\(\int (a+b \log (c (d (e+f x)^m)^n))^2 \, dx\) [406]

Optimal result

Integrand size = 20, antiderivative size = 78 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=-2 a b m n x+2 b^2 m^2 n^2 x-\frac {2 b^2 m n (e+f x) \log \left (c \left (d (e+f x)^m\right )^n\right )}{f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}{f} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2436, 2333, 2332, 2495} \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}{f}-2 a b m n x-\frac {2 b^2 m n (e+f x) \log \left (c \left (d (e+f x)^m\right )^n\right )}{f}+2 b^2 m^2 n^2 x \]

[In]

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

[Out]

-2*a*b*m*n*x + 2*b^2*m^2*n^2*x - (2*b^2*m*n*(e + f*x)*Log[c*(d*(e + f*x)^m)^n])/f + ((e + f*x)*(a + b*Log[c*(d
*(e + f*x)^m)^n])^2)/f

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[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]

Rule 2495

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]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2 \, dx,c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right ) \\ & = \text {Subst}\left (\frac {\text {Subst}\left (\int \left (a+b \log \left (c d^n x^{m n}\right )\right )^2 \, dx,x,e+f x\right )}{f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right ) \\ & = \frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}{f}-\text {Subst}\left (\frac {(2 b m n) \text {Subst}\left (\int \left (a+b \log \left (c d^n x^{m n}\right )\right ) \, dx,x,e+f x\right )}{f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right ) \\ & = -2 a b m n x+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}{f}-\text {Subst}\left (\frac {\left (2 b^2 m n\right ) \text {Subst}\left (\int \log \left (c d^n x^{m n}\right ) \, dx,x,e+f x\right )}{f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right ) \\ & = -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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}{f}-2 b m n \left (a x-b m n x+\frac {b (e+f x) \log \left (c \left (d (e+f x)^m\right )^n\right )}{f}\right ) \]

[In]

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

[Out]

((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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(164\) vs. \(2(78)=156\).

Time = 0.42 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.12

method	result	size
parallelrisch	\(\frac {-2 \ln \left (f x +e \right ) b^{2} e^{2} m^{2} n^{2}+2 x \,b^{2} e f \,m^{2} n^{2}-2 x \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right ) b^{2} e f m n +2 \ln \left (f x +e \right ) a b \,e^{2} m n +x {\ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )}^{2} b^{2} e f -2 x a b e f m n +2 x \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right ) a b e f +{\ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )}^{2} b^{2} e^{2}+e \,a^{2} f x}{e f}\)	\(165\)

[In]

int((a+b*ln(c*(d*(f*x+e)^m)^n))^2,x,method=_RETURNVERBOSE)

[Out]

(-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+e*a^2*f*x)/e/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (78) = 156\).

Time = 0.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.96 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {b^{2} f n^{2} x \log \left (d\right )^{2} + b^{2} f x \log \left (c\right )^{2} + {\left (b^{2} f m^{2} n^{2} x + b^{2} e m^{2} n^{2}\right )} \log \left (f x + e\right )^{2} - 2 \, {\left (b^{2} f m n - a b f\right )} x \log \left (c\right ) + {\left (2 \, b^{2} f m^{2} n^{2} - 2 \, a b f m n + a^{2} f\right )} x - 2 \, {\left (b^{2} e m^{2} n^{2} - a b e m n + {\left (b^{2} f m^{2} n^{2} - a b f m n\right )} x - {\left (b^{2} f m n x + b^{2} e m n\right )} \log \left (c\right ) - {\left (b^{2} f m n^{2} x + b^{2} e m n^{2}\right )} \log \left (d\right )\right )} \log \left (f x + e\right ) + 2 \, {\left (b^{2} f n x \log \left (c\right ) - {\left (b^{2} f m n^{2} - a b f n\right )} x\right )} \log \left (d\right )}{f} \]

[In]

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

[Out]

(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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (76) = 152\).

Time = 0.52 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.28 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\begin {cases} a^{2} x + \frac {2 a b e \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}}{f} - 2 a b m n x + 2 a b x \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )} - \frac {2 b^{2} e m n \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}}{f} + \frac {b^{2} e \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}^{2}}{f} + 2 b^{2} m^{2} n^{2} x - 2 b^{2} m n x \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )} + b^{2} x \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}^{2} & \text {for}\: f \neq 0 \\x \left (a + b \log {\left (c \left (d e^{m}\right )^{n} \right )}\right )^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.90 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=-2 \, a b f m n {\left (\frac {x}{f} - \frac {e \log \left (f x + e\right )}{f^{2}}\right )} + b^{2} x \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right )^{2} + 2 \, a b x \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right ) - {\left (2 \, f m n {\left (\frac {x}{f} - \frac {e \log \left (f x + e\right )}{f^{2}}\right )} \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right ) + \frac {{\left (e \log \left (f x + e\right )^{2} - 2 \, f x + 2 \, e \log \left (f x + e\right )\right )} m^{2} n^{2}}{f}\right )} b^{2} + a^{2} x \]

[In]

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

[Out]

-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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (78) = 156\).

Time = 0.33 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.63 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx=\frac {{\left (f x + e\right )} b^{2} m^{2} n^{2} \log \left (f x + e\right )^{2}}{f} - \frac {2 \, {\left (f x + e\right )} b^{2} m^{2} n^{2} \log \left (f x + e\right )}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} m n^{2} \log \left (f x + e\right ) \log \left (d\right )}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} m^{2} n^{2}}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} m n \log \left (f x + e\right ) \log \left (c\right )}{f} - \frac {2 \, {\left (f x + e\right )} b^{2} m n^{2} \log \left (d\right )}{f} + \frac {{\left (f x + e\right )} b^{2} n^{2} \log \left (d\right )^{2}}{f} + \frac {2 \, {\left (f x + e\right )} a b m n \log \left (f x + e\right )}{f} - \frac {2 \, {\left (f x + e\right )} b^{2} m n \log \left (c\right )}{f} + \frac {2 \, {\left (f x + e\right )} b^{2} n \log \left (c\right ) \log \left (d\right )}{f} - \frac {2 \, {\left (f x + e\right )} a b m n}{f} + \frac {{\left (f x + e\right )} b^{2} \log \left (c\right )^{2}}{f} + \frac {2 \, {\left (f x + e\right )} a b n \log \left (d\right )}{f} + \frac {2 \, {\left (f x + e\right )} a b \log \left (c\right )}{f} + \frac {{\left (f x + e\right )} a^{2}}{f} \]

[In]

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

[Out]

(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

Mupad [B] (verification not implemented)

Time = 1.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.42 \[ \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2 \, dx={\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )}^2\,\left (b^2\,x+\frac {b^2\,e}{f}\right )+x\,\left (a^2-2\,a\,b\,m\,n+2\,b^2\,m^2\,n^2\right )-\frac {\ln \left (e+f\,x\right )\,\left (2\,b^2\,e\,m^2\,n^2-2\,a\,b\,e\,m\,n\right )}{f}+2\,b\,x\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )\,\left (a-b\,m\,n\right ) \]

[In]

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

[Out]

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)