3.1.100 \(\int (a g+b g x) (A+B \log (\frac {e (a+b x)}{c+d x}))^2 \, dx\) [100]

Optimal. Leaf size=180 \[ -\frac {B (b c-a d) g (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b d}+\frac {g (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{2 b}-\frac {B (b c-a d)^2 g \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b d^2}-\frac {B^2 (b c-a d)^2 g \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{b d^2} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2550, 2381, 2384, 2354, 2438} \begin {gather*} -\frac {B^2 g (b c-a d)^2 \text {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d^2}-\frac {B g (b c-a d)^2 \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A+B\right )}{b d^2}-\frac {B g (a+b x) (b c-a d) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b d}+\frac {g (a+b x)^2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.12, size = 203, normalized size = 1.13 \begin {gather*} \frac {g \left ((a+b x)^2 \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 x+2 B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )-2 B (b c-a d) \log (c+d x)-2 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)+B (b c-a d) \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 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{d^2}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (b g x +a g \right ) \left (A +B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (180) = 360\).
time = 0.36, size = 571, normalized size = 3.17 \begin {gather*} \frac {1}{2} \, A^{2} b g x^{2} + 2 \, {\left (x \log \left (\frac {b x e}{d x + c} + \frac {a e}{d x + c}\right ) + \frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} A B a g + {\left (x^{2} \log \left (\frac {b x e}{d x + c} + \frac {a e}{d x + c}\right ) - \frac {a^{2} \log \left (b x + a\right )}{b^{2}} + \frac {c^{2} \log \left (d x + c\right )}{d^{2}} - \frac {{\left (b c - a d\right )} x}{b d}\right )} A B b g + A^{2} a g x + \frac {{\left (2 \, b c^{2} g - 3 \, a c d g\right )} B^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b^{2} c^{2} g - 2 \, a b c d g + a^{2} d^{2} g\right )} {\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} B^{2}}{b d^{2}} + \frac {B^{2} b^{2} d^{2} g x^{2} - 2 \, {\left (b^{2} c d g - 2 \, a b d^{2} g\right )} B^{2} x + {\left (B^{2} b^{2} d^{2} g x^{2} + 2 \, B^{2} a b d^{2} g x + B^{2} a^{2} d^{2} g\right )} \log \left (b x + a\right )^{2} + {\left (B^{2} b^{2} d^{2} g x^{2} + 2 \, B^{2} a b d^{2} g x - {\left (b^{2} c^{2} g - 2 \, a b c d g\right )} B^{2}\right )} \log \left (d x + c\right )^{2} + 2 \, {\left (B^{2} b^{2} d^{2} g x^{2} - {\left (b^{2} c d g - 3 \, a b d^{2} g\right )} B^{2} x - {\left (a b c d g - 2 \, a^{2} d^{2} g\right )} B^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left (B^{2} b^{2} d^{2} g x^{2} - {\left (b^{2} c d g - 3 \, a b d^{2} g\right )} B^{2} x + {\left (B^{2} b^{2} d^{2} g x^{2} + 2 \, B^{2} a b d^{2} g x + B^{2} a^{2} d^{2} g\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, b d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a\,g+b\,g\,x\right )\,{\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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