\(\int (f x)^m \log (c (d+e x^3)^p) \, dx\) [55]

Optimal result

Integrand size = 18, antiderivative size = 81 \[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=-\frac {3 e p (f x)^{4+m} \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{3},\frac {7+m}{3},-\frac {e x^3}{d}\right )}{d f^4 (1+m) (4+m)}+\frac {(f x)^{1+m} \log \left (c \left (d+e x^3\right )^p\right )}{f (1+m)} \] Output:

-3*e*p*(f*x)^(4+m)*hypergeom([1, 4/3+1/3*m],[7/3+1/3*m],-e*x^3/d)/d/f^4/(1 
+m)/(4+m)+(f*x)^(1+m)*ln(c*(e*x^3+d)^p)/f/(1+m)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {x (f x)^m \left (-3 e p x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{3},\frac {7+m}{3},-\frac {e x^3}{d}\right )+d (4+m) \log \left (c \left (d+e x^3\right )^p\right )\right )}{d (1+m) (4+m)} \] Input:

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

Output:

(x*(f*x)^m*(-3*e*p*x^3*Hypergeometric2F1[1, (4 + m)/3, (7 + m)/3, -((e*x^3 
)/d)] + d*(4 + m)*Log[c*(d + e*x^3)^p]))/(d*(1 + m)*(4 + m))
 

Rubi [A] (verified)

Time = 0.39 (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, 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.

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

\(\Big \downarrow \) 2905

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

\(\Big \downarrow \) 8

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

\(\Big \downarrow \) 888

\(\displaystyle \frac {(f x)^{m+1} \log \left (c \left (d+e x^3\right )^p\right )}{f (m+1)}-\frac {3 e p (f x)^{m+4} \operatorname {Hypergeometric2F1}\left (1,\frac {m+4}{3},\frac {m+7}{3},-\frac {e x^3}{d}\right )}{d f^4 (m+1) (m+4)}\)

Input:

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

Output:

(-3*e*p*(f*x)^(4 + m)*Hypergeometric2F1[1, (4 + m)/3, (7 + m)/3, -((e*x^3) 
/d)])/(d*f^4*(1 + m)*(4 + 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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int (f x)^m \log \left (c \left (d+e x^3\right )^p\right ) \, dx=\frac {f^{m} \left (x^{m} \mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right ) m x +x^{m} \mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right ) x -3 x^{m} p x +3 \left (\int \frac {x^{m}}{e m \,x^{3}+e \,x^{3}+d m +d}d x \right ) d \,m^{2} p +6 \left (\int \frac {x^{m}}{e m \,x^{3}+e \,x^{3}+d m +d}d x \right ) d m p +3 \left (\int \frac {x^{m}}{e m \,x^{3}+e \,x^{3}+d m +d}d x \right ) d p \right )}{m^{2}+2 m +1} \] Input:

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

Output:

(f**m*(x**m*log((d + e*x**3)**p*c)*m*x + x**m*log((d + e*x**3)**p*c)*x - 3 
*x**m*p*x + 3*int(x**m/(d*m + d + e*m*x**3 + e*x**3),x)*d*m**2*p + 6*int(x 
**m/(d*m + d + e*m*x**3 + e*x**3),x)*d*m*p + 3*int(x**m/(d*m + d + e*m*x** 
3 + e*x**3),x)*d*p))/(m**2 + 2*m + 1)