3.346 \(\int \frac {\tan ^{-1}(a x)^2}{x^4 (c+a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=397 \[ -\frac {a^2 \sqrt {a^2 c x^2+c}}{3 c^2 x}+\frac {5 a^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 c^2 x}-\frac {a \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 c^2 x^2}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 c^2 x^3}-\frac {2 a^4 x}{c \sqrt {a^2 c x^2+c}}+\frac {a^4 x \tan ^{-1}(a x)^2}{c \sqrt {a^2 c x^2+c}}-\frac {11 i a^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {a^2 c x^2+c}}+\frac {11 i a^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {a^2 c x^2+c}}+\frac {2 a^3 \tan ^{-1}(a x)}{c \sqrt {a^2 c x^2+c}}+\frac {22 a^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {a^2 c x^2+c}} \]

[Out]

-2*a^4*x/c/(a^2*c*x^2+c)^(1/2)+2*a^3*arctan(a*x)/c/(a^2*c*x^2+c)^(1/2)+a^4*x*arctan(a*x)^2/c/(a^2*c*x^2+c)^(1/
2)+22/3*a^3*arctan(a*x)*arctanh((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)-11/3*
I*a^3*polylog(2,-(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)+11/3*I*a^3*polylog(2
,(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)-1/3*a^2*(a^2*c*x^2+c)^(1/2)/c^2/x-1/
3*a*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/c^2/x^2-1/3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/c^2/x^3+5/3*a^2*arctan(a*x)^
2*(a^2*c*x^2+c)^(1/2)/c^2/x

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Rubi [A]  time = 1.20, antiderivative size = 397, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4966, 4962, 264, 4958, 4954, 4944, 4898, 191} \[ -\frac {11 i a^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {a^2 c x^2+c}}+\frac {11 i a^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {a^2 c x^2+c}}-\frac {a^2 \sqrt {a^2 c x^2+c}}{3 c^2 x}+\frac {5 a^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 c^2 x}-\frac {a \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 c^2 x^2}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 c^2 x^3}-\frac {2 a^4 x}{c \sqrt {a^2 c x^2+c}}+\frac {a^4 x \tan ^{-1}(a x)^2}{c \sqrt {a^2 c x^2+c}}+\frac {2 a^3 \tan ^{-1}(a x)}{c \sqrt {a^2 c x^2+c}}+\frac {22 a^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]^2/(x^4*(c + a^2*c*x^2)^(3/2)),x]

[Out]

(-2*a^4*x)/(c*Sqrt[c + a^2*c*x^2]) - (a^2*Sqrt[c + a^2*c*x^2])/(3*c^2*x) + (2*a^3*ArcTan[a*x])/(c*Sqrt[c + a^2
*c*x^2]) - (a*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(3*c^2*x^2) + (a^4*x*ArcTan[a*x]^2)/(c*Sqrt[c + a^2*c*x^2]) - (
Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(3*c^2*x^3) + (5*a^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(3*c^2*x) + (22*a^3
*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(3*c*Sqrt[c + a^2*c*x^2]) - (((11*I)/
3)*a^3*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x])])/(c*Sqrt[c + a^2*c*x^2]) + (((11*I)/3)
*a^3*Sqrt[1 + a^2*x^2]*PolyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(c*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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -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 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 4962

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

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^4 \left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )^{3/2}} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^2}{x^4 \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 c^2 x^3}+a^4 \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx+\frac {(2 a) \int \frac {\tan ^{-1}(a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx}{3 c}-\frac {\left (2 a^2\right ) \int \frac {\tan ^{-1}(a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{3 c}-\frac {a^2 \int \frac {\tan ^{-1}(a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=\frac {2 a^3 \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c^2 x^2}+\frac {a^4 x \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 c^2 x^3}+\frac {5 a^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 c^2 x}-\left (2 a^4\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx+\frac {a^2 \int \frac {1}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{3 c}-\frac {a^3 \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{3 c}-\frac {\left (4 a^3\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{3 c}-\frac {\left (2 a^3\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=-\frac {2 a^4 x}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 \sqrt {c+a^2 c x^2}}{3 c^2 x}+\frac {2 a^3 \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c^2 x^2}+\frac {a^4 x \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 c^2 x^3}+\frac {5 a^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 c^2 x}-\frac {\left (a^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{3 c \sqrt {c+a^2 c x^2}}-\frac {\left (4 a^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{3 c \sqrt {c+a^2 c x^2}}-\frac {\left (2 a^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{c \sqrt {c+a^2 c x^2}}\\ &=-\frac {2 a^4 x}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 \sqrt {c+a^2 c x^2}}{3 c^2 x}+\frac {2 a^3 \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {a \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c^2 x^2}+\frac {a^4 x \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 c^2 x^3}+\frac {5 a^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 c^2 x}+\frac {22 a^3 \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 )}{3 c \sqrt {c+a^2 c x^2}}-\frac {11 i a^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {c+a^2 c x^2}}+\frac {11 i a^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 3.60, size = 270, normalized size = 0.68 \[ \frac {a^3 \sqrt {a^2 x^2+1} \left (\frac {\left (a^2 x^2+1\right )^{3/2} \left (\tan ^{-1}(a x) \left (\frac {66 a x \left (\log \left (1+e^{i \tan ^{-1}(a x)}\right )-\log \left (1-e^{i \tan ^{-1}(a x)}\right )\right )}{\sqrt {a^2 x^2+1}}+8 \sin \left (2 \tan ^{-1}(a x)\right )-6 \sin \left (4 \tan ^{-1}(a x)\right )+22 \left (\log \left (1-e^{i \tan ^{-1}(a x)}\right )-\log \left (1+e^{i \tan ^{-1}(a x)}\right )\right ) \sin \left (3 \tan ^{-1}(a x)\right )\right )+\frac {88 i a^3 x^3 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\left (a^2 x^2+1\right )^{3/2}}+28 \cos \left (2 \tan ^{-1}(a x)\right )-6 \cos \left (4 \tan ^{-1}(a x)\right )+\tan ^{-1}(a x)^2 \left (-36 \cos \left (2 \tan ^{-1}(a x)\right )+3 \cos \left (4 \tan ^{-1}(a x)\right )+25\right )-22\right )}{a^3 x^3}-88 i \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )\right )}{24 c \sqrt {a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a*x]^2/(x^4*(c + a^2*c*x^2)^(3/2)),x]

[Out]

(a^3*Sqrt[1 + a^2*x^2]*((-88*I)*PolyLog[2, -E^(I*ArcTan[a*x])] + ((1 + a^2*x^2)^(3/2)*(-22 + 28*Cos[2*ArcTan[a
*x]] - 6*Cos[4*ArcTan[a*x]] + ArcTan[a*x]^2*(25 - 36*Cos[2*ArcTan[a*x]] + 3*Cos[4*ArcTan[a*x]]) + ((88*I)*a^3*
x^3*PolyLog[2, E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(3/2) + ArcTan[a*x]*((66*a*x*(-Log[1 - E^(I*ArcTan[a*x])] + L
og[1 + E^(I*ArcTan[a*x])]))/Sqrt[1 + a^2*x^2] + 8*Sin[2*ArcTan[a*x]] + 22*(Log[1 - E^(I*ArcTan[a*x])] - Log[1
+ E^(I*ArcTan[a*x])])*Sin[3*ArcTan[a*x]] - 6*Sin[4*ArcTan[a*x]])))/(a^3*x^3)))/(24*c*Sqrt[c + a^2*c*x^2])

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}}{a^{4} c^{2} x^{8} + 2 \, a^{2} c^{2} x^{6} + c^{2} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^4/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

sage0*x

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maple [A]  time = 2.44, size = 318, normalized size = 0.80 \[ \frac {a^{3} \left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right ) a^{3}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}+\frac {\left (5 \arctan \left (a x \right )^{2} x^{2} a^{2}-a^{2} x^{2}-\arctan \left (a x \right ) x a -\arctan \left (a x \right )^{2}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{3 x^{3} c^{2}}+\frac {11 i a^{3} \left (-i \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+i \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )+\polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{3 \sqrt {a^{2} x^{2}+1}\, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^2/x^4/(a^2*c*x^2+c)^(3/2),x)

[Out]

1/2*a^3*(arctan(a*x)^2-2+2*I*arctan(a*x))*(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)^2-2-2*I*arctan(a*x))*a^3/(a^2*x^2+1)/c^2+1/3*(5*arctan(a*x)^2*x^2*a^2-a^2*x^2
-arctan(a*x)*x*a-arctan(a*x)^2)*(c*(a*x-I)*(I+a*x))^(1/2)/x^3/c^2+11/3*I*a^3*(-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))-polylo
g(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^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^4/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(arctan(a*x)^2/((a^2*c*x^2 + c)^(3/2)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^4\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)^2/(x^4*(c + a^2*c*x^2)^(3/2)),x)

[Out]

int(atan(a*x)^2/(x^4*(c + a^2*c*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{x^{4} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)**2/x**4/(a**2*c*x**2+c)**(3/2),x)

[Out]

Integral(atan(a*x)**2/(x**4*(c*(a**2*x**2 + 1))**(3/2)), x)

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