\(\int \frac {\arctan (a x)^2}{x^2 (c+a^2 c x^2)^{5/2}} \, dx\) [354]

Optimal result

Integrand size = 24, antiderivative size = 381 \[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {2 a^2 x}{27 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {94 a^2 x}{27 c^2 \sqrt {c+a^2 c x^2}}-\frac {2 a \arctan (a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {10 a \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)^2}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c^3 x}-\frac {4 a \sqrt {1+a^2 x^2} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {c+a^2 c x^2}}+\frac {2 i a \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {c+a^2 c x^2}}-\frac {2 i a \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

2/27*a^2*x/c/(a^2*c*x^2+c)^(3/2)-2/9*a*arctan(a*x)/c/(a^2*c*x^2+c)^(3/2)-1/3*a^2*x*arctan(a*x)^2/c/(a^2*c*x^2+
c)^(3/2)+94/27*a^2*x/c^2/(a^2*c*x^2+c)^(1/2)-10/3*a*arctan(a*x)/c^2/(a^2*c*x^2+c)^(1/2)-5/3*a^2*x*arctan(a*x)^
2/c^2/(a^2*c*x^2+c)^(1/2)-4*a*arctan(a*x)*arctanh((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/c^2/(a^2*
c*x^2+c)^(1/2)+2*I*a*polylog(2,-(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/c^2/(a^2*c*x^2+c)^(1/2)-2*I
*a*polylog(2,(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/c^2/(a^2*c*x^2+c)^(1/2)-arctan(a*x)^2*(a^2*c*x
^2+c)^(1/2)/c^3/x

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5086, 5064, 5078, 5074, 5018, 197, 5020, 198} \[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {4 a \sqrt {a^2 x^2+1} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^2 \sqrt {a^2 c x^2+c}}{c^3 x}-\frac {5 a^2 x \arctan (a x)^2}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {10 a \arctan (a x)}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {a^2 x \arctan (a x)^2}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 a \arctan (a x)}{9 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 i a \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}-\frac {2 i a \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}+\frac {94 a^2 x}{27 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 a^2 x}{27 c \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

Int[ArcTan[a*x]^2/(x^2*(c + a^2*c*x^2)^(5/2)),x]

[Out]

(2*a^2*x)/(27*c*(c + a^2*c*x^2)^(3/2)) + (94*a^2*x)/(27*c^2*Sqrt[c + a^2*c*x^2]) - (2*a*ArcTan[a*x])/(9*c*(c +
a^2*c*x^2)^(3/2)) - (10*a*ArcTan[a*x])/(3*c^2*Sqrt[c + a^2*c*x^2]) - (a^2*x*ArcTan[a*x]^2)/(3*c*(c + a^2*c*x^
2)^(3/2)) - (5*a^2*x*ArcTan[a*x]^2)/(3*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(c^3*x)
- (4*a*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(c^2*Sqrt[c + a^2*c*x^2]) + ((2
*I)*a*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x])])/(c^2*Sqrt[c + a^2*c*x^2]) - ((2*I)*a*S
qrt[1 + a^2*x^2]*PolyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(c^2*Sqrt[c + a^2*c*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 5018

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 5020

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[b*p*(d + e*x^2)^(q +
1)*((a + b*ArcTan[c*x])^(p - 1)/(4*c*d*(q + 1)^2)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q +
1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[b^2*p*((p - 1)/(4*(q + 1)^2)), Int[(d + e*x^2)^q*(a + b*ArcTan[c*x])^(
p - 2), x], x] - Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e
}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && GtQ[p, 1] && NeQ[q, -3/2]

Rule 5064

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 5074

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

Rule 5078

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 5086

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]

Rubi steps \begin{align*} \text {integral}& = -\left (a^2 \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx\right )+\frac {\int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{3/2}} \, dx}{c} \\ & = -\frac {2 a \arctan (a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {a^2 x \arctan (a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {1}{9} \left (2 a^2\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2}} \, dx+\frac {\int \frac {\arctan (a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{c^2}-\frac {\left (2 a^2\right ) \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 c}-\frac {a^2 \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{c} \\ & = \frac {2 a^2 x}{27 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 a \arctan (a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {10 a \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)^2}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c^3 x}+\frac {(2 a) \int \frac {\arctan (a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{c^2}+\frac {\left (4 a^2\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{27 c}+\frac {\left (4 a^2\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 c}+\frac {\left (2 a^2\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{c} \\ & = \frac {2 a^2 x}{27 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {94 a^2 x}{27 c^2 \sqrt {c+a^2 c x^2}}-\frac {2 a \arctan (a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {10 a \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)^2}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c^3 x}+\frac {\left (2 a \sqrt {1+a^2 x^2}\right ) \int \frac {\arctan (a x)}{x \sqrt {1+a^2 x^2}} \, dx}{c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {2 a^2 x}{27 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {94 a^2 x}{27 c^2 \sqrt {c+a^2 c x^2}}-\frac {2 a \arctan (a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {10 a \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)^2}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c^3 x}-\frac {4 a \sqrt {1+a^2 x^2} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {c+a^2 c x^2}}+\frac {2 i a \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {c+a^2 c x^2}}-\frac {2 i a \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 296, normalized size of antiderivative = 0.78 \[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {a \left (-378 a x+378 \arctan (a x)+189 a x \arctan (a x)^2+6 \sqrt {1+a^2 x^2} \arctan (a x) \cos (3 \arctan (a x))+27 a x \arctan (a x)^2 \csc ^2\left (\frac {1}{2} \arctan (a x)\right )-216 \sqrt {1+a^2 x^2} \arctan (a x) \log \left (1-e^{i \arctan (a x)}\right )+216 \sqrt {1+a^2 x^2} \arctan (a x) \log \left (1+e^{i \arctan (a x)}\right )-216 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-e^{i \arctan (a x)}\right )+216 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,e^{i \arctan (a x)}\right )-2 \sqrt {1+a^2 x^2} \sin (3 \arctan (a x))+9 \sqrt {1+a^2 x^2} \arctan (a x)^2 \sin (3 \arctan (a x))+54 \sqrt {1+a^2 x^2} \arctan (a x)^2 \tan \left (\frac {1}{2} \arctan (a x)\right )\right )}{108 c^2 \sqrt {c+a^2 c x^2}} \]

[In]

Integrate[ArcTan[a*x]^2/(x^2*(c + a^2*c*x^2)^(5/2)),x]

[Out]

-1/108*(a*(-378*a*x + 378*ArcTan[a*x] + 189*a*x*ArcTan[a*x]^2 + 6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Cos[3*ArcTan[a
*x]] + 27*a*x*ArcTan[a*x]^2*Csc[ArcTan[a*x]/2]^2 - 216*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])
] + 216*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Log[1 + E^(I*ArcTan[a*x])] - (216*I)*Sqrt[1 + a^2*x^2]*PolyLog[2, -E^(I*
ArcTan[a*x])] + (216*I)*Sqrt[1 + a^2*x^2]*PolyLog[2, E^(I*ArcTan[a*x])] - 2*Sqrt[1 + a^2*x^2]*Sin[3*ArcTan[a*x
]] + 9*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*Sin[3*ArcTan[a*x]] + 54*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*Tan[ArcTan[a*x]
/2]))/(c^2*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.52

method	result	size
default	\(-\frac {\left (54 \arctan \left (a x \right ) \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right ) a^{5} x^{5}-54 \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{5} x^{5}+108 i \operatorname {polylog}\left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{3} x^{3}-54 i \operatorname {polylog}\left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a x +72 \arctan \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a^{4} x^{4}-94 \sqrt {a^{2} x^{2}+1}\, a^{4} x^{4}+90 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+108 \arctan \left (a x \right ) \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right ) a^{3} x^{3}-108 \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{3} x^{3}+54 i \operatorname {polylog}\left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{5} x^{5}-108 i \operatorname {polylog}\left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{3} x^{3}+108 \arctan \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}-96 a^{2} x^{2} \sqrt {a^{2} x^{2}+1}+96 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x +54 \arctan \left (a x \right ) \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right ) a x -54 \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a x -54 i \operatorname {polylog}\left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{5} x^{5}+54 i \operatorname {polylog}\left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a x +27 \arctan \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{27 \sqrt {a^{2} x^{2}+1}\, x \,c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(579\)

[In]

int(arctan(a*x)^2/x^2/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/27*(54*arctan(a*x)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)*a^5*x^5-54*arctan(a*x)*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2
))*a^5*x^5+108*I*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))*a^3*x^3-54*I*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))*a
*x+72*arctan(a*x)^2*(a^2*x^2+1)^(1/2)*a^4*x^4-94*(a^2*x^2+1)^(1/2)*a^4*x^4+90*arctan(a*x)*(a^2*x^2+1)^(1/2)*a^
3*x^3+108*arctan(a*x)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)*a^3*x^3-108*arctan(a*x)*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/
2))*a^3*x^3+54*I*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))*a^5*x^5-108*I*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))*
a^3*x^3+108*arctan(a*x)^2*(a^2*x^2+1)^(1/2)*a^2*x^2-96*a^2*x^2*(a^2*x^2+1)^(1/2)+96*arctan(a*x)*(a^2*x^2+1)^(1
/2)*a*x+54*arctan(a*x)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)*a*x-54*arctan(a*x)*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))*
a*x-54*I*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))*a^5*x^5+54*I*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))*a*x+27*ar
ctan(a*x)^2*(a^2*x^2+1)^(1/2))/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(I+a*x))^(1/2)/x/c^3/(a^4*x^4+2*a^2*x^2+1)

Fricas [F]

\[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2}} \,d x } \]

[In]

integrate(arctan(a*x)^2/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)^2/(a^6*c^3*x^8 + 3*a^4*c^3*x^6 + 3*a^2*c^3*x^4 + c^3*x^2), x)

Sympy [F]

\[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{x^{2} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(atan(a*x)**2/x**2/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(atan(a*x)**2/(x**2*(c*(a**2*x**2 + 1))**(5/2)), x)

Maxima [F]

\[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2}} \,d x } \]

[In]

integrate(arctan(a*x)^2/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(arctan(a*x)^2/((a^2*c*x^2 + c)^(5/2)*x^2), x)

Giac [F]

\[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2}} \,d x } \]

[In]

integrate(arctan(a*x)^2/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^2\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

[In]

int(atan(a*x)^2/(x^2*(c + a^2*c*x^2)^(5/2)),x)

[Out]

int(atan(a*x)^2/(x^2*(c + a^2*c*x^2)^(5/2)), x)