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

Optimal. Leaf size=67 \[ \frac{(e x)^{m+1} \left (a+b \log \left (c \log ^p(d x)\right )\right )}{e (m+1)}-\frac{b p (d x)^{-m-1} (e x)^{m+1} \text{Ei}((m+1) \log (d x))}{e (m+1)} \]

[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))

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Rubi [A]  time = 0.0572072, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2522, 2310, 2178} \[ \frac{(e x)^{m+1} \left (a+b \log \left (c \log ^p(d x)\right )\right )}{e (m+1)}-\frac{b p (d x)^{-m-1} (e x)^{m+1} \text{Ei}((m+1) \log (d x))}{e (m+1)} \]

Antiderivative was successfully verified.

[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 2522

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]

Rule 2310

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)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2178

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

Rubi steps

\begin{align*} \int (e x)^m \left (a+b \log \left (c \log ^p(d x)\right )\right ) \, dx &=\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 ) \operatorname{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]  time = 0.13092, size = 56, normalized size = 0.84 \[ \frac{(d x)^{-m} (e x)^m \left (d x (d x)^m \left (a+b \log \left (c \log ^p(d x)\right )\right )-b p \text{Ei}((m+1) \log (d x))\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( a+b\ln \left ( c \left ( \ln \left ( dx \right ) \right ) ^{p} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05197, size = 204, normalized size = 3.04 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \left (a + b \log{\left (c \log{\left (d x \right )}^{p} \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.31031, size = 112, normalized size = 1.67 \begin{align*} \frac{b p x x^{m} e^{m} \log \left (\log \left (d\right ) + \log \left (x\right )\right )}{m + 1} + \frac{b x x^{m} e^{m} \log \left (c\right )}{m + 1} + \frac{a x x^{m} e^{m}}{m + 1} - \frac{b p{\rm Ei}\left (m \log \left (d\right ) + m \log \left (x\right ) + \log \left (d\right ) + \log \left (x\right )\right ) e^{m}}{d d^{m} m + d d^{m}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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