e)^(1/2)*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*I*b*Pi*cs
gn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^2/x-1/2*b*e/d^2*x/(e*x^2+d)*ln(x^n)-3/2*b*e/d^2/(d*e)^(1/2)*arctan(x*e/(d*
e)^(1/2))*ln(x^n)-b*ln(c)/d^2/x-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e/d^2*x/(e*x^2+d)-1/4*I*b*Pi*csgn(I*c*x
^n)^2*csgn(I*c)*e/d^2*x/(e*x^2+d)-b*ln(x^n)/d^2/x-b*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^2/(e*x^2+d)^2,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^{2} x^{6} + 2 \, d e x^{4} + d^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(c*x^n) + a)/(e^2*x^6 + 2*d*e*x^4 + d^2*x^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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 )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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