3.312 \(\int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x^2} \, dx\)

Optimal. Leaf size=458 \[ \frac {2 i a c \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i a c \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}+\frac {2 i a c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i a c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 a c \sqrt {a^2 x^2+1} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {2 a c \sqrt {a^2 x^2+1} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i a c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{\sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{x}-\frac {4 a c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}} \]

[Out]

-2*I*a*c*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)^2*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-4*a*c*arctan(
a*x)*arctanh((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)+2*I*a*c*arctan(a*x)*polylo
g(2,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-2*I*a*c*arctan(a*x)*polylog(2,I*(1+I
*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)+2*I*a*c*polylog(2,-(1+I*a*x)^(1/2)/(1-I*a*x)^(1
/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-2*I*a*c*polylog(2,(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2
)/(a^2*c*x^2+c)^(1/2)-2*a*c*polylog(3,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)+2*
a*c*polylog(3,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-arctan(a*x)^2*(a^2*c*x^2+c)
^(1/2)/x

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Rubi [A]  time = 0.53, antiderivative size = 458, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4950, 4944, 4958, 4954, 4890, 4888, 4181, 2531, 2282, 6589} \[ \frac {2 i a c \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i a c \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}+\frac {2 i a c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i a c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 a c \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {2 a c \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i a c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{\sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{x}-\frac {4 a c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/x^2,x]

[Out]

-((Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/x) - ((2*I)*a*c*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^
2)/Sqrt[c + a^2*c*x^2] - (4*a*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c
+ a^2*c*x^2] + ((2*I)*a*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^
2] - ((2*I)*a*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] + ((2*I)*a*
c*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x])])/Sqrt[c + a^2*c*x^2] - ((2*I)*a*c*Sqrt[1 +
a^2*x^2]*PolyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (2*a*c*Sqrt[1 + a^2*x^2]*PolyLog[3
, (-I)*E^(I*ArcTan[a*x])])/Sqrt[c + a^2*c*x^2] + (2*a*c*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[a*x])])/Sqr
t[c + a^2*c*x^2]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4888

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

Rule 4890

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

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 4950

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

Rule 4954

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

Rule 4958

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

Rule 6589

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

Rubi steps

\begin {align*} \int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x^2} \, dx &=c \int \frac {\tan ^{-1}(a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx+\left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x}+(2 a c) \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx+\frac {\left (a^2 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x}+\frac {\left (a c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (2 a c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x}-\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}}-\frac {4 a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (2 a c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (2 a c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x}-\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}}-\frac {4 a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (2 i a c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (2 i a c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x}-\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}}-\frac {4 a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (2 a c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (2 a c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x}-\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}}-\frac {4 a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 a c \sqrt {1+a^2 x^2} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 a c \sqrt {1+a^2 x^2} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.80, size = 265, normalized size = 0.58 \[ -\frac {a \sqrt {c \left (a^2 x^2+1\right )} \left (\frac {\sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2}{a x}-2 i \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+2 i \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )-2 i \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )+2 i \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )+2 \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )-2 \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )+\tan ^{-1}(a x)^2 \left (-\log \left (1-i e^{i \tan ^{-1}(a x)}\right )\right )+\tan ^{-1}(a x)^2 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )-2 \tan ^{-1}(a x) \log \left (1-e^{i \tan ^{-1}(a x)}\right )+2 \tan ^{-1}(a x) \log \left (1+e^{i \tan ^{-1}(a x)}\right )\right )}{\sqrt {a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/x^2,x]

[Out]

-((a*Sqrt[c*(1 + a^2*x^2)]*((Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2)/(a*x) - 2*ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])]
- ArcTan[a*x]^2*Log[1 - I*E^(I*ArcTan[a*x])] + ArcTan[a*x]^2*Log[1 + I*E^(I*ArcTan[a*x])] + 2*ArcTan[a*x]*Log
[1 + E^(I*ArcTan[a*x])] - (2*I)*PolyLog[2, -E^(I*ArcTan[a*x])] - (2*I)*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan
[a*x])] + (2*I)*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])] + (2*I)*PolyLog[2, E^(I*ArcTan[a*x])] + 2*PolyLog[
3, (-I)*E^(I*ArcTan[a*x])] - 2*PolyLog[3, I*E^(I*ArcTan[a*x])]))/Sqrt[1 + a^2*x^2])

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/x^2,x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)^2/x^2, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/x^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.94, size = 309, normalized size = 0.67 \[ -\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (a x \right )^{2}}{x}-\frac {a \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (-\arctan \left (a x \right )^{2} \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \arctan \left (a x \right ) \polylog \left (2, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+\arctan \left (a x \right )^{2} \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \arctan \left (a x \right ) \polylog \left (2, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \dilog \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \dilog \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 \polylog \left (3, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 \polylog \left (3, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{\sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/x^2,x)

[Out]

-(c*(a*x-I)*(I+a*x))^(1/2)*arctan(a*x)^2/x-a*(c*(a*x-I)*(I+a*x))^(1/2)*(-arctan(a*x)^2*ln(1-I*(1+I*a*x)/(a^2*x
^2+1)^(1/2))+2*I*arctan(a*x)*polylog(2,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+arctan(a*x)^2*ln(1+I*(1+I*a*x)/(a^2*x^2+
1)^(1/2))-2*I*arctan(a*x)*polylog(2,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+2*arctan(a*x)*ln(1+(1+I*a*x)/(a^2*x^2+1)^(
1/2))-2*I*dilog((1+I*a*x)/(a^2*x^2+1)^(1/2))-2*I*dilog(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))-2*polylog(3,I*(1+I*a*x)/
(a^2*x^2+1)^(1/2))+2*polylog(3,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))/(a^2*x^2+1)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/x^2,x, algorithm="maxima")

[Out]

integrate(sqrt(a^2*c*x^2 + c)*arctan(a*x)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(a*x)^2*(c + a^2*c*x^2)^(1/2))/x^2,x)

[Out]

int((atan(a*x)^2*(c + a^2*c*x^2)^(1/2))/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {atan}^{2}{\left (a x \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)**2*(a**2*c*x**2+c)**(1/2)/x**2,x)

[Out]

Integral(sqrt(c*(a**2*x**2 + 1))*atan(a*x)**2/x**2, x)

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