\(\int \frac {(A+B \log (e (\frac {a+b x}{c+d x})^n))^2}{(c i+d i x)^2} \, dx\) [197]

Optimal result

Integrand size = 35, antiderivative size = 163 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx=-\frac {2 A B n (a+b x)}{(b c-a d) i^2 (c+d x)}+\frac {2 B^2 n^2 (a+b x)}{(b c-a d) i^2 (c+d x)}-\frac {2 B^2 n (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(b c-a d) i^2 (c+d x)}+\frac {(a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) i^2 (c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2551, 2333, 2332} \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx=\frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{i^2 (c+d x) (b c-a d)}-\frac {2 A B n (a+b x)}{i^2 (c+d x) (b c-a d)}-\frac {2 B^2 n (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{i^2 (c+d x) (b c-a d)}+\frac {2 B^2 n^2 (a+b x)}{i^2 (c+d x) (b c-a d)} \]

[In]

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

[Out]

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

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 2551

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/d)^m, Subst[Int[(A + B*Log[e*x^n])^p/(b - d*x)^(m + 2), x], x, (
a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[m, p] &&
EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.23 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.03 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx=\frac {-\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2+\frac {B n \left (2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+2 b (c+d x) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 b (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)-2 B n (b c-a d+b (c+d x) \log (a+b x)-b (c+d x) \log (c+d x))-b B n (c+d x) \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+b B n (c+d x) \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{b c-a d}}{d i^2 (c+d x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.80

method	result	size
parallelrisch	\(-\frac {2 B^{2} a b \,d^{3} n^{3}-2 B^{2} b^{2} c \,d^{2} n^{3}+A^{2} a b \,d^{3} n -A^{2} b^{2} c \,d^{2} n +2 A B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} d^{3} n +2 A B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a b \,d^{3} n -2 A B a b \,d^{3} n^{2}+2 A B \,b^{2} c \,d^{2} n^{2}+B^{2} x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} b^{2} d^{3} n -2 B^{2} x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} d^{3} n^{2}+B^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} a b \,d^{3} n -2 B^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a b \,d^{3} n^{2}}{i^{2} \left (d x +c \right ) b \,d^{3} n \left (a d -c b \right )}\)	\(294\)

[In]

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

[Out]

-(2*B^2*a*b*d^3*n^3-2*B^2*b^2*c*d^2*n^3+A^2*a*b*d^3*n-A^2*b^2*c*d^2*n+2*A*B*x*ln(e*((b*x+a)/(d*x+c))^n)*b^2*d^
3*n+2*A*B*ln(e*((b*x+a)/(d*x+c))^n)*a*b*d^3*n-2*A*B*a*b*d^3*n^2+2*A*B*b^2*c*d^2*n^2+B^2*x*ln(e*((b*x+a)/(d*x+c
))^n)^2*b^2*d^3*n-2*B^2*x*ln(e*((b*x+a)/(d*x+c))^n)*b^2*d^3*n^2+B^2*ln(e*((b*x+a)/(d*x+c))^n)^2*a*b*d^3*n-2*B^
2*ln(e*((b*x+a)/(d*x+c))^n)*a*b*d^3*n^2)/i^2/(d*x+c)/b/d^3/n/(a*d-b*c)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.61 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx=-\frac {A^{2} b c - A^{2} a d + 2 \, {\left (B^{2} b c - B^{2} a d\right )} n^{2} + {\left (B^{2} b c - B^{2} a d\right )} \log \left (e\right )^{2} - {\left (B^{2} b d n^{2} x + B^{2} a d n^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} - 2 \, {\left (A B b c - A B a d\right )} n + 2 \, {\left (A B b c - A B a d - {\left (B^{2} b c - B^{2} a d\right )} n - {\left (B^{2} b d n x + B^{2} a d n\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \log \left (e\right ) + 2 \, {\left (B^{2} a d n^{2} - A B a d n + {\left (B^{2} b d n^{2} - A B b d n\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{{\left (b c d^{2} - a d^{3}\right )} i^{2} x + {\left (b c^{2} d - a c d^{2}\right )} i^{2}} \]

[In]

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

[Out]

-(A^2*b*c - A^2*a*d + 2*(B^2*b*c - B^2*a*d)*n^2 + (B^2*b*c - B^2*a*d)*log(e)^2 - (B^2*b*d*n^2*x + B^2*a*d*n^2)
*log((b*x + a)/(d*x + c))^2 - 2*(A*B*b*c - A*B*a*d)*n + 2*(A*B*b*c - A*B*a*d - (B^2*b*c - B^2*a*d)*n - (B^2*b*
d*n*x + B^2*a*d*n)*log((b*x + a)/(d*x + c)))*log(e) + 2*(B^2*a*d*n^2 - A*B*a*d*n + (B^2*b*d*n^2 - A*B*b*d*n)*x
)*log((b*x + a)/(d*x + c)))/((b*c*d^2 - a*d^3)*i^2*x + (b*c^2*d - a*c*d^2)*i^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (163) = 326\).

Time = 0.21 (sec) , antiderivative size = 428, normalized size of antiderivative = 2.63 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx=2 \, A B n {\left (\frac {1}{d^{2} i^{2} x + c d i^{2}} + \frac {b \log \left (b x + a\right )}{{\left (b c d - a d^{2}\right )} i^{2}} - \frac {b \log \left (d x + c\right )}{{\left (b c d - a d^{2}\right )} i^{2}}\right )} + {\left (2 \, n {\left (\frac {1}{d^{2} i^{2} x + c d i^{2}} + \frac {b \log \left (b x + a\right )}{{\left (b c d - a d^{2}\right )} i^{2}} - \frac {b \log \left (d x + c\right )}{{\left (b c d - a d^{2}\right )} i^{2}}\right )} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) - \frac {{\left ({\left (b d x + b c\right )} \log \left (b x + a\right )^{2} + {\left (b d x + b c\right )} \log \left (d x + c\right )^{2} + 2 \, b c - 2 \, a d + 2 \, {\left (b d x + b c\right )} \log \left (b x + a\right ) - 2 \, {\left (b d x + b c + {\left (b d x + b c\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )\right )} n^{2}}{b c^{2} d i^{2} - a c d^{2} i^{2} + {\left (b c d^{2} i^{2} - a d^{3} i^{2}\right )} x}\right )} B^{2} - \frac {B^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )^{2}}{d^{2} i^{2} x + c d i^{2}} - \frac {2 \, A B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{d^{2} i^{2} x + c d i^{2}} - \frac {A^{2}}{d^{2} i^{2} x + c d i^{2}} \]

[In]

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

[Out]

2*A*B*n*(1/(d^2*i^2*x + c*d*i^2) + b*log(b*x + a)/((b*c*d - a*d^2)*i^2) - b*log(d*x + c)/((b*c*d - a*d^2)*i^2)
) + (2*n*(1/(d^2*i^2*x + c*d*i^2) + b*log(b*x + a)/((b*c*d - a*d^2)*i^2) - b*log(d*x + c)/((b*c*d - a*d^2)*i^2
))*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) - ((b*d*x + b*c)*log(b*x + a)^2 + (b*d*x + b*c)*log(d*x + c)^2 + 2*b
*c - 2*a*d + 2*(b*d*x + b*c)*log(b*x + a) - 2*(b*d*x + b*c + (b*d*x + b*c)*log(b*x + a))*log(d*x + c))*n^2/(b*
c^2*d*i^2 - a*c*d^2*i^2 + (b*c*d^2*i^2 - a*d^3*i^2)*x))*B^2 - B^2*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)^2/(d^
2*i^2*x + c*d*i^2) - 2*A*B*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(d^2*i^2*x + c*d*i^2) - A^2/(d^2*i^2*x + c*d
*i^2)

Giac [A] (verification not implemented)

none

Time = 0.99 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.07 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx={\left (\frac {{\left (b x + a\right )} B^{2} n^{2} \log \left (\frac {b x + a}{d x + c}\right )^{2}}{{\left (d x + c\right )} i^{2}} - \frac {2 \, {\left (B^{2} n^{2} - B^{2} n \log \left (e\right ) - A B n\right )} {\left (b x + a\right )} \log \left (\frac {b x + a}{d x + c}\right )}{{\left (d x + c\right )} i^{2}} + \frac {{\left (2 \, B^{2} n^{2} - 2 \, B^{2} n \log \left (e\right ) + B^{2} \log \left (e\right )^{2} - 2 \, A B n + 2 \, A B \log \left (e\right ) + A^{2}\right )} {\left (b x + a\right )}}{{\left (d x + c\right )} i^{2}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

[In]

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

[Out]

((b*x + a)*B^2*n^2*log((b*x + a)/(d*x + c))^2/((d*x + c)*i^2) - 2*(B^2*n^2 - B^2*n*log(e) - A*B*n)*(b*x + a)*l
og((b*x + a)/(d*x + c))/((d*x + c)*i^2) + (2*B^2*n^2 - 2*B^2*n*log(e) + B^2*log(e)^2 - 2*A*B*n + 2*A*B*log(e)
+ A^2)*(b*x + a)/((d*x + c)*i^2))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)

Mupad [B] (verification not implemented)

Time = 1.99 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.45 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx=\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {2\,B^2\,n}{x\,d^2\,i^2+c\,d\,i^2}-\frac {2\,A\,B}{x\,d^2\,i^2+c\,d\,i^2}\right )-{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2\,\left (\frac {B^2}{d\,\left (c\,i^2+d\,i^2\,x\right )}+\frac {B^2\,b}{d\,i^2\,\left (a\,d-b\,c\right )}\right )-\frac {A^2-2\,A\,B\,n+2\,B^2\,n^2}{x\,d^2\,i^2+c\,d\,i^2}+\frac {B\,b\,n\,\mathrm {atan}\left (\frac {\left (2\,b\,d\,x+\frac {a\,d^2\,i^2+b\,c\,d\,i^2}{d\,i^2}\right )\,1{}\mathrm {i}}{a\,d-b\,c}\right )\,\left (A-B\,n\right )\,4{}\mathrm {i}}{d\,i^2\,\left (a\,d-b\,c\right )} \]

[In]

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

[Out]

log(e*((a + b*x)/(c + d*x))^n)*((2*B^2*n)/(d^2*i^2*x + c*d*i^2) - (2*A*B)/(d^2*i^2*x + c*d*i^2)) - log(e*((a +
b*x)/(c + d*x))^n)^2*(B^2/(d*(c*i^2 + d*i^2*x)) + (B^2*b)/(d*i^2*(a*d - b*c))) - (A^2 + 2*B^2*n^2 - 2*A*B*n)/
(d^2*i^2*x + c*d*i^2) + (B*b*n*atan(((2*b*d*x + (a*d^2*i^2 + b*c*d*i^2)/(d*i^2))*1i)/(a*d - b*c))*(A - B*n)*4i
)/(d*i^2*(a*d - b*c))