3.5.52 \(\int \frac {1}{(a+b \log (c (d (e+f x)^p)^q))^2} \, dx\) [452]

Optimal. Leaf size=123 \[ \frac {e^{-\frac {a}{b p q}} (e+f x) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b^2 f p^2 q^2}-\frac {e+f x}{b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \]

[Out]

(f*x+e)*Ei((a+b*ln(c*(d*(f*x+e)^p)^q))/b/p/q)/b^2/exp(a/b/p/q)/f/p^2/q^2/((c*(d*(f*x+e)^p)^q)^(1/p/q))+(-f*x-e
)/b/f/p/q/(a+b*ln(c*(d*(f*x+e)^p)^q))

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Rubi [A]
time = 0.12, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2436, 2334, 2337, 2209, 2495} \begin {gather*} \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^2 f p^2 q^2}-\frac {e+f x}{b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])^(-2),x]

[Out]

((e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(b^2*E^(a/(b*p*q))*f*p^2*q^2*(c*(d*(e + f*
x)^p)^q)^(1/(p*q))) - (e + f*x)/(b*f*p*q*(a + b*Log[c*(d*(e + f*x)^p)^q]))

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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2337

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 2436

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 2495

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}{\left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx &=\text {Subst}\left (\int \frac {1}{\left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right )^2} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\text {Subst}\left (\frac {\text {Subst}\left (\int \frac {1}{\left (a+b \log \left (c d^q x^{p q}\right )\right )^2} \, 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 )\\ &=-\frac {e+f x}{b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}+\text {Subst}\left (\frac {\text {Subst}\left (\int \frac {1}{a+b \log \left (c d^q x^{p q}\right )} \, dx,x,e+f x\right )}{b f p q},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {e+f x}{b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}+\text {Subst}\left (\frac {\left ((e+f x) \left (c d^q (e+f x)^{p q}\right )^{-\frac {1}{p q}}\right ) \text {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 )}{b f p^2 q^2},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^2 f p^2 q^2}-\frac {e+f x}{b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 163, normalized size = 1.33 \begin {gather*} -\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}} \left (b e^{\frac {a}{b p q}} 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 ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )\right )}{b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])^(-2),x]

[Out]

-(((e + f*x)*(b*E^(a/(b*p*q))*p*q*(c*(d*(e + f*x)^p)^q)^(1/(p*q)) - ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)
^q])/(b*p*q)]*(a + b*Log[c*(d*(e + f*x)^p)^q])))/(b^2*E^(a/(b*p*q))*f*p^2*q^2*(c*(d*(e + f*x)^p)^q)^(1/(p*q))*
(a + b*Log[c*(d*(e + f*x)^p)^q])))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*ln(c*(d*(f*x+e)^p)^q))^2,x)

[Out]

int(1/(a+b*ln(c*(d*(f*x+e)^p)^q))^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 175, normalized size = 1.42 \begin {gather*} -\frac {{\left ({\left (b f p q x + b p q e\right )} e^{\left (\frac {b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )} - {\left (b p q \log \left (f x + e\right ) + b q \log \left (d\right ) + b \log \left (c\right ) + a\right )} \operatorname {log\_integral}\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 )\right )} e^{\left (-\frac {b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )}}{b^{3} f p^{3} q^{3} \log \left (f x + e\right ) + b^{3} f p^{2} q^{3} \log \left (d\right ) + b^{3} f p^{2} q^{2} \log \left (c\right ) + a b^{2} f p^{2} q^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="fricas")

[Out]

-((b*f*p*q*x + b*p*q*e)*e^((b*q*log(d) + b*log(c) + a)/(b*p*q)) - (b*p*q*log(f*x + e) + b*q*log(d) + b*log(c)
+ a)*log_integral((f*x + e)*e^((b*q*log(d) + b*log(c) + a)/(b*p*q))))*e^(-(b*q*log(d) + b*log(c) + a)/(b*p*q))
/(b^3*f*p^3*q^3*log(f*x + e) + b^3*f*p^2*q^3*log(d) + b^3*f*p^2*q^2*log(c) + a*b^2*f*p^2*q^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*ln(c*(d*(f*x+e)**p)**q))**2,x)

[Out]

Integral((a + b*log(c*(d*(e + f*x)**p)**q))**(-2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 593 vs. \(2 (128) = 256\).
time = 4.28, size = 593, normalized size = 4.82 \begin {gather*} -\frac {{\left (f x + e\right )} b p q}{b^{3} f p^{3} q^{3} \log \left (f x + e\right ) + b^{3} f p^{2} q^{3} \log \left (d\right ) + b^{3} f p^{2} q^{2} \log \left (c\right ) + a b^{2} f p^{2} q^{2}} + \frac {b p q {\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 )} \log \left (f x + e\right )}{{\left (b^{3} f p^{3} q^{3} \log \left (f x + e\right ) + b^{3} f p^{2} q^{3} \log \left (d\right ) + b^{3} f p^{2} q^{2} \log \left (c\right ) + a b^{2} f p^{2} q^{2}\right )} c^{\frac {1}{p q}} d^{\left (\frac {1}{p}\right )}} + \frac {b q {\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 )} \log \left (d\right )}{{\left (b^{3} f p^{3} q^{3} \log \left (f x + e\right ) + b^{3} f p^{2} q^{3} \log \left (d\right ) + b^{3} f p^{2} q^{2} \log \left (c\right ) + a b^{2} f p^{2} q^{2}\right )} c^{\frac {1}{p q}} d^{\left (\frac {1}{p}\right )}} + \frac {b {\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 )} \log \left (c\right )}{{\left (b^{3} f p^{3} q^{3} \log \left (f x + e\right ) + b^{3} f p^{2} q^{3} \log \left (d\right ) + b^{3} f p^{2} q^{2} \log \left (c\right ) + a b^{2} f p^{2} q^{2}\right )} c^{\frac {1}{p q}} d^{\left (\frac {1}{p}\right )}} + \frac {a {\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 )}}{{\left (b^{3} f p^{3} q^{3} \log \left (f x + e\right ) + b^{3} f p^{2} q^{3} \log \left (d\right ) + b^{3} f p^{2} q^{2} \log \left (c\right ) + a b^{2} f p^{2} q^{2}\right )} c^{\frac {1}{p q}} d^{\left (\frac {1}{p}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="giac")

[Out]

-(f*x + e)*b*p*q/(b^3*f*p^3*q^3*log(f*x + e) + b^3*f*p^2*q^3*log(d) + b^3*f*p^2*q^2*log(c) + a*b^2*f*p^2*q^2)
+ b*p*q*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(f*x + e)/((b^3*f*p^3*q^3*log
(f*x + e) + b^3*f*p^2*q^3*log(d) + b^3*f*p^2*q^2*log(c) + a*b^2*f*p^2*q^2)*c^(1/(p*q))*d^(1/p)) + b*q*Ei(log(d
)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(d)/((b^3*f*p^3*q^3*log(f*x + e) + b^3*f*p^2*
q^3*log(d) + b^3*f*p^2*q^2*log(c) + a*b^2*f*p^2*q^2)*c^(1/(p*q))*d^(1/p)) + b*Ei(log(d)/p + log(c)/(p*q) + a/(
b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(c)/((b^3*f*p^3*q^3*log(f*x + e) + b^3*f*p^2*q^3*log(d) + b^3*f*p^2*q
^2*log(c) + a*b^2*f*p^2*q^2)*c^(1/(p*q))*d^(1/p)) + a*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e
^(-a/(b*p*q))/((b^3*f*p^3*q^3*log(f*x + e) + b^3*f*p^2*q^3*log(d) + b^3*f*p^2*q^2*log(c) + a*b^2*f*p^2*q^2)*c^
(1/(p*q))*d^(1/p))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*log(c*(d*(e + f*x)^p)^q))^2,x)

[Out]

int(1/(a + b*log(c*(d*(e + f*x)^p)^q))^2, x)

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