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

Optimal. Leaf size=76 \[ -\frac {2 e^{\frac {2 a}{b n}} \left (c x^n\right )^{2/n} \text {Ei}\left (-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b^2 n^2 x^2}-\frac {1}{b n x^2 \left (a+b \log \left (c x^n\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2306, 2310, 2178} \[ -\frac {2 e^{\frac {2 a}{b n}} \left (c x^n\right )^{2/n} \text {Ei}\left (-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b^2 n^2 x^2}-\frac {1}{b n x^2 \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*Log[c*x^n])^2),x]

[Out]

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

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Mathematica [A]  time = 0.09, size = 80, normalized size = 1.05 \[ -\frac {2 e^{\frac {2 a}{b n}} \left (c x^n\right )^{2/n} \left (a+b \log \left (c x^n\right )\right ) \text {Ei}\left (-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )+b n}{b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 102, normalized size = 1.34 \[ -\frac {2 \, {\left (b n x^{2} \log \relax (x) + b x^{2} \log \relax (c) + a x^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \relax (c) + a\right )}}{b n}\right )} \operatorname {log\_integral}\left (\frac {e^{\left (-\frac {2 \, {\left (b \log \relax (c) + a\right )}}{b n}\right )}}{x^{2}}\right ) + b n}{b^{3} n^{3} x^{2} \log \relax (x) + b^{3} n^{2} x^{2} \log \relax (c) + a b^{2} n^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.98, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \ln \left (c \,x^{n}\right )+a \right )^{2} x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*log(c*x^n))^2),x)

[Out]

int(1/(x^3*(a + b*log(c*x^n))^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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