Integrand size = 20, antiderivative size = 78 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=-2 a b m n x+2 b^2 m^2 n^2 x-\frac {2 b^2 m n (e+f x) \log \left (c \left (d (e+f x)^m\right )^n\right )}{f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}{f} \] Output:
-2*a*b*m*n*x+2*b^2*m^2*n^2*x-2*b^2*m*n*(f*x+e)*ln(c*(d*(f*x+e)^m)^n)/f+(f* x+e)*(a+b*ln(c*(d*(f*x+e)^m)^n))^2/f
Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}{f}-2 b m n \left (a x-b m n x+\frac {b (e+f x) \log \left (c \left (d (e+f x)^m\right )^n\right )}{f}\right ) \] Input:
Integrate[(a + b*Log[c*(d*(e + f*x)^m)^n])^2,x]
Output:
((e + f*x)*(a + b*Log[c*(d*(e + f*x)^m)^n])^2)/f - 2*b*m*n*(a*x - b*m*n*x + (b*(e + f*x)*Log[c*(d*(e + f*x)^m)^n])/f)
Time = 0.59 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2895, 2836, 2733, 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.
\(\displaystyle \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 2895 |
\(\displaystyle \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2dx\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle \frac {\int \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2d(e+f x)}{f}\) |
\(\Big \downarrow \) 2733 |
\(\displaystyle \frac {(e+f x) \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2-2 b m n \int \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )d(e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(e+f x) \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2-2 b m n \left (a (e+f x)+b (e+f x) \log \left (c d^n (e+f x)^{m n}\right )-b m n (e+f x)\right )}{f}\) |
Input:
Int[(a + b*Log[c*(d*(e + f*x)^m)^n])^2,x]
Output:
((e + f*x)*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^2 - 2*b*m*n*(a*(e + f*x) - b *m*n*(e + f*x) + b*(e + f*x)*Log[c*d^n*(e + f*x)^(m*n)]))/f
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b *Log[c*x^n])^p, x] - Simp[b*n*p Int[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : > Simp[1/e Subst[Int[(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{ a, b, c, d, e, n, p}, x]
Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_. )*(u_.), x_Symbol] :> Subst[Int[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[n] && !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[ IntHide[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(164\) vs. \(2(78)=156\).
Time = 0.51 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.12
method | result | size |
parallelrisch | \(\frac {-2 \ln \left (f x +e \right ) b^{2} e^{2} m^{2} n^{2}+2 x \,b^{2} e f \,m^{2} n^{2}-2 x \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right ) b^{2} e f m n +2 \ln \left (f x +e \right ) a b \,e^{2} m n +x {\ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )}^{2} b^{2} e f -2 x a b e f m n +2 x \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right ) a b e f +{\ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )}^{2} b^{2} e^{2}+x \,a^{2} e f}{e f}\) | \(165\) |
Input:
int((a+b*ln(c*(d*(f*x+e)^m)^n))^2,x,method=_RETURNVERBOSE)
Output:
(-2*ln(f*x+e)*b^2*e^2*m^2*n^2+2*x*b^2*e*f*m^2*n^2-2*x*ln(c*(d*(f*x+e)^m)^n )*b^2*e*f*m*n+2*ln(f*x+e)*a*b*e^2*m*n+x*ln(c*(d*(f*x+e)^m)^n)^2*b^2*e*f-2* x*a*b*e*f*m*n+2*x*ln(c*(d*(f*x+e)^m)^n)*a*b*e*f+ln(c*(d*(f*x+e)^m)^n)^2*b^ 2*e^2+x*a^2*e*f)/e/f
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (78) = 156\).
Time = 0.10 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.96 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {b^{2} f n^{2} x \log \left (d\right )^{2} + b^{2} f x \log \left (c\right )^{2} + {\left (b^{2} f m^{2} n^{2} x + b^{2} e m^{2} n^{2}\right )} \log \left (f x + e\right )^{2} - 2 \, {\left (b^{2} f m n - a b f\right )} x \log \left (c\right ) + {\left (2 \, b^{2} f m^{2} n^{2} - 2 \, a b f m n + a^{2} f\right )} x - 2 \, {\left (b^{2} e m^{2} n^{2} - a b e m n + {\left (b^{2} f m^{2} n^{2} - a b f m n\right )} x - {\left (b^{2} f m n x + b^{2} e m n\right )} \log \left (c\right ) - {\left (b^{2} f m n^{2} x + b^{2} e m n^{2}\right )} \log \left (d\right )\right )} \log \left (f x + e\right ) + 2 \, {\left (b^{2} f n x \log \left (c\right ) - {\left (b^{2} f m n^{2} - a b f n\right )} x\right )} \log \left (d\right )}{f} \] Input:
integrate((a+b*log(c*(d*(f*x+e)^m)^n))^2,x, algorithm="fricas")
Output:
(b^2*f*n^2*x*log(d)^2 + b^2*f*x*log(c)^2 + (b^2*f*m^2*n^2*x + b^2*e*m^2*n^ 2)*log(f*x + e)^2 - 2*(b^2*f*m*n - a*b*f)*x*log(c) + (2*b^2*f*m^2*n^2 - 2* a*b*f*m*n + a^2*f)*x - 2*(b^2*e*m^2*n^2 - a*b*e*m*n + (b^2*f*m^2*n^2 - a*b *f*m*n)*x - (b^2*f*m*n*x + b^2*e*m*n)*log(c) - (b^2*f*m*n^2*x + b^2*e*m*n^ 2)*log(d))*log(f*x + e) + 2*(b^2*f*n*x*log(c) - (b^2*f*m*n^2 - a*b*f*n)*x) *log(d))/f
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (76) = 152\).
Time = 0.55 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.28 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\begin {cases} a^{2} x + \frac {2 a b e \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}}{f} - 2 a b m n x + 2 a b x \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )} - \frac {2 b^{2} e m n \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}}{f} + \frac {b^{2} e \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}^{2}}{f} + 2 b^{2} m^{2} n^{2} x - 2 b^{2} m n x \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )} + b^{2} x \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}^{2} & \text {for}\: f \neq 0 \\x \left (a + b \log {\left (c \left (d e^{m}\right )^{n} \right )}\right )^{2} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*ln(c*(d*(f*x+e)**m)**n))**2,x)
Output:
Piecewise((a**2*x + 2*a*b*e*log(c*(d*(e + f*x)**m)**n)/f - 2*a*b*m*n*x + 2 *a*b*x*log(c*(d*(e + f*x)**m)**n) - 2*b**2*e*m*n*log(c*(d*(e + f*x)**m)**n )/f + b**2*e*log(c*(d*(e + f*x)**m)**n)**2/f + 2*b**2*m**2*n**2*x - 2*b**2 *m*n*x*log(c*(d*(e + f*x)**m)**n) + b**2*x*log(c*(d*(e + f*x)**m)**n)**2, Ne(f, 0)), (x*(a + b*log(c*(d*e**m)**n))**2, True))
Time = 0.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.90 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=-2 \, a b f m n {\left (\frac {x}{f} - \frac {e \log \left (f x + e\right )}{f^{2}}\right )} + b^{2} x \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right )^{2} + 2 \, a b x \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right ) - {\left (2 \, f m n {\left (\frac {x}{f} - \frac {e \log \left (f x + e\right )}{f^{2}}\right )} \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right ) + \frac {{\left (e \log \left (f x + e\right )^{2} - 2 \, f x + 2 \, e \log \left (f x + e\right )\right )} m^{2} n^{2}}{f}\right )} b^{2} + a^{2} x \] Input:
integrate((a+b*log(c*(d*(f*x+e)^m)^n))^2,x, algorithm="maxima")
Output:
-2*a*b*f*m*n*(x/f - e*log(f*x + e)/f^2) + b^2*x*log(((f*x + e)^m*d)^n*c)^2 + 2*a*b*x*log(((f*x + e)^m*d)^n*c) - (2*f*m*n*(x/f - e*log(f*x + e)/f^2)* log(((f*x + e)^m*d)^n*c) + (e*log(f*x + e)^2 - 2*f*x + 2*e*log(f*x + e))*m ^2*n^2/f)*b^2 + a^2*x
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (78) = 156\).
Time = 0.13 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.63 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {{\left (f x + e\right )} b^{2} m^{2} n^{2} \log \left (f x + e\right )^{2}}{f} - \frac {2 \, {\left (f x + e\right )} b^{2} m^{2} n^{2} \log \left (f x + e\right )}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} m n^{2} \log \left (f x + e\right ) \log \left (d\right )}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} m^{2} n^{2}}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} m n \log \left (f x + e\right ) \log \left (c\right )}{f} - \frac {2 \, {\left (f x + e\right )} b^{2} m n^{2} \log \left (d\right )}{f} + \frac {{\left (f x + e\right )} b^{2} n^{2} \log \left (d\right )^{2}}{f} + \frac {2 \, {\left (f x + e\right )} a b m n \log \left (f x + e\right )}{f} - \frac {2 \, {\left (f x + e\right )} b^{2} m n \log \left (c\right )}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} n \log \left (c\right ) \log \left (d\right )}{f} - \frac {2 \, {\left (f x + e\right )} a b m n}{f} + \frac {{\left (f x + e\right )} b^{2} \log \left (c\right )^{2}}{f} + \frac {2 \, {\left (f x + e\right )} a b n \log \left (d\right )}{f} + \frac {2 \, {\left (f x + e\right )} a b \log \left (c\right )}{f} + \frac {{\left (f x + e\right )} a^{2}}{f} \] Input:
integrate((a+b*log(c*(d*(f*x+e)^m)^n))^2,x, algorithm="giac")
Output:
(f*x + e)*b^2*m^2*n^2*log(f*x + e)^2/f - 2*(f*x + e)*b^2*m^2*n^2*log(f*x + e)/f + 2*(f*x + e)*b^2*m*n^2*log(f*x + e)*log(d)/f + 2*(f*x + e)*b^2*m^2* n^2/f + 2*(f*x + e)*b^2*m*n*log(f*x + e)*log(c)/f - 2*(f*x + e)*b^2*m*n^2* log(d)/f + (f*x + e)*b^2*n^2*log(d)^2/f + 2*(f*x + e)*a*b*m*n*log(f*x + e) /f - 2*(f*x + e)*b^2*m*n*log(c)/f + 2*(f*x + e)*b^2*n*log(c)*log(d)/f - 2* (f*x + e)*a*b*m*n/f + (f*x + e)*b^2*log(c)^2/f + 2*(f*x + e)*a*b*n*log(d)/ f + 2*(f*x + e)*a*b*log(c)/f + (f*x + e)*a^2/f
Time = 25.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.42 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx={\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )}^2\,\left (b^2\,x+\frac {b^2\,e}{f}\right )+x\,\left (a^2-2\,a\,b\,m\,n+2\,b^2\,m^2\,n^2\right )-\frac {\ln \left (e+f\,x\right )\,\left (2\,b^2\,e\,m^2\,n^2-2\,a\,b\,e\,m\,n\right )}{f}+2\,b\,x\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )\,\left (a-b\,m\,n\right ) \] Input:
int((a + b*log(c*(d*(e + f*x)^m)^n))^2,x)
Output:
log(c*(d*(e + f*x)^m)^n)^2*(b^2*x + (b^2*e)/f) + x*(a^2 + 2*b^2*m^2*n^2 - 2*a*b*m*n) - (log(e + f*x)*(2*b^2*e*m^2*n^2 - 2*a*b*e*m*n))/f + 2*b*x*log( c*(d*(e + f*x)^m)^n)*(a - b*m*n)
Time = 0.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.12 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {\mathrm {log}\left (d^{n} \left (f x +e \right )^{m n} c \right )^{2} b^{2} e +\mathrm {log}\left (d^{n} \left (f x +e \right )^{m n} c \right )^{2} b^{2} f x +2 \,\mathrm {log}\left (d^{n} \left (f x +e \right )^{m n} c \right ) a b e +2 \,\mathrm {log}\left (d^{n} \left (f x +e \right )^{m n} c \right ) a b f x -2 \,\mathrm {log}\left (d^{n} \left (f x +e \right )^{m n} c \right ) b^{2} e m n -2 \,\mathrm {log}\left (d^{n} \left (f x +e \right )^{m n} c \right ) b^{2} f m n x +a^{2} f x -2 a b f m n x +2 b^{2} f \,m^{2} n^{2} x}{f} \] Input:
int((a+b*log(c*(d*(f*x+e)^m)^n))^2,x)
Output:
(log(d**n*(e + f*x)**(m*n)*c)**2*b**2*e + log(d**n*(e + f*x)**(m*n)*c)**2* b**2*f*x + 2*log(d**n*(e + f*x)**(m*n)*c)*a*b*e + 2*log(d**n*(e + f*x)**(m *n)*c)*a*b*f*x - 2*log(d**n*(e + f*x)**(m*n)*c)*b**2*e*m*n - 2*log(d**n*(e + f*x)**(m*n)*c)*b**2*f*m*n*x + a**2*f*x - 2*a*b*f*m*n*x + 2*b**2*f*m**2* n**2*x)/f