\(\int (A+B \log (e (a+b x)^n (c+d x)^{-n}))^2 \, dx\) [305]

Optimal result

Integrand size = 25, antiderivative size = 137 \[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\frac {2 B (b c-a d) n \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{b}+\frac {2 B^2 (b c-a d) n^2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \]

[Out]

2*B*(-a*d+b*c)*n*ln((-a*d+b*c)/b/(d*x+c))*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/b/d+(b*x+a)*(A+B*ln(e*(b*x+a)^n/((
d*x+c)^n)))^2/b+2*B^2*(-a*d+b*c)*n^2*polylog(2,d*(b*x+a)/b/(d*x+c))/b/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2536, 2542, 2458, 2378, 2370, 2352} \[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\frac {2 B n (b c-a d) \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{b d}+\frac {(a+b x) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{b}+\frac {2 B^2 n^2 (b c-a d) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \]

[In]

Int[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2,x]

[Out]

(2*B*(b*c - a*d)*n*Log[(b*c - a*d)/(b*(c + d*x))]*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n]))/(b*d) + ((a + b*x)
*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2)/b + (2*B^2*(b*c - a*d)*n^2*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))
])/(b*d)

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2370

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

Rule 2378

Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Dist[1/n, Subst[Int[(a
+ b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2536

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.), x_Symbol] :> Simp[
(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Dist[B*n*p*((b*c - a*d)/b), Int[(A + B*Log[e*((
a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] &&
NeQ[b*c - a*d, 0] && IGtQ[p, 0]

Rule 2542

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))/((f_.) + (g_.)*(x_)), x_S
ymbol] :> Simp[(-Log[-(b*c - a*d)/(d*(a + b*x))])*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/g), x] + Dist[B*n*
((b*c - a*d)/g), Int[Log[-(b*c - a*d)/(d*(a + b*x))]/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f,
g, A, B, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{b}-\frac {(2 B (b c-a d) n) \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{c+d x} \, dx}{b} \\ & = \frac {2 B (b c-a d) n \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{b}-\frac {\left (2 B^2 (b c-a d)^2 n^2\right ) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{b d} \\ & = \frac {2 B (b c-a d) n \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{b}-\frac {\left (2 B^2 (b c-a d)^2 n^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {b c-a d}{b x}\right )}{x \left (\frac {-b c+a d}{d}+\frac {b x}{d}\right )} \, dx,x,c+d x\right )}{b d^2} \\ & = \frac {2 B (b c-a d) n \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{b}+\frac {\left (2 B^2 (b c-a d)^2 n^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {(b c-a d) x}{b}\right )}{\left (\frac {-b c+a d}{d}+\frac {b}{d x}\right ) x} \, dx,x,\frac {1}{c+d x}\right )}{b d^2} \\ & = \frac {2 B (b c-a d) n \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{b}+\frac {\left (2 B^2 (b c-a d)^2 n^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {(b c-a d) x}{b}\right )}{\frac {b}{d}+\frac {(-b c+a d) x}{d}} \, dx,x,\frac {1}{c+d x}\right )}{b d^2} \\ & = \frac {2 B (b c-a d) n \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{b}+\frac {2 B^2 (b c-a d) n^2 \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.58 \[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\frac {A^2 b d x-2 A B (b c-a d) n \log (c+d x)+2 A B d (a+b x) \log \left (e (a+b x)^n (c+d x)^{-n}\right )+B^2 d (a+b x) \log ^2\left (e (a+b x)^n (c+d x)^{-n}\right )+B^2 (b c-a d) n \left (-\log \left (\frac {b c-a d}{b c+b d x}\right ) \left (2 n \log \left (\frac {d (a+b x)}{-b c+a d}\right )-2 \log \left (e (a+b x)^n (c+d x)^{-n}\right )+n \log \left (\frac {b c-a d}{b c+b d x}\right )\right )+2 n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )}{b d} \]

[In]

Integrate[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2,x]

[Out]

(A^2*b*d*x - 2*A*B*(b*c - a*d)*n*Log[c + d*x] + 2*A*B*d*(a + b*x)*Log[(e*(a + b*x)^n)/(c + d*x)^n] + B^2*d*(a
+ b*x)*Log[(e*(a + b*x)^n)/(c + d*x)^n]^2 + B^2*(b*c - a*d)*n*(-(Log[(b*c - a*d)/(b*c + b*d*x)]*(2*n*Log[(d*(a
+ b*x))/(-(b*c) + a*d)] - 2*Log[(e*(a + b*x)^n)/(c + d*x)^n] + n*Log[(b*c - a*d)/(b*c + b*d*x)])) + 2*n*PolyL
og[2, (b*(c + d*x))/(b*c - a*d)]))/(b*d)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.85 (sec) , antiderivative size = 2240, normalized size of antiderivative = 16.35

method	result	size
risch	\(\text {Expression too large to display}\)	\(2240\)

[In]

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2,x,method=_RETURNVERBOSE)

[Out]

-2*n^2*B^2*c/d+(-2*x*B^2*ln((b*x+a)^n)-B*(-I*B*Pi*b*d*x*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x
+c)^n)*(b*x+a)^n)+I*B*Pi*b*d*x*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*b*d*x*csgn(I*(b*x+a)^n)*csgn
(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))+I*B*Pi*b*d*x*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I
*B*Pi*b*d*x*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-I*B*Pi*b*d*x*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+I
*B*Pi*b*d*x*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*b*d*x*csgn(I*e/((d*x+c)^n)*
(b*x+a)^n)^3+2*B*ln(e)*b*d*x-2*B*ln(d*x+c)*b*c*n+2*B*a*d*n*ln(b*x+a)+2*A*b*d*x)/b/d)*ln((d*x+c)^n)+B^2/b*ln((b
*x+a)^n)^2*a+x*B^2*ln((d*x+c)^n)^2-2*B^2/b*ln((b*x+a)^n)*a*n+1/4*x*(-I*B*Pi*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c
)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)+I*B*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*csgn(I*(b*x+a)
^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))+I*B*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2
+I*B*Pi*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-I*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+I*B*Pi*csgn
(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3+2*B*ln(e)
+2*A)^2+x*B^2*ln((b*x+a)^n)^2+B*(-I*B*Pi*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^
n)+I*B*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x
+a)^n/((d*x+c)^n))+I*B*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*B*Pi*csgn(I/((d*x+c)^n))*csgn(I*
(b*x+a)^n/((d*x+c)^n))^2-I*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+I*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/
((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3+2*B*ln(e)+2*A)*(ln((b*x+a)^n)*x-b*n*(x/b-a/b
^2*ln(b*x+a)))+2*B^2*a*n^2/b+2*B^2*ln(e)*x*n-I*n*x*B^2*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)
^n/((d*x+c)^n))-n^2*B^2*c/d*ln(d*x+c)^2-2/d*n*B*c*ln(d*x+c)*A+2*n^2*B^2*a/b*ln(b*x+a)*ln((-a*d+c*b+d*(b*x+a))/
(-a*d+b*c))-I*n*x*B^2*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-I*n*x*B^2*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3+I/d*n*
c*ln(d*x+c)*B^2*Pi*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)+I/d*n*c*ln(d*x+c)*B
^2*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))+2/d*n^2*B^2*c*dilog((a*d-c*b+b*(d*x+
c))/(a*d-b*c))+2*n^2*B^2/b*a*ln(a*d-c*b+b*(d*x+c))+2*n^2*B^2*a/b*dilog((-a*d+c*b+d*(b*x+a))/(-a*d+b*c))+I/d*n*
c*ln(d*x+c)*B^2*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+I/d*n*c*ln(d*x+c)*B^2*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-
I*n*x*B^2*Pi*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)+2/d*n^2*B^2*c*ln(d*x+c)*l
n((a*d-c*b+b*(d*x+c))/(a*d-b*c))-2/d*n*B^2*ln((b*x+a)^n)*c*ln(d*x+c)-2/d*n*c*ln(d*x+c)*B^2*ln(e)+I*n*x*B^2*Pi*
csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+I*n*x*B^2*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*n
*x*B^2*Pi*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*n*x*B^2*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(
I*e/((d*x+c)^n)*(b*x+a)^n)^2+2*B*A*x*n-I/d*n*c*ln(d*x+c)*B^2*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I/
d*n*c*ln(d*x+c)*B^2*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-I/d*n*c*ln(d*x+c)*B^2*Pi*csgn(I/((d*x
+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-I/d*n*c*ln(d*x+c)*B^2*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+
c)^n)*(b*x+a)^n)^2

Fricas [F]

\[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2} \,d x } \]

[In]

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

[Out]

integral(B^2*log((b*x + a)^n*e/(d*x + c)^n)^2 + 2*A*B*log((b*x + a)^n*e/(d*x + c)^n) + A^2, x)

Sympy [F(-2)]

Exception generated. \[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))**2,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2} \,d x } \]

[In]

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

[Out]

2*A*B*x*log((b*x + a)^n*e/(d*x + c)^n) + A^2*x + B^2*((2*b*c*n^2*log(b*x + a)*log(d*x + c) - b*c*n^2*log(d*x +
c)^2 + b*d*x*log((b*x + a)^n)^2 + b*d*x*log((d*x + c)^n)^2 + 2*(a*d*n*log(b*x + a) - b*c*n*log(d*x + c) + b*d
*x*log(e))*log((b*x + a)^n) - 2*(a*d*n*log(b*x + a) - b*c*n*log(d*x + c) + b*d*x*log((b*x + a)^n) + b*d*x*log(
e))*log((d*x + c)^n))/(b*d) - integrate(-(b^2*d*x^2*log(e)^2 + a*b*c*log(e)^2 - ((2*n*log(e) - log(e)^2)*b^2*c
- (2*n*log(e) + log(e)^2)*a*b*d)*x - 2*(b^2*c*n^2*x + 2*a*b*c*n^2 - a^2*d*n^2)*log(b*x + a))/(b^2*d*x^2 + a*b
*c + (b^2*c + a*b*d)*x), x)) + 2*(a*e*n*log(b*x + a)/b - c*e*n*log(d*x + c)/d)*A*B/e

Giac [F]

\[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2} \,d x } \]

[In]

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

[Out]

integrate((B*log((b*x + a)^n*e/(d*x + c)^n) + A)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \, dx=\int {\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^2 \,d x \]

[In]

int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^2,x)

[Out]

int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^2, x)