\(\int (f x)^m \log (c (d+\frac {e}{x^3})^p) \, dx\) [60]

Optimal result

Integrand size = 18, antiderivative size = 85 \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=-\frac {3 e f^2 p (f x)^{-2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2-m}{3},\frac {5-m}{3},-\frac {e}{d x^3}\right )}{d \left (2+m-m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (1+m)} \] Output:

-3*e*f^2*p*(f*x)^(-2+m)*hypergeom([1, 2/3-1/3*m],[5/3-1/3*m],-e/d/x^3)/d/( 
-m^2+m+2)+(f*x)^(1+m)*ln(c*(d+e/x^3)^p)/f/(1+m)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=\frac {(f x)^m \left (3 e p \operatorname {Hypergeometric2F1}\left (1,\frac {2}{3}-\frac {m}{3},\frac {5}{3}-\frac {m}{3},-\frac {e}{d x^3}\right )+d (-2+m) x^3 \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )\right )}{d (-2+m) (1+m) x^2} \] Input:

Integrate[(f*x)^m*Log[c*(d + e/x^3)^p],x]
 

Output:

((f*x)^m*(3*e*p*Hypergeometric2F1[1, 2/3 - m/3, 5/3 - m/3, -(e/(d*x^3))] + 
 d*(-2 + m)*x^3*Log[c*(d + e/x^3)^p]))/(d*(-2 + m)*(1 + m)*x^2)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2905, 8, 862, 888}

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.

\(\displaystyle \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx\)

\(\Big \downarrow \) 2905

\(\displaystyle \frac {3 e p \int \frac {(f x)^{m+1}}{\left (d+\frac {e}{x^3}\right ) x^4}dx}{f (m+1)}+\frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (m+1)}\)

\(\Big \downarrow \) 8

\(\displaystyle \frac {3 e f^3 p \int \frac {(f x)^{m-3}}{d+\frac {e}{x^3}}dx}{m+1}+\frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (m+1)}\)

\(\Big \downarrow \) 862

\(\displaystyle \frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (m+1)}-\frac {3 e f^2 p \left (\frac {1}{x}\right )^{m-2} (f x)^{m-2} \int \frac {\left (\frac {1}{x}\right )^{1-m}}{d+\frac {e}{x^3}}d\frac {1}{x}}{m+1}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{x^3}\right )^p\right )}{f (m+1)}-\frac {3 e f^2 p (f x)^{m-2} \operatorname {Hypergeometric2F1}\left (1,\frac {2-m}{3},\frac {5-m}{3},-\frac {e}{d x^3}\right )}{d (2-m) (m+1)}\)

Input:

Int[(f*x)^m*Log[c*(d + e/x^3)^p],x]
 

Output:

(-3*e*f^2*p*(f*x)^(-2 + m)*Hypergeometric2F1[1, (2 - m)/3, (5 - m)/3, -(e/ 
(d*x^3))])/(d*(2 - m)*(1 + m)) + ((f*x)^(1 + m)*Log[c*(d + e/x^3)^p])/(f*( 
1 + m))
 

Defintions of rubi rules used

rule 8

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m   Int[u*(a* 
x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
 

rule 862

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-c^ 
(-1))*(c*x)^(m + 1)*(1/x)^(m + 1)   Subst[Int[(a + b/x^n)^p/x^(m + 2), x], 
x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 2905

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ 
(m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m 
+ 1))), x] - Simp[b*e*n*(p/(f*(m + 1)))   Int[x^(n - 1)*((f*x)^(m + 1)/(d + 
 e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
 

Maple [F]

\[\int \left (f x \right )^{m} \ln \left (c \left (d +\frac {e}{x^{3}}\right )^{p}\right )d x\]

Input:

int((f*x)^m*ln(c*(d+e/x^3)^p),x)
 

Output:

int((f*x)^m*ln(c*(d+e/x^3)^p),x)
 

Fricas [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{x^{3}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^3)^p),x, algorithm="fricas")
 

Output:

integral((f*x)^m*log(c*((d*x^3 + e)/x^3)^p), x)
 

Sympy [F(-1)]

Timed out. \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=\text {Timed out} \] Input:

integrate((f*x)**m*ln(c*(d+e/x**3)**p),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{x^{3}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^3)^p),x, algorithm="maxima")
 

Output:

(f^m*x*x^m*log((d*x^3 + e)^p) - 3*f^m*x*x^m*log(x^p))/(m + 1) + integrate( 
(d*f^m*(m + 1)*x^3*log(c) + e*f^m*(m + 1)*log(c) + 3*e*f^m*p)*x^m/(d*(m + 
1)*x^3 + e*(m + 1)), x)
 

Giac [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{x^{3}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^3)^p),x, algorithm="giac")
 

Output:

integrate((f*x)^m*log(c*(d + e/x^3)^p), x)
 

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+\frac {e}{x^3}\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \] Input:

int(log(c*(d + e/x^3)^p)*(f*x)^m,x)
 

Output:

int(log(c*(d + e/x^3)^p)*(f*x)^m, x)
 

Reduce [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^3}\right )^p\right ) \, dx=f^{m} \left (\int x^{m} \mathrm {log}\left (\frac {\left (d \,x^{3}+e \right )^{p} c}{x^{3 p}}\right )d x \right ) \] Input:

int((f*x)^m*log(c*(d+e/x^3)^p),x)
 

Output:

f**m*int(x**m*log(((d*x**3 + e)**p*c)/x**(3*p)),x)