3.409 \(\int \frac {1}{(a+b \log (c (d (e+f x)^m)^n))^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.17, 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 = {2389, 2297, 2300, 2178, 2445} \[ \frac {(e+f x) e^{-\frac {a}{b m n}} \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right )}{b^2 f m^2 n^2}-\frac {e+f x}{b f m n \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2297

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

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Mathematica [A]  time = 0.12, size = 163, normalized size = 1.33 \[ -\frac {(e+f x) e^{-\frac {a}{b m n}} \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \left (b m n e^{\frac {a}{b m n}} \left (c \left (d (e+f x)^m\right )^n\right )^{\frac {1}{m n}}-\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right ) \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right )\right )}{b^2 f m^2 n^2 \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 171, normalized size = 1.39 \[ -\frac {{\left ({\left (b f m n x + b e m n\right )} e^{\left (\frac {b n \log \relax (d) + b \log \relax (c) + a}{b m n}\right )} - {\left (b m n \log \left (f x + e\right ) + b n \log \relax (d) + b \log \relax (c) + a\right )} \operatorname {log\_integral}\left ({\left (f x + e\right )} e^{\left (\frac {b n \log \relax (d) + b \log \relax (c) + a}{b m n}\right )}\right )\right )} e^{\left (-\frac {b n \log \relax (d) + b \log \relax (c) + a}{b m n}\right )}}{b^{3} f m^{3} n^{3} \log \left (f x + e\right ) + b^{3} f m^{2} n^{3} \log \relax (d) + b^{3} f m^{2} n^{2} \log \relax (c) + a b^{2} f m^{2} n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.25, size = 593, normalized size = 4.82 \[ -\frac {{\left (f x + e\right )} b m n}{b^{3} f m^{3} n^{3} \log \left (f x + e\right ) + b^{3} f m^{2} n^{3} \log \relax (d) + b^{3} f m^{2} n^{2} \log \relax (c) + a b^{2} f m^{2} n^{2}} + \frac {b m n {\rm Ei}\left (\frac {\log \relax (d)}{m} + \frac {\log \relax (c)}{m n} + \frac {a}{b m n} + \log \left (f x + e\right )\right ) e^{\left (-\frac {a}{b m n}\right )} \log \left (f x + e\right )}{{\left (b^{3} f m^{3} n^{3} \log \left (f x + e\right ) + b^{3} f m^{2} n^{3} \log \relax (d) + b^{3} f m^{2} n^{2} \log \relax (c) + a b^{2} f m^{2} n^{2}\right )} c^{\frac {1}{m n}} d^{\left (\frac {1}{m}\right )}} + \frac {b n {\rm Ei}\left (\frac {\log \relax (d)}{m} + \frac {\log \relax (c)}{m n} + \frac {a}{b m n} + \log \left (f x + e\right )\right ) e^{\left (-\frac {a}{b m n}\right )} \log \relax (d)}{{\left (b^{3} f m^{3} n^{3} \log \left (f x + e\right ) + b^{3} f m^{2} n^{3} \log \relax (d) + b^{3} f m^{2} n^{2} \log \relax (c) + a b^{2} f m^{2} n^{2}\right )} c^{\frac {1}{m n}} d^{\left (\frac {1}{m}\right )}} + \frac {b {\rm Ei}\left (\frac {\log \relax (d)}{m} + \frac {\log \relax (c)}{m n} + \frac {a}{b m n} + \log \left (f x + e\right )\right ) e^{\left (-\frac {a}{b m n}\right )} \log \relax (c)}{{\left (b^{3} f m^{3} n^{3} \log \left (f x + e\right ) + b^{3} f m^{2} n^{3} \log \relax (d) + b^{3} f m^{2} n^{2} \log \relax (c) + a b^{2} f m^{2} n^{2}\right )} c^{\frac {1}{m n}} d^{\left (\frac {1}{m}\right )}} + \frac {a {\rm Ei}\left (\frac {\log \relax (d)}{m} + \frac {\log \relax (c)}{m n} + \frac {a}{b m n} + \log \left (f x + e\right )\right ) e^{\left (-\frac {a}{b m n}\right )}}{{\left (b^{3} f m^{3} n^{3} \log \left (f x + e\right ) + b^{3} f m^{2} n^{3} \log \relax (d) + b^{3} f m^{2} n^{2} \log \relax (c) + a b^{2} f m^{2} n^{2}\right )} c^{\frac {1}{m n}} d^{\left (\frac {1}{m}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {f x + e}{b^{2} f m n \log \left ({\left ({\left (f x + e\right )}^{m}\right )}^{n}\right ) + a b f m n + {\left (f m n^{2} \log \relax (d) + f m n \log \relax (c)\right )} b^{2}} + \int \frac {1}{b^{2} m n \log \left ({\left ({\left (f x + e\right )}^{m}\right )}^{n}\right ) + a b m n + {\left (m n^{2} \log \relax (d) + m n \log \relax (c)\right )} b^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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