3.234 \(\int \frac{a+b \log (c x^n)}{x (d+e x^2)^3} \, dx\)

Optimal. Leaf size=115 \[ \frac{b n \text{PolyLog}\left (2,-\frac{d}{e x^2}\right )}{4 d^3}-\frac{\log \left (\frac{d}{e x^2}+1\right ) \left (4 a+4 b \log \left (c x^n\right )-3 b n\right )}{8 d^3}+\frac{4 a+4 b \log \left (c x^n\right )-b n}{8 d^2 \left (d+e x^2\right )}+\frac{a+b \log \left (c x^n\right )}{4 d \left (d+e x^2\right )^2} \]

[Out]

(a + b*Log[c*x^n])/(4*d*(d + e*x^2)^2) - (Log[1 + d/(e*x^2)]*(4*a - 3*b*n + 4*b*Log[c*x^n]))/(8*d^3) + (4*a -
b*n + 4*b*Log[c*x^n])/(8*d^2*(d + e*x^2)) + (b*n*PolyLog[2, -(d/(e*x^2))])/(4*d^3)

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Rubi [A]  time = 0.217559, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2340, 2345, 2391} \[ \frac{b n \text{PolyLog}\left (2,-\frac{d}{e x^2}\right )}{4 d^3}-\frac{\log \left (\frac{d}{e x^2}+1\right ) \left (4 a+4 b \log \left (c x^n\right )-3 b n\right )}{8 d^3}+\frac{4 a+4 b \log \left (c x^n\right )-b n}{8 d^2 \left (d+e x^2\right )}+\frac{a+b \log \left (c x^n\right )}{4 d \left (d+e x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x*(d + e*x^2)^3),x]

[Out]

(a + b*Log[c*x^n])/(4*d*(d + e*x^2)^2) - (Log[1 + d/(e*x^2)]*(4*a - 3*b*n + 4*b*Log[c*x^n]))/(8*d^3) + (4*a -
b*n + 4*b*Log[c*x^n])/(8*d^2*(d + e*x^2)) + (b*n*PolyLog[2, -(d/(e*x^2))])/(4*d^3)

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 )^3} \, dx &=\frac{a+b \log \left (c x^n\right )}{4 d \left (d+e x^2\right )^2}-\frac{\int \frac{-4 a+b n-4 b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx}{4 d}\\ &=\frac{a+b \log \left (c x^n\right )}{4 d \left (d+e x^2\right )^2}+\frac{4 a-b n+4 b \log \left (c x^n\right )}{8 d^2 \left (d+e x^2\right )}+\frac{\int \frac{-4 b n-2 (-4 a+b n)+8 b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx}{8 d^2}\\ &=\frac{a+b \log \left (c x^n\right )}{4 d \left (d+e x^2\right )^2}-\frac{\log \left (1+\frac{d}{e x^2}\right ) \left (4 a-3 b n+4 b \log \left (c x^n\right )\right )}{8 d^3}+\frac{4 a-b n+4 b \log \left (c x^n\right )}{8 d^2 \left (d+e x^2\right )}+\frac{(b n) \int \frac{\log \left (1+\frac{d}{e x^2}\right )}{x} \, dx}{2 d^3}\\ &=\frac{a+b \log \left (c x^n\right )}{4 d \left (d+e x^2\right )^2}-\frac{\log \left (1+\frac{d}{e x^2}\right ) \left (4 a-3 b n+4 b \log \left (c x^n\right )\right )}{8 d^3}+\frac{4 a-b n+4 b \log \left (c x^n\right )}{8 d^2 \left (d+e x^2\right )}+\frac{b n \text{Li}_2\left (-\frac{d}{e x^2}\right )}{4 d^3}\\ \end{align*}

Mathematica [C]  time = 0.916985, size = 396, normalized size = 3.44 \[ \frac{-b n \left (8 \text{PolyLog}\left (2,-\frac{i \sqrt{e} x}{\sqrt{d}}\right )+8 \text{PolyLog}\left (2,\frac{i \sqrt{e} x}{\sqrt{d}}\right )+\frac{d}{d-i \sqrt{d} \sqrt{e} x}+\frac{d}{d+i \sqrt{d} \sqrt{e} x}+8 \log (x) \log \left (1-\frac{i \sqrt{e} x}{\sqrt{d}}\right )+8 \log (x) \log \left (1+\frac{i \sqrt{e} x}{\sqrt{d}}\right )+\frac{5 \sqrt{e} x \log (x)}{\sqrt{e} x-i \sqrt{d}}+\frac{5 \sqrt{e} x \log (x)}{\sqrt{e} x+i \sqrt{d}}-\frac{d \log (x)}{\left (\sqrt{d}-i \sqrt{e} x\right )^2}-\frac{d \log (x)}{\left (\sqrt{d}+i \sqrt{e} x\right )^2}-6 \log \left (-\sqrt{e} x+i \sqrt{d}\right )-6 \log \left (\sqrt{e} x+i \sqrt{d}\right )-8 \log ^2(x)+2 \log (x)\right )+\frac{4 d^2 \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{\left (d+e x^2\right )^2}+\frac{8 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d+e x^2}-8 \log \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )+16 \log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{16 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x*(d + e*x^2)^3),x]

[Out]

((4*d^2*(a - b*n*Log[x] + b*Log[c*x^n]))/(d + e*x^2)^2 + (8*d*(a - b*n*Log[x] + b*Log[c*x^n]))/(d + e*x^2) + 1
6*Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]) - 8*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + e*x^2] - b*n*(d/(d - I*Sq
rt[d]*Sqrt[e]*x) + d/(d + I*Sqrt[d]*Sqrt[e]*x) + 2*Log[x] - (d*Log[x])/(Sqrt[d] - I*Sqrt[e]*x)^2 - (d*Log[x])/
(Sqrt[d] + I*Sqrt[e]*x)^2 + (5*Sqrt[e]*x*Log[x])/((-I)*Sqrt[d] + Sqrt[e]*x) + (5*Sqrt[e]*x*Log[x])/(I*Sqrt[d]
+ Sqrt[e]*x) - 8*Log[x]^2 - 6*Log[I*Sqrt[d] - Sqrt[e]*x] - 6*Log[I*Sqrt[d] + Sqrt[e]*x] + 8*Log[x]*Log[1 - (I*
Sqrt[e]*x)/Sqrt[d]] + 8*Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + 8*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]] + 8*Pol
yLog[2, (I*Sqrt[e]*x)/Sqrt[d]]))/(16*d^3)

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Maple [C]  time = 0.166, size = 841, normalized size = 7.3 \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/(e*x^2+d)^3,x)

[Out]

b*ln(c)/d^3*ln(x)+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^3*ln(e*x^2+d)-1/2*a/d^3*ln(e*x^2+d)+1/4*a/d
/(e*x^2+d)^2+1/2*a/d^2/(e*x^2+d)-1/2*b*n/d^3*ln(x)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3/d^3*ln(x)-1/2*b*n/d^3*ln(x)*ln
((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/4*I*b*Pi*csgn(I*c*x^n)^3/d^3*ln(e*x^2+d)-1/8*I*b*Pi*csgn(I*x^n)*csgn(I*c*x
^n)*csgn(I*c)/d/(e*x^2+d)^2-1/2*b*n/d^3*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n/d^3*ln(x)*ln(e*x^2+
d)-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^2/(e*x^2+d)+b*ln(x^n)/d^3*ln(x)-1/2*b*ln(x^n)/d^3*ln(e*x^2
+d)+1/4*b*ln(x^n)/d/(e*x^2+d)^2+1/2*b*ln(x^n)/d^2/(e*x^2+d)+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^3*ln(x)-1/2
*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^3*ln(x)-1/2*b*ln(c)/d^3*ln(e*x^2+d)+1/4*b*ln(c)/d/(e*x^2+d)^2+1/
2*b*ln(c)/d^2/(e*x^2+d)-1/2*b*n/d^3*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n/d^3*dilog((e*x+(-d*e)^(1/2
))/(-d*e)^(1/2))+3/8*b*n/d^3*ln(e*x^2+d)-1/8*b*n/d^2/(e*x^2+d)+a/d^3*ln(x)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n
)^2/d^3*ln(x)-1/4*I*b*Pi*csgn(I*c*x^n)^3/d^2/(e*x^2+d)-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^3*ln(e*x^2+d)-
1/8*I*b*Pi*csgn(I*c*x^n)^3/d/(e*x^2+d)^2+1/8*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d/(e*x^2+d)^2+1/4*I*b*Pi*csgn(I*
c*x^n)^2*csgn(I*c)/d^2/(e*x^2+d)+1/8*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d/(e*x^2+d)^2+1/4*I*b*Pi*csgn(I*x^n)*c
sgn(I*c*x^n)^2/d^2/(e*x^2+d)-1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^3*ln(e*x^2+d)-3/4*b*n*ln(x)/d^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a{\left (\frac{2 \, e x^{2} + 3 \, d}{d^{2} e^{2} x^{4} + 2 \, d^{3} e x^{2} + d^{4}} - \frac{2 \, \log \left (e x^{2} + d\right )}{d^{3}} + \frac{4 \, \log \left (x\right )}{d^{3}}\right )} + b \int \frac{\log \left (c\right ) + \log \left (x^{n}\right )}{e^{3} x^{7} + 3 \, d e^{2} x^{5} + 3 \, d^{2} e x^{3} + d^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

1/4*a*((2*e*x^2 + 3*d)/(d^2*e^2*x^4 + 2*d^3*e*x^2 + d^4) - 2*log(e*x^2 + d)/d^3 + 4*log(x)/d^3) + b*integrate(
(log(c) + log(x^n))/(e^3*x^7 + 3*d*e^2*x^5 + 3*d^2*e*x^3 + d^3*x), x)

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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^{3} x^{7} + 3 \, d e^{2} x^{5} + 3 \, d^{2} e x^{3} + d^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)/(e^3*x^7 + 3*d*e^2*x^5 + 3*d^2*e*x^3 + d^3*x), 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/(e*x**2+d)**3,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 )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^3,x, algorithm="giac")

[Out]

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