\(\int \frac {(a+b \log (c \sqrt {x}))^p}{x^2} \, dx\) [188]

Optimal result

Integrand size = 18, antiderivative size = 73 \[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=-2^{-p} c^2 e^{\frac {2 a}{b}} \Gamma \left (1+p,\frac {2 \left (a+b \log \left (c \sqrt {x}\right )\right )}{b}\right ) \left (a+b \log \left (c \sqrt {x}\right )\right )^p \left (\frac {a+b \log \left (c \sqrt {x}\right )}{b}\right )^{-p} \]

[Out]

-c^2*exp(2*a/b)*GAMMA(p+1,2*(a+b*ln(c*x^(1/2)))/b)*(a+b*ln(c*x^(1/2)))^p/(2^p)/(((a+b*ln(c*x^(1/2)))/b)^p)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2347, 2212} \[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=c^2 \left (-2^{-p}\right ) e^{\frac {2 a}{b}} \left (a+b \log \left (c \sqrt {x}\right )\right )^p \left (\frac {a+b \log \left (c \sqrt {x}\right )}{b}\right )^{-p} \Gamma \left (p+1,\frac {2 \left (a+b \log \left (c \sqrt {x}\right )\right )}{b}\right ) \]

[In]

Int[(a + b*Log[c*Sqrt[x]])^p/x^2,x]

[Out]

-((c^2*E^((2*a)/b)*Gamma[1 + p, (2*(a + b*Log[c*Sqrt[x]]))/b]*(a + b*Log[c*Sqrt[x]])^p)/(2^p*((a + b*Log[c*Sqr
t[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 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]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=-2^{-p} c^2 e^{\frac {2 a}{b}} \Gamma \left (1+p,\frac {2 \left (a+b \log \left (c \sqrt {x}\right )\right )}{b}\right ) \left (a+b \log \left (c \sqrt {x}\right )\right )^p \left (\frac {a+b \log \left (c \sqrt {x}\right )}{b}\right )^{-p} \]

[In]

Integrate[(a + b*Log[c*Sqrt[x]])^p/x^2,x]

[Out]

-((c^2*E^((2*a)/b)*Gamma[1 + p, (2*(a + b*Log[c*Sqrt[x]]))/b]*(a + b*Log[c*Sqrt[x]])^p)/(2^p*((a + b*Log[c*Sqr
t[x]])/b)^p))

Maple [F]

\[\int \frac {\left (a +b \ln \left (c \sqrt {x}\right )\right )^{p}}{x^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c \sqrt {x}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]

[In]

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

[Out]

integral((b*log(c*sqrt(x)) + a)^p/x^2, x)

Sympy [F]

\[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=\int \frac {\left (a + b \log {\left (c \sqrt {x} \right )}\right )^{p}}{x^{2}}\, dx \]

[In]

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

[Out]

Integral((a + b*log(c*sqrt(x)))**p/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=-\frac {2 \, {\left (b \log \left (c \sqrt {x}\right ) + a\right )}^{p + 1} c^{2} e^{\left (\frac {2 \, a}{b}\right )} E_{-p}\left (\frac {2 \, {\left (b \log \left (c \sqrt {x}\right ) + a\right )}}{b}\right )}{b} \]

[In]

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

[Out]

-2*(b*log(c*sqrt(x)) + a)^(p + 1)*c^2*e^(2*a/b)*exp_integral_e(-p, 2*(b*log(c*sqrt(x)) + a)/b)/b

Giac [F]

\[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c \sqrt {x}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate((b*log(c*sqrt(x)) + a)^p/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c \sqrt {x}\right )\right )^p}{x^2} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,\sqrt {x}\right )\right )}^p}{x^2} \,d x \]

[In]

int((a + b*log(c*x^(1/2)))^p/x^2,x)

[Out]

int((a + b*log(c*x^(1/2)))^p/x^2, x)