3.447 \(\int \frac{1}{a+b \log (c (d (e+f x)^p)^q)} \, dx\)

Optimal. Leaf size=83 \[ \frac{(e+f x) e^{-\frac{a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f p q} \]

[Out]

((e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(b*E^(a/(b*p*q))*f*p*q*(c*(d*(e + f*x)^p)^
q)^(1/(p*q)))

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Rubi [A]  time = 0.119018, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2389, 2300, 2178, 2445} \[ \frac{(e+f x) e^{-\frac{a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f p q} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])^(-1),x]

[Out]

((e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(b*E^(a/(b*p*q))*f*p*q*(c*(d*(e + f*x)^p)^
q)^(1/(p*q)))

Rule 2389

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

Rule 2300

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

Rule 2445

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

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

Mathematica [A]  time = 0.0619316, size = 83, normalized size = 1. \[ \frac{(e+f x) e^{-\frac{a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac{1}{p q}} \text{Ei}\left (\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f p q} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])^(-1),x]

[Out]

((e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(b*E^(a/(b*p*q))*f*p*q*(c*(d*(e + f*x)^p)^
q)^(1/(p*q)))

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Maple [F]  time = 0.271, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) \right ) ^{-1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*ln(c*(d*(f*x+e)^p)^q)),x)

[Out]

int(1/(a+b*ln(c*(d*(f*x+e)^p)^q)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="maxima")

[Out]

integrate(1/(b*log(((f*x + e)^p*d)^q*c) + a), x)

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Fricas [A]  time = 1.94225, size = 157, normalized size = 1.89 \begin{align*} \frac{e^{\left (-\frac{b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )} \logintegral \left ({\left (f x + e\right )} e^{\left (\frac{b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )}\right )}{b f p q} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="fricas")

[Out]

e^(-(b*q*log(d) + b*log(c) + a)/(b*p*q))*log_integral((f*x + e)*e^((b*q*log(d) + b*log(c) + a)/(b*p*q)))/(b*f*
p*q)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \log{\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*ln(c*(d*(f*x+e)**p)**q)),x)

[Out]

Integral(1/(a + b*log(c*(d*(e + f*x)**p)**q)), x)

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Giac [A]  time = 1.30398, size = 107, normalized size = 1.29 \begin{align*} \frac{{\rm Ei}\left (\frac{\log \left (d\right )}{p} + \frac{\log \left (c\right )}{p q} + \frac{a}{b p q} + \log \left (f x + e\right )\right ) e^{\left (-\frac{a}{b p q}\right )}}{b c^{\frac{1}{p q}} d^{\left (\frac{1}{p}\right )} f p q} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="giac")

[Out]

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