\(\int \frac {a+b \log (c \log ^p(d x^n))}{x^3} \, dx\) [53]

Optimal result

Integrand size = 19, antiderivative size = 55 \[ \int \frac {a+b \log \left (c \log ^p\left (d x^n\right )\right )}{x^3} \, dx=\frac {b p \left (d x^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (-\frac {2 \log \left (d x^n\right )}{n}\right )}{2 x^2}-\frac {a+b \log \left (c \log ^p\left (d x^n\right )\right )}{2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2602, 2347, 2209} \[ \int \frac {a+b \log \left (c \log ^p\left (d x^n\right )\right )}{x^3} \, dx=\frac {b p \left (d x^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (-\frac {2 \log \left (d x^n\right )}{n}\right )}{2 x^2}-\frac {a+b \log \left (c \log ^p\left (d x^n\right )\right )}{2 x^2} \]

[In]

Int[(a + b*Log[c*Log[d*x^n]^p])/x^3,x]

[Out]

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

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 2347

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)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2602

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \log \left (c \log ^p\left (d x^n\right )\right )}{x^3} \, dx=-\frac {a-b p \left (d x^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (-\frac {2 \log \left (d x^n\right )}{n}\right )+b \log \left (c \log ^p\left (d x^n\right )\right )}{2 x^2} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {a +b \ln \left (c \ln \left (d \,x^{n}\right )^{p}\right )}{x^{3}}d x\]

[In]

int((a+b*ln(c*ln(d*x^n)^p))/x^3,x)

[Out]

int((a+b*ln(c*ln(d*x^n)^p))/x^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \log \left (c \log ^p\left (d x^n\right )\right )}{x^3} \, dx=\frac {b d^{\frac {2}{n}} p x^{2} \operatorname {log\_integral}\left (\frac {1}{d^{\frac {2}{n}} x^{2}}\right ) - b p \log \left (n \log \left (x\right ) + \log \left (d\right )\right ) - b \log \left (c\right ) - a}{2 \, x^{2}} \]

[In]

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

[Out]

1/2*(b*d^(2/n)*p*x^2*log_integral(1/(d^(2/n)*x^2)) - b*p*log(n*log(x) + log(d)) - b*log(c) - a)/x^2

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*log(c*log(d*x**n)**p))/x**3, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((b*log(c*log(d*x^n)^p) + a)/x^3, x)

Mupad [F(-1)]

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

[In]

int((a + b*log(c*log(d*x^n)^p))/x^3,x)

[Out]

int((a + b*log(c*log(d*x^n)^p))/x^3, x)