3.5.49 \(\int (f x)^m (d+e x^r)^2 (a+b \log (c x^n))^p \, dx\) [449]

3.5.49.1 Optimal result

Integrand size = 27, antiderivative size = 350 \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\frac {d^2 e^{-\frac {a (1+m)}{b n}} (f x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \Gamma \left (1+p,-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{f (1+m)}+\frac {2 d e e^{-\frac {a (1+m+r)}{b n}} x^{1+r} (f x)^m \left (c x^n\right )^{-\frac {1+m+r}{n}} \Gamma \left (1+p,-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+r}+\frac {e^2 e^{-\frac {a (1+m+2 r)}{b n}} x^{1+2 r} (f x)^m \left (c x^n\right )^{-\frac {1+m+2 r}{n}} \Gamma \left (1+p,-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+2 r} \]

output

d^2*(f*x)^(1+m)*GAMMA(p+1,-(1+m)*(a+b*ln(c*x^n))/b/n)*(a+b*ln(c*x^n))^p/ex 
p(a*(1+m)/b/n)/f/(1+m)/((c*x^n)^((1+m)/n))/((-(1+m)*(a+b*ln(c*x^n))/b/n)^p 
)+2*d*e*x^(1+r)*(f*x)^m*GAMMA(p+1,-(1+m+r)*(a+b*ln(c*x^n))/b/n)*(a+b*ln(c* 
x^n))^p/exp(a*(1+m+r)/b/n)/(1+m+r)/((c*x^n)^((1+m+r)/n))/((-(1+m+r)*(a+b*l 
n(c*x^n))/b/n)^p)+e^2*x^(1+2*r)*(f*x)^m*GAMMA(p+1,-(1+m+2*r)*(a+b*ln(c*x^n 
))/b/n)*(a+b*ln(c*x^n))^p/exp(a*(1+m+2*r)/b/n)/(1+m+2*r)/((c*x^n)^((1+m+2* 
r)/n))/((-(1+m+2*r)*(a+b*ln(c*x^n))/b/n)^p)
 

3.5.49.2 Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.87 \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=x^{-m} (f x)^m \left (a+b \log \left (c x^n\right )\right )^p \left (\frac {d^2 e^{-\frac {(1+m) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} \Gamma \left (1+p,-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m}+e \left (\frac {2 d e^{-\frac {(1+m+r) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} \Gamma \left (1+p,-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+r}+\frac {e e^{-\frac {(1+m+2 r) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} \Gamma \left (1+p,-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+2 r}\right )\right ) \]

input

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

output

((f*x)^m*(a + b*Log[c*x^n])^p*((d^2*Gamma[1 + p, -(((1 + m)*(a + b*Log[c*x 
^n]))/(b*n))])/(E^(((1 + m)*(a - b*n*Log[x] + b*Log[c*x^n]))/(b*n))*(1 + m 
)*(-(((1 + m)*(a + b*Log[c*x^n]))/(b*n)))^p) + e*((2*d*Gamma[1 + p, -(((1 
+ m + r)*(a + b*Log[c*x^n]))/(b*n))])/(E^(((1 + m + r)*(a - b*n*Log[x] + b 
*Log[c*x^n]))/(b*n))*(1 + m + r)*(-(((1 + m + r)*(a + b*Log[c*x^n]))/(b*n) 
))^p) + (e*Gamma[1 + p, -(((1 + m + 2*r)*(a + b*Log[c*x^n]))/(b*n))])/(E^( 
((1 + m + 2*r)*(a - b*n*Log[x] + b*Log[c*x^n]))/(b*n))*(1 + m + 2*r)*(-((( 
1 + m + 2*r)*(a + b*Log[c*x^n]))/(b*n)))^p))))/x^m
 

3.5.49.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2795, 2009}

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.

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

\(\Big \downarrow \) 2795

\(\displaystyle \int \left (d^2 (f x)^m \left (a+b \log \left (c x^n\right )\right )^p+2 d e x^r (f x)^m \left (a+b \log \left (c x^n\right )\right )^p+e^2 x^{2 r} (f x)^m \left (a+b \log \left (c x^n\right )\right )^p\right )dx\)

\(\Big \downarrow \) 2009

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

input

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

output

(d^2*(f*x)^(1 + m)*Gamma[1 + p, -(((1 + m)*(a + b*Log[c*x^n]))/(b*n))]*(a 
+ b*Log[c*x^n])^p)/(E^((a*(1 + m))/(b*n))*f*(1 + m)*(c*x^n)^((1 + m)/n)*(- 
(((1 + m)*(a + b*Log[c*x^n]))/(b*n)))^p) + (2*d*e*x^(1 + r)*(f*x)^m*Gamma[ 
1 + p, -(((1 + m + r)*(a + b*Log[c*x^n]))/(b*n))]*(a + b*Log[c*x^n])^p)/(E 
^((a*(1 + m + r))/(b*n))*(1 + m + r)*(c*x^n)^((1 + m + r)/n)*(-(((1 + m + 
r)*(a + b*Log[c*x^n]))/(b*n)))^p) + (e^2*x^(1 + 2*r)*(f*x)^m*Gamma[1 + p, 
-(((1 + m + 2*r)*(a + b*Log[c*x^n]))/(b*n))]*(a + b*Log[c*x^n])^p)/(E^((a* 
(1 + m + 2*r))/(b*n))*(1 + m + 2*r)*(c*x^n)^((1 + m + 2*r)/n)*(-(((1 + m + 
 2*r)*(a + b*Log[c*x^n]))/(b*n)))^p)
 

3.5.49.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2795

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + 
(e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[ 
c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b 
, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0 
] && IntegerQ[m] && IntegerQ[r]))
 

3.5.49.4 Maple [F]

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

input

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

output

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

3.5.49.5 Fricas [F]

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

input

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

output

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

3.5.49.6 Sympy [F(-1)]

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

input

integrate((f*x)**m*(d+e*x**r)**2*(a+b*ln(c*x**n))**p,x)
 

output

Timed out
 

3.5.49.7 Maxima [F(-2)]

Exception generated. \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\text {Exception raised: RuntimeError} \]

input

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

output

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

3.5.49.8 Giac [F]

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

input

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

output

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

3.5.49.9 Mupad [F(-1)]

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

input

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

output

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