Integrand size = 24, antiderivative size = 216 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx=\frac {2^{1+p} d e^{-\frac {a}{2 b}} \left (d+\frac {e}{\sqrt {x}}\right ) \Gamma \left (1+p,\frac {-a-b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{2 b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p}}{e^2 \sqrt {c \left (d+\frac {e}{\sqrt {x}}\right )^2}}-\frac {e^{-\frac {a}{b}} \Gamma \left (1+p,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p}}{c e^2} \]
-GAMMA(p+1,(-a-b*ln(c*(d+e/x^(1/2))^2))/b)*(a+b*ln(c*(d+e/x^(1/2))^2))^p/c /e^2/exp(a/b)/(((-a-b*ln(c*(d+e/x^(1/2))^2))/b)^p)+2^(p+1)*d*GAMMA(p+1,1/2 *(-a-b*ln(c*(d+e/x^(1/2))^2))/b)*(a+b*ln(c*(d+e/x^(1/2))^2))^p*(d+e/x^(1/2 ))/e^2/exp(1/2*a/b)/(((-a-b*ln(c*(d+e/x^(1/2))^2))/b)^p)/(c*(d+e/x^(1/2))^ 2)^(1/2)
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx=\int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx \]
Time = 0.52 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.
\(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -2 \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{\sqrt {x}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 2848 |
\(\displaystyle -2 \int \left (\frac {\left (d+\frac {e}{\sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{e}-\frac {d \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{e}\right )d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\frac {e^{-\frac {a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )}{2 c e^2}-\frac {d 2^p e^{-\frac {a}{2 b}} \left (d+\frac {e}{\sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{2 b}\right )}{e^2 \sqrt {c \left (d+\frac {e}{\sqrt {x}}\right )^2}}\right )\) |
-2*((Gamma[1 + p, -((a + b*Log[c*(d + e/Sqrt[x])^2])/b)]*(a + b*Log[c*(d + e/Sqrt[x])^2])^p)/(2*c*e^2*E^(a/b)*(-((a + b*Log[c*(d + e/Sqrt[x])^2])/b) )^p) - (2^p*d*(d + e/Sqrt[x])*Gamma[1 + p, -1/2*(a + b*Log[c*(d + e/Sqrt[x ])^2])/b]*(a + b*Log[c*(d + e/Sqrt[x])^2])^p)/(e^2*E^(a/(2*b))*Sqrt[c*(d + e/Sqrt[x])^2]*(-((a + b*Log[c*(d + e/Sqrt[x])^2])/b))^p))
3.6.53.3.1 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])
\[\int \frac {{\left (a +b \ln \left (c \left (d +\frac {e}{\sqrt {x}}\right )^{2}\right )\right )}^{p}}{x^{2}}d x\]
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{2}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{2}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{2}\right ) + a\right )}^{p}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^2\right )\right )}^p}{x^2} \,d x \]