3.337 \(\int \frac {\tan ^{-1}(a x)^2}{x^3 \sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=328 \[ -\frac {i a^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {i a^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {a^2 \sqrt {a^2 x^2+1} \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {a^2 \sqrt {a^2 x^2+1} \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {a \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 c x^2}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{\sqrt {c}}+\frac {a^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.48, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {4962, 4944, 266, 63, 208, 4958, 4956, 4183, 2531, 2282, 6589} \[ -\frac {i a^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {i a^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {a^2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {a^2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {a \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 c x^2}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{\sqrt {c}}+\frac {a^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]^2/(x^3*Sqrt[c + a^2*c*x^2]),x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 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 4956

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[1/Sqrt[d], Sub
st[Int[(a + b*x)^p*Csc[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 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 4962

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

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 {\tan ^{-1}(a x)^2}{x^3 \sqrt {c+a^2 c x^2}} \, dx &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 c x^2}+a \int \frac {\tan ^{-1}(a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx-\frac {1}{2} a^2 \int \frac {\tan ^{-1}(a x)^2}{x \sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 c x^2}+a^2 \int \frac {1}{x \sqrt {c+a^2 c x^2}} \, dx-\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^2}{x \sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 c x^2}+\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )-\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 c x^2}+\frac {a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c+a^2 c x^2}\right )}{c}+\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 c x^2}+\frac {a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {i a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (i a^2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (i a^2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 c x^2}+\frac {a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {i a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c x}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 c x^2}+\frac {a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {i a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {a^2 \sqrt {1+a^2 x^2} \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {a^2 \sqrt {1+a^2 x^2} \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.23, size = 231, normalized size = 0.70 \[ \frac {a^2 \sqrt {a^2 x^2+1} \left (-8 i \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )+8 i \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )+8 \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )-8 \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )-4 \tan \left (\frac {1}{2} \tan ^{-1}(a x)\right ) \tan ^{-1}(a x)-4 \tan ^{-1}(a x)^2 \log \left (1-e^{i \tan ^{-1}(a x)}\right )+4 \tan ^{-1}(a x)^2 \log \left (1+e^{i \tan ^{-1}(a x)}\right )+8 \log \left (\tan \left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )-4 \tan ^{-1}(a x) \cot \left (\frac {1}{2} \tan ^{-1}(a x)\right )+\tan ^{-1}(a x)^2 \left (-\csc ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )+\tan ^{-1}(a x)^2 \sec ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )}{8 \sqrt {c \left (a^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a*x]^2/(x^3*Sqrt[c + a^2*c*x^2]),x]

[Out]

(a^2*Sqrt[1 + a^2*x^2]*(-4*ArcTan[a*x]*Cot[ArcTan[a*x]/2] - ArcTan[a*x]^2*Csc[ArcTan[a*x]/2]^2 - 4*ArcTan[a*x]
^2*Log[1 - E^(I*ArcTan[a*x])] + 4*ArcTan[a*x]^2*Log[1 + E^(I*ArcTan[a*x])] + 8*Log[Tan[ArcTan[a*x]/2]] - (8*I)
*ArcTan[a*x]*PolyLog[2, -E^(I*ArcTan[a*x])] + (8*I)*ArcTan[a*x]*PolyLog[2, E^(I*ArcTan[a*x])] + 8*PolyLog[3, -
E^(I*ArcTan[a*x])] - 8*PolyLog[3, E^(I*ArcTan[a*x])] + ArcTan[a*x]^2*Sec[ArcTan[a*x]/2]^2 - 4*ArcTan[a*x]*Tan[
ArcTan[a*x]/2]))/(8*Sqrt[c*(1 + a^2*x^2)])

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fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}}{a^{2} c x^{5} + c x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^3/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)^2/(a^2*c*x^5 + c*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^3/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 1.30, size = 261, normalized size = 0.80 \[ -\frac {\left (2 a x +\arctan \left (a x \right )\right ) \arctan \left (a x \right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 x^{2} c}-\frac {a^{2} \left (-2 i \arctan \left (a x \right ) \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \arctan \left (a x \right ) \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+\arctan \left (a x \right )^{2} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right )^{2} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+4 \arctanh \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^2/x^3/(a^2*c*x^2+c)^(1/2),x)

[Out]

-1/2*(2*a*x+arctan(a*x))*arctan(a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/x^2/c-1/2*a^2*(-2*I*arctan(a*x)*polylog(2,(1+I*
a*x)/(a^2*x^2+1)^(1/2))+2*I*arctan(a*x)*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+arctan(a*x)^2*ln(1-(1+I*a*x)/(
a^2*x^2+1)^(1/2))-arctan(a*x)^2*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+2*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))-2*p
olylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+4*arctanh((1+I*a*x)/(a^2*x^2+1)^(1/2)))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*
x^2+1)^(1/2)/c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{2}}{\sqrt {a^{2} c x^{2} + c} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^3/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(arctan(a*x)^2/(sqrt(a^2*c*x^2 + c)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^3\,\sqrt {c\,a^2\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)^2/(x^3*(c + a^2*c*x^2)^(1/2)),x)

[Out]

int(atan(a*x)^2/(x^3*(c + a^2*c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{x^{3} \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)**2/x**3/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(atan(a*x)**2/(x**3*sqrt(c*(a**2*x**2 + 1))), x)

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