3.278 \(\int \frac {(A+B \log (\frac {e (a+b x)^2}{(c+d x)^2}))^2}{(f+g x)^3} \, dx\)

Optimal. Leaf size=381 \[ \frac {b^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{2 g (b f-a g)^2}+\frac {2 B g (a+b x) (b c-a d) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{(f+g x) (b f-a g)^2 (d f-c g)}+\frac {2 B (b c-a d) (-a d g-b c g+2 b d f) \log \left (1-\frac {(a+b x) (d f-c g)}{(c+d x) (b f-a g)}\right ) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{(b f-a g)^2 (d f-c g)^2}-\frac {\left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{2 g (f+g x)^2}+\frac {4 B^2 (b c-a d) (-a d g-b c g+2 b d f) \text {Li}_2\left (\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g)^2 (d f-c g)^2}+\frac {4 B^2 g (b c-a d)^2 \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g)^2 (d f-c g)^2} \]

[Out]

2*B*(-a*d+b*c)*g*(b*x+a)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/(-a*g+b*f)^2/(-c*g+d*f)/(g*x+f)+1/2*b^2*(A+B*ln(e*(b*
x+a)^2/(d*x+c)^2))^2/g/(-a*g+b*f)^2-1/2*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2/g/(g*x+f)^2+4*B^2*(-a*d+b*c)^2*g*ln(
(g*x+f)/(d*x+c))/(-a*g+b*f)^2/(-c*g+d*f)^2+2*B*(-a*d+b*c)*(-a*d*g-b*c*g+2*b*d*f)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2
))*ln(1-(-c*g+d*f)*(b*x+a)/(-a*g+b*f)/(d*x+c))/(-a*g+b*f)^2/(-c*g+d*f)^2+4*B^2*(-a*d+b*c)*(-a*d*g-b*c*g+2*b*d*
f)*polylog(2,(-c*g+d*f)*(b*x+a)/(-a*g+b*f)/(d*x+c))/(-a*g+b*f)^2/(-c*g+d*f)^2

________________________________________________________________________________________

Rubi [B]  time = 1.64, antiderivative size = 899, normalized size of antiderivative = 2.36, number of steps used = 36, number of rules used = 11, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {2525, 12, 2528, 2524, 2418, 2390, 2301, 2394, 2393, 2391, 72} \[ -\frac {2 B^2 \log ^2(a+b x) b^2}{g (b f-a g)^2}+\frac {2 B \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) b^2}{g (b f-a g)^2}+\frac {4 B^2 \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right ) b^2}{g (b f-a g)^2}+\frac {4 B^2 \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right ) b^2}{g (b f-a g)^2}+\frac {4 B^2 (b c-a d) \log (a+b x) b}{(b f-a g)^2 (d f-c g)}-\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{2 g (f+g x)^2}-\frac {2 B^2 d^2 \log ^2(c+d x)}{g (d f-c g)^2}-\frac {2 B (b c-a d) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g) (d f-c g) (f+g x)}+\frac {4 B^2 d^2 \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{g (d f-c g)^2}-\frac {2 B d^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)}{g (d f-c g)^2}-\frac {4 B^2 d (b c-a d) \log (c+d x)}{(b f-a g) (d f-c g)^2}-\frac {4 B^2 (b c-a d) (2 b d f-b c g-a d g) \log \left (-\frac {g (a+b x)}{b f-a g}\right ) \log (f+g x)}{(b f-a g)^2 (d f-c g)^2}+\frac {2 B (b c-a d) (2 b d f-b c g-a d g) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (f+g x)}{(b f-a g)^2 (d f-c g)^2}+\frac {4 B^2 (b c-a d) (2 b d f-b c g-a d g) \log \left (-\frac {g (c+d x)}{d f-c g}\right ) \log (f+g x)}{(b f-a g)^2 (d f-c g)^2}+\frac {4 B^2 (b c-a d)^2 g \log (f+g x)}{(b f-a g)^2 (d f-c g)^2}+\frac {4 B^2 d^2 \text {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{g (d f-c g)^2}-\frac {4 B^2 (b c-a d) (2 b d f-b c g-a d g) \text {PolyLog}\left (2,\frac {b (f+g x)}{b f-a g}\right )}{(b f-a g)^2 (d f-c g)^2}+\frac {4 B^2 (b c-a d) (2 b d f-b c g-a d g) \text {PolyLog}\left (2,\frac {d (f+g x)}{d f-c g}\right )}{(b f-a g)^2 (d f-c g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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 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 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]

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 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 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 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 2525

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

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]

Rubi steps

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

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Mathematica [A]  time = 1.58, size = 603, normalized size = 1.58 \[ -\frac {\frac {4 B (f+g x) \left (-b^2 (f+g x) \log (a+b x) (d f-c g)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )+d^2 (f+g x) (b f-a g)^2 \log (c+d x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )+g (b c-a d) (b f-a g) (d f-c g) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )+g (f+g x) (b c-a d) \log (f+g x) (a d g+b c g-2 b d f) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )+b^2 B (f+g x) (d f-c g)^2 \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )\right )-B d^2 (f+g x) (b f-a g)^2 \left (2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )+\log (c+d x) \left (2 \log \left (\frac {d (a+b x)}{a d-b c}\right )-\log (c+d x)\right )\right )-2 B g (f+g x) (b c-a d) (a d g+b c g-2 b d f) \left (\log (f+g x) \left (\log \left (\frac {g (a+b x)}{a g-b f}\right )-\log \left (\frac {g (c+d x)}{c g-d f}\right )\right )+\text {Li}_2\left (\frac {b (f+g x)}{b f-a g}\right )-\text {Li}_2\left (\frac {d (f+g x)}{d f-c g}\right )\right )-2 B g (f+g x) (b c-a d) (b \log (a+b x) (d f-c g)+\log (c+d x) (a d g-b d f)+g (b c-a d) \log (f+g x))\right )}{(b f-a g)^2 (d f-c g)^2}+\left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{2 g (f+g x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2 + (4*B*(f + g*x)*((b*c - a*d)*g*(b*f - a*g)*(d*f - c*g)*(A +
B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) - b^2*(d*f - c*g)^2*(f + g*x)*Log[a + b*x]*(A + B*Log[(e*(a + b*x)^2)/(c +
d*x)^2]) + d^2*(b*f - a*g)^2*(f + g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[c + d*x] + (b*c - a*d)*g*
(-2*b*d*f + b*c*g + a*d*g)*(f + g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[f + g*x] - 2*B*(b*c - a*d)*g
*(f + g*x)*(b*(d*f - c*g)*Log[a + b*x] + (-(b*d*f) + a*d*g)*Log[c + d*x] + (b*c - a*d)*g*Log[f + g*x]) + b^2*B
*(d*f - c*g)^2*(f + g*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*d^2*(b*f - a*g)^2*(f + g*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)]) - 2*B*(b*c - a*d)*g*(-2*b*d*f + b*c*g + a*d*g)*(f + g
*x)*((Log[(g*(a + b*x))/(-(b*f) + a*g)] - Log[(g*(c + d*x))/(-(d*f) + c*g)])*Log[f + g*x] + PolyLog[2, (b*(f +
g*x))/(b*f - a*g)] - PolyLog[2, (d*(f + g*x))/(d*f - c*g)])))/((b*f - a*g)^2*(d*f - c*g)^2))/(g*(f + g*x)^2)

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B^{2} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} + 2 \, A B \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + A^{2}}{g^{3} x^{3} + 3 \, f g^{2} x^{2} + 3 \, f^{2} g x + f^{3}}, x\right ) \]

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/(g*x+f)^3,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)/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{3}}\,{d x} \]

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/(g*x+f)^3,x, algorithm="giac")

[Out]

integrate((B*log((b*x + a)^2*e/(d*x + c)^2) + A)^2/(g*x + f)^3, x)

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maple [F]  time = 1.75, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \ln \left (\frac {\left (b x +a \right )^{2} e}{\left (d x +c \right )^{2}}\right )+A \right )^{2}}{\left (g x +f \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*ln((b*x+a)^2/(d*x+c)^2*e)+A)^2/(g*x+f)^3,x)

[Out]

int((B*ln((b*x+a)^2/(d*x+c)^2*e)+A)^2/(g*x+f)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (\frac {2 \, b^{2} \log \left (b x + a\right )}{b^{2} f^{2} g - 2 \, a b f g^{2} + a^{2} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{d^{2} f^{2} g - 2 \, c d f g^{2} + c^{2} g^{3}} + \frac {2 \, {\left (2 \, {\left (b^{2} c d - a b d^{2}\right )} f - {\left (b^{2} c^{2} - a^{2} d^{2}\right )} g\right )} \log \left (g x + f\right )}{b^{2} d^{2} f^{4} + a^{2} c^{2} g^{4} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{3} g + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{2} g^{2} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f g^{3}} - \frac {2 \, {\left (b c - a d\right )}}{b d f^{3} + a c f g^{2} - {\left (b c + a d\right )} f^{2} g + {\left (b d f^{2} g + a c g^{3} - {\left (b c + a d\right )} f g^{2}\right )} x} - \frac {\log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g}\right )} A B - B^{2} {\left (\frac {2 \, \log \left (d x + c\right )^{2}}{g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g} + \int -\frac {d g x \log \relax (e)^{2} + c g \log \relax (e)^{2} + 4 \, {\left (d g x + c g\right )} \log \left (b x + a\right )^{2} + 4 \, {\left (d g x \log \relax (e) + c g \log \relax (e)\right )} \log \left (b x + a\right ) - 4 \, {\left ({\left (g \log \relax (e) - g\right )} d x + c g \log \relax (e) - d f + 2 \, {\left (d g x + c g\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{d g^{4} x^{4} + c f^{3} g + {\left (3 \, d f g^{3} + c g^{4}\right )} x^{3} + 3 \, {\left (d f^{2} g^{2} + c f g^{3}\right )} x^{2} + {\left (d f^{3} g + 3 \, c f^{2} g^{2}\right )} x}\,{d x}\right )} - \frac {A^{2}}{2 \, {\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} \]

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/(g*x+f)^3,x, algorithm="maxima")

[Out]

(2*b^2*log(b*x + a)/(b^2*f^2*g - 2*a*b*f*g^2 + a^2*g^3) - 2*d^2*log(d*x + c)/(d^2*f^2*g - 2*c*d*f*g^2 + c^2*g^
3) + 2*(2*(b^2*c*d - a*b*d^2)*f - (b^2*c^2 - a^2*d^2)*g)*log(g*x + f)/(b^2*d^2*f^4 + a^2*c^2*g^4 - 2*(b^2*c*d
+ a*b*d^2)*f^3*g + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*f^2*g^2 - 2*(a*b*c^2 + a^2*c*d)*f*g^3) - 2*(b*c - a*d)/(b*d
*f^3 + a*c*f*g^2 - (b*c + a*d)*f^2*g + (b*d*f^2*g + a*c*g^3 - (b*c + a*d)*f*g^2)*x) - log(b^2*e*x^2/(d^2*x^2 +
2*c*d*x + c^2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) + a^2*e/(d^2*x^2 + 2*c*d*x + c^2))/(g^3*x^2 + 2*f*g^2*x
+ f^2*g))*A*B - B^2*(2*log(d*x + c)^2/(g^3*x^2 + 2*f*g^2*x + f^2*g) + integrate(-(d*g*x*log(e)^2 + c*g*log(e)^
2 + 4*(d*g*x + c*g)*log(b*x + a)^2 + 4*(d*g*x*log(e) + c*g*log(e))*log(b*x + a) - 4*((g*log(e) - g)*d*x + c*g*
log(e) - d*f + 2*(d*g*x + c*g)*log(b*x + a))*log(d*x + c))/(d*g^4*x^4 + c*f^3*g + (3*d*f*g^3 + c*g^4)*x^3 + 3*
(d*f^2*g^2 + c*f*g^3)*x^2 + (d*f^3*g + 3*c*f^2*g^2)*x), x)) - 1/2*A^2/(g^3*x^2 + 2*f*g^2*x + f^2*g)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\right )}^2}{{\left (f+g\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log((e*(a + b*x)^2)/(c + d*x)^2))^2/(f + g*x)^3,x)

[Out]

int((A + B*log((e*(a + b*x)^2)/(c + d*x)^2))^2/(f + g*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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/(g*x+f)**3,x)

[Out]

Timed out

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