3.354 \(\int \frac {\tan ^{-1}(a x)^2}{x^2 (c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=381 \[ -\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{c^3 x}+\frac {2 i a \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}-\frac {2 i a \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}+\frac {94 a^2 x}{27 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 a^2 x \tan ^{-1}(a x)^2}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {10 a \tan ^{-1}(a x)}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {4 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}+\frac {2 a^2 x}{27 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 a \tan ^{-1}(a x)}{9 c \left (a^2 c x^2+c\right )^{3/2}} \]

[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

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Rubi [A]  time = 0.69, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4966, 4944, 4958, 4954, 4898, 191, 4900, 192} \[ \frac {2 i a \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}-\frac {2 i a \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}+\frac {94 a^2 x}{27 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 a^2 x \tan ^{-1}(a x)^2}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {10 a \tan ^{-1}(a x)}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{c^3 x}-\frac {4 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {a^2 c x^2+c}}+\frac {2 a^2 x}{27 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 a \tan ^{-1}(a x)}{9 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[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 191

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 192

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 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 4900

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 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 4954

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

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )^{3/2}} \, dx}{c}\\ &=-\frac {2 a \tan ^{-1}(a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {a^2 x \tan ^{-1}(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 {\tan ^{-1}(a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{c^2}-\frac {\left (2 a^2\right ) \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 c}-\frac {a^2 \int \frac {\tan ^{-1}(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 \tan ^{-1}(a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {10 a \tan ^{-1}(a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \tan ^{-1}(a x)^2}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{c^3 x}+\frac {(2 a) \int \frac {\tan ^{-1}(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 \tan ^{-1}(a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {10 a \tan ^{-1}(a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \tan ^{-1}(a x)^2}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{c^3 x}+\frac {\left (2 a \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(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 \tan ^{-1}(a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {10 a \tan ^{-1}(a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \tan ^{-1}(a x)^2}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{c^3 x}-\frac {4 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\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} \text {Li}_2\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} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.70, size = 296, normalized size = 0.78 \[ -\frac {a \left (-216 i \sqrt {a^2 x^2+1} \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )+216 i \sqrt {a^2 x^2+1} \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )+54 \sqrt {a^2 x^2+1} \tan \left (\frac {1}{2} \tan ^{-1}(a x)\right ) \tan ^{-1}(a x)^2-216 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \log \left (1-e^{i \tan ^{-1}(a x)}\right )+216 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \log \left (1+e^{i \tan ^{-1}(a x)}\right )+9 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \sin \left (3 \tan ^{-1}(a x)\right )-2 \sqrt {a^2 x^2+1} \sin \left (3 \tan ^{-1}(a x)\right )+6 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \cos \left (3 \tan ^{-1}(a x)\right )-378 a x+189 a x \tan ^{-1}(a x)^2+378 \tan ^{-1}(a x)+27 a x \tan ^{-1}(a x)^2 \csc ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )}{108 c^2 \sqrt {a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[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])

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}}{a^{6} c^{3} x^{8} + 3 \, a^{4} c^{3} x^{6} + 3 \, a^{2} c^{3} x^{4} + c^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

sage0*x

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maple [A]  time = 0.80, size = 433, normalized size = 1.14 \[ \frac {a \left (6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \left (a^{3} x^{3}-3 i x^{2} a^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{216 \left (a^{2} x^{2}+1\right )^{2} c^{3}}-\frac {7 a \left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} \left (a^{2} x^{2}+1\right )}-\frac {7 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right ) a}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i x^{2} a^{2}-3 a x -i\right ) \left (-6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) a}{216 c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{x \,c^{3}}-\frac {2 i a \left (-i \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+i \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )+\polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*a*(6*I*arctan(a*x)+9*arctan(a*x)^2-2)*(a^3*x^3-3*I*x^2*a^2-3*a*x+I)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1
)^2/c^3-7/8*a*(arctan(a*x)^2-2+2*I*arctan(a*x))*(a*x-I)*(c*(a*x-I)*(I+a*x))^(1/2)/c^3/(a^2*x^2+1)-7/8*(c*(a*x-
I)*(I+a*x))^(1/2)*(I+a*x)*(arctan(a*x)^2-2-2*I*arctan(a*x))*a/c^3/(a^2*x^2+1)+1/216*(c*(a*x-I)*(I+a*x))^(1/2)*
(a^3*x^3+3*I*x^2*a^2-3*a*x-I)*(-6*I*arctan(a*x)+9*arctan(a*x)^2-2)*a/c^3/(a^4*x^4+2*a^2*x^2+1)-arctan(a*x)^2*(
c*(a*x-I)*(I+a*x))^(1/2)/x/c^3-2*I*a*(-I*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)+I*ln(1-(1+I*a*x)/(a^2*x
^2+1)^(1/2))*arctan(a*x)+polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))-polylog(2,-(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^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^2\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{x^{2} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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