\(\int \frac {1}{\log ^4(c (d+e x))} \, dx\) [8]

Optimal result

Integrand size = 10, antiderivative size = 85 \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=-\frac {d+e x}{3 e \log ^3(c (d+e x))}-\frac {d+e x}{6 e \log ^2(c (d+e x))}-\frac {d+e x}{6 e \log (c (d+e x))}+\frac {\operatorname {LogIntegral}(c (d+e x))}{6 c e} \]

[Out]

1/6*Li(c*(e*x+d))/c/e+1/3*(-e*x-d)/e/ln(c*(e*x+d))^3+1/6*(-e*x-d)/e/ln(c*(e*x+d))^2+1/6*(-e*x-d)/e/ln(c*(e*x+d
))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2436, 2334, 2335} \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=\frac {\operatorname {LogIntegral}(c (d+e x))}{6 c e}-\frac {d+e x}{3 e \log ^3(c (d+e x))}-\frac {d+e x}{6 e \log ^2(c (d+e x))}-\frac {d+e x}{6 e \log (c (d+e x))} \]

[In]

Int[Log[c*(d + e*x)]^(-4),x]

[Out]

-1/3*(d + e*x)/(e*Log[c*(d + e*x)]^3) - (d + e*x)/(6*e*Log[c*(d + e*x)]^2) - (d + e*x)/(6*e*Log[c*(d + e*x)])
+ LogIntegral[c*(d + e*x)]/(6*c*e)

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 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\log ^4(c x)} \, dx,x,d+e x\right )}{e} \\ & = -\frac {d+e x}{3 e \log ^3(c (d+e x))}+\frac {\text {Subst}\left (\int \frac {1}{\log ^3(c x)} \, dx,x,d+e x\right )}{3 e} \\ & = -\frac {d+e x}{3 e \log ^3(c (d+e x))}-\frac {d+e x}{6 e \log ^2(c (d+e x))}+\frac {\text {Subst}\left (\int \frac {1}{\log ^2(c x)} \, dx,x,d+e x\right )}{6 e} \\ & = -\frac {d+e x}{3 e \log ^3(c (d+e x))}-\frac {d+e x}{6 e \log ^2(c (d+e x))}-\frac {d+e x}{6 e \log (c (d+e x))}+\frac {\text {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,d+e x\right )}{6 e} \\ & = -\frac {d+e x}{3 e \log ^3(c (d+e x))}-\frac {d+e x}{6 e \log ^2(c (d+e x))}-\frac {d+e x}{6 e \log (c (d+e x))}+\frac {\text {li}(c (d+e x))}{6 c e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=\frac {-\frac {(d+e x) \left (2+\log (c (d+e x))+\log ^2(c (d+e x))\right )}{\log ^3(c (d+e x))}+\frac {\operatorname {LogIntegral}(c (d+e x))}{c}}{6 e} \]

[In]

Integrate[Log[c*(d + e*x)]^(-4),x]

[Out]

(-(((d + e*x)*(2 + Log[c*(d + e*x)] + Log[c*(d + e*x)]^2))/Log[c*(d + e*x)]^3) + LogIntegral[c*(d + e*x)]/c)/(
6*e)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {-\frac {c e x +c d}{3 \ln \left (c e x +c d \right )^{3}}-\frac {c e x +c d}{6 \ln \left (c e x +c d \right )^{2}}-\frac {c e x +c d}{6 \ln \left (c e x +c d \right )}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{6}}{c e}\)	\(87\)
default	\(\frac {-\frac {c e x +c d}{3 \ln \left (c e x +c d \right )^{3}}-\frac {c e x +c d}{6 \ln \left (c e x +c d \right )^{2}}-\frac {c e x +c d}{6 \ln \left (c e x +c d \right )}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{6}}{c e}\)	\(87\)
risch	\(-\frac {x \ln \left (c \left (e x +d \right )\right )^{2} e +\ln \left (c \left (e x +d \right )\right )^{2} d +\ln \left (c \left (e x +d \right )\right ) x e +d \ln \left (c \left (e x +d \right )\right )+2 e x +2 d}{6 e \ln \left (c \left (e x +d \right )\right )^{3}}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c e x +c d \right )\right )}{6 c e}\)	\(92\)

[In]

int(1/ln(c*(e*x+d))^4,x,method=_RETURNVERBOSE)

[Out]

1/c/e*(-1/3*(c*e*x+c*d)/ln(c*e*x+c*d)^3-1/6*(c*e*x+c*d)/ln(c*e*x+c*d)^2-1/6*(c*e*x+c*d)/ln(c*e*x+c*d)-1/6*Ei(1
,-ln(c*e*x+c*d)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=\frac {\log \left (c e x + c d\right )^{3} \operatorname {log\_integral}\left (c e x + c d\right ) - 2 \, c e x - {\left (c e x + c d\right )} \log \left (c e x + c d\right )^{2} - 2 \, c d - {\left (c e x + c d\right )} \log \left (c e x + c d\right )}{6 \, c e \log \left (c e x + c d\right )^{3}} \]

[In]

integrate(1/log(c*(e*x+d))^4,x, algorithm="fricas")

[Out]

1/6*(log(c*e*x + c*d)^3*log_integral(c*e*x + c*d) - 2*c*e*x - (c*e*x + c*d)*log(c*e*x + c*d)^2 - 2*c*d - (c*e*
x + c*d)*log(c*e*x + c*d))/(c*e*log(c*e*x + c*d)^3)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=\frac {- d - e x + \left (- \frac {d}{2} - \frac {e x}{2}\right ) \log {\left (c \left (d + e x\right ) \right )}^{2} + \left (- \frac {d}{2} - \frac {e x}{2}\right ) \log {\left (c \left (d + e x\right ) \right )}}{3 e \log {\left (c \left (d + e x\right ) \right )}^{3}} + \frac {\operatorname {li}{\left (c d + c e x \right )}}{6 c e} \]

[In]

integrate(1/ln(c*(e*x+d))**4,x)

[Out]

(-d - e*x + (-d/2 - e*x/2)*log(c*(d + e*x))**2 + (-d/2 - e*x/2)*log(c*(d + e*x)))/(3*e*log(c*(d + e*x))**3) +
li(c*d + c*e*x)/(6*c*e)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.24 \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=\frac {\Gamma \left (-3, -\log \left (c e x + c d\right )\right )}{c e} \]

[In]

integrate(1/log(c*(e*x+d))^4,x, algorithm="maxima")

[Out]

gamma(-3, -log(c*e*x + c*d))/(c*e)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=\frac {{\rm Ei}\left (\log \left ({\left (e x + d\right )} c\right )\right )}{6 \, c e} - \frac {e x + d}{6 \, e \log \left ({\left (e x + d\right )} c\right )} - \frac {e x + d}{6 \, e \log \left ({\left (e x + d\right )} c\right )^{2}} - \frac {e x + d}{3 \, e \log \left ({\left (e x + d\right )} c\right )^{3}} \]

[In]

integrate(1/log(c*(e*x+d))^4,x, algorithm="giac")

[Out]

1/6*Ei(log((e*x + d)*c))/(c*e) - 1/6*(e*x + d)/(e*log((e*x + d)*c)) - 1/6*(e*x + d)/(e*log((e*x + d)*c)^2) - 1
/3*(e*x + d)/(e*log((e*x + d)*c)^3)

Mupad [B] (verification not implemented)

Time = 1.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\log ^4(c (d+e x))} \, dx=-\frac {\left (d+e\,x\right )\,\left (\frac {1}{6\,\ln \left (c\,\left (d+e\,x\right )\right )}+\frac {1}{6\,{\ln \left (c\,\left (d+e\,x\right )\right )}^2}+\frac {1}{3\,{\ln \left (c\,\left (d+e\,x\right )\right )}^3}\right )}{e}-\frac {\mathrm {expint}\left (-\ln \left (c\,\left (d+e\,x\right )\right )\right )}{6\,c\,e} \]

[In]

int(1/log(c*(d + e*x))^4,x)

[Out]

- ((d + e*x)*(1/(6*log(c*(d + e*x))) + 1/(6*log(c*(d + e*x))^2) + 1/(3*log(c*(d + e*x))^3)))/e - expint(-log(c
*(d + e*x)))/(6*c*e)