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

Optimal. Leaf size=107 \[ \frac{b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^3}-\frac{x \left (2 a+2 b \log \left (c x^n\right )+b n\right )}{2 e^2 (d+e x)}+\frac{\log \left (\frac{e x}{d}+1\right ) \left (2 a+2 b \log \left (c x^n\right )+3 b n\right )}{2 e^3}-\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.182732, antiderivative size = 132, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {43, 2351, 2319, 44, 2314, 31, 2317, 2391} \[ \frac{b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^3}-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^3 (d+e x)^2}-\frac{2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}+\frac{\log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac{b d n}{2 e^3 (d+e x)}+\frac{3 b n \log (d+e x)}{2 e^3}+\frac{b n \log (x)}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2314

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

Rule 31

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

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, 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{x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx &=\int \left (\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)^3}-\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)^2}+\frac{a+b \log \left (c x^n\right )}{e^2 (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^2}-\frac{(2 d) \int \frac{a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^2}+\frac{d^2 \int \frac{a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{e^2}\\ &=-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^3 (d+e x)^2}-\frac{2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}+\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{e^3}-\frac{(b n) \int \frac{\log \left (1+\frac{e x}{d}\right )}{x} \, dx}{e^3}+\frac{\left (b d^2 n\right ) \int \frac{1}{x (d+e x)^2} \, dx}{2 e^3}+\frac{(2 b n) \int \frac{1}{d+e x} \, dx}{e^2}\\ &=-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^3 (d+e x)^2}-\frac{2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}+\frac{2 b n \log (d+e x)}{e^3}+\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{e^3}+\frac{b n \text{Li}_2\left (-\frac{e x}{d}\right )}{e^3}+\frac{\left (b d^2 n\right ) \int \left (\frac{1}{d^2 x}-\frac{e}{d (d+e x)^2}-\frac{e}{d^2 (d+e x)}\right ) \, dx}{2 e^3}\\ &=\frac{b d n}{2 e^3 (d+e x)}+\frac{b n \log (x)}{2 e^3}-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^3 (d+e x)^2}-\frac{2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}+\frac{3 b n \log (d+e x)}{2 e^3}+\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{e^3}+\frac{b n \text{Li}_2\left (-\frac{e x}{d}\right )}{e^3}\\ \end{align*}

Mathematica [A]  time = 0.109, size = 122, normalized size = 1.14 \[ \frac{2 b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac{4 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}+2 \log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )-4 b n (\log (x)-\log (d+e x))+b n \left (\frac{d}{d+e x}-\log (d+e x)+\log (x)\right )}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.148, size = 596, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{2} \log \left (c x^{n}\right ) + a x^{2}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 49.5046, size = 328, normalized size = 3.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/e**2 - 2*a*d*Piecewise((x/d**2, Eq(e, 0)),
(-1/(d*e + e**2*x), True))/e**2 + a*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**2 - b*d**2*n*Piecew
ise((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - log(x)/(2*d**2*e) + log(d/e + x)/(2*d**2*e), True))/e**2
+ b*d**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/e**2 + 2*b*d*n*Piecewise((x
/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), True))/e**2 - 2*b*d*Piecewise((x/d**2, Eq(e, 0)), (-1/(
d*e + e**2*x), True))*log(c*x**n)/e**2 - b*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((log(d)*log(x) - polylog(2,
e*x*exp_polar(I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/d), 1/Abs(x) < 1), (-
meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x
*exp_polar(I*pi)/d), True))/e, True))/e**2 + b*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/
e**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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