\(\int \frac {(a+b \log (c x))^p}{x^4} \, dx\) [182]

Optimal result

Integrand size = 14, antiderivative size = 63 \[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=-3^{-1-p} c^3 e^{\frac {3 a}{b}} \Gamma \left (1+p,\frac {3 (a+b \log (c x))}{b}\right ) (a+b \log (c x))^p \left (\frac {a+b \log (c x)}{b}\right )^{-p} \]

[Out]

-3^(-1-p)*c^3*exp(3*a/b)*GAMMA(p+1,3*(a+b*ln(c*x))/b)*(a+b*ln(c*x))^p/(((a+b*ln(c*x))/b)^p)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2346, 2212} \[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=c^3 \left (-3^{-p-1}\right ) e^{\frac {3 a}{b}} (a+b \log (c x))^p \left (\frac {a+b \log (c x)}{b}\right )^{-p} \Gamma \left (p+1,\frac {3 (a+b \log (c x))}{b}\right ) \]

[In]

Int[(a + b*Log[c*x])^p/x^4,x]

[Out]

-((3^(-1 - p)*c^3*E^((3*a)/b)*Gamma[1 + p, (3*(a + b*Log[c*x]))/b]*(a + b*Log[c*x])^p)/((a + b*Log[c*x])/b)^p)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2346

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=-3^{-1-p} c^3 e^{\frac {3 a}{b}} \Gamma \left (1+p,\frac {3 (a+b \log (c x))}{b}\right ) (a+b \log (c x))^p \left (\frac {a+b \log (c x)}{b}\right )^{-p} \]

[In]

Integrate[(a + b*Log[c*x])^p/x^4,x]

[Out]

-((3^(-1 - p)*c^3*E^((3*a)/b)*Gamma[1 + p, (3*(a + b*Log[c*x]))/b]*(a + b*Log[c*x])^p)/((a + b*Log[c*x])/b)^p)

Maple [F]

\[\int \frac {\left (a +b \ln \left (x c \right )\right )^{p}}{x^{4}}d x\]

[In]

int((a+b*ln(x*c))^p/x^4,x)

[Out]

int((a+b*ln(x*c))^p/x^4,x)

Fricas [F]

\[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x\right ) + a\right )}^{p}}{x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*x))^p/x^4,x, algorithm="fricas")

[Out]

integral((b*log(c*x) + a)^p/x^4, x)

Sympy [F]

\[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=\int \frac {\left (a + b \log {\left (c x \right )}\right )^{p}}{x^{4}}\, dx \]

[In]

integrate((a+b*ln(c*x))**p/x**4,x)

[Out]

Integral((a + b*log(c*x))**p/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=-\frac {{\left (b \log \left (c x\right ) + a\right )}^{p + 1} c^{3} e^{\left (\frac {3 \, a}{b}\right )} E_{-p}\left (\frac {3 \, {\left (b \log \left (c x\right ) + a\right )}}{b}\right )}{b} \]

[In]

integrate((a+b*log(c*x))^p/x^4,x, algorithm="maxima")

[Out]

-(b*log(c*x) + a)^(p + 1)*c^3*e^(3*a/b)*exp_integral_e(-p, 3*(b*log(c*x) + a)/b)/b

Giac [F]

\[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x\right ) + a\right )}^{p}}{x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*x))^p/x^4,x, algorithm="giac")

[Out]

integrate((b*log(c*x) + a)^p/x^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \log (c x))^p}{x^4} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,x\right )\right )}^p}{x^4} \,d x \]

[In]

int((a + b*log(c*x))^p/x^4,x)

[Out]

int((a + b*log(c*x))^p/x^4, x)