\(\int (e x)^m (a+b \log (c \log ^p(d x))) \, dx\) [46]

Optimal result

Integrand size = 19, antiderivative size = 67 \[ \int (e x)^m \left (a+b \log \left (c \log ^p(d x)\right )\right ) \, dx=-\frac {b p (d x)^{-1-m} (e x)^{1+m} \operatorname {ExpIntegralEi}((1+m) \log (d x))}{e (1+m)}+\frac {(e x)^{1+m} \left (a+b \log \left (c \log ^p(d x)\right )\right )}{e (1+m)} \]

[Out]

-b*p*(d*x)^(-1-m)*(e*x)^(1+m)*Ei((1+m)*ln(d*x))/e/(1+m)+(e*x)^(1+m)*(a+b*ln(c*ln(d*x)^p))/e/(1+m)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2602, 2347, 2209} \[ \int (e x)^m \left (a+b \log \left (c \log ^p(d x)\right )\right ) \, dx=\frac {(e x)^{m+1} \left (a+b \log \left (c \log ^p(d x)\right )\right )}{e (m+1)}-\frac {b p (d x)^{-m-1} (e x)^{m+1} \operatorname {ExpIntegralEi}((m+1) \log (d x))}{e (m+1)} \]

[In]

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

[Out]

-((b*p*(d*x)^(-1 - m)*(e*x)^(1 + m)*ExpIntegralEi[(1 + m)*Log[d*x]])/(e*(1 + m))) + ((e*x)^(1 + m)*(a + b*Log[
c*Log[d*x]^p]))/(e*(1 + m))

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2602

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

Rubi steps \begin{align*} \text {integral}& = \frac {(e x)^{1+m} \left (a+b \log \left (c \log ^p(d x)\right )\right )}{e (1+m)}-\frac {(b p) \int \frac {(e x)^m}{\log (d x)} \, dx}{1+m} \\ & = \frac {(e x)^{1+m} \left (a+b \log \left (c \log ^p(d x)\right )\right )}{e (1+m)}-\frac {\left (b p (d x)^{-1-m} (e x)^{1+m}\right ) \text {Subst}\left (\int \frac {e^{(1+m) x}}{x} \, dx,x,\log (d x)\right )}{e (1+m)} \\ & = -\frac {b p (d x)^{-1-m} (e x)^{1+m} \text {Ei}((1+m) \log (d x))}{e (1+m)}+\frac {(e x)^{1+m} \left (a+b \log \left (c \log ^p(d x)\right )\right )}{e (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int (e x)^m \left (a+b \log \left (c \log ^p(d x)\right )\right ) \, dx=\frac {(d x)^{-m} (e x)^m \left (-b p \operatorname {ExpIntegralEi}((1+m) \log (d x))+d x (d x)^m \left (a+b \log \left (c \log ^p(d x)\right )\right )\right )}{d (1+m)} \]

[In]

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

[Out]

((e*x)^m*(-(b*p*ExpIntegralEi[(1 + m)*Log[d*x]]) + d*x*(d*x)^m*(a + b*Log[c*Log[d*x]^p])))/(d*(1 + m)*(d*x)^m)

Maple [F]

\[\int \left (e x \right )^{m} \left (a +b \ln \left (c \ln \left (d x \right )^{p}\right )\right )d x\]

[In]

int((e*x)^m*(a+b*ln(c*ln(d*x)^p)),x)

[Out]

int((e*x)^m*(a+b*ln(c*ln(d*x)^p)),x)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24 \[ \int (e x)^m \left (a+b \log \left (c \log ^p(d x)\right )\right ) \, dx=\frac {b d p x e^{\left (m \log \left (d x\right ) + m \log \left (\frac {e}{d}\right )\right )} \log \left (\log \left (d x\right )\right ) - b p \left (\frac {e}{d}\right )^{m} {\rm Ei}\left ({\left (m + 1\right )} \log \left (d x\right )\right ) + {\left (b d x \log \left (c\right ) + a d x\right )} e^{\left (m \log \left (d x\right ) + m \log \left (\frac {e}{d}\right )\right )}}{d m + d} \]

[In]

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

[Out]

(b*d*p*x*e^(m*log(d*x) + m*log(e/d))*log(log(d*x)) - b*p*(e/d)^m*Ei((m + 1)*log(d*x)) + (b*d*x*log(c) + a*d*x)
*e^(m*log(d*x) + m*log(e/d)))/(d*m + d)

Sympy [F]

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

[In]

integrate((e*x)**m*(a+b*ln(c*ln(d*x)**p)),x)

[Out]

Integral((e*x)**m*(a + b*log(c*log(d*x)**p)), x)

Maxima [F]

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

[In]

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

[Out]

-(e^m*p*integrate(x^m/((m^2 + 2*m + 1)*log(d)^2 + 2*(m^2 + 2*m + 1)*log(d)*log(x) + (m^2 + 2*m + 1)*log(x)^2),
x) - ((e^m*(m + 1)*x*log(d) + e^m*(m + 1)*x*log(x))*x^m*log((log(d) + log(x))^p) + (e^m*(m + 1)*x*log(c)*log(
x) + (e^m*(m + 1)*log(c)*log(d) - e^m*p)*x)*x^m)/((m^2 + 2*m + 1)*log(d) + (m^2 + 2*m + 1)*log(x)))*b + (e*x)^
(m + 1)*a/(e*(m + 1))

Giac [F]

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

[In]

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

[Out]

integrate((b*log(c*log(d*x)^p) + a)*(e*x)^m, x)

Mupad [F(-1)]

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

[In]

int((a + b*log(c*log(d*x)^p))*(e*x)^m,x)

[Out]

int((a + b*log(c*log(d*x)^p))*(e*x)^m, x)