\(\int (a g+b g x)^2 (A+B \log (\frac {e (a+b x)}{c+d x}))^2 \, dx\) [99]

Optimal result

Integrand size = 32, antiderivative size = 253 \[ \int (a g+b g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=-\frac {B (b c-a d) g^2 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {g^2 (a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 b}+\frac {B (b c-a d)^2 g^2 (a+b x) \left (2 A+B+2 B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d^2}+\frac {B (b c-a d)^3 g^2 \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (2 A+3 B+2 B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d^3}+\frac {2 B^2 (b c-a d)^3 g^2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{3 b d^3} \]

[Out]

-1/3*B*(-a*d+b*c)*g^2*(b*x+a)^2*(A+B*ln(e*(b*x+a)/(d*x+c)))/b/d+1/3*g^2*(b*x+a)^3*(A+B*ln(e*(b*x+a)/(d*x+c)))^
2/b+1/3*B*(-a*d+b*c)^2*g^2*(b*x+a)*(2*A+B+2*B*ln(e*(b*x+a)/(d*x+c)))/b/d^2+1/3*B*(-a*d+b*c)^3*g^2*ln((-a*d+b*c
)/b/(d*x+c))*(2*A+3*B+2*B*ln(e*(b*x+a)/(d*x+c)))/b/d^3+2/3*B^2*(-a*d+b*c)^3*g^2*polylog(2,d*(b*x+a)/b/(d*x+c))
/b/d^3

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2550, 2381, 2384, 2354, 2438} \[ \int (a g+b g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\frac {B g^2 (b c-a d)^3 \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (2 B \log \left (\frac {e (a+b x)}{c+d x}\right )+2 A+3 B\right )}{3 b d^3}+\frac {B g^2 (a+b x) (b c-a d)^2 \left (2 B \log \left (\frac {e (a+b x)}{c+d x}\right )+2 A+B\right )}{3 b d^2}-\frac {B g^2 (a+b x)^2 (b c-a d) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{3 b d}+\frac {g^2 (a+b x)^3 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{3 b}+\frac {2 B^2 g^2 (b c-a d)^3 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{3 b d^3} \]

[In]

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

[Out]

-1/3*(B*(b*c - a*d)*g^2*(a + b*x)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(b*d) + (g^2*(a + b*x)^3*(A + B*Log[
(e*(a + b*x))/(c + d*x)])^2)/(3*b) + (B*(b*c - a*d)^2*g^2*(a + b*x)*(2*A + B + 2*B*Log[(e*(a + b*x))/(c + d*x)
]))/(3*b*d^2) + (B*(b*c - a*d)^3*g^2*Log[(b*c - a*d)/(b*(c + d*x))]*(2*A + 3*B + 2*B*Log[(e*(a + b*x))/(c + d*
x)]))/(3*b*d^3) + (2*B^2*(b*c - a*d)^3*g^2*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))])/(3*b*d^3)

Rule 2354

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

Rule 2381

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_), x_Symbol] :> Simp
[(-(f*x)^(m + 1))*(d + e*x)^(q + 1)*((a + b*Log[c*x^n])^p/(d*f*(q + 1))), x] + Dist[b*n*(p/(d*(q + 1))), Int[(
f*x)^m*(d + e*x)^(q + 1)*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && EqQ[m
+ q + 2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2384

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] - Dist[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 2438

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

Rule 2550

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_)
)^(m_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/b)^m, Subst[Int[x^m*((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] && EqQ[n + mn, 0] && IGtQ[n, 0]
&& NeQ[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rubi steps \begin{align*} \text {integral}& = \left ((b c-a d)^3 g^2\right ) \text {Subst}\left (\int \frac {x^2 (A+B \log (e x))^2}{(b-d x)^4} \, dx,x,\frac {a+b x}{c+d x}\right ) \\ & = \frac {g^2 (a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 b}-\frac {\left (2 B (b c-a d)^3 g^2\right ) \text {Subst}\left (\int \frac {x^2 (A+B \log (e x))}{(b-d x)^3} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b} \\ & = -\frac {B (b c-a d) g^2 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {g^2 (a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 b}+\frac {\left (B (b c-a d)^3 g^2\right ) \text {Subst}\left (\int \frac {x (2 A+B+2 B \log (e x))}{(b-d x)^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b d} \\ & = -\frac {B (b c-a d) g^2 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {g^2 (a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 b}+\frac {B (b c-a d)^2 g^2 (a+b x) \left (2 A+B+2 B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d^2}-\frac {\left (B (b c-a d)^3 g^2\right ) \text {Subst}\left (\int \frac {2 A+3 B+2 B \log (e x)}{b-d x} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b d^2} \\ & = -\frac {B (b c-a d) g^2 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {g^2 (a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 b}+\frac {B (b c-a d)^2 g^2 (a+b x) \left (2 A+B+2 B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d^2}+\frac {B (b c-a d)^3 g^2 \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (2 A+3 B+2 B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d^3}-\frac {\left (2 B^2 (b c-a d)^3 g^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {d x}{b}\right )}{x} \, dx,x,\frac {a+b x}{c+d x}\right )}{3 b d^3} \\ & = -\frac {B (b c-a d) g^2 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d}+\frac {g^2 (a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{3 b}+\frac {B (b c-a d)^2 g^2 (a+b x) \left (2 A+B+2 B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d^2}+\frac {B (b c-a d)^3 g^2 \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (2 A+3 B+2 B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 b d^3}+\frac {2 B^2 (b c-a d)^3 g^2 \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{3 b d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.13 \[ \int (a g+b g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\frac {g^2 \left ((a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2+\frac {B (b c-a d) \left (2 A b d (b c-a d) x+2 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )-d^2 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-2 B (b c-a d)^2 \log (c+d x)-2 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)+B (b c-a d) (b d x+(-b c+a d) \log (c+d x))+B (b c-a d)^2 \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{d^3}\right )}{3 b} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1165 vs. \(2 (242) = 484\).

Time = 0.29 (sec) , antiderivative size = 1165, normalized size of antiderivative = 4.60 \[ \int (a g+b g x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/3*A^2*b^2*g^2*x^3 + A^2*a*b*g^2*x^2 + 2*(x*log(b*e*x/(d*x + c) + a*e/(d*x + c)) + a*log(b*x + a)/b - c*log(d
*x + c)/d)*A*B*a^2*g^2 + 2*(x^2*log(b*e*x/(d*x + c) + a*e/(d*x + c)) - a^2*log(b*x + a)/b^2 + c^2*log(d*x + c)
/d^2 - (b*c - a*d)*x/(b*d))*A*B*a*b*g^2 + 1/3*(2*x^3*log(b*e*x/(d*x + c) + a*e/(d*x + c)) + 2*a^3*log(b*x + a)
/b^3 - 2*c^3*log(d*x + c)/d^3 - ((b^2*c*d - a*b*d^2)*x^2 - 2*(b^2*c^2 - a^2*d^2)*x)/(b^2*d^2))*A*B*b^2*g^2 + A
^2*a^2*g^2*x - 1/3*((2*g^2*log(e) + 3*g^2)*b^2*c^3 - (6*g^2*log(e) + 7*g^2)*a*b*c^2*d + 2*(3*g^2*log(e) + 2*g^
2)*a^2*c*d^2)*B^2*log(d*x + c)/d^3 - 2/3*(b^3*c^3*g^2 - 3*a*b^2*c^2*d*g^2 + 3*a^2*b*c*d^2*g^2 - a^3*d^3*g^2)*(
log(b*x + a)*log((b*d*x + a*d)/(b*c - a*d) + 1) + dilog(-(b*d*x + a*d)/(b*c - a*d)))*B^2/(b*d^3) + 1/3*(B^2*b^
3*d^3*g^2*x^3*log(e)^2 - (b^3*c*d^2*g^2*log(e) - (3*g^2*log(e)^2 + g^2*log(e))*a*b^2*d^3)*B^2*x^2 + ((2*g^2*lo
g(e) + g^2)*b^3*c^2*d - 2*(3*g^2*log(e) + g^2)*a*b^2*c*d^2 + (3*g^2*log(e)^2 + 4*g^2*log(e) + g^2)*a^2*b*d^3)*
B^2*x + (B^2*b^3*d^3*g^2*x^3 + 3*B^2*a*b^2*d^3*g^2*x^2 + 3*B^2*a^2*b*d^3*g^2*x + B^2*a^3*d^3*g^2)*log(b*x + a)
^2 + (B^2*b^3*d^3*g^2*x^3 + 3*B^2*a*b^2*d^3*g^2*x^2 + 3*B^2*a^2*b*d^3*g^2*x + (b^3*c^3*g^2 - 3*a*b^2*c^2*d*g^2
+ 3*a^2*b*c*d^2*g^2)*B^2)*log(d*x + c)^2 + (2*B^2*b^3*d^3*g^2*x^3*log(e) - (b^3*c*d^2*g^2 - (6*g^2*log(e) + g
^2)*a*b^2*d^3)*B^2*x^2 + 2*(b^3*c^2*d*g^2 - 3*a*b^2*c*d^2*g^2 + (3*g^2*log(e) + 2*g^2)*a^2*b*d^3)*B^2*x + (2*a
*b^2*c^2*d*g^2 - 5*a^2*b*c*d^2*g^2 + (2*g^2*log(e) + 3*g^2)*a^3*d^3)*B^2)*log(b*x + a) - (2*B^2*b^3*d^3*g^2*x^
3*log(e) - (b^3*c*d^2*g^2 - (6*g^2*log(e) + g^2)*a*b^2*d^3)*B^2*x^2 + 2*(b^3*c^2*d*g^2 - 3*a*b^2*c*d^2*g^2 + (
3*g^2*log(e) + 2*g^2)*a^2*b*d^3)*B^2*x + 2*(B^2*b^3*d^3*g^2*x^3 + 3*B^2*a*b^2*d^3*g^2*x^2 + 3*B^2*a^2*b*d^3*g^
2*x + B^2*a^3*d^3*g^2)*log(b*x + a))*log(d*x + c))/(b*d^3)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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