\(\int \frac {\arctan (a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx\) [336]

Optimal result

Integrand size = 24, antiderivative size = 208 \[ \int \frac {\arctan (a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx=-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c x}-\frac {4 a \sqrt {1+a^2 x^2} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i a \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i a \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}} \]

[Out]

-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)/(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)/(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)/(a^2*c*x^2+c)^(1/2)-arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/c/x

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5064, 5078, 5074} \[ \int \frac {\arctan (a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx=-\frac {4 a \sqrt {a^2 x^2+1} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^2 \sqrt {a^2 c x^2+c}}{c x}+\frac {2 i a \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i a \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

-((Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(c*x)) - (4*a*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/Sqrt
[1 - I*a*x]])/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])])
/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]])/Sqrt[c + a^2*c*
x^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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c x}+(2 a) \int \frac {\arctan (a x)}{x \sqrt {c+a^2 c x^2}} \, dx \\ & = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c 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}{\sqrt {c+a^2 c x^2}} \\ & = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{c 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 )}{\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 )}{\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 )}{\sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.62 \[ \int \frac {\arctan (a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx=-\frac {a \sqrt {1+a^2 x^2} \left (\arctan (a x) \left (\frac {\sqrt {1+a^2 x^2} \arctan (a x)}{a x}-2 \log \left (1-e^{i \arctan (a x)}\right )+2 \log \left (1+e^{i \arctan (a x)}\right )\right )-2 i \operatorname {PolyLog}\left (2,-e^{i \arctan (a x)}\right )+2 i \operatorname {PolyLog}\left (2,e^{i \arctan (a x)}\right )\right )}{\sqrt {c \left (1+a^2 x^2\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {\arctan \left (a x \right )^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{c x}+\frac {2 i a \left (i \arctan \left (a x \right ) \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+\operatorname {polylog}\left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\operatorname {polylog}\left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c}\)	\(171\)

[In]

int(arctan(a*x)^2/x^2/(a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-arctan(a*x)^2*(c*(a*x-I)*(I+a*x))^(1/2)/c/x+2*I*a*(I*arctan(a*x)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)-I*arctan(a
*x)*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))+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)))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/c

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(atan(a*x)^2/(x^2*(c + a^2*c*x^2)^(1/2)),x)

[Out]

int(atan(a*x)^2/(x^2*(c + a^2*c*x^2)^(1/2)), x)