\(\int \frac {\arctan (a x)}{x^3 (c+a^2 c x^2)^{3/2}} \, dx\) [238]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 300 \[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {a^3 x}{c \sqrt {c+a^2 c x^2}}-\frac {a \sqrt {c+a^2 c x^2}}{2 c^2 x}-\frac {a^2 \arctan (a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{2 c^2 x^2}+\frac {3 a^2 \sqrt {1+a^2 x^2} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {3 i a^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 c \sqrt {c+a^2 c x^2}}+\frac {3 i a^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 c \sqrt {c+a^2 c x^2}} \]

[Out]

a^3*x/c/(a^2*c*x^2+c)^(1/2)-a^2*arctan(a*x)/c/(a^2*c*x^2+c)^(1/2)+3*a^2*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)-3/2*I*a^2*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)+3/2*I*a^2*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)-1/2*a*(a^2*c*x^2+c)^(1/2)/c^2/x-1/2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/c^2/x^2

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5086, 5082, 270, 5078, 5074, 5050, 197} \[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {3 a^2 \sqrt {a^2 x^2+1} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{2 c^2 x^2}-\frac {a^2 \arctan (a x)}{c \sqrt {a^2 c x^2+c}}-\frac {a \sqrt {a^2 c x^2+c}}{2 c^2 x}-\frac {3 i a^2 \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 c \sqrt {a^2 c x^2+c}}+\frac {3 i a^2 \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 c \sqrt {a^2 c x^2+c}}+\frac {a^3 x}{c \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

(a^3*x)/(c*Sqrt[c + a^2*c*x^2]) - (a*Sqrt[c + a^2*c*x^2])/(2*c^2*x) - (a^2*ArcTan[a*x])/(c*Sqrt[c + a^2*c*x^2]
) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(2*c^2*x^2) + (3*a^2*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*
x]/Sqrt[1 - I*a*x]])/(c*Sqrt[c + a^2*c*x^2]) - (((3*I)/2)*a^2*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a*x]/S
qrt[1 - I*a*x])])/(c*Sqrt[c + a^2*c*x^2]) + (((3*I)/2)*a^2*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 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 270

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 5050

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(d + e*x^2)^(
q + 1)*((a + b*ArcTan[c*x])^p/(2*e*(q + 1))), x] - Dist[b*(p/(2*c*(q + 1))), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -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 5082

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 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)}{x \left (c+a^2 c x^2\right )^{3/2}} \, dx\right )+\frac {\int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx}{c} \\ & = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{2 c^2 x^2}+a^4 \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx+\frac {a \int \frac {1}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{2 c}-\frac {a^2 \int \frac {\arctan (a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{2 c}-\frac {a^2 \int \frac {\arctan (a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{c} \\ & = -\frac {a \sqrt {c+a^2 c x^2}}{2 c^2 x}-\frac {a^2 \arctan (a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{2 c^2 x^2}+a^3 \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx-\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \int \frac {\arctan (a x)}{x \sqrt {1+a^2 x^2}} \, dx}{2 c \sqrt {c+a^2 c x^2}}-\frac {\left (a^2 \sqrt {1+a^2 x^2}\right ) \int \frac {\arctan (a x)}{x \sqrt {1+a^2 x^2}} \, dx}{c \sqrt {c+a^2 c x^2}} \\ & = \frac {a^3 x}{c \sqrt {c+a^2 c x^2}}-\frac {a \sqrt {c+a^2 c x^2}}{2 c^2 x}-\frac {a^2 \arctan (a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{2 c^2 x^2}+\frac {3 a^2 \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 \sqrt {c+a^2 c x^2}}-\frac {3 i a^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 c \sqrt {c+a^2 c x^2}}+\frac {3 i a^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.86 \[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {a^2 \left (-8 a x+8 \arctan (a x)+a x \csc ^2\left (\frac {1}{2} \arctan (a x)\right )+\sqrt {1+a^2 x^2} \arctan (a x) \csc ^2\left (\frac {1}{2} \arctan (a x)\right )+12 \sqrt {1+a^2 x^2} \arctan (a x) \log \left (1-e^{i \arctan (a x)}\right )-12 \sqrt {1+a^2 x^2} \arctan (a x) \log \left (1+e^{i \arctan (a x)}\right )+12 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-e^{i \arctan (a x)}\right )-12 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,e^{i \arctan (a x)}\right )-\sqrt {1+a^2 x^2} \arctan (a x) \sec ^2\left (\frac {1}{2} \arctan (a x)\right )+2 \sqrt {1+a^2 x^2} \tan \left (\frac {1}{2} \arctan (a x)\right )\right )}{8 c \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

-1/8*(a^2*(-8*a*x + 8*ArcTan[a*x] + a*x*Csc[ArcTan[a*x]/2]^2 + Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Csc[ArcTan[a*x]/2
]^2 + 12*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])] - 12*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Log[1 + E
^(I*ArcTan[a*x])] + (12*I)*Sqrt[1 + a^2*x^2]*PolyLog[2, -E^(I*ArcTan[a*x])] - (12*I)*Sqrt[1 + a^2*x^2]*PolyLog
[2, E^(I*ArcTan[a*x])] - Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Sec[ArcTan[a*x]/2]^2 + 2*Sqrt[1 + a^2*x^2]*Tan[ArcTan[a
*x]/2]))/(c*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.91

method result size
default \(-\frac {a^{2} \left (\arctan \left (a x \right )+i\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )-i\right ) a^{2}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\left (a x +\arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 c^{2} x^{2}}-\frac {3 i a^{2} \left (i \arctan \left (a x \right ) \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+\operatorname {polylog}\left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\operatorname {polylog}\left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{2}}\) \(273\)

[In]

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

[Out]

-1/2*a^2*(arctan(a*x)+I)*(1+I*a*x)*(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-1)*(arctan(a*x)-I)*a^2/(a^2*x^2+1)/c^2-1/2*(a*x+arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)/c^2/x^2-3/2*I*a^2*(
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^2

Fricas [F]

\[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{3}} \,d x } \]

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)/(a^4*c^2*x^7 + 2*a^2*c^2*x^5 + c^2*x^3), x)

Sympy [F]

\[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{3} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{3}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{3}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a x)}{x^3 \left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )}{x^3\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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