\(\int \frac {\log ^{-1+q}(c x^n) (a x^m+b \log ^q(c x^n))^2}{x} \, dx\) [3]

Optimal result

Integrand size = 32, antiderivative size = 156 \[ \int \frac {\log ^{-1+q}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx=\frac {b^2 \log ^{3 q}\left (c x^n\right )}{3 n q}-\frac {2 a b x^m \left (c x^n\right )^{-\frac {m}{n}} \Gamma \left (2 q,-\frac {m \log \left (c x^n\right )}{n}\right ) \log ^{2 q}\left (c x^n\right ) \left (-\frac {m \log \left (c x^n\right )}{n}\right )^{-2 q}}{n}-\frac {2^{-q} a^2 x^{2 m} \left (c x^n\right )^{-\frac {2 m}{n}} \Gamma \left (q,-\frac {2 m \log \left (c x^n\right )}{n}\right ) \log ^q\left (c x^n\right ) \left (-\frac {m \log \left (c x^n\right )}{n}\right )^{-q}}{n} \] Output:

1/3*b^2*ln(c*x^n)^(3*q)/n/q-2*a*b*x^m*GAMMA(2*q,-m*ln(c*x^n)/n)*ln(c*x^n)^ 
(2*q)/n/((c*x^n)^(m/n))/((-m*ln(c*x^n)/n)^(2*q))-a^2*x^(2*m)*GAMMA(q,-2*m* 
ln(c*x^n)/n)*ln(c*x^n)^q/(2^q)/n/((c*x^n)^(2*m/n))/((-m*ln(c*x^n)/n)^q)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96 \[ \int \frac {\log ^{-1+q}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx=\frac {\log ^q\left (c x^n\right ) \left (\frac {b^2 \log ^{2 q}\left (c x^n\right )}{q}-6 a b x^m \left (c x^n\right )^{-\frac {m}{n}} \Gamma \left (2 q,-\frac {m \log \left (c x^n\right )}{n}\right ) \log ^q\left (c x^n\right ) \left (-\frac {m \log \left (c x^n\right )}{n}\right )^{-2 q}-3\ 2^{-q} a^2 x^{2 m} \left (c x^n\right )^{-\frac {2 m}{n}} \Gamma \left (q,-\frac {2 m \log \left (c x^n\right )}{n}\right ) \left (-\frac {m \log \left (c x^n\right )}{n}\right )^{-q}\right )}{3 n} \] Input:

Integrate[(Log[c*x^n]^(-1 + q)*(a*x^m + b*Log[c*x^n]^q)^2)/x,x]
 

Output:

(Log[c*x^n]^q*((b^2*Log[c*x^n]^(2*q))/q - (6*a*b*x^m*Gamma[2*q, -((m*Log[c 
*x^n])/n)]*Log[c*x^n]^q)/((c*x^n)^(m/n)*(-((m*Log[c*x^n])/n))^(2*q)) - (3* 
a^2*x^(2*m)*Gamma[q, (-2*m*Log[c*x^n])/n])/(2^q*(c*x^n)^((2*m)/n)*(-((m*Lo 
g[c*x^n])/n))^q)))/(3*n)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {3019, 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.

\(\displaystyle \int \frac {\log ^{q-1}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx\)

\(\Big \downarrow \) 3019

\(\displaystyle \int \left (a^2 x^{2 m-1} \log ^{q-1}\left (c x^n\right )+2 a b x^{m-1} \log ^{2 q-1}\left (c x^n\right )+\frac {b^2 \log ^{3 q-1}\left (c x^n\right )}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 2^{-q} x^{2 m} \left (c x^n\right )^{-\frac {2 m}{n}} \log ^q\left (c x^n\right ) \left (-\frac {m \log \left (c x^n\right )}{n}\right )^{-q} \Gamma \left (q,-\frac {2 m \log \left (c x^n\right )}{n}\right )}{n}-\frac {2 a b x^m \left (c x^n\right )^{-\frac {m}{n}} \log ^{2 q}\left (c x^n\right ) \left (-\frac {m \log \left (c x^n\right )}{n}\right )^{-2 q} \Gamma \left (2 q,-\frac {m \log \left (c x^n\right )}{n}\right )}{n}+\frac {b^2 \log ^{3 q}\left (c x^n\right )}{3 n q}\)

Input:

Int[(Log[c*x^n]^(-1 + q)*(a*x^m + b*Log[c*x^n]^q)^2)/x,x]
 

Output:

(b^2*Log[c*x^n]^(3*q))/(3*n*q) - (2*a*b*x^m*Gamma[2*q, -((m*Log[c*x^n])/n) 
]*Log[c*x^n]^(2*q))/(n*(c*x^n)^(m/n)*(-((m*Log[c*x^n])/n))^(2*q)) - (a^2*x 
^(2*m)*Gamma[q, (-2*m*Log[c*x^n])/n]*Log[c*x^n]^q)/(2^q*n*(c*x^n)^((2*m)/n 
)*(-((m*Log[c*x^n])/n))^q)
 

Defintions of rubi rules used

rule 2009

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

rule 3019

Int[(Log[(c_.)*(x_)^(n_.)]^(r_.)*(Log[(c_.)*(x_)^(n_.)]^(q_)*(b_.) + (a_.)* 
(x_)^(m_.))^(p_.))/(x_), x_Symbol] :> Int[ExpandIntegrand[Log[c*x^n]^r/x, ( 
a*x^m + b*Log[c*x^n]^q)^p, x], x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && 
EqQ[r, q - 1] && IGtQ[p, 0]
 

Maple [F]

\[\int \frac {\ln \left (c \,x^{n}\right )^{-1+q} \left (a \,x^{m}+b \ln \left (c \,x^{n}\right )^{q}\right )^{2}}{x}d x\]

Input:

int(ln(c*x^n)^(-1+q)*(a*x^m+b*ln(c*x^n)^q)^2/x,x)
 

Output:

int(ln(c*x^n)^(-1+q)*(a*x^m+b*ln(c*x^n)^q)^2/x,x)
 

Fricas [F]

\[ \int \frac {\log ^{-1+q}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx=\int { \frac {{\left (a x^{m} + b \log \left (c x^{n}\right )^{q}\right )}^{2} \log \left (c x^{n}\right )^{q - 1}}{x} \,d x } \] Input:

integrate(log(c*x^n)^(-1+q)*(a*x^m+b*log(c*x^n)^q)^2/x,x, algorithm="frica 
s")
 

Output:

integral((2*a*b*x^m*log(c*x^n)^(q - 1)*log(c*x^n)^q + a^2*x^(2*m)*log(c*x^ 
n)^(q - 1) + b^2*log(c*x^n)^(2*q)*log(c*x^n)^(q - 1))/x, x)
 

Sympy [F]

\[ \int \frac {\log ^{-1+q}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx=\int \frac {\left (a x^{m} + b \log {\left (c x^{n} \right )}^{q}\right )^{2} \log {\left (c x^{n} \right )}^{q - 1}}{x}\, dx \] Input:

integrate(ln(c*x**n)**(-1+q)*(a*x**m+b*ln(c*x**n)**q)**2/x,x)
 

Output:

Integral((a*x**m + b*log(c*x**n)**q)**2*log(c*x**n)**(q - 1)/x, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\log ^{-1+q}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(log(c*x^n)^(-1+q)*(a*x^m+b*log(c*x^n)^q)^2/x,x, algorithm="maxim 
a")
 

Output:

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

Giac [F]

\[ \int \frac {\log ^{-1+q}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx=\int { \frac {{\left (a x^{m} + b \log \left (c x^{n}\right )^{q}\right )}^{2} \log \left (c x^{n}\right )^{q - 1}}{x} \,d x } \] Input:

integrate(log(c*x^n)^(-1+q)*(a*x^m+b*log(c*x^n)^q)^2/x,x, algorithm="giac" 
)
 

Output:

integrate((a*x^m + b*log(c*x^n)^q)^2*log(c*x^n)^(q - 1)/x, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^{-1+q}\left (c x^n\right ) \left (a x^m+b \log ^q\left (c x^n\right )\right )^2}{x} \, dx=\int \frac {{\ln \left (c\,x^n\right )}^{q-1}\,{\left (a\,x^m+b\,{\ln \left (c\,x^n\right )}^q\right )}^2}{x} \,d x \] Input:

int((log(c*x^n)^(q - 1)*(a*x^m + b*log(c*x^n)^q)^2)/x,x)
 

Output:

int((log(c*x^n)^(q - 1)*(a*x^m + b*log(c*x^n)^q)^2)/x, x)
 

Reduce [F]

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

int(log(c*x^n)^(-1+q)*(a*x^m+b*log(c*x^n)^q)^2/x,x)
 

Output:

(log(x**n*c)**(3*q)*b**2 + 3*int((x**(2*m)*log(x**n*c)**q)/(log(x**n*c)*x) 
,x)*a**2*n*q + 6*int((x**m*log(x**n*c)**(2*q))/(log(x**n*c)*x),x)*a*b*n*q) 
/(3*n*q)