Integrand size = 21, antiderivative size = 103 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=-\frac {b^2 d n^2}{4 x^2}-\frac {2 b^2 e n^2}{x}-\frac {b d n \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {2 b e n \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 x^2}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{x} \] Output:
-1/4*b^2*d*n^2/x^2-2*b^2*e*n^2/x-1/2*b*d*n*(a+b*ln(c*x^n))/x^2-2*b*e*n*(a+ b*ln(c*x^n))/x-1/2*d*(a+b*ln(c*x^n))^2/x^2-e*(a+b*ln(c*x^n))^2/x
Time = 0.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=-\frac {2 a^2 (d+2 e x)+2 a b n (d+4 e x)+b^2 n^2 (d+8 e x)+2 b (2 a (d+2 e x)+b n (d+4 e x)) \log \left (c x^n\right )+2 b^2 (d+2 e x) \log ^2\left (c x^n\right )}{4 x^2} \] Input:
Integrate[((d + e*x)*(a + b*Log[c*x^n])^2)/x^3,x]
Output:
-1/4*(2*a^2*(d + 2*e*x) + 2*a*b*n*(d + 4*e*x) + b^2*n^2*(d + 8*e*x) + 2*b* (2*a*(d + 2*e*x) + b*n*(d + 4*e*x))*Log[c*x^n] + 2*b^2*(d + 2*e*x)*Log[c*x ^n]^2)/x^2
Time = 0.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
\(\displaystyle \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx\) |
\(\Big \downarrow \) 2795 |
\(\displaystyle \int \left (\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b d n \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 x^2}-\frac {2 b e n \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{x}-\frac {b^2 d n^2}{4 x^2}-\frac {2 b^2 e n^2}{x}\) |
Input:
Int[((d + e*x)*(a + b*Log[c*x^n])^2)/x^3,x]
Output:
-1/4*(b^2*d*n^2)/x^2 - (2*b^2*e*n^2)/x - (b*d*n*(a + b*Log[c*x^n]))/(2*x^2 ) - (2*b*e*n*(a + b*Log[c*x^n]))/x - (d*(a + b*Log[c*x^n])^2)/(2*x^2) - (e *(a + b*Log[c*x^n])^2)/x
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]))
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.26
method | result | size |
parallelrisch | \(-\frac {4 b^{2} \ln \left (c \,x^{n}\right )^{2} e x +8 b^{2} e n x \ln \left (c \,x^{n}\right )+8 b^{2} e \,n^{2} x +8 a b \ln \left (c \,x^{n}\right ) e x +8 a b e n x +2 b^{2} \ln \left (c \,x^{n}\right )^{2} d +2 \ln \left (c \,x^{n}\right ) n \,b^{2} d +b^{2} d \,n^{2}+4 a^{2} e x +4 a b \ln \left (c \,x^{n}\right ) d +2 d a b n +2 a^{2} d}{4 x^{2}}\) | \(130\) |
risch | \(\text {Expression too large to display}\) | \(1483\) |
Input:
int((e*x+d)*(a+b*ln(c*x^n))^2/x^3,x,method=_RETURNVERBOSE)
Output:
-1/4/x^2*(4*b^2*ln(c*x^n)^2*e*x+8*b^2*e*n*x*ln(c*x^n)+8*b^2*e*n^2*x+8*a*b* ln(c*x^n)*e*x+8*a*b*e*n*x+2*b^2*ln(c*x^n)^2*d+2*ln(c*x^n)*n*b^2*d+b^2*d*n^ 2+4*a^2*e*x+4*a*b*ln(c*x^n)*d+2*d*a*b*n+2*a^2*d)
Time = 0.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.74 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=-\frac {b^{2} d n^{2} + 2 \, a b d n + 2 \, a^{2} d + 2 \, {\left (2 \, b^{2} e x + b^{2} d\right )} \log \left (c\right )^{2} + 2 \, {\left (2 \, b^{2} e n^{2} x + b^{2} d n^{2}\right )} \log \left (x\right )^{2} + 4 \, {\left (2 \, b^{2} e n^{2} + 2 \, a b e n + a^{2} e\right )} x + 2 \, {\left (b^{2} d n + 2 \, a b d + 4 \, {\left (b^{2} e n + a b e\right )} x\right )} \log \left (c\right ) + 2 \, {\left (b^{2} d n^{2} + 2 \, a b d n + 4 \, {\left (b^{2} e n^{2} + a b e n\right )} x + 2 \, {\left (2 \, b^{2} e n x + b^{2} d n\right )} \log \left (c\right )\right )} \log \left (x\right )}{4 \, x^{2}} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))^2/x^3,x, algorithm="fricas")
Output:
-1/4*(b^2*d*n^2 + 2*a*b*d*n + 2*a^2*d + 2*(2*b^2*e*x + b^2*d)*log(c)^2 + 2 *(2*b^2*e*n^2*x + b^2*d*n^2)*log(x)^2 + 4*(2*b^2*e*n^2 + 2*a*b*e*n + a^2*e )*x + 2*(b^2*d*n + 2*a*b*d + 4*(b^2*e*n + a*b*e)*x)*log(c) + 2*(b^2*d*n^2 + 2*a*b*d*n + 4*(b^2*e*n^2 + a*b*e*n)*x + 2*(2*b^2*e*n*x + b^2*d*n)*log(c) )*log(x))/x^2
Time = 0.37 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.60 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=- \frac {a^{2} d}{2 x^{2}} - \frac {a^{2} e}{x} - \frac {a b d n}{2 x^{2}} - \frac {a b d \log {\left (c x^{n} \right )}}{x^{2}} - \frac {2 a b e n}{x} - \frac {2 a b e \log {\left (c x^{n} \right )}}{x} - \frac {b^{2} d n^{2}}{4 x^{2}} - \frac {b^{2} d n \log {\left (c x^{n} \right )}}{2 x^{2}} - \frac {b^{2} d \log {\left (c x^{n} \right )}^{2}}{2 x^{2}} - \frac {2 b^{2} e n^{2}}{x} - \frac {2 b^{2} e n \log {\left (c x^{n} \right )}}{x} - \frac {b^{2} e \log {\left (c x^{n} \right )}^{2}}{x} \] Input:
integrate((e*x+d)*(a+b*ln(c*x**n))**2/x**3,x)
Output:
-a**2*d/(2*x**2) - a**2*e/x - a*b*d*n/(2*x**2) - a*b*d*log(c*x**n)/x**2 - 2*a*b*e*n/x - 2*a*b*e*log(c*x**n)/x - b**2*d*n**2/(4*x**2) - b**2*d*n*log( c*x**n)/(2*x**2) - b**2*d*log(c*x**n)**2/(2*x**2) - 2*b**2*e*n**2/x - 2*b* *2*e*n*log(c*x**n)/x - b**2*e*log(c*x**n)**2/x
Time = 0.03 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.46 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=-2 \, b^{2} e {\left (\frac {n^{2}}{x} + \frac {n \log \left (c x^{n}\right )}{x}\right )} - \frac {1}{4} \, b^{2} d {\left (\frac {n^{2}}{x^{2}} + \frac {2 \, n \log \left (c x^{n}\right )}{x^{2}}\right )} - \frac {b^{2} e \log \left (c x^{n}\right )^{2}}{x} - \frac {2 \, a b e n}{x} - \frac {2 \, a b e \log \left (c x^{n}\right )}{x} - \frac {b^{2} d \log \left (c x^{n}\right )^{2}}{2 \, x^{2}} - \frac {a b d n}{2 \, x^{2}} - \frac {a^{2} e}{x} - \frac {a b d \log \left (c x^{n}\right )}{x^{2}} - \frac {a^{2} d}{2 \, x^{2}} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))^2/x^3,x, algorithm="maxima")
Output:
-2*b^2*e*(n^2/x + n*log(c*x^n)/x) - 1/4*b^2*d*(n^2/x^2 + 2*n*log(c*x^n)/x^ 2) - b^2*e*log(c*x^n)^2/x - 2*a*b*e*n/x - 2*a*b*e*log(c*x^n)/x - 1/2*b^2*d *log(c*x^n)^2/x^2 - 1/2*a*b*d*n/x^2 - a^2*e/x - a*b*d*log(c*x^n)/x^2 - 1/2 *a^2*d/x^2
Time = 0.12 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.86 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=-\frac {{\left (2 \, b^{2} e n^{2} x + b^{2} d n^{2}\right )} \log \left (x\right )^{2}}{2 \, x^{2}} - \frac {{\left (4 \, b^{2} e n^{2} x + 4 \, b^{2} e n x \log \left (c\right ) + b^{2} d n^{2} + 4 \, a b e n x + 2 \, b^{2} d n \log \left (c\right ) + 2 \, a b d n\right )} \log \left (x\right )}{2 \, x^{2}} - \frac {8 \, b^{2} e n^{2} x + 8 \, b^{2} e n x \log \left (c\right ) + 4 \, b^{2} e x \log \left (c\right )^{2} + b^{2} d n^{2} + 8 \, a b e n x + 2 \, b^{2} d n \log \left (c\right ) + 8 \, a b e x \log \left (c\right ) + 2 \, b^{2} d \log \left (c\right )^{2} + 2 \, a b d n + 4 \, a^{2} e x + 4 \, a b d \log \left (c\right ) + 2 \, a^{2} d}{4 \, x^{2}} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))^2/x^3,x, algorithm="giac")
Output:
-1/2*(2*b^2*e*n^2*x + b^2*d*n^2)*log(x)^2/x^2 - 1/2*(4*b^2*e*n^2*x + 4*b^2 *e*n*x*log(c) + b^2*d*n^2 + 4*a*b*e*n*x + 2*b^2*d*n*log(c) + 2*a*b*d*n)*lo g(x)/x^2 - 1/4*(8*b^2*e*n^2*x + 8*b^2*e*n*x*log(c) + 4*b^2*e*x*log(c)^2 + b^2*d*n^2 + 8*a*b*e*n*x + 2*b^2*d*n*log(c) + 8*a*b*e*x*log(c) + 2*b^2*d*lo g(c)^2 + 2*a*b*d*n + 4*a^2*e*x + 4*a*b*d*log(c) + 2*a^2*d)/x^2
Time = 27.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=-\frac {x\,\left (2\,e\,a^2+4\,e\,a\,b\,n+4\,e\,b^2\,n^2\right )+a^2\,d+\frac {b^2\,d\,n^2}{2}+a\,b\,d\,n}{2\,x^2}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d\,\left (2\,a+b\,n\right )}{2}+2\,b\,e\,x\,\left (a+b\,n\right )\right )}{x^2}-\frac {{\ln \left (c\,x^n\right )}^2\,\left (\frac {b^2\,d}{2}+b^2\,e\,x\right )}{x^2} \] Input:
int(((a + b*log(c*x^n))^2*(d + e*x))/x^3,x)
Output:
- (x*(2*a^2*e + 4*b^2*e*n^2 + 4*a*b*e*n) + a^2*d + (b^2*d*n^2)/2 + a*b*d*n )/(2*x^2) - (log(c*x^n)*((b*d*(2*a + b*n))/2 + 2*b*e*x*(a + b*n)))/x^2 - ( log(c*x^n)^2*((b^2*d)/2 + b^2*e*x))/x^2
Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )^2}{x^3} \, dx=\frac {-2 \mathrm {log}\left (x^{n} c \right )^{2} b^{2} d -4 \mathrm {log}\left (x^{n} c \right )^{2} b^{2} e x -4 \,\mathrm {log}\left (x^{n} c \right ) a b d -8 \,\mathrm {log}\left (x^{n} c \right ) a b e x -2 \,\mathrm {log}\left (x^{n} c \right ) b^{2} d n -8 \,\mathrm {log}\left (x^{n} c \right ) b^{2} e n x -2 a^{2} d -4 a^{2} e x -2 a b d n -8 a b e n x -b^{2} d \,n^{2}-8 b^{2} e \,n^{2} x}{4 x^{2}} \] Input:
int((e*x+d)*(a+b*log(c*x^n))^2/x^3,x)
Output:
( - 2*log(x**n*c)**2*b**2*d - 4*log(x**n*c)**2*b**2*e*x - 4*log(x**n*c)*a* b*d - 8*log(x**n*c)*a*b*e*x - 2*log(x**n*c)*b**2*d*n - 8*log(x**n*c)*b**2* e*n*x - 2*a**2*d - 4*a**2*e*x - 2*a*b*d*n - 8*a*b*e*n*x - b**2*d*n**2 - 8* b**2*e*n**2*x)/(4*x**2)