\(\int x \log (d+e (f^{c (a+b x)})^n) \, dx\) [125]

Optimal result

Integrand size = 18, antiderivative size = 118 \[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{2} x^2 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{2} x^2 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x \operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {\operatorname {PolyLog}\left (3,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)} \]

[Out]

1/2*x^2*ln(d+e*(f^(c*(b*x+a)))^n)-1/2*x^2*ln(1+e*(f^(c*(b*x+a)))^n/d)-x*polylog(2,-e*(f^(c*(b*x+a)))^n/d)/b/c/
n/ln(f)+polylog(3,-e*(f^(c*(b*x+a)))^n/d)/b^2/c^2/n^2/ln(f)^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2612, 2611, 2320, 6724} \[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {\operatorname {PolyLog}\left (3,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {1}{2} x^2 \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-\frac {1}{2} x^2 \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right ) \]

[In]

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

[Out]

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

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2612

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

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{2} x^2 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )+\int x \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right ) \, dx \\ & = \frac {1}{2} x^2 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{2} x^2 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {\int \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right ) \, dx}{b c n \log (f)} \\ & = \frac {1}{2} x^2 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{2} x^2 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {\text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {e x^n}{d}\right )}{x} \, dx,x,f^{c (a+b x)}\right )}{b^2 c^2 n \log ^2(f)} \\ & = \frac {1}{2} x^2 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{2} x^2 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {\text {Li}_3\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00 \[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{2} x^2 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{2} x^2 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x \operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {\operatorname {PolyLog}\left (3,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(557\) vs. \(2(114)=228\).

Time = 0.58 (sec) , antiderivative size = 558, normalized size of antiderivative = 4.73

method	result	size
risch	\(\frac {x^{2} \ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right )}{2}-\frac {\ln \left (f^{c \left (b x +a \right )}\right )^{2} \ln \left (1+\frac {e \,f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{2 \ln \left (f \right )^{2} b^{2} c^{2}}-\frac {\ln \left (f^{c \left (b x +a \right )}\right ) \operatorname {Li}_{2}\left (-\frac {e \,f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{n \ln \left (f \right )^{2} b^{2} c^{2}}+\frac {\operatorname {Li}_{3}\left (-\frac {e \,f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{n^{2} \ln \left (f \right )^{2} b^{2} c^{2}}-\frac {\ln \left (d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) x^{2}}{2}+\frac {\ln \left (d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right ) x}{\ln \left (f \right ) b c}-\frac {\ln \left (d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right )^{2}}{2 \ln \left (f \right )^{2} b^{2} c^{2}}-\frac {\operatorname {dilog}\left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right ) x}{n \ln \left (f \right ) b c}+\frac {\operatorname {dilog}\left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right ) \ln \left (f^{c \left (b x +a \right )}\right )}{n \ln \left (f \right )^{2} b^{2} c^{2}}-\frac {\ln \left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right ) x \ln \left (f^{c \left (b x +a \right )}\right )}{c b \ln \left (f \right )}+\frac {\ln \left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right ) \ln \left (f^{c \left (b x +a \right )}\right )^{2}}{c^{2} b^{2} \ln \left (f \right )^{2}}\)	\(558\)

[In]

int(x*ln(d+e*(f^(c*(b*x+a)))^n),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2*ln(d+e*(f^(c*(b*x+a)))^n)-1/2/ln(f)^2/b^2/c^2*ln(f^(c*(b*x+a)))^2*ln(1+e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(
c*(b*x+a)))^n/d)-1/n/ln(f)^2/b^2/c^2*ln(f^(c*(b*x+a)))*polylog(2,-e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n
/d)+1/n^2/ln(f)^2/b^2/c^2*polylog(3,-e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)-1/2*ln(d+f^(x*b*c*n)*f^(-
x*b*c*n)*(f^(c*(b*x+a)))^n*e)*x^2+1/ln(f)/b/c*ln(d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln(f^(c*(b*x+
a)))*x-1/2/ln(f)^2/b^2/c^2*ln(d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln(f^(c*(b*x+a)))^2-1/n/ln(f)/b/
c*dilog((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*x+1/n/ln(f)^2/b^2/c^2*dilog((d+f^(x*b*c*n)*f^(-x*b
*c*n)*(f^(c*(b*x+a)))^n*e)/d)*ln(f^(c*(b*x+a)))-1/c/b/ln(f)*ln((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e
)/d)*x*ln(f^(c*(b*x+a)))+1/c^2/b^2/ln(f)^2*ln((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*ln(f^(c*(b*x
+a)))^2

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.43 \[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {2 \, b c n x {\rm Li}_2\left (-\frac {e f^{b c n x + a c n} + d}{d} + 1\right ) \log \left (f\right ) - {\left (b^{2} c^{2} n^{2} x^{2} - a^{2} c^{2} n^{2}\right )} \log \left (e f^{b c n x + a c n} + d\right ) \log \left (f\right )^{2} + {\left (b^{2} c^{2} n^{2} x^{2} - a^{2} c^{2} n^{2}\right )} \log \left (f\right )^{2} \log \left (\frac {e f^{b c n x + a c n} + d}{d}\right ) - 2 \, {\rm polylog}\left (3, -\frac {e f^{b c n x + a c n}}{d}\right )}{2 \, b^{2} c^{2} n^{2} \log \left (f\right )^{2}} \]

[In]

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

[Out]

-1/2*(2*b*c*n*x*dilog(-(e*f^(b*c*n*x + a*c*n) + d)/d + 1)*log(f) - (b^2*c^2*n^2*x^2 - a^2*c^2*n^2)*log(e*f^(b*
c*n*x + a*c*n) + d)*log(f)^2 + (b^2*c^2*n^2*x^2 - a^2*c^2*n^2)*log(f)^2*log((e*f^(b*c*n*x + a*c*n) + d)/d) - 2
*polylog(3, -e*f^(b*c*n*x + a*c*n)/d))/(b^2*c^2*n^2*log(f)^2)

Sympy [F]

\[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x \log {\left (d + e \left (f^{a c + b c x}\right )^{n} \right )}\, dx \]

[In]

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

[Out]

Integral(x*log(d + e*(f**(a*c + b*c*x))**n), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.07 \[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (e f^{{\left (b x + a\right )} c n} + d\right ) - \frac {b^{2} c^{2} n^{2} x^{2} \log \left (\frac {e f^{b c n x} f^{a c n}}{d} + 1\right ) \log \left (f\right )^{2} + 2 \, b c n x {\rm Li}_2\left (-\frac {e f^{b c n x} f^{a c n}}{d}\right ) \log \left (f\right ) - 2 \, {\rm Li}_{3}(-\frac {e f^{b c n x} f^{a c n}}{d})}{2 \, b^{2} c^{2} n^{2} \log \left (f\right )^{2}} \]

[In]

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

[Out]

1/2*x^2*log(e*f^((b*x + a)*c*n) + d) - 1/2*(b^2*c^2*n^2*x^2*log(e*f^(b*c*n*x)*f^(a*c*n)/d + 1)*log(f)^2 + 2*b*
c*n*x*dilog(-e*f^(b*c*n*x)*f^(a*c*n)/d)*log(f) - 2*polylog(3, -e*f^(b*c*n*x)*f^(a*c*n)/d))/(b^2*c^2*n^2*log(f)
^2)

Giac [F]

\[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int { x \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + d\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x\,\ln \left (d+e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n\right ) \,d x \]

[In]

int(x*log(d + e*(f^(c*(a + b*x)))^n),x)

[Out]

int(x*log(d + e*(f^(c*(a + b*x)))^n), x)