\(\int (f x)^m \log (c (d+e x^2)^p) \, dx\) [56]

Optimal result

Integrand size = 18, antiderivative size = 81 \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 e p (f x)^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},-\frac {e x^2}{d}\right )}{d f^3 (1+m) (3+m)}+\frac {(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {x (f x)^m \left (-2 e p x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},-\frac {e x^2}{d}\right )+d (3+m) \log \left (c \left (d+e x^2\right )^p\right )\right )}{d (1+m) (3+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2905, 8, 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.

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

\(\Big \downarrow \) 2905

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

\(\Big \downarrow \) 8

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 34.92 (sec) , antiderivative size = 377, normalized size of antiderivative = 4.65 \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

-2*e*p*Piecewise((0**m*sqrt(-d/e**3)*log(-e*sqrt(-d/e**3) + x)/2 - 0**m*sq 
rt(-d/e**3)*log(e*sqrt(-d/e**3) + x)/2 + 0**m*x/e, Eq(f, 0) | (Eq(f, 0) & 
Ne(m, -1))), (f**(m + 1)*m*x**(m + 3)*lerchphi(e*x**2*exp_polar(I*pi)/d, 1 
, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*d*f*m*gamma(m/2 + 5/2) + 4*d*f*gamma(m/2 
+ 5/2)) + 3*f**(m + 1)*x**(m + 3)*lerchphi(e*x**2*exp_polar(I*pi)/d, 1, m/ 
2 + 3/2)*gamma(m/2 + 3/2)/(4*d*f*m*gamma(m/2 + 5/2) + 4*d*f*gamma(m/2 + 5/ 
2)), (m > -oo) & (m < oo) & Ne(m, -1)), (-Piecewise((-polylog(2, e*x**2*ex 
p_polar(I*pi)/d)/2, (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) - polyl 
og(2, e*x**2*exp_polar(I*pi)/d)/2, Abs(x) < 1), (-log(d)*log(1/x) - polylo 
g(2, e*x**2*exp_polar(I*pi)/d)/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ( 
(0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - p 
olylog(2, e*x**2*exp_polar(I*pi)/d)/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]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {f^{m} \left (x^{m} \mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) m x +x^{m} \mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) x -2 x^{m} p x +2 \left (\int \frac {x^{m}}{e m \,x^{2}+e \,x^{2}+d m +d}d x \right ) d \,m^{2} p +4 \left (\int \frac {x^{m}}{e m \,x^{2}+e \,x^{2}+d m +d}d x \right ) d m p +2 \left (\int \frac {x^{m}}{e m \,x^{2}+e \,x^{2}+d m +d}d x \right ) d p \right )}{m^{2}+2 m +1} \] Input:

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

Output:

(f**m*(x**m*log((d + e*x**2)**p*c)*m*x + x**m*log((d + e*x**2)**p*c)*x - 2 
*x**m*p*x + 2*int(x**m/(d*m + d + e*m*x**2 + e*x**2),x)*d*m**2*p + 4*int(x 
**m/(d*m + d + e*m*x**2 + e*x**2),x)*d*m*p + 2*int(x**m/(d*m + d + e*m*x** 
2 + e*x**2),x)*d*p))/(m**2 + 2*m + 1)