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

Optimal. Leaf size=129 \[ \frac{8 B^2 (b c-a d) \text{PolyLog}\left (2,\frac{d (a+b x)}{b (c+d x)}\right )}{b d}+\frac{4 B (b c-a d) \log \left (\frac{b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b d}+\frac{(a+b x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{b} \]

[Out]

((a + b*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2)/b + (4*B*(b*c - a*d)*(A + B*Log[(e*(a + b*x)^2)/(c + d*
x)^2])*Log[(b*c - a*d)/(b*(c + d*x))])/(b*d) + (8*B^2*(b*c - a*d)*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))])/(b*
d)

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Rubi [A]  time = 0.773903, antiderivative size = 252, normalized size of antiderivative = 1.95, number of steps used = 22, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {2523, 12, 2528, 2524, 2418, 2390, 2301, 2394, 2393, 2391} \[ \frac{8 a B^2 \text{PolyLog}\left (2,-\frac{d (a+b x)}{b c-a d}\right )}{b}+\frac{8 B^2 c \text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )}{d}+\frac{4 a B \log (a+b x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b}-\frac{4 B c \log (c+d x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{d}+x \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2+\frac{8 B^2 c \log (c+d x) \log \left (-\frac{d (a+b x)}{b c-a d}\right )}{d}+\frac{8 a B^2 \log (a+b x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b}-\frac{4 a B^2 \log ^2(a+b x)}{b}-\frac{4 B^2 c \log ^2(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a*B^2*Log[a + b*x]^2)/b + (4*a*B*Log[a + b*x]*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/b + x*(A + B*Log[(
e*(a + b*x)^2)/(c + d*x)^2])^2 + (8*B^2*c*Log[-((d*(a + b*x))/(b*c - a*d))]*Log[c + d*x])/d - (4*B*c*(A + B*Lo
g[(e*(a + b*x)^2)/(c + d*x)^2])*Log[c + d*x])/d - (4*B^2*c*Log[c + d*x]^2)/d + (8*a*B^2*Log[a + b*x]*Log[(b*(c
+ d*x))/(b*c - a*d)])/b + (8*a*B^2*PolyLog[2, -((d*(a + b*x))/(b*c - a*d))])/b + (8*B^2*c*PolyLog[2, (b*(c +
d*x))/(b*c - a*d)])/d

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rule 2524

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

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2390

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2394

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

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2391

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]

Rubi steps

\begin{align*} \int \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx &=x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2-(2 B) \int \frac{2 (b c-a d) x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{(a+b x) (c+d x)} \, dx\\ &=x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2-(4 B (b c-a d)) \int \frac{x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{(a+b x) (c+d x)} \, dx\\ &=x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2-(4 B (b c-a d)) \int \left (-\frac{a \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b c-a d) (a+b x)}+\frac{c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b c-a d) (c+d x)}\right ) \, dx\\ &=x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2+(4 a B) \int \frac{A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )}{a+b x} \, dx-(4 B c) \int \frac{A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )}{c+d x} \, dx\\ &=\frac{4 a B \log (a+b x) \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{b}+x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2-\frac{4 B c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{d}-\frac{\left (4 a B^2\right ) \int \frac{(c+d x)^2 \left (-\frac{2 d e (a+b x)^2}{(c+d x)^3}+\frac{2 b e (a+b x)}{(c+d x)^2}\right ) \log (a+b x)}{e (a+b x)^2} \, dx}{b}+\frac{\left (4 B^2 c\right ) \int \frac{(c+d x)^2 \left (-\frac{2 d e (a+b x)^2}{(c+d x)^3}+\frac{2 b e (a+b x)}{(c+d x)^2}\right ) \log (c+d x)}{e (a+b x)^2} \, dx}{d}\\ &=\frac{4 a B \log (a+b x) \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{b}+x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2-\frac{4 B c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{d}-\frac{\left (4 a B^2\right ) \int \frac{(c+d x)^2 \left (-\frac{2 d e (a+b x)^2}{(c+d x)^3}+\frac{2 b e (a+b x)}{(c+d x)^2}\right ) \log (a+b x)}{(a+b x)^2} \, dx}{b e}+\frac{\left (4 B^2 c\right ) \int \frac{(c+d x)^2 \left (-\frac{2 d e (a+b x)^2}{(c+d x)^3}+\frac{2 b e (a+b x)}{(c+d x)^2}\right ) \log (c+d x)}{(a+b x)^2} \, dx}{d e}\\ &=\frac{4 a B \log (a+b x) \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{b}+x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2-\frac{4 B c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{d}-\frac{\left (4 a B^2\right ) \int \left (\frac{2 b e \log (a+b x)}{a+b x}-\frac{2 d e \log (a+b x)}{c+d x}\right ) \, dx}{b e}+\frac{\left (4 B^2 c\right ) \int \left (\frac{2 b e \log (c+d x)}{a+b x}-\frac{2 d e \log (c+d x)}{c+d x}\right ) \, dx}{d e}\\ &=\frac{4 a B \log (a+b x) \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{b}+x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2-\frac{4 B c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{d}-\left (8 a B^2\right ) \int \frac{\log (a+b x)}{a+b x} \, dx-\left (8 B^2 c\right ) \int \frac{\log (c+d x)}{c+d x} \, dx+\frac{\left (8 b B^2 c\right ) \int \frac{\log (c+d x)}{a+b x} \, dx}{d}+\frac{\left (8 a B^2 d\right ) \int \frac{\log (a+b x)}{c+d x} \, dx}{b}\\ &=\frac{4 a B \log (a+b x) \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{b}+x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2+\frac{8 B^2 c \log \left (-\frac{d (a+b x)}{b c-a d}\right ) \log (c+d x)}{d}-\frac{4 B c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{d}+\frac{8 a B^2 \log (a+b x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b}-\left (8 a B^2\right ) \int \frac{\log \left (\frac{b (c+d x)}{b c-a d}\right )}{a+b x} \, dx-\frac{\left (8 a B^2\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,a+b x\right )}{b}-\left (8 B^2 c\right ) \int \frac{\log \left (\frac{d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx-\frac{\left (8 B^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,c+d x\right )}{d}\\ &=-\frac{4 a B^2 \log ^2(a+b x)}{b}+\frac{4 a B \log (a+b x) \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{b}+x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2+\frac{8 B^2 c \log \left (-\frac{d (a+b x)}{b c-a d}\right ) \log (c+d x)}{d}-\frac{4 B c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{d}-\frac{4 B^2 c \log ^2(c+d x)}{d}+\frac{8 a B^2 \log (a+b x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b}-\frac{\left (8 a B^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{d x}{b c-a d}\right )}{x} \, dx,x,a+b x\right )}{b}-\frac{\left (8 B^2 c\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{d}\\ &=-\frac{4 a B^2 \log ^2(a+b x)}{b}+\frac{4 a B \log (a+b x) \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{b}+x \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )^2+\frac{8 B^2 c \log \left (-\frac{d (a+b x)}{b c-a d}\right ) \log (c+d x)}{d}-\frac{4 B c \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{d}-\frac{4 B^2 c \log ^2(c+d x)}{d}+\frac{8 a B^2 \log (a+b x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b}+\frac{8 a B^2 \text{Li}_2\left (-\frac{d (a+b x)}{b c-a d}\right )}{b}+\frac{8 B^2 c \text{Li}_2\left (\frac{b (c+d x)}{b c-a d}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.162735, size = 220, normalized size = 1.71 \[ \frac{4 B \left (-a B d \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac{b (c+d x)}{b c-a d}\right )\right )-2 \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )\right )+b B c \left (2 \text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )+\log (c+d x) \left (2 \log \left (\frac{d (a+b x)}{a d-b c}\right )-\log (c+d x)\right )\right )+a d \log (a+b x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )-b c \log (c+d x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )\right )}{b d}+x \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2 + (4*B*(a*d*Log[a + b*x]*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])
- b*c*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[c + d*x] - a*B*d*(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*c*((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*d)

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Maple [F]  time = 1.319, size = 0, normalized size = 0. \begin{align*} \int \left ( A+B\ln \left ({\frac{e \left ( bx+a \right ) ^{2}}{ \left ( dx+c \right ) ^{2}}} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \,{\left (x \log \left (\frac{{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + \frac{2 \,{\left (\frac{a e \log \left (b x + a\right )}{b} - \frac{c e \log \left (d x + c\right )}{d}\right )}}{e}\right )} A B + A^{2} x + B^{2}{\left (\frac{4 \,{\left (b d x \log \left (b x + a\right )^{2} +{\left (b d x + b c\right )} \log \left (d x + c\right )^{2} -{\left (b d x \log \left (e\right ) + 2 \,{\left (b d x + a d\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )\right )}}{b d} + \int \frac{{\left (\log \left (e\right )^{2} + 4 \, \log \left (e\right )\right )} b^{2} d x^{2} + a b c \log \left (e\right )^{2} +{\left (b^{2} c \log \left (e\right )^{2} +{\left (\log \left (e\right )^{2} + 4 \, \log \left (e\right )\right )} a b d\right )} x + 4 \,{\left (b^{2} d x^{2} \log \left (e\right ) + a b c \log \left (e\right ) + 2 \, a^{2} d +{\left (a b d{\left (\log \left (e\right ) + 4\right )} + b^{2} c{\left (\log \left (e\right ) - 2\right )}\right )} x\right )} \log \left (b x + a\right )}{b^{2} d x^{2} + a b c +{\left (b^{2} c + a b d\right )} x}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(x*log((b*x + a)^2*e/(d*x + c)^2) + 2*(a*e*log(b*x + a)/b - c*e*log(d*x + c)/d)/e)*A*B + A^2*x + B^2*(4*(b*d
*x*log(b*x + a)^2 + (b*d*x + b*c)*log(d*x + c)^2 - (b*d*x*log(e) + 2*(b*d*x + a*d)*log(b*x + a))*log(d*x + c))
/(b*d) + integrate(((log(e)^2 + 4*log(e))*b^2*d*x^2 + a*b*c*log(e)^2 + (b^2*c*log(e)^2 + (log(e)^2 + 4*log(e))
*a*b*d)*x + 4*(b^2*d*x^2*log(e) + a*b*c*log(e) + 2*a^2*d + (a*b*d*(log(e) + 4) + b^2*c*(log(e) - 2))*x)*log(b*
x + a))/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x), x))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (B^{2} \log \left (\frac{b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} + 2 \, A B \log \left (\frac{b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + A^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \log \left (\frac{{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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