3.4.32 \(\int \frac {x^2 \text {ArcTan}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx\) [332]

Optimal. Leaf size=344 \[ -\frac {\sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}{a^3 c}+\frac {x \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2}{2 a^2 c}+\frac {i \sqrt {1+a^2 x^2} \text {ArcTan}\left (e^{i \text {ArcTan}(a x)}\right ) \text {ArcTan}(a x)^2}{a^3 \sqrt {c+a^2 c x^2}}+\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^3 \sqrt {c}}-\frac {i \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \text {PolyLog}\left (3,-i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \text {PolyLog}\left (3,i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {c+a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 344, 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 = {5072, 5050, 223, 212, 5010, 5008, 4266, 2611, 2320, 6724} \begin {gather*} \frac {x \text {ArcTan}(a x)^2 \sqrt {a^2 c x^2+c}}{2 a^2 c}-\frac {i \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {Li}_2\left (-i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {Li}_2\left (i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \text {Li}_3\left (-i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} \text {Li}_3\left (i e^{i \text {ArcTan}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \text {ArcTan}\left (e^{i \text {ArcTan}(a x)}\right ) \text {ArcTan}(a x)^2}{a^3 \sqrt {a^2 c x^2+c}}-\frac {\text {ArcTan}(a x) \sqrt {a^2 c x^2+c}}{a^3 c}+\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^3 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 2320

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 2611

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 4266

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 5008

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 5010

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 5050

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

Rule 5072

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*((a + b*ArcTan[c*x])^p/(c^2*d*m)), x] + (-Dist[b*f*(p/(c*m)), Int[(f*x)^(m - 1
)*((a + b*ArcTan[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] - Dist[f^2*((m - 1)/(c^2*m)), 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] && GtQ[p, 0] && Gt
Q[m, 1]

Rule 6724

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

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Mathematica [A]
time = 0.88, size = 505, normalized size = 1.47 \begin {gather*} \frac {\sqrt {c+a^2 c x^2} \left (\text {ArcTan}(a x) (-2+a x \text {ArcTan}(a x))+\frac {-\text {ArcTan}(a x)^2 \log \left (1-i e^{i \text {ArcTan}(a x)}\right )-\pi \text {ArcTan}(a x) \log \left (\frac {1}{2} \sqrt [4]{-1} e^{-\frac {1}{2} i \text {ArcTan}(a x)} \left (1-i e^{i \text {ArcTan}(a x)}\right )\right )+\text {ArcTan}(a x)^2 \log \left (1+i e^{i \text {ArcTan}(a x)}\right )+\text {ArcTan}(a x)^2 \log \left (\left (\frac {1}{2}+\frac {i}{2}\right ) e^{-\frac {1}{2} i \text {ArcTan}(a x)} \left (-i+e^{i \text {ArcTan}(a x)}\right )\right )-\pi \text {ArcTan}(a x) \log \left (-\frac {1}{2} \sqrt [4]{-1} e^{-\frac {1}{2} i \text {ArcTan}(a x)} \left (-i+e^{i \text {ArcTan}(a x)}\right )\right )-\text {ArcTan}(a x)^2 \log \left (\frac {1}{2} e^{-\frac {1}{2} i \text {ArcTan}(a x)} \left ((1+i)+(1-i) e^{i \text {ArcTan}(a x)}\right )\right )+\pi \text {ArcTan}(a x) \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcTan}(a x))\right )\right )-2 \log \left (\cos \left (\frac {1}{2} \text {ArcTan}(a x)\right )-\sin \left (\frac {1}{2} \text {ArcTan}(a x)\right )\right )-\text {ArcTan}(a x)^2 \log \left (\cos \left (\frac {1}{2} \text {ArcTan}(a x)\right )-\sin \left (\frac {1}{2} \text {ArcTan}(a x)\right )\right )+2 \log \left (\cos \left (\frac {1}{2} \text {ArcTan}(a x)\right )+\sin \left (\frac {1}{2} \text {ArcTan}(a x)\right )\right )+\text {ArcTan}(a x)^2 \log \left (\cos \left (\frac {1}{2} \text {ArcTan}(a x)\right )+\sin \left (\frac {1}{2} \text {ArcTan}(a x)\right )\right )+\pi \text {ArcTan}(a x) \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcTan}(a x))\right )\right )-2 i \text {ArcTan}(a x) \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a x)}\right )+2 i \text {ArcTan}(a x) \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a x)}\right )+2 \text {PolyLog}\left (3,-i e^{i \text {ArcTan}(a x)}\right )-2 \text {PolyLog}\left (3,i e^{i \text {ArcTan}(a x)}\right )}{\sqrt {1+a^2 x^2}}\right )}{2 a^3 c} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c + a^2*c*x^2]*(ArcTan[a*x]*(-2 + a*x*ArcTan[a*x]) + (-(ArcTan[a*x]^2*Log[1 - I*E^(I*ArcTan[a*x])]) - Pi
*ArcTan[a*x]*Log[((-1)^(1/4)*(1 - I*E^(I*ArcTan[a*x])))/(2*E^((I/2)*ArcTan[a*x]))] + ArcTan[a*x]^2*Log[1 + I*E
^(I*ArcTan[a*x])] + ArcTan[a*x]^2*Log[((1/2 + I/2)*(-I + E^(I*ArcTan[a*x])))/E^((I/2)*ArcTan[a*x])] - Pi*ArcTa
n[a*x]*Log[-1/2*((-1)^(1/4)*(-I + E^(I*ArcTan[a*x])))/E^((I/2)*ArcTan[a*x])] - ArcTan[a*x]^2*Log[((1 + I) + (1
- I)*E^(I*ArcTan[a*x]))/(2*E^((I/2)*ArcTan[a*x]))] + Pi*ArcTan[a*x]*Log[-Cos[(Pi + 2*ArcTan[a*x])/4]] - 2*Log
[Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2]] - ArcTan[a*x]^2*Log[Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2]] + 2*Log
[Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2]] + ArcTan[a*x]^2*Log[Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2]] + Pi*Ar
cTan[a*x]*Log[Sin[(Pi + 2*ArcTan[a*x])/4]] - (2*I)*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] + (2*I)*ArcT
an[a*x]*PolyLog[2, I*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]))/(2*a^3*c)

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Maple [A]
time = 1.14, size = 271, normalized size = 0.79

method	result	size
default	\(\frac {\left (\arctan \left (a x \right ) a x -2\right ) \arctan \left (a x \right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 c \,a^{3}}+\frac {\left (\arctan \left (a x \right )^{2} \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right )^{2} \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \arctan \left (a x \right ) \polylog \left (2, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \arctan \left (a x \right ) \polylog \left (2, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-4 i \arctan \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 \polylog \left (3, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 \polylog \left (3, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, a^{3} c}\)	\(271\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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