\(\int \frac {1}{(a+b \log (c x^n))^3} \, dx\) [84]

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

Optimal result

Integrand size = 12, antiderivative size = 98 \[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=\frac {e^{-\frac {a}{b n}} x \left (c x^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c x^n\right )}{b n}\right )}{2 b^3 n^3}-\frac {x}{2 b n \left (a+b \log \left (c x^n\right )\right )^2}-\frac {x}{2 b^2 n^2 \left (a+b \log \left (c x^n\right )\right )} \]

[Out]

1/2*x*Ei((a+b*ln(c*x^n))/b/n)/b^3/exp(a/b/n)/n^3/((c*x^n)^(1/n))-1/2*x/b/n/(a+b*ln(c*x^n))^2-1/2*x/b^2/n^2/(a+
b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2334, 2337, 2209} \[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=\frac {x e^{-\frac {a}{b n}} \left (c x^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c x^n\right )}{b n}\right )}{2 b^3 n^3}-\frac {x}{2 b^2 n^2 \left (a+b \log \left (c x^n\right )\right )}-\frac {x}{2 b n \left (a+b \log \left (c x^n\right )\right )^2} \]

[In]

Int[(a + b*Log[c*x^n])^(-3),x]

[Out]

(x*ExpIntegralEi[(a + b*Log[c*x^n])/(b*n)])/(2*b^3*E^(a/(b*n))*n^3*(c*x^n)^n^(-1)) - x/(2*b*n*(a + b*Log[c*x^n
])^2) - x/(2*b^2*n^2*(a + b*Log[c*x^n]))

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {x}{2 b n \left (a+b \log \left (c x^n\right )\right )^2}+\frac {\int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx}{2 b n} \\ & = -\frac {x}{2 b n \left (a+b \log \left (c x^n\right )\right )^2}-\frac {x}{2 b^2 n^2 \left (a+b \log \left (c x^n\right )\right )}+\frac {\int \frac {1}{a+b \log \left (c x^n\right )} \, dx}{2 b^2 n^2} \\ & = -\frac {x}{2 b n \left (a+b \log \left (c x^n\right )\right )^2}-\frac {x}{2 b^2 n^2 \left (a+b \log \left (c x^n\right )\right )}+\frac {\left (x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{n}}}{a+b x} \, dx,x,\log \left (c x^n\right )\right )}{2 b^2 n^3} \\ & = \frac {e^{-\frac {a}{b n}} x \left (c x^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c x^n\right )}{b n}\right )}{2 b^3 n^3}-\frac {x}{2 b n \left (a+b \log \left (c x^n\right )\right )^2}-\frac {x}{2 b^2 n^2 \left (a+b \log \left (c x^n\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=\frac {x \left (e^{-\frac {a}{b n}} \left (c x^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c x^n\right )}{b n}\right )-\frac {b n \left (a+b n+b \log \left (c x^n\right )\right )}{\left (a+b \log \left (c x^n\right )\right )^2}\right )}{2 b^3 n^3} \]

[In]

Integrate[(a + b*Log[c*x^n])^(-3),x]

[Out]

(x*(ExpIntegralEi[(a + b*Log[c*x^n])/(b*n)]/(E^(a/(b*n))*(c*x^n)^n^(-1)) - (b*n*(a + b*n + b*Log[c*x^n]))/(a +
 b*Log[c*x^n])^2))/(2*b^3*n^3)

Maple [C] (warning: unable to verify)

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

Time = 0.39 (sec) , antiderivative size = 459, normalized size of antiderivative = 4.68

method result size
risch \(-\frac {-i \pi b x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i \pi b x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b x \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 \ln \left (c \right ) b x +2 b x \ln \left (x^{n}\right )+2 a x +2 b n x}{b^{2} n^{2} {\left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 \ln \left (x^{n}\right ) b +2 a \right )}^{2}}-\frac {x \,c^{-\frac {1}{n}} \left (x^{n}\right )^{-\frac {1}{n}} {\mathrm e}^{-\frac {-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 a}{2 b n}} \operatorname {Ei}_{1}\left (-\ln \left (x \right )-\frac {-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+2 a}{2 b n}\right )}{2 b^{3} n^{3}}\) \(459\)

[In]

int(1/(a+b*ln(c*x^n))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (91) = 182\).

Time = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.02 \[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=-\frac {{\left ({\left (b^{2} n^{2} x \log \left (x\right ) + b^{2} n x \log \left (c\right ) + {\left (b^{2} n^{2} + a b n\right )} x\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} - {\left (b^{2} n^{2} \log \left (x\right )^{2} + b^{2} \log \left (c\right )^{2} + 2 \, a b \log \left (c\right ) + a^{2} + 2 \, {\left (b^{2} n \log \left (c\right ) + a b n\right )} \log \left (x\right )\right )} \operatorname {log\_integral}\left (x e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right )\right )} e^{\left (-\frac {b \log \left (c\right ) + a}{b n}\right )}}{2 \, {\left (b^{5} n^{5} \log \left (x\right )^{2} + b^{5} n^{3} \log \left (c\right )^{2} + 2 \, a b^{4} n^{3} \log \left (c\right ) + a^{2} b^{3} n^{3} + 2 \, {\left (b^{5} n^{4} \log \left (c\right ) + a b^{4} n^{4}\right )} \log \left (x\right )\right )}} \]

[In]

integrate(1/(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

-1/2*((b^2*n^2*x*log(x) + b^2*n*x*log(c) + (b^2*n^2 + a*b*n)*x)*e^((b*log(c) + a)/(b*n)) - (b^2*n^2*log(x)^2 +
 b^2*log(c)^2 + 2*a*b*log(c) + a^2 + 2*(b^2*n*log(c) + a*b*n)*log(x))*log_integral(x*e^((b*log(c) + a)/(b*n)))
)*e^(-(b*log(c) + a)/(b*n))/(b^5*n^5*log(x)^2 + b^5*n^3*log(c)^2 + 2*a*b^4*n^3*log(c) + a^2*b^3*n^3 + 2*(b^5*n
^4*log(c) + a*b^4*n^4)*log(x))

Sympy [F]

\[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=\int \frac {1}{\left (a + b \log {\left (c x^{n} \right )}\right )^{3}}\, dx \]

[In]

integrate(1/(a+b*ln(c*x**n))**3,x)

[Out]

Integral((a + b*log(c*x**n))**(-3), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate(1/(a+b*log(c*x^n))^3,x, algorithm="maxima")

[Out]

-1/2*(b*x*log(x^n) + (b*(n + log(c)) + a)*x)/(b^4*n^2*log(c)^2 + b^4*n^2*log(x^n)^2 + 2*a*b^3*n^2*log(c) + a^2
*b^2*n^2 + 2*(b^4*n^2*log(c) + a*b^3*n^2)*log(x^n)) + integrate(1/2/(b^3*n^2*log(c) + b^3*n^2*log(x^n) + a*b^2
*n^2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 982 vs. \(2 (91) = 182\).

Time = 0.42 (sec) , antiderivative size = 982, normalized size of antiderivative = 10.02 \[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

1/2*b^2*n^2*Ei(log(c)/n + a/(b*n) + log(x))*e^(-a/(b*n))*log(x)^2/((b^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log(x)
 + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a^2*b^3*n^3)*c^(1/n)) - 1/2*b^2*n^2*x*log(x)/(
b^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a^2*
b^3*n^3) + b^2*n*Ei(log(c)/n + a/(b*n) + log(x))*e^(-a/(b*n))*log(c)*log(x)/((b^5*n^5*log(x)^2 + 2*b^5*n^4*log
(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a^2*b^3*n^3)*c^(1/n)) - 1/2*b^2*n^2*
x/(b^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a
^2*b^3*n^3) - 1/2*b^2*n*x*log(c)/(b^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*
log(x) + 2*a*b^4*n^3*log(c) + a^2*b^3*n^3) + 1/2*b^2*Ei(log(c)/n + a/(b*n) + log(x))*e^(-a/(b*n))*log(c)^2/((b
^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a^2*b
^3*n^3)*c^(1/n)) + a*b*n*Ei(log(c)/n + a/(b*n) + log(x))*e^(-a/(b*n))*log(x)/((b^5*n^5*log(x)^2 + 2*b^5*n^4*lo
g(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a^2*b^3*n^3)*c^(1/n)) - 1/2*a*b*n*x
/(b^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a^
2*b^3*n^3) + a*b*Ei(log(c)/n + a/(b*n) + log(x))*e^(-a/(b*n))*log(c)/((b^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log
(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^4*log(x) + 2*a*b^4*n^3*log(c) + a^2*b^3*n^3)*c^(1/n)) + 1/2*a^2*Ei(log(c)/n
 + a/(b*n) + log(x))*e^(-a/(b*n))/((b^5*n^5*log(x)^2 + 2*b^5*n^4*log(c)*log(x) + b^5*n^3*log(c)^2 + 2*a*b^4*n^
4*log(x) + 2*a*b^4*n^3*log(c) + a^2*b^3*n^3)*c^(1/n))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^3} \, dx=\int \frac {1}{{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3} \,d x \]

[In]

int(1/(a + b*log(c*x^n))^3,x)

[Out]

int(1/(a + b*log(c*x^n))^3, x)