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

Optimal. Leaf size=377 \[ -\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{c^2 x}+\frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {6 a \sqrt {a^2 x^2+1} \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}+\frac {6 a \sqrt {a^2 x^2+1} \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}+\frac {6 a}{c \sqrt {a^2 c x^2+c}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {a^2 c x^2+c}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {a^2 c x^2+c}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {a^2 c x^2+c}}-\frac {6 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}} \]

[Out]

6*a/c/(a^2*c*x^2+c)^(1/2)+6*a^2*x*arctan(a*x)/c/(a^2*c*x^2+c)^(1/2)-3*a*arctan(a*x)^2/c/(a^2*c*x^2+c)^(1/2)-a^
2*x*arctan(a*x)^3/c/(a^2*c*x^2+c)^(1/2)-6*a*arctan(a*x)^2*arctanh((1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/
2)/c/(a^2*c*x^2+c)^(1/2)+6*I*a*arctan(a*x)*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*
x^2+c)^(1/2)-6*I*a*arctan(a*x)*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)-
6*a*polylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)+6*a*polylog(3,(1+I*a*x)/(a
^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)-arctan(a*x)^3*(a^2*c*x^2+c)^(1/2)/c^2/x

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Rubi [A]  time = 0.58, antiderivative size = 377, 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 = {4966, 4944, 4958, 4956, 4183, 2531, 2282, 6589, 4898, 4894} \[ \frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {6 a \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}+\frac {6 a \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{c^2 x}+\frac {6 a}{c \sqrt {a^2 c x^2+c}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {a^2 c x^2+c}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {a^2 c x^2+c}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {a^2 c x^2+c}}-\frac {6 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a)/(c*Sqrt[c + a^2*c*x^2]) + (6*a^2*x*ArcTan[a*x])/(c*Sqrt[c + a^2*c*x^2]) - (3*a*ArcTan[a*x]^2)/(c*Sqrt[c
+ a^2*c*x^2]) - (a^2*x*ArcTan[a*x]^3)/(c*Sqrt[c + a^2*c*x^2]) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^3)/(c^2*x) -
(6*a*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*ArcTanh[E^(I*ArcTan[a*x])])/(c*Sqrt[c + a^2*c*x^2]) + ((6*I)*a*Sqrt[1 + a
^2*x^2]*ArcTan[a*x]*PolyLog[2, -E^(I*ArcTan[a*x])])/(c*Sqrt[c + a^2*c*x^2]) - ((6*I)*a*Sqrt[1 + a^2*x^2]*ArcTa
n[a*x]*PolyLog[2, E^(I*ArcTan[a*x])])/(c*Sqrt[c + a^2*c*x^2]) - (6*a*Sqrt[1 + a^2*x^2]*PolyLog[3, -E^(I*ArcTan
[a*x])])/(c*Sqrt[c + a^2*c*x^2]) + (6*a*Sqrt[1 + a^2*x^2]*PolyLog[3, E^(I*ArcTan[a*x])])/(c*Sqrt[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 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 4894

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[b/(c*d*Sqrt[d + e*x^2]),
x] + Simp[(x*(a + b*ArcTan[c*x]))/(d*Sqrt[d + e*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

Rule 4898

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

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 4966

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int[
x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTan[c*
x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && ILtQ[m, 0] &
& NeQ[p, -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 {\tan ^{-1}(a x)^3}{x^2 \left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^3}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c^2 x}+\left (6 a^2\right ) \int \frac {\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx+\frac {(3 a) \int \frac {\tan ^{-1}(a x)^2}{x \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=\frac {6 a}{c \sqrt {c+a^2 c x^2}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c^2 x}+\frac {\left (3 a \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^2}{x \sqrt {1+a^2 x^2}} \, dx}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 a}{c \sqrt {c+a^2 c x^2}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c^2 x}+\frac {\left (3 a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 a}{c \sqrt {c+a^2 c x^2}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c^2 x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {\left (6 a \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 )}{c \sqrt {c+a^2 c x^2}}+\frac {\left (6 a \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 )}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 a}{c \sqrt {c+a^2 c x^2}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c^2 x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}+\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {\left (6 i a \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 )}{c \sqrt {c+a^2 c x^2}}+\frac {\left (6 i a \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 )}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 a}{c \sqrt {c+a^2 c x^2}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c^2 x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}+\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {\left (6 a \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 )}{c \sqrt {c+a^2 c x^2}}+\frac {\left (6 a \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 )}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {6 a}{c \sqrt {c+a^2 c x^2}}+\frac {6 a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {3 a \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^3}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c^2 x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}+\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {6 a \sqrt {1+a^2 x^2} \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}+\frac {6 a \sqrt {1+a^2 x^2} \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.60, size = 301, normalized size = 0.80 \[ \frac {a \left (12 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )-12 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )-12 \sqrt {a^2 x^2+1} \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )+12 \sqrt {a^2 x^2+1} \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )+6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \log \left (1-e^{i \tan ^{-1}(a x)}\right )-6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \log \left (1+e^{i \tan ^{-1}(a x)}\right )-\frac {2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^3 \sin ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )}{a x}-2 a x \tan ^{-1}(a x)^3-6 \tan ^{-1}(a x)^2+12 a x \tan ^{-1}(a x)-\frac {1}{2} a x \tan ^{-1}(a x)^3 \csc ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )+12\right )}{2 c \sqrt {a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(12 + 12*a*x*ArcTan[a*x] - 6*ArcTan[a*x]^2 - 2*a*x*ArcTan[a*x]^3 - (a*x*ArcTan[a*x]^3*Csc[ArcTan[a*x]/2]^2)
/2 + 6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*Log[1 - E^(I*ArcTan[a*x])] - 6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*Log[1 +
E^(I*ArcTan[a*x])] + (12*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, -E^(I*ArcTan[a*x])] - (12*I)*Sqrt[1 + a^2
*x^2]*ArcTan[a*x]*PolyLog[2, E^(I*ArcTan[a*x])] - 12*Sqrt[1 + a^2*x^2]*PolyLog[3, -E^(I*ArcTan[a*x])] + 12*Sqr
t[1 + a^2*x^2]*PolyLog[3, E^(I*ArcTan[a*x])] - (2*(1 + a^2*x^2)*ArcTan[a*x]^3*Sin[ArcTan[a*x]/2]^2)/(a*x)))/(2
*c*Sqrt[c + a^2*c*x^2])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{3}}{a^{4} c^{2} x^{6} + 2 \, a^{2} c^{2} x^{4} + c^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)^3/(a^4*c^2*x^6 + 2*a^2*c^2*x^4 + c^2*x^2), 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)^3/x^2/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.99, size = 356, normalized size = 0.94 \[ -\frac {a \left (\arctan \left (a x \right )^{3}-6 \arctan \left (a x \right )+3 i \arctan \left (a x \right )^{2}-6 i\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{3}-6 \arctan \left (a x \right )-3 i \arctan \left (a x \right )^{2}+6 i\right ) a}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\arctan \left (a x \right )^{3} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{x \,c^{2}}+\frac {3 a \left (\arctan \left (a x \right )^{2} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )-\arctan \left (a x \right )^{2} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )+2 \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(arctan(a*x)^3-6*arctan(a*x)+3*I*arctan(a*x)^2-6*I)*(a*x-I)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)/c^2-1
/2*(c*(a*x-I)*(I+a*x))^(1/2)*(I+a*x)*(arctan(a*x)^3-6*arctan(a*x)-3*I*arctan(a*x)^2+6*I)*a/(a^2*x^2+1)/c^2-arc
tan(a*x)^3*(c*(a*x-I)*(I+a*x))^(1/2)/x/c^2+3*a*(arctan(a*x)^2*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))-2*I*polylog(2,
(1+I*a*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)-arctan(a*x)^2*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+2*I*polylog(2,-(1+I*a
*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)+2*polylog(3,(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)))/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(I+a*x))^(1/2)/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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