3.2.44 \(\int \frac {(a g+b g x)^2 (A+B \log (e (\frac {a+b x}{c+d x})^n))}{(c i+d i x)^2} \, dx\) [144]

3.2.44.1 Optimal result

Integrand size = 43, antiderivative size = 275 \[ \int \frac {(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c i+d i x)^2} \, dx=-\frac {2 B (b c-a d) g^2 n (a+b x)}{d^2 i^2 (c+d x)}+\frac {(b c-a d) g^2 (2 A+B n) (a+b x)}{d^2 i^2 (c+d x)}+\frac {2 B (b c-a d) g^2 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d^2 i^2 (c+d x)}+\frac {g^2 (a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{d i^2 (c+d x)}+\frac {b (b c-a d) g^2 \left (2 A+B n+2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{d^3 i^2}+\frac {2 b B (b c-a d) g^2 n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d^3 i^2} \]

output

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

3.2.44.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.92 \[ \int \frac {(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c i+d i x)^2} \, dx=\frac {g^2 \left (A b^2 d x+\frac {B (b c-a d)^2 n}{c+d x}+b B (b c-a d) n \log (a+b x)+b B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-\frac {(b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{c+d x}-2 b B (b c-a d) n \log (c+d x)-2 b (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)+b B (b c-a d) n \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{d^3 i^2} \]

input

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

output

(g^2*(A*b^2*d*x + (B*(b*c - a*d)^2*n)/(c + d*x) + b*B*(b*c - a*d)*n*Log[a 
+ b*x] + b*B*d*(a + b*x)*Log[e*((a + b*x)/(c + d*x))^n] - ((b*c - a*d)^2*( 
A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(c + d*x) - 2*b*B*(b*c - a*d)*n*Log 
[c + d*x] - 2*b*(b*c - a*d)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[c + 
 d*x] + b*B*(b*c - a*d)*n*((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)])))/(d^3*i^2)
 

3.2.44.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {2961, 2784, 2793, 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.

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

\(\Big \downarrow \) 2961

\(\displaystyle \frac {g^2 (b c-a d) \int \frac {(a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c+d x)^2 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}d\frac {a+b x}{c+d x}}{i^2}\)

\(\Big \downarrow \) 2784

\(\displaystyle \frac {g^2 (b c-a d) \left (\frac {(a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d (c+d x)^2 \left (b-\frac {d (a+b x)}{c+d x}\right )}-\frac {\int \frac {(a+b x) \left (2 A+B n+2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )}d\frac {a+b x}{c+d x}}{d}\right )}{i^2}\)

\(\Big \downarrow \) 2793

\(\displaystyle \frac {g^2 (b c-a d) \left (\frac {(a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d (c+d x)^2 \left (b-\frac {d (a+b x)}{c+d x}\right )}-\frac {\int \left (-\frac {2 A+B n+2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d}-\frac {b \left (2 A+B n+2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{d \left (\frac {d (a+b x)}{c+d x}-b\right )}\right )d\frac {a+b x}{c+d x}}{d}\right )}{i^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {g^2 (b c-a d) \left (\frac {(a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d (c+d x)^2 \left (b-\frac {d (a+b x)}{c+d x}\right )}-\frac {-\frac {b \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right ) \left (2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+2 A+B n\right )}{d^2}-\frac {(a+b x) (2 A+B n)}{d (c+d x)}-\frac {2 b B n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d^2}-\frac {2 B (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d (c+d x)}+\frac {2 B n (a+b x)}{d (c+d x)}}{d}\right )}{i^2}\)

input

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

output

((b*c - a*d)*g^2*(((a + b*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(d* 
(c + d*x)^2*(b - (d*(a + b*x))/(c + d*x))) - ((2*B*n*(a + b*x))/(d*(c + d* 
x)) - ((2*A + B*n)*(a + b*x))/(d*(c + d*x)) - (2*B*(a + b*x)*Log[e*((a + b 
*x)/(c + d*x))^n])/(d*(c + d*x)) - (b*(2*A + B*n + 2*B*Log[e*((a + b*x)/(c 
 + d*x))^n])*Log[1 - (d*(a + b*x))/(b*(c + d*x))])/d^2 - (2*b*B*n*PolyLog[ 
2, (d*(a + b*x))/(b*(c + d*x))])/d^2)/d))/i^2
 

3.2.44.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2784

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* 
(x_))^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] 
)/(e*(q + 1))), x] - Simp[f/(e*(q + 1))   Int[(f*x)^(m - 1)*(d + e*x)^(q + 
1)*(a*m + b*n + b*m*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, 
x] && ILtQ[q, -1] && GtQ[m, 0]
 

rule 2793

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* 
(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], 
 (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, 
 f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer 
Q[r]))
 

rule 2961

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + B*L 
og[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, i, A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]
 

3.2.44.4 Maple [F]

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

input

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

output

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

3.2.44.5 Fricas [F]

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

input

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

output

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

3.2.44.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.2.44.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1273 vs. \(2 (274) = 548\).

Time = 0.48 (sec) , antiderivative size = 1273, normalized size of antiderivative = 4.63 \[ \int \frac {(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c i+d i x)^2} \, dx=\text {Too large to display} \]

input

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

output

B*a^2*g^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)) - A*b^2*(c^2/(d^4*i^2*x + c*d^3* 
i^2) - x/(d^2*i^2) + 2*c*log(d*x + c)/(d^3*i^2))*g^2 + 2*A*a*b*g^2*(c/(d^3 
*i^2*x + c*d^2*i^2) + log(d*x + c)/(d^2*i^2)) - B*a^2*g^2*log(e*(b*x/(d*x 
+ c) + a/(d*x + c))^n)/(d^2*i^2*x + c*d*i^2) - A*a^2*g^2/(d^2*i^2*x + c*d* 
i^2) - (2*a^2*b*d^2*g^2*log(e) + 2*(g^2*n + g^2*log(e))*b^3*c^2 - (3*g^2*n 
 + 4*g^2*log(e))*a*b^2*c*d)*B*log(d*x + c)/(b*c*d^3*i^2 - a*d^4*i^2) + ((b 
^3*c*d^2*g^2*log(e) - a*b^2*d^3*g^2*log(e))*B*x^2 + (b^3*c^2*d*g^2*log(e) 
- a*b^2*c*d^2*g^2*log(e))*B*x + 2*((b^3*c^2*d*g^2*n - 2*a*b^2*c*d^2*g^2*n 
+ a^2*b*d^3*g^2*n)*B*x + (b^3*c^3*g^2*n - 2*a*b^2*c^2*d*g^2*n + a^2*b*c*d^ 
2*g^2*n)*B)*log(b*x + a)*log(d*x + c) - ((b^3*c^2*d*g^2*n - 2*a*b^2*c*d^2* 
g^2*n + a^2*b*d^3*g^2*n)*B*x + (b^3*c^3*g^2*n - 2*a*b^2*c^2*d*g^2*n + a^2* 
b*c*d^2*g^2*n)*B)*log(d*x + c)^2 + ((g^2*n - g^2*log(e))*b^3*c^3 - 3*(g^2* 
n - g^2*log(e))*a*b^2*c^2*d + 2*(g^2*n - g^2*log(e))*a^2*b*c*d^2)*B + ((b^ 
3*c^2*d*g^2*n - a*b^2*c*d^2*g^2*n - a^2*b*d^3*g^2*n)*B*x + (b^3*c^3*g^2*n 
- a*b^2*c^2*d*g^2*n - a^2*b*c*d^2*g^2*n)*B)*log(b*x + a) + ((b^3*c*d^2*g^2 
 - a*b^2*d^3*g^2)*B*x^2 + (b^3*c^2*d*g^2 - a*b^2*c*d^2*g^2)*B*x - (b^3*c^3 
*g^2 - 3*a*b^2*c^2*d*g^2 + 2*a^2*b*c*d^2*g^2)*B - 2*((b^3*c^2*d*g^2 - 2*a* 
b^2*c*d^2*g^2 + a^2*b*d^3*g^2)*B*x + (b^3*c^3*g^2 - 2*a*b^2*c^2*d*g^2 + a^ 
2*b*c*d^2*g^2)*B)*log(d*x + c))*log((b*x + a)^n) - ((b^3*c*d^2*g^2 - a*...
 

3.2.44.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1898 vs. \(2 (274) = 548\).

Time = 221.56 (sec) , antiderivative size = 1898, normalized size of antiderivative = 6.90 \[ \int \frac {(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c i+d i x)^2} \, dx=\text {Too large to display} \]

input

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

output

1/6*(2*(B*b^6*c^4*g^2*n - 4*B*a*b^5*c^3*d*g^2*n - 3*(b*x + a)*B*b^5*c^4*d* 
g^2*n/(d*x + c) + 6*B*a^2*b^4*c^2*d^2*g^2*n + 12*(b*x + a)*B*a*b^4*c^3*d^2 
*g^2*n/(d*x + c) + 3*(b*x + a)^2*B*b^4*c^4*d^2*g^2*n/(d*x + c)^2 - 4*B*a^3 
*b^3*c*d^3*g^2*n - 18*(b*x + a)*B*a^2*b^3*c^2*d^3*g^2*n/(d*x + c) - 12*(b* 
x + a)^2*B*a*b^3*c^3*d^3*g^2*n/(d*x + c)^2 + B*a^4*b^2*d^4*g^2*n + 12*(b*x 
 + a)*B*a^3*b^2*c*d^4*g^2*n/(d*x + c) + 18*(b*x + a)^2*B*a^2*b^2*c^2*d^4*g 
^2*n/(d*x + c)^2 - 3*(b*x + a)*B*a^4*b*d^5*g^2*n/(d*x + c) - 12*(b*x + a)^ 
2*B*a^3*b*c*d^5*g^2*n/(d*x + c)^2 + 3*(b*x + a)^2*B*a^4*d^6*g^2*n/(d*x + c 
)^2)*log((b*x + a)/(d*x + c))/(b^3*d^3*i^2 - 3*(b*x + a)*b^2*d^4*i^2/(d*x 
+ c) + 3*(b*x + a)^2*b*d^5*i^2/(d*x + c)^2 - (b*x + a)^3*d^6*i^2/(d*x + c) 
^3) + (3*B*b^6*c^4*g^2*n - 12*B*a*b^5*c^3*d*g^2*n - 7*(b*x + a)*B*b^5*c^4* 
d*g^2*n/(d*x + c) + 18*B*a^2*b^4*c^2*d^2*g^2*n + 28*(b*x + a)*B*a*b^4*c^3* 
d^2*g^2*n/(d*x + c) + 4*(b*x + a)^2*B*b^4*c^4*d^2*g^2*n/(d*x + c)^2 - 12*B 
*a^3*b^3*c*d^3*g^2*n - 42*(b*x + a)*B*a^2*b^3*c^2*d^3*g^2*n/(d*x + c) - 16 
*(b*x + a)^2*B*a*b^3*c^3*d^3*g^2*n/(d*x + c)^2 + 3*B*a^4*b^2*d^4*g^2*n + 2 
8*(b*x + a)*B*a^3*b^2*c*d^4*g^2*n/(d*x + c) + 24*(b*x + a)^2*B*a^2*b^2*c^2 
*d^4*g^2*n/(d*x + c)^2 - 7*(b*x + a)*B*a^4*b*d^5*g^2*n/(d*x + c) - 16*(b*x 
 + a)^2*B*a^3*b*c*d^5*g^2*n/(d*x + c)^2 + 4*(b*x + a)^2*B*a^4*d^6*g^2*n/(d 
*x + c)^2 + 2*B*b^6*c^4*g^2*log(e) - 8*B*a*b^5*c^3*d*g^2*log(e) - 6*(b*x + 
 a)*B*b^5*c^4*d*g^2*log(e)/(d*x + c) + 12*B*a^2*b^4*c^2*d^2*g^2*log(e) ...
 

3.2.44.9 Mupad [F(-1)]

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

input

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

output

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