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

Optimal. Leaf size=344 \[ \frac {x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a^2 c}-\frac {i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^3 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^3 c}+\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^3 \sqrt {c}} \]

[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 = 0.33, 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 = {4952, 4930, 217, 206, 4890, 4888, 4181, 2531, 2282, 6589} \[ -\frac {i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^3 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^3 \sqrt {a^2 c x^2+c}}+\frac {x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a^2 c}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^3 c}+\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^3 \sqrt {c}} \]

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 206

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

Rule 217

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 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 4930

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 4952

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 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 {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 {\operatorname {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} \operatorname {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} \operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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} \operatorname {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} \operatorname {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 = 0.36, size = 175, normalized size = 0.51 \[ \frac {\sqrt {a^2 c x^2+c} \left (\tan ^{-1}(a x) \left (a x \tan ^{-1}(a x)-2\right )+\frac {2 \left (\tanh ^{-1}\left (\frac {a x}{\sqrt {a^2 x^2+1}}\right )-i \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+i \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+\text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )-\text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )+i \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2\right )}{\sqrt {a^2 x^2+1}}\right )}{2 a^3 c} \]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2} \arctan \left (a x\right )^{2}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

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

[Out]

Timed out

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maple [A]  time = 2.91, size = 271, normalized size = 0.79 \[ \frac {\left (\arctan \left (a x \right ) x a -2\right ) \arctan \left (a x \right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 c \,a^{3}}-\frac {\left (\arctan \left (a x \right )^{2} \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right )^{2} \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \arctan \left (a x \right ) \polylog \left (2, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \arctan \left (a x \right ) \polylog \left (2, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+4 i \arctan \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 \polylog \left (3, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 \polylog \left (3, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, a^{3} c} \]

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)

[Out]

1/2*(arctan(a*x)*x*a-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 \[ \int \frac {x^{2} \arctan \left (a x\right )^{2}}{\sqrt {a^{2} c x^{2} + c}}\,{d x} \]

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

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

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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