3.35 \(\int \frac {(a+b \tan ^{-1}(c+d x))^2}{(e+f x)^2} \, dx\)

Optimal. Leaf size=568 \[ \frac {2 a b d \log (e+f x)}{(d e-c f)^2+f^2}-\frac {a b d \log \left ((c+d x)^2+1\right )}{(d e-c f)^2+f^2}+\frac {2 a b d (d e-c f) \tan ^{-1}(c+d x)}{f \left ((d e-c f)^2+f^2\right )}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {i b^2 d \text {Li}_2\left (1-\frac {2}{1-i (c+d x)}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {i b^2 d \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+i f) (1-i (c+d x))}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {i b^2 d \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {i b^2 d \tan ^{-1}(c+d x)^2}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {b^2 d (d e-c f) \tan ^{-1}(c+d x)^2}{f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac {2 b^2 d \log \left (\frac {2}{1-i (c+d x)}\right ) \tan ^{-1}(c+d x)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {2 b^2 d \tan ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {2 b^2 d \log \left (\frac {2}{1+i (c+d x)}\right ) \tan ^{-1}(c+d x)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2} \]

[Out]

2*a*b*d*(-c*f+d*e)*arctan(d*x+c)/f/(f^2+(-c*f+d*e)^2)+I*b^2*d*arctan(d*x+c)^2/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)+
b^2*d*(-c*f+d*e)*arctan(d*x+c)^2/f/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)-(a+b*arctan(d*x+c))^2/f/(f*x+e)+2*a*b*d*ln(
f*x+e)/(f^2+(-c*f+d*e)^2)-2*b^2*d*arctan(d*x+c)*ln(2/(1-I*(d*x+c)))/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)+2*b^2*d*ar
ctan(d*x+c)*ln(2*d*(f*x+e)/(d*e+I*f-c*f)/(1-I*(d*x+c)))/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)+2*b^2*d*arctan(d*x+c)*
ln(2/(1+I*(d*x+c)))/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)-a*b*d*ln(1+(d*x+c)^2)/(f^2+(-c*f+d*e)^2)+I*b^2*d*polylog(2
,1-2/(1-I*(d*x+c)))/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)-I*b^2*d*polylog(2,1-2*d*(f*x+e)/(d*e+I*f-c*f)/(1-I*(d*x+c)
))/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)+I*b^2*d*polylog(2,1-2/(1+I*(d*x+c)))/(d^2*e^2-2*c*d*e*f+(c^2+1)*f^2)

________________________________________________________________________________________

Rubi [A]  time = 1.35, antiderivative size = 568, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 25, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {5045, 1982, 705, 31, 634, 618, 204, 628, 6741, 5057, 706, 635, 203, 260, 6688, 12, 6725, 4856, 2402, 2315, 2447, 4984, 4884, 4920, 4854} \[ \frac {i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {i b^2 d \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(1-i (c+d x)) (d e+(-c+i) f)}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {2 a b d \log (e+f x)}{(d e-c f)^2+f^2}-\frac {a b d \log \left ((c+d x)^2+1\right )}{(d e-c f)^2+f^2}+\frac {2 a b d (d e-c f) \tan ^{-1}(c+d x)}{f \left ((d e-c f)^2+f^2\right )}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {i b^2 d \tan ^{-1}(c+d x)^2}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {b^2 d (d e-c f) \tan ^{-1}(c+d x)^2}{f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac {2 b^2 d \log \left (\frac {2}{1-i (c+d x)}\right ) \tan ^{-1}(c+d x)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {2 b^2 d \tan ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (d e+(-c+i) f)}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {2 b^2 d \log \left (\frac {2}{1+i (c+d x)}\right ) \tan ^{-1}(c+d x)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^2/(e + f*x)^2,x]

[Out]

(2*a*b*d*(d*e - c*f)*ArcTan[c + d*x])/(f*(f^2 + (d*e - c*f)^2)) + (I*b^2*d*ArcTan[c + d*x]^2)/(d^2*e^2 - 2*c*d
*e*f + (1 + c^2)*f^2) + (b^2*d*(d*e - c*f)*ArcTan[c + d*x]^2)/(f*(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2)) - (a +
b*ArcTan[c + d*x])^2/(f*(e + f*x)) + (2*a*b*d*Log[e + f*x])/(f^2 + (d*e - c*f)^2) - (2*b^2*d*ArcTan[c + d*x]*
Log[2/(1 - I*(c + d*x))])/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2) + (2*b^2*d*ArcTan[c + d*x]*Log[(2*d*(e + f*x))
/((d*e + (I - c)*f)*(1 - I*(c + d*x)))])/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2) + (2*b^2*d*ArcTan[c + d*x]*Log[
2/(1 + I*(c + d*x))])/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2) - (a*b*d*Log[1 + (c + d*x)^2])/(f^2 + (d*e - c*f)^
2) + (I*b^2*d*PolyLog[2, 1 - 2/(1 - I*(c + d*x))])/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2) - (I*b^2*d*PolyLog[2,
1 - (2*d*(e + f*x))/((d*e + (I - c)*f)*(1 - I*(c + d*x)))])/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2) + (I*b^2*d*
PolyLog[2, 1 - 2/(1 + I*(c + d*x))])/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 1982

Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^p, x] /; FreeQ[{m, p}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 4984

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> I
nt[ExpandIntegrand[(a + b*ArcTan[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& IGtQ[p, 0] && EqQ[e, c^2*d] && IGtQ[m, 0]

Rule 5045

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

Rule 5057

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_
)^2)^(q_.), x_Symbol] :> Dist[1/d, Subst[Int[((d*e - c*f)/d + (f*x)/d)^m*(C/d^2 + (C*x^2)/d^2)^q*(a + b*ArcTan
[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0]
&& EqQ[2*c*C - B*d, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

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

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Mathematica [A]  time = 7.49, size = 419, normalized size = 0.74 \[ \frac {-\frac {a^2}{f}+\frac {2 a b \left (d (e+f x) \log \left (\frac {d (e+f x)}{\sqrt {(c+d x)^2+1}}\right )-\tan ^{-1}(c+d x) \left (c^2 f-c d e+c d f x-d^2 e x+f\right )\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {b^2 d (e+f x) \left (-\frac {(d e-c f) \left (i \text {Li}_2\left (\exp \left (2 i \left (\tan ^{-1}\left (\frac {d e-c f}{f}\right )+\tan ^{-1}(c+d x)\right )\right )\right )-2 \left (\tan ^{-1}\left (\frac {d e-c f}{f}\right )+\tan ^{-1}(c+d x)\right ) \log \left (1-\exp \left (2 i \left (\tan ^{-1}\left (\frac {d e-c f}{f}\right )+\tan ^{-1}(c+d x)\right )\right )\right )-i \tan ^{-1}(c+d x) \left (\pi -2 \tan ^{-1}\left (\frac {d e-c f}{f}\right )\right )+2 \tan ^{-1}\left (\frac {d e-c f}{f}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac {d e-c f}{f}\right )+\tan ^{-1}(c+d x)\right )\right )+\pi \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )-\pi \log \left (1+e^{-2 i \tan ^{-1}(c+d x)}\right )\right )}{f^2 \left (\frac {(d e-c f)^2}{f^2}+1\right )}-\frac {\tan ^{-1}(c+d x)^2 e^{i \tan ^{-1}\left (\frac {d e-c f}{f}\right )}}{f \sqrt {\frac {(d e-c f)^2}{f^2}+1}}+\frac {(c+d x) \tan ^{-1}(c+d x)^2}{d (e+f x)}\right )}{d e-c f}}{e+f x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c + d*x])^2/(e + f*x)^2,x]

[Out]

(-(a^2/f) + (2*a*b*(-((-(c*d*e) + f + c^2*f - d^2*e*x + c*d*f*x)*ArcTan[c + d*x]) + d*(e + f*x)*Log[(d*(e + f*
x))/Sqrt[1 + (c + d*x)^2]]))/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2) + (b^2*d*(e + f*x)*(-((E^(I*ArcTan[(d*e - c
*f)/f])*ArcTan[c + d*x]^2)/(f*Sqrt[1 + (d*e - c*f)^2/f^2])) + ((c + d*x)*ArcTan[c + d*x]^2)/(d*(e + f*x)) - ((
d*e - c*f)*((-I)*(Pi - 2*ArcTan[(d*e - c*f)/f])*ArcTan[c + d*x] - Pi*Log[1 + E^((-2*I)*ArcTan[c + d*x])] - 2*(
ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x])*Log[1 - E^((2*I)*(ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x]))] + Pi*Log
[1/Sqrt[1 + (c + d*x)^2]] + 2*ArcTan[(d*e - c*f)/f]*Log[Sin[ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x]]] + I*Poly
Log[2, E^((2*I)*(ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x]))]))/(f^2*(1 + (d*e - c*f)^2/f^2))))/(d*e - c*f))/(e
+ f*x)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (d x + c\right )^{2} + 2 \, a b \arctan \left (d x + c\right ) + a^{2}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(f*x+e)^2,x, algorithm="fricas")

[Out]

integral((b^2*arctan(d*x + c)^2 + 2*a*b*arctan(d*x + c) + a^2)/(f^2*x^2 + 2*e*f*x + e^2), x)

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giac [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*arctan(d*x+c))^2/(f*x+e)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.15, size = 1087, normalized size = 1.91 \[ -\frac {d \,a^{2}}{\left (d f x +d e \right ) f}-\frac {d \,b^{2} \arctan \left (d x +c \right )^{2}}{\left (d f x +d e \right ) f}+\frac {2 d \,b^{2} \arctan \left (d x +c \right ) \ln \left (f \left (d x +c \right )-c f +d e \right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {d \,b^{2} \arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {d \,b^{2} \arctan \left (d x +c \right )^{2} c}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}+\frac {d^{2} b^{2} \arctan \left (d x +c \right )^{2} e}{f \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )}+\frac {i d \,b^{2} \ln \left (d x +c -i\right )^{2}}{4 c^{2} f^{2}-8 c d e f +4 d^{2} e^{2}+4 f^{2}}-\frac {i d \,b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )}+\frac {i d \,b^{2} \ln \left (f \left (d x +c \right )-c f +d e \right ) \ln \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {i d \,b^{2} \ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )}-\frac {i d \,b^{2} \dilog \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}+\frac {i d \,b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{2 c^{2} f^{2}-4 c d e f +2 d^{2} e^{2}+2 f^{2}}+\frac {i d \,b^{2} \ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{2 c^{2} f^{2}-4 c d e f +2 d^{2} e^{2}+2 f^{2}}-\frac {i d \,b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right )}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )}-\frac {i d \,b^{2} \ln \left (f \left (d x +c \right )-c f +d e \right ) \ln \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {i d \,b^{2} \ln \left (d x +c +i\right )^{2}}{4 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )}+\frac {i d \,b^{2} \ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 c^{2} f^{2}-4 c d e f +2 d^{2} e^{2}+2 f^{2}}+\frac {i d \,b^{2} \dilog \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {2 d a b \arctan \left (d x +c \right )}{\left (d f x +d e \right ) f}+\frac {2 d a b \ln \left (f \left (d x +c \right )-c f +d e \right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {d a b \ln \left (1+\left (d x +c \right )^{2}\right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {2 d a b \arctan \left (d x +c \right ) c}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}+\frac {2 d^{2} a b \arctan \left (d x +c \right ) e}{f \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^2/(f*x+e)^2,x)

[Out]

-d*a^2/(d*f*x+d*e)/f-d*b^2/(d*f*x+d*e)/f*arctan(d*x+c)^2+2*d*b^2*arctan(d*x+c)/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)
*ln(f*(d*x+c)-c*f+d*e)-d*b^2*arctan(d*x+c)/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(1+(d*x+c)^2)-d*b^2/(c^2*f^2-2*c*
d*e*f+d^2*e^2+f^2)*arctan(d*x+c)^2*c+d^2*b^2/f/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*arctan(d*x+c)^2*e+1/4*I*d*b^2/(
c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(d*x+c-I)^2-1/2*I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*dilog(1/2*I*(d*x+c-I)
)-I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*dilog((I*f+f*(d*x+c))/(I*f+c*f-d*e))+1/2*I*d*b^2/(c^2*f^2-2*c*d*e*f+
d^2*e^2+f^2)*dilog(-1/2*I*(I+d*x+c))+I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*dilog((I*f-f*(d*x+c))/(d*e+I*f-c*
f))-1/2*I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(I+d*x+c)*ln(1/2*I*(d*x+c-I))+I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2
*e^2+f^2)*ln(f*(d*x+c)-c*f+d*e)*ln((I*f-f*(d*x+c))/(d*e+I*f-c*f))-1/2*I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*
ln(d*x+c-I)*ln(1+(d*x+c)^2)+1/2*I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(d*x+c-I)*ln(-1/2*I*(I+d*x+c))-I*d*b
^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(f*(d*x+c)-c*f+d*e)*ln((I*f+f*(d*x+c))/(I*f+c*f-d*e))-1/4*I*d*b^2/(c^2*f^
2-2*c*d*e*f+d^2*e^2+f^2)*ln(I+d*x+c)^2+1/2*I*d*b^2/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(I+d*x+c)*ln(1+(d*x+c)^2)
-2*d*a*b/(d*f*x+d*e)/f*arctan(d*x+c)+2*d*a*b/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(f*(d*x+c)-c*f+d*e)-d*a*b/(c^2*
f^2-2*c*d*e*f+d^2*e^2+f^2)*ln(1+(d*x+c)^2)-2*d*a*b/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*arctan(d*x+c)*c+2*d^2*a*b/f
/(c^2*f^2-2*c*d*e*f+d^2*e^2+f^2)*arctan(d*x+c)*e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (d {\left (\frac {2 \, {\left (d^{2} e - c d f\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{{\left (d^{2} e^{2} f - 2 \, c d e f^{2} + {\left (c^{2} + 1\right )} f^{3}\right )} d} - \frac {\log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} + 1\right )} f^{2}} + \frac {2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} + 1\right )} f^{2}}\right )} - \frac {2 \, \arctan \left (d x + c\right )}{f^{2} x + e f}\right )} a b - \frac {\frac {1}{4} \, {\left (28 \, \arctan \left (d x + c\right )^{2} - 4 \, {\left (f^{2} x + e f\right )} \int \frac {36 \, {\left (d^{2} f x^{2} + 2 \, c d f x + {\left (c^{2} + 1\right )} f\right )} \arctan \left (d x + c\right )^{2} + 3 \, {\left (d^{2} f x^{2} + 2 \, c d f x + {\left (c^{2} + 1\right )} f\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2} + 56 \, {\left (d f x + d e\right )} \arctan \left (d x + c\right ) - 12 \, {\left (d^{2} f x^{2} + c d e + {\left (d^{2} e + c d f\right )} x\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{4 \, {\left (d^{2} f^{3} x^{4} + {\left (c^{2} + 1\right )} e^{2} f + 2 \, {\left (d^{2} e f^{2} + c d f^{3}\right )} x^{3} + {\left (d^{2} e^{2} f + 4 \, c d e f^{2} + {\left (c^{2} + 1\right )} f^{3}\right )} x^{2} + 2 \, {\left (c d e^{2} f + {\left (c^{2} + 1\right )} e f^{2}\right )} x\right )}}\,{d x} - 3 \, \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2}\right )} b^{2}}{16 \, {\left (f^{2} x + e f\right )}} - \frac {a^{2}}{f^{2} x + e f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(f*x+e)^2,x, algorithm="maxima")

[Out]

(d*(2*(d^2*e - c*d*f)*arctan((d^2*x + c*d)/d)/((d^2*e^2*f - 2*c*d*e*f^2 + (c^2 + 1)*f^3)*d) - log(d^2*x^2 + 2*
c*d*x + c^2 + 1)/(d^2*e^2 - 2*c*d*e*f + (c^2 + 1)*f^2) + 2*log(f*x + e)/(d^2*e^2 - 2*c*d*e*f + (c^2 + 1)*f^2))
- 2*arctan(d*x + c)/(f^2*x + e*f))*a*b - 1/16*(4*arctan(d*x + c)^2 - 16*(f^2*x + e*f)*integrate(1/16*(12*(d^2
*f*x^2 + 2*c*d*f*x + (c^2 + 1)*f)*arctan(d*x + c)^2 + (d^2*f*x^2 + 2*c*d*f*x + (c^2 + 1)*f)*log(d^2*x^2 + 2*c*
d*x + c^2 + 1)^2 + 8*(d*f*x + d*e)*arctan(d*x + c) - 4*(d^2*f*x^2 + c*d*e + (d^2*e + c*d*f)*x)*log(d^2*x^2 + 2
*c*d*x + c^2 + 1))/(d^2*f^3*x^4 + (c^2 + 1)*e^2*f + 2*(d^2*e*f^2 + c*d*f^3)*x^3 + (d^2*e^2*f + 4*c*d*e*f^2 + (
c^2 + 1)*f^3)*x^2 + 2*(c*d*e^2*f + (c^2 + 1)*e*f^2)*x), x) - log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2)*b^2/(f^2*x +
e*f) - a^2/(f^2*x + e*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^2/(e + f*x)^2,x)

[Out]

int((a + b*atan(c + d*x))^2/(e + f*x)^2, 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*atan(d*x+c))**2/(f*x+e)**2,x)

[Out]

Timed out

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