Integrand size = 22, antiderivative size = 267 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=-\frac {3^{-p} e^{-\frac {3 a}{b}} \Gamma \left (1+p,-\frac {3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )}{b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )^{-p}}{c^3 e^3}+\frac {3\ 2^{-p} d e^{-\frac {2 a}{b}} \Gamma \left (1+p,-\frac {2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )}{b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )^{-p}}{c^2 e^3}-\frac {3 d^2 e^{-\frac {a}{b}} \Gamma \left (1+p,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )^{-p}}{c e^3} \] Output:
-GAMMA(p+1,(-3*a-3*b*ln(c*(d+e/x^(1/3))))/b)*(a+b*ln(c*(d+e/x^(1/3))))^p/( 3^p)/c^3/e^3/exp(3*a/b)/((-(a+b*ln(c*(d+e/x^(1/3))))/b)^p)+3*d*GAMMA(p+1,( -2*a-2*b*ln(c*(d+e/x^(1/3))))/b)*(a+b*ln(c*(d+e/x^(1/3))))^p/(2^p)/c^2/e^3 /exp(2*a/b)/((-(a+b*ln(c*(d+e/x^(1/3))))/b)^p)-3*d^2*GAMMA(p+1,-(a+b*ln(c* (d+e/x^(1/3))))/b)*(a+b*ln(c*(d+e/x^(1/3))))^p/c/e^3/exp(a/b)/((-(a+b*ln(c *(d+e/x^(1/3))))/b)^p)
Time = 0.43 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.66 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=-\frac {6^{-p} e^{-\frac {3 a}{b}} \left (2^p \Gamma \left (1+p,-\frac {3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )}{b}\right )+3^{1+p} c d e^{a/b} \left (-\Gamma \left (1+p,-\frac {2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )}{b}\right )+2^p c d e^{a/b} \Gamma \left (1+p,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )\right )\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )^{-p}}{c^3 e^3} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(1/3))])^p/x^2,x]
Output:
-(((2^p*Gamma[1 + p, (-3*(a + b*Log[c*(d + e/x^(1/3))]))/b] + 3^(1 + p)*c* d*E^(a/b)*(-Gamma[1 + p, (-2*(a + b*Log[c*(d + e/x^(1/3))]))/b] + 2^p*c*d* E^(a/b)*Gamma[1 + p, -((a + b*Log[c*(d + e/x^(1/3))])/b)]))*(a + b*Log[c*( d + e/x^(1/3))])^p)/(6^p*c^3*e^3*E^((3*a)/b)*(-((a + b*Log[c*(d + e/x^(1/3 ))])/b))^p))
Time = 1.05 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx\) |
| \(\Big \downarrow \) 2904 |
| \(\displaystyle -3 \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^{2/3}}d\frac {1}{\sqrt [3]{x}}\) |
| \(\Big \downarrow \) 2848 |
| \(\displaystyle -3 \int \left (\frac {\left (d+\frac {e}{\sqrt [3]{x}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{e^2}-\frac {2 d \left (d+\frac {e}{\sqrt [3]{x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{e^2}+\frac {d^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{e^2}\right )d\frac {1}{\sqrt [3]{x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -3 \left (\frac {3^{-p-1} e^{-\frac {3 a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )}{b}\right )}{c^3 e^3}-\frac {d 2^{-p} e^{-\frac {2 a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )}{b}\right )}{c^2 e^3}+\frac {d^2 e^{-\frac {a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )}{b}\right )}{c e^3}\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(1/3))])^p/x^2,x]
Output:
-3*((3^(-1 - p)*Gamma[1 + p, (-3*(a + b*Log[c*(d + e/x^(1/3))]))/b]*(a + b *Log[c*(d + e/x^(1/3))])^p)/(c^3*e^3*E^((3*a)/b)*(-((a + b*Log[c*(d + e/x^ (1/3))])/b))^p) - (d*Gamma[1 + p, (-2*(a + b*Log[c*(d + e/x^(1/3))]))/b]*( a + b*Log[c*(d + e/x^(1/3))])^p)/(2^p*c^2*e^3*E^((2*a)/b)*(-((a + b*Log[c* (d + e/x^(1/3))])/b))^p) + (d^2*Gamma[1 + p, -((a + b*Log[c*(d + e/x^(1/3) )])/b)]*(a + b*Log[c*(d + e/x^(1/3))])^p)/(c*e^3*E^(a/b)*(-((a + b*Log[c*( d + e/x^(1/3))])/b))^p))
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])
\[\int \frac {{\left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )\right )\right )}^{p}}{x^{2}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(1/3))))^p/x^2,x)
Output:
int((a+b*ln(c*(d+e/x^(1/3))))^p/x^2,x)
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}\right ) + a\right )}^{p}}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))))^p/x^2,x, algorithm="fricas")
Output:
integral((b*log((c*d*x + c*e*x^(2/3))/x) + a)^p/x^2, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(1/3))))**p/x**2,x)
Output:
Timed out
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}\right ) + a\right )}^{p}}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))))^p/x^2,x, algorithm="maxima")
Output:
integrate((b*log(c*(d + e/x^(1/3))) + a)^p/x^2, x)
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}\right ) + a\right )}^{p}}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))))^p/x^2,x, algorithm="giac")
Output:
integrate((b*log(c*(d + e/x^(1/3))) + a)^p/x^2, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,\left (d+\frac {e}{x^{1/3}}\right )\right )\right )}^p}{x^2} \,d x \] Input:
int((a + b*log(c*(d + e/x^(1/3))))^p/x^2,x)
Output:
int((a + b*log(c*(d + e/x^(1/3))))^p/x^2, x)
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\right )^p}{x^2} \, dx=\text {too large to display} \] Input:
int((a+b*log(c*(d+e/x^(1/3))))^p/x^2,x)
Output:
(6*x**(2/3)*(log((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p*b*d**2*e*p**2 + 6*x**(2/3)*(log((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p*b*d**2*e*p - 3*x* *(1/3)*(log((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p*b*d*e**2*p**2 - 3*x** (1/3)*(log((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p*b*d*e**2*p - 6*(log((x **(1/3)*c*d + c*e)/x**(1/3))*b + a)**p*log((x**(1/3)*c*d + c*e)/x**(1/3))* b*d**3*p*x - 6*(log((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p*a*d**3*p*x - 6*(log((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p*a*e**3*p - 6*(log((x**(1/3 )*c*d + c*e)/x**(1/3))*b + a)**p*a*e**3 + 6*int((log((x**(1/3)*c*d + c*e)/ x**(1/3))*b + a)**p/(3*x**(2/3)*log((x**(1/3)*c*d + c*e)/x**(1/3))*a*b*d*x + x**(2/3)*log((x**(1/3)*c*d + c*e)/x**(1/3))*b**2*d*p*x + 3*x**(2/3)*a** 2*d*x + x**(2/3)*a*b*d*p*x + 3*x**(1/3)*log((x**(1/3)*c*d + c*e)/x**(1/3)) *a*b*e*x + x**(1/3)*log((x**(1/3)*c*d + c*e)/x**(1/3))*b**2*e*p*x + 3*x**( 1/3)*a**2*e*x + x**(1/3)*a*b*e*p*x),x)*a*b**2*d**2*e**2*p**3*x + 6*int((lo g((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p/(3*x**(2/3)*log((x**(1/3)*c*d + c*e)/x**(1/3))*a*b*d*x + x**(2/3)*log((x**(1/3)*c*d + c*e)/x**(1/3))*b**2 *d*p*x + 3*x**(2/3)*a**2*d*x + x**(2/3)*a*b*d*p*x + 3*x**(1/3)*log((x**(1/ 3)*c*d + c*e)/x**(1/3))*a*b*e*x + x**(1/3)*log((x**(1/3)*c*d + c*e)/x**(1/ 3))*b**2*e*p*x + 3*x**(1/3)*a**2*e*x + x**(1/3)*a*b*e*p*x),x)*a*b**2*d**2* e**2*p**2*x + 2*int((log((x**(1/3)*c*d + c*e)/x**(1/3))*b + a)**p/(3*x**(2 /3)*log((x**(1/3)*c*d + c*e)/x**(1/3))*a*b*d*x + x**(2/3)*log((x**(1/3)...