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]

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

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Rubi [A]  time = 0.12, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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 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*}

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Mathematica [A]  time = 0.06, size = 83, normalized size = 1.00 \[ \frac {(e+f x) e^{-\frac {a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f p q} \]

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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fricas [A]  time = 0.45, size = 65, normalized size = 0.78 \[ \frac {e^{\left (-\frac {b q \log \relax (d) + b \log \relax (c) + a}{b p q}\right )} \operatorname {log\_integral}\left ({\left (f x + e\right )} e^{\left (\frac {b q \log \relax (d) + b \log \relax (c) + a}{b p q}\right )}\right )}{b f p q} \]

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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giac [A]  time = 0.17, size = 79, normalized size = 0.95 \[ \frac {{\rm Ei}\left (\frac {\log \relax (d)}{p} + \frac {\log \relax (c)}{p q} + \frac {a}{b p q} + \log \left (f x + e\right )\right ) e^{\left (-\frac {a}{b p q}\right )}}{b c^{\frac {1}{p q}} d^{\left (\frac {1}{p}\right )} f p q} \]

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)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}}\, dx \]

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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