\(\int \frac {1}{(a+b \log (c (d+e x)^n))^2} \, dx\) [97]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 16, antiderivative size = 96 \[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 e n^2}-\frac {d+e x}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=-\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \left (b e^{\frac {a}{b n}} n \left (c (d+e x)^n\right )^{\frac {1}{n}}-\operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2836, 2734, 2737, 2609}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx\)

\(\Big \downarrow \) 2836

\(\displaystyle \frac {\int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2}d(d+e x)}{e}\)

\(\Big \downarrow \) 2734

\(\displaystyle \frac {\frac {\int \frac {1}{a+b \log \left (c (d+e x)^n\right )}d(d+e x)}{b n}-\frac {d+e x}{b n \left (a+b \log \left (c (d+e x)^n\right )\right )}}{e}\)

\(\Big \downarrow \) 2737

\(\displaystyle \frac {\frac {(d+e x) \left (c (d+e x)^n\right )^{-1/n} \int \frac {\left (c (d+e x)^n\right )^{\frac {1}{n}}}{a+b \log \left (c (d+e x)^n\right )}d\log \left (c (d+e x)^n\right )}{b n^2}-\frac {d+e x}{b n \left (a+b \log \left (c (d+e x)^n\right )\right )}}{e}\)

\(\Big \downarrow \) 2609

\(\displaystyle \frac {\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 n^2}-\frac {d+e x}{b n \left (a+b \log \left (c (d+e x)^n\right )\right )}}{e}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2609
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si 
mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F 
reeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2734
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] - Simp[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] && Int 
egerQ[2*p]
 

rule 2737
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[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 2836
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : 
> Simp[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]
 
Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.42 (sec) , antiderivative size = 456, normalized size of antiderivative = 4.75

method result size
risch \(-\frac {2 \left (e x +d \right )}{\left (i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (\left (e x +d \right )^{n}\right ) b +2 b \ln \left (c \right )+2 a \right ) b n e}-\frac {\left (e x +d \right ) \left (\left (e x +d \right )^{n}\right )^{-\frac {1}{n}} c^{-\frac {1}{n}} {\mathrm e}^{-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 a}{2 n b}} \operatorname {expIntegral}_{1}\left (-\ln \left (e x +d \right )-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 b \ln \left (c \right )+2 b \left (\ln \left (\left (e x +d \right )^{n}\right )-n \ln \left (e x +d \right )\right )+2 a}{2 n b}\right )}{b^{2} n^{2} e}\) \(456\)

Input:

int(1/(a+b*ln(c*(e*x+d)^n))^2,x,method=_RETURNVERBOSE)
 

Output:

-2/(I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*(e*x+d)^n 
)*csgn(I*c*(e*x+d)^n)*csgn(I*c)-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+I*b*Pi*csgn(I 
*c*(e*x+d)^n)^2*csgn(I*c)+2*ln((e*x+d)^n)*b+2*b*ln(c)+2*a)/b/n/e*(e*x+d)-1 
/b^2/n^2/e*(e*x+d)*c^(-1/n)*((e*x+d)^n)^(-1/n)*exp(-1/2*(I*b*Pi*csgn(I*(e* 
x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n) 
*csgn(I*c)-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csgn( 
I*c)+2*a)/n/b)*Ei(1,-ln(e*x+d)-1/2*(I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x 
+d)^n)^2-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csgn(I*c)-I*b*Pi*csg 
n(I*c*(e*x+d)^n)^3+I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csgn(I*c)+2*b*ln(c)+2*b*(l 
n((e*x+d)^n)-n*ln(e*x+d))+2*a)/n/b)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=-\frac {{\left ({\left (b e n x + b d n\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} - {\left (b n \log \left (e x + d\right ) + b \log \left (c\right ) + a\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right )\right )} e^{\left (-\frac {b \log \left (c\right ) + a}{b n}\right )}}{b^{3} e n^{3} \log \left (e x + d\right ) + b^{3} e n^{2} \log \left (c\right ) + a b^{2} e n^{2}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\int \frac {1}{\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\int { \frac {1}{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

-(e*x + d)/(b^2*e*n*log((e*x + d)^n) + b^2*e*n*log(c) + a*b*e*n) + integra 
te(1/(b^2*n*log((e*x + d)^n) + b^2*n*log(c) + a*b*n), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (95) = 190\).

Time = 0.12 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.98 \[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\frac {b n {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )} \log \left (e x + d\right )}{{\left (b^{3} e n^{3} \log \left (e x + d\right ) + b^{3} e n^{2} \log \left (c\right ) + a b^{2} e n^{2}\right )} c^{\left (\frac {1}{n}\right )}} - \frac {{\left (e x + d\right )} b n}{b^{3} e n^{3} \log \left (e x + d\right ) + b^{3} e n^{2} \log \left (c\right ) + a b^{2} e n^{2}} + \frac {b {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )} \log \left (c\right )}{{\left (b^{3} e n^{3} \log \left (e x + d\right ) + b^{3} e n^{2} \log \left (c\right ) + a b^{2} e n^{2}\right )} c^{\left (\frac {1}{n}\right )}} + \frac {a {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{{\left (b^{3} e n^{3} \log \left (e x + d\right ) + b^{3} e n^{2} \log \left (c\right ) + a b^{2} e n^{2}\right )} c^{\left (\frac {1}{n}\right )}} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\frac {\left (\int \frac {x}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b^{2} d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b^{2} e x +2 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) a b d +2 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) a b e x +a^{2} d +a^{2} e x}d x \right ) \mathrm {log}\left (\left (e x +d \right )^{n} c \right ) a b \,e^{2} n +\left (\int \frac {x}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b^{2} d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b^{2} e x +2 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) a b d +2 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) a b e x +a^{2} d +a^{2} e x}d x \right ) a^{2} e^{2} n +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) d}{a e n \left (\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b +a \right )} \] Input:

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

Output:

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