\(\int (f x)^m \log (c (d+e x^n)^p) \, dx\) [63]

Optimal result

Integrand size = 18, antiderivative size = 87 \[ \int (f x)^m \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e n p x^{1+n} (f x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+n}{n},\frac {1+m+2 n}{n},-\frac {e x^n}{d}\right )}{d (1+m) (1+m+n)}+\frac {(f x)^{1+m} \log \left (c \left (d+e x^n\right )^p\right )}{f (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89 \[ \int (f x)^m \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x (f x)^m \left (-e n p x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+n}{n},\frac {1+m+2 n}{n},-\frac {e x^n}{d}\right )+d (1+m+n) \log \left (c \left (d+e x^n\right )^p\right )\right )}{d (1+m) (1+m+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 87, 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, 30, 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.

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

\(\Big \downarrow \) 2905

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

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((f*x)**m*log(c*(d + e*x**n)**p), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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