\(\int (f x)^m \log (c (d+\frac {e}{x^2})^p) \, dx\) [59]

Optimal result

Integrand size = 18, antiderivative size = 82 \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx=-\frac {2 e f p (f x)^{-1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1-m}{2},\frac {3-m}{2},-\frac {e}{d x^2}\right )}{d \left (1-m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )}{f (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93 \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx=\frac {(f x)^m \left (2 e p \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {m}{2},\frac {3}{2}-\frac {m}{2},-\frac {e}{d x^2}\right )+d (-1+m) x^2 \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )\right )}{d (-1+m) (1+m) x} \] Input:

Integrate[(f*x)^m*Log[c*(d + e/x^2)^p],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04, 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, 278}

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.

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

\(\Big \downarrow \) 2905

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

\(\Big \downarrow \) 8

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

\(\Big \downarrow \) 862

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

\(\Big \downarrow \) 278

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

Input:

Int[(f*x)^m*Log[c*(d + e/x^2)^p],x]
 

Output:

(-2*e*f*p*(f*x)^(-1 + m)*Hypergeometric2F1[1, (1 - m)/2, (3 - m)/2, -(e/(d 
*x^2))])/(d*(1 - m)*(1 + m)) + ((f*x)^(1 + m)*Log[c*(d + e/x^2)^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 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

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 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]

\[\int \left (f x \right )^{m} \ln \left (c \left (d +\frac {e}{x^{2}}\right )^{p}\right )d x\]

Input:

int((f*x)^m*ln(c*(d+e/x^2)^p),x)
 

Output:

int((f*x)^m*ln(c*(d+e/x^2)^p),x)
 

Fricas [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{x^{2}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^2)^p),x, algorithm="fricas")
 

Output:

integral((f*x)^m*log(c*((d*x^2 + e)/x^2)^p), x)
 

Sympy [A] (verification not implemented)

Time = 25.77 (sec) , antiderivative size = 366, normalized size of antiderivative = 4.46 \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx =\text {Too large to display} \] Input:

integrate((f*x)**m*ln(c*(d+e/x**2)**p),x)
 

Output:

2*e*p*Piecewise((-0**m*sqrt(-1/(d*e))*log(-e*sqrt(-1/(d*e)) + x)/2 + 0**m* 
sqrt(-1/(d*e))*log(e*sqrt(-1/(d*e)) + x)/2, Eq(f, 0) | (Eq(f, 0) & Ne(m, - 
1))), (f**(m + 1)*m*x**(m - 1)*lerchphi(e*exp_polar(I*pi)/(d*x**2), 1, 1/2 
 - m/2)*gamma(1/2 - m/2)/(4*d*f*m*gamma(3/2 - m/2) + 4*d*f*gamma(3/2 - m/2 
)) - f**(m + 1)*x**(m - 1)*lerchphi(e*exp_polar(I*pi)/(d*x**2), 1, 1/2 - m 
/2)*gamma(1/2 - m/2)/(4*d*f*m*gamma(3/2 - m/2) + 4*d*f*gamma(3/2 - m/2)), 
(m > -oo) & (m < oo) & Ne(m, -1)), (Piecewise((polylog(2, e*exp_polar(I*pi 
)/(d*x**2))/2, (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) + polylog(2, 
 e*exp_polar(I*pi)/(d*x**2))/2, Abs(x) < 1), (-log(d)*log(1/x) + polylog(2 
, e*exp_polar(I*pi)/(d*x**2))/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), (( 
0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) + po 
lylog(2, e*exp_polar(I*pi)/(d*x**2))/2, True))/(2*e*f) - log(f*x)*log(d + 
e/x**2)/(2*e*f), True)) + Piecewise((0**m*x, Eq(f, 0)), (Piecewise(((f*x)* 
*(m + 1)/(m + 1), Ne(m, -1)), (log(f*x), True))/f, True))*log(c*(d + e/x** 
2)**p)
 

Maxima [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{x^{2}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^2)^p),x, algorithm="maxima")
 

Output:

(f^m*x*x^m*log((d*x^2 + e)^p) - 2*f^m*x*x^m*log(x^p))/(m + 1) + integrate( 
(d*f^m*(m + 1)*x^2*log(c) + e*f^m*(m + 1)*log(c) + 2*e*f^m*p)*x^m/(d*(m + 
1)*x^2 + e*(m + 1)), x)
 

Giac [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{x^{2}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^2)^p),x, algorithm="giac")
 

Output:

integrate((f*x)^m*log(c*(d + e/x^2)^p), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+\frac {e}{x^2}\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \] Input:

int(log(c*(d + e/x^2)^p)*(f*x)^m,x)
 

Output:

int(log(c*(d + e/x^2)^p)*(f*x)^m, x)
 

Reduce [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx=f^{m} \left (\int x^{m} \mathrm {log}\left (\frac {\left (d \,x^{2}+e \right )^{p} c}{x^{2 p}}\right )d x \right ) \] Input:

int((f*x)^m*log(c*(d+e/x^2)^p),x)
 

Output:

f**m*int(x**m*log(((d*x**2 + e)**p*c)/x**(2*p)),x)