Integrand size = 22, antiderivative size = 175 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx=-\frac {2^{-p} e^{-\frac {2 a}{b}} \Gamma \left (1+p,-\frac {2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )}{b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right )^{-p}}{c^2 e^2}+\frac {2 d e^{-\frac {a}{b}} \Gamma \left (1+p,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right )^{-p}}{c e^2} \]
-GAMMA(p+1,-2*(a+b*ln(c*(d+e/x^(1/2))))/b)*(a+b*ln(c*(d+e/x^(1/2))))^p/(2^ p)/c^2/e^2/exp(2*a/b)/(((-a-b*ln(c*(d+e/x^(1/2))))/b)^p)+2*d*GAMMA(p+1,(-a -b*ln(c*(d+e/x^(1/2))))/b)*(a+b*ln(c*(d+e/x^(1/2))))^p/c/e^2/exp(a/b)/(((- a-b*ln(c*(d+e/x^(1/2))))/b)^p)
Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx=\frac {2^{-p} e^{-\frac {2 a}{b}} \left (-\Gamma \left (1+p,-\frac {2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )}{b}\right )+2^{1+p} c d e^{a/b} \Gamma \left (1+p,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right )\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right )^{-p}}{c^2 e^2} \]
((-Gamma[1 + p, (-2*(a + b*Log[c*(d + e/Sqrt[x])]))/b] + 2^(1 + p)*c*d*E^( a/b)*Gamma[1 + p, -((a + b*Log[c*(d + e/Sqrt[x])])/b)])*(a + b*Log[c*(d + e/Sqrt[x])])^p)/(2^p*c^2*e^2*E^((2*a)/b)*(-((a + b*Log[c*(d + e/Sqrt[x])]) /b))^p)
Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2904, 2848, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -2 \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{\sqrt {x}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 2848 |
\(\displaystyle -2 \int \left (\frac {\left (d+\frac {e}{\sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{e}-\frac {d \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{e}\right )d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\frac {2^{-p-1} e^{-\frac {2 a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )}{b}\right )}{c^2 e^2}-\frac {d e^{-\frac {a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )}{b}\right )}{c e^2}\right )\) |
-2*((2^(-1 - p)*Gamma[1 + p, (-2*(a + b*Log[c*(d + e/Sqrt[x])]))/b]*(a + b *Log[c*(d + e/Sqrt[x])])^p)/(c^2*e^2*E^((2*a)/b)*(-((a + b*Log[c*(d + e/Sq rt[x])])/b))^p) - (d*Gamma[1 + p, -((a + b*Log[c*(d + e/Sqrt[x])])/b)]*(a + b*Log[c*(d + e/Sqrt[x])])^p)/(c*e^2*E^(a/b)*(-((a + b*Log[c*(d + e/Sqrt[ x])])/b))^p))
3.6.47.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - d*g, 0] && IGtQ[q, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
\[\int \frac {\left (a +b \ln \left (c \left (d +\frac {e}{\sqrt {x}}\right )\right )\right )^{p}}{x^{2}}d x\]
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )\right )\right )^p}{x^2} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,\left (d+\frac {e}{\sqrt {x}}\right )\right )\right )}^p}{x^2} \,d x \]