3.361 \(\int \frac{(f+g x^{2 n}) \log (c (d+e x^n)^p)}{x} \, dx\)

Optimal. Leaf size=124 \[ \frac{f p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{f \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac{d^2 g p \log \left (d+e x^n\right )}{2 e^2 n}+\frac{d g p x^n}{2 e n}-\frac{g p x^{2 n}}{4 n} \]

[Out]

(d*g*p*x^n)/(2*e*n) - (g*p*x^(2*n))/(4*n) - (d^2*g*p*Log[d + e*x^n])/(2*e^2*n) + (g*x^(2*n)*Log[c*(d + e*x^n)^
p])/(2*n) + (f*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/n + (f*p*PolyLog[2, 1 + (e*x^n)/d])/n

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Rubi [A]  time = 0.142098, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2475, 14, 2416, 2394, 2315, 2395, 43} \[ \frac{f p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{f \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac{d^2 g p \log \left (d+e x^n\right )}{2 e^2 n}+\frac{d g p x^n}{2 e n}-\frac{g p x^{2 n}}{4 n} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x^(2*n))*Log[c*(d + e*x^n)^p])/x,x]

[Out]

(d*g*p*x^n)/(2*e*n) - (g*p*x^(2*n))/(4*n) - (d^2*g*p*Log[d + e*x^n])/(2*e^2*n) + (g*x^(2*n)*Log[c*(d + e*x^n)^
p])/(2*n) + (f*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/n + (f*p*PolyLog[2, 1 + (e*x^n)/d])/n

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2395

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

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])

Rubi steps

\begin{align*} \int \frac{\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (f+g x^2\right ) \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{f \log \left (c (d+e x)^p\right )}{x}+g x \log \left (c (d+e x)^p\right )\right ) \, dx,x,x^n\right )}{n}\\ &=\frac{f \operatorname{Subst}\left (\int \frac{\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}+\frac{g \operatorname{Subst}\left (\int x \log \left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}\\ &=\frac{g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac{f \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac{(e f p) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n}-\frac{(e g p) \operatorname{Subst}\left (\int \frac{x^2}{d+e x} \, dx,x,x^n\right )}{2 n}\\ &=\frac{g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac{f \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{f p \text{Li}_2\left (1+\frac{e x^n}{d}\right )}{n}-\frac{(e g p) \operatorname{Subst}\left (\int \left (-\frac{d}{e^2}+\frac{x}{e}+\frac{d^2}{e^2 (d+e x)}\right ) \, dx,x,x^n\right )}{2 n}\\ &=\frac{d g p x^n}{2 e n}-\frac{g p x^{2 n}}{4 n}-\frac{d^2 g p \log \left (d+e x^n\right )}{2 e^2 n}+\frac{g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac{f \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{f p \text{Li}_2\left (1+\frac{e x^n}{d}\right )}{n}\\ \end{align*}

Mathematica [A]  time = 0.111868, size = 100, normalized size = 0.81 \[ \frac{4 e^2 f p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )+2 e^2 \log \left (c \left (d+e x^n\right )^p\right ) \left (2 f \log \left (-\frac{e x^n}{d}\right )+g x^{2 n}\right )-2 d^2 g p \log \left (d+e x^n\right )-e g p x^n \left (e x^n-2 d\right )}{4 e^2 n} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x^(2*n))*Log[c*(d + e*x^n)^p])/x,x]

[Out]

(-(e*g*p*x^n*(-2*d + e*x^n)) - 2*d^2*g*p*Log[d + e*x^n] + 2*e^2*(g*x^(2*n) + 2*f*Log[-((e*x^n)/d)])*Log[c*(d +
e*x^n)^p] + 4*e^2*f*p*PolyLog[2, 1 + (e*x^n)/d])/(4*e^2*n)

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Maple [C]  time = 4.243, size = 410, normalized size = 3.3 \begin{align*}{\frac{ \left ( 2\,f\ln \left ( x \right ) n+g \left ({x}^{n} \right ) ^{2} \right ) \ln \left ( \left ( d+e{x}^{n} \right ) ^{p} \right ) }{2\,n}}-{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{3}f\ln \left ( x \right ) -{\frac{{\frac{i}{4}}\pi \, \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{3}g \left ({x}^{n} \right ) ^{2}}{n}}+{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( d+e{x}^{n} \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{2}f\ln \left ( x \right ) -{\frac{{\frac{i}{4}}\pi \,{\it csgn} \left ( i \left ( d+e{x}^{n} \right ) ^{p} \right ){\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ){\it csgn} \left ( ic \right ) g \left ({x}^{n} \right ) ^{2}}{n}}+{\frac{{\frac{i}{4}}\pi \, \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) g \left ({x}^{n} \right ) ^{2}}{n}}+{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) f\ln \left ( x \right ) +{\frac{{\frac{i}{4}}\pi \,{\it csgn} \left ( i \left ( d+e{x}^{n} \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{2}g \left ({x}^{n} \right ) ^{2}}{n}}-{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( d+e{x}^{n} \right ) ^{p} \right ){\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ){\it csgn} \left ( ic \right ) f\ln \left ( x \right ) +\ln \left ( c \right ) f\ln \left ( x \right ) +{\frac{g\ln \left ( c \right ) \left ({x}^{n} \right ) ^{2}}{2\,n}}-{\frac{gp \left ({x}^{n} \right ) ^{2}}{4\,n}}+{\frac{dgp{x}^{n}}{2\,en}}-{\frac{{d}^{2}gp\ln \left ( d+e{x}^{n} \right ) }{2\,{e}^{2}n}}-{\frac{pf}{n}{\it dilog} \left ({\frac{d+e{x}^{n}}{d}} \right ) }-pf\ln \left ( x \right ) \ln \left ({\frac{d+e{x}^{n}}{d}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f+g*x^(2*n))*ln(c*(d+e*x^n)^p)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g*x^(2*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="maxima")

[Out]

-1/4*(2*e^2*f*n^2*p*log(x)^2 - 2*d*e*g*p*x^n + (e^2*g*p - 2*e^2*g*log(c))*x^(2*n) - 2*(2*e^2*f*n*log(x) + e^2*
g*x^(2*n))*log((e*x^n + d)^p) + 2*(d^2*g*n*p - 2*e^2*f*n*log(c))*log(x))/(e^2*n) + integrate(1/2*(2*d*e^2*f*n*
p*log(x) + d^3*g*p)/(e^3*x*x^n + d*e^2*x), x)

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Fricas [A]  time = 2.08481, size = 323, normalized size = 2.6 \begin{align*} -\frac{4 \, e^{2} f n p \log \left (x\right ) \log \left (\frac{e x^{n} + d}{d}\right ) - 4 \, e^{2} f n \log \left (c\right ) \log \left (x\right ) - 2 \, d e g p x^{n} + 4 \, e^{2} f p{\rm Li}_2\left (-\frac{e x^{n} + d}{d} + 1\right ) +{\left (e^{2} g p - 2 \, e^{2} g \log \left (c\right )\right )} x^{2 \, n} - 2 \,{\left (2 \, e^{2} f n p \log \left (x\right ) + e^{2} g p x^{2 \, n} - d^{2} g p\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g*x^(2*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="fricas")

[Out]

-1/4*(4*e^2*f*n*p*log(x)*log((e*x^n + d)/d) - 4*e^2*f*n*log(c)*log(x) - 2*d*e*g*p*x^n + 4*e^2*f*p*dilog(-(e*x^
n + d)/d + 1) + (e^2*g*p - 2*e^2*g*log(c))*x^(2*n) - 2*(2*e^2*f*n*p*log(x) + e^2*g*p*x^(2*n) - d^2*g*p)*log(e*
x^n + d))/(e^2*n)

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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((f+g*x**(2*n))*ln(c*(d+e*x**n)**p)/x,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x^{2 \, n} + f\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g*x^(2*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="giac")

[Out]

integrate((g*x^(2*n) + f)*log((e*x^n + d)^p*c)/x, x)