3.86 \(\int \frac {1}{x^2 (a+b \log (c x^n))^3} \, dx\)

Optimal. Leaf size=102 \[ \frac {e^{\frac {a}{b n}} \left (c x^n\right )^{\frac {1}{n}} \text {Ei}\left (-\frac {a+b \log \left (c x^n\right )}{b n}\right )}{2 b^3 n^3 x}+\frac {1}{2 b^2 n^2 x \left (a+b \log \left (c x^n\right )\right )}-\frac {1}{2 b n x \left (a+b \log \left (c x^n\right )\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2306, 2310, 2178} \[ \frac {e^{\frac {a}{b n}} \left (c x^n\right )^{\frac {1}{n}} \text {Ei}\left (-\frac {a+b \log \left (c x^n\right )}{b n}\right )}{2 b^3 n^3 x}+\frac {1}{2 b^2 n^2 x \left (a+b \log \left (c x^n\right )\right )}-\frac {1}{2 b n x \left (a+b \log \left (c x^n\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*Log[c*x^n])^3),x]

[Out]

(E^(a/(b*n))*(c*x^n)^n^(-1)*ExpIntegralEi[-((a + b*Log[c*x^n])/(b*n))])/(2*b^3*n^3*x) - 1/(2*b*n*x*(a + b*Log[
c*x^n])^2) + 1/(2*b^2*n^2*x*(a + b*Log[c*x^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 2306

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

Rule 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 94, normalized size = 0.92 \[ \frac {e^{\frac {a}{b n}} \left (c x^n\right )^{\frac {1}{n}} \left (a+b \log \left (c x^n\right )\right )^2 \text {Ei}\left (-\frac {a+b \log \left (c x^n\right )}{b n}\right )+b n \left (a+b \log \left (c x^n\right )-b n\right )}{2 b^3 n^3 x \left (a+b \log \left (c x^n\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*Log[c*x^n])^3),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.44, size = 192, normalized size = 1.88 \[ \frac {b^{2} n^{2} \log \relax (x) - b^{2} n^{2} + b^{2} n \log \relax (c) + a b n + {\left (b^{2} n^{2} x \log \relax (x)^{2} + b^{2} x \log \relax (c)^{2} + 2 \, a b x \log \relax (c) + a^{2} x + 2 \, {\left (b^{2} n x \log \relax (c) + a b n x\right )} \log \relax (x)\right )} e^{\left (\frac {b \log \relax (c) + a}{b n}\right )} \operatorname {log\_integral}\left (\frac {e^{\left (-\frac {b \log \relax (c) + a}{b n}\right )}}{x}\right )}{2 \, {\left (b^{5} n^{5} x \log \relax (x)^{2} + b^{5} n^{3} x \log \relax (c)^{2} + 2 \, a b^{4} n^{3} x \log \relax (c) + a^{2} b^{3} n^{3} x + 2 \, {\left (b^{5} n^{4} x \log \relax (c) + a b^{4} n^{4} x\right )} \log \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.95, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \ln \left (c \,x^{n}\right )+a \right )^{3} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*ln(c*x^n)+a)^3,x)

[Out]

int(1/x^2/(b*ln(c*x^n)+a)^3,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b {\left (n - \log \relax (c)\right )} - b \log \left (x^{n}\right ) - a}{2 \, {\left (b^{4} n^{2} x \log \left (x^{n}\right )^{2} + 2 \, {\left (b^{4} n^{2} \log \relax (c) + a b^{3} n^{2}\right )} x \log \left (x^{n}\right ) + {\left (b^{4} n^{2} \log \relax (c)^{2} + 2 \, a b^{3} n^{2} \log \relax (c) + a^{2} b^{2} n^{2}\right )} x\right )}} + \int \frac {1}{2 \, {\left (b^{3} n^{2} x^{2} \log \left (x^{n}\right ) + {\left (b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}\right )} x^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*log(c*x^n))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*log(c*x^n))^3),x)

[Out]

int(1/(x^2*(a + b*log(c*x^n))^3), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \left (a + b \log {\left (c x^{n} \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(a+b*ln(c*x**n))**3,x)

[Out]

Integral(1/(x**2*(a + b*log(c*x**n))**3), x)

________________________________________________________________________________________