3.221 \(\int \frac{x^5 (a+b \log (c x^n))}{(d+e x^2)^2} \, dx\)

Optimal. Leaf size=129 \[ -\frac{b d n \text{PolyLog}\left (2,-\frac{e x^2}{d}\right )}{2 e^3}+\frac{d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2 \left (d+e x^2\right )}-\frac{d \log \left (\frac{e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac{b d n \log \left (d+e x^2\right )}{4 e^3}-\frac{b n x^2}{4 e^2} \]

[Out]

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

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Rubi [A]  time = 0.220363, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {266, 43, 2351, 2304, 2335, 260, 2337, 2391} \[ -\frac{b d n \text{PolyLog}\left (2,-\frac{e x^2}{d}\right )}{2 e^3}+\frac{d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2 \left (d+e x^2\right )}-\frac{d \log \left (\frac{e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac{b d n \log \left (d+e x^2\right )}{4 e^3}-\frac{b n x^2}{4 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*Log[c*x^n]))/(d + e*x^2)^2,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

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

Mathematica [C]  time = 0.467416, size = 287, normalized size = 2.22 \[ \frac{b n \left (-4 d \left (\text{PolyLog}\left (2,-\frac{i \sqrt{e} x}{\sqrt{d}}\right )+\log (x) \log \left (1+\frac{i \sqrt{e} x}{\sqrt{d}}\right )\right )-4 d \left (\text{PolyLog}\left (2,\frac{i \sqrt{e} x}{\sqrt{d}}\right )+\log (x) \log \left (1-\frac{i \sqrt{e} x}{\sqrt{d}}\right )\right )+\frac{d \sqrt{e} x \log (x)}{\sqrt{e} x-i \sqrt{d}}+\frac{d \sqrt{e} x \log (x)}{\sqrt{e} x+i \sqrt{d}}-d \log \left (-\sqrt{e} x+i \sqrt{d}\right )-d \log \left (\sqrt{e} x+i \sqrt{d}\right )+e x^2 (2 \log (x)-1)\right )-\frac{2 d^2 \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d+e x^2}-4 d \log \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )+2 e x^2 \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{4 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*Log[c*x^n]))/(d + e*x^2)^2,x]

[Out]

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

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Maple [C]  time = 0.19, size = 687, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*ln(c*x^n))/(e*x^2+d)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{5} \log \left (c x^{n}\right ) + a x^{5}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*x^5*log(c*x^n) + a*x^5)/(e^2*x^4 + 2*d*e*x^2 + d^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(x**5*(a+b*ln(c*x**n))/(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{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^5/(e*x^2 + d)^2, x)