\(\int \frac {\arctan (a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx\) [229]

Optimal result

Integrand size = 22, antiderivative size = 56 \[ \int \frac {\arctan (a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx=-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c x}-\frac {a \text {arctanh}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5064, 272, 65, 214} \[ \int \frac {\arctan (a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx=-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{c x}-\frac {a \text {arctanh}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{\sqrt {c}} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c x}+a \int \frac {1}{x \sqrt {c+a^2 c x^2}} \, dx \\ & = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c x}+\frac {1}{2} a \text {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c x}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c+a^2 c x^2}\right )}{a c} \\ & = -\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c x}-\frac {a \text {arctanh}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.11 \[ \int \frac {\arctan (a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx=-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c x}+\frac {a \left (\log (x)-\log \left (c+\sqrt {c} \sqrt {c+a^2 c x^2}\right )\right )}{\sqrt {c}} \]

[In]

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

[Out]

-((Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(c*x)) + (a*(Log[x] - Log[c + Sqrt[c]*Sqrt[c + a^2*c*x^2]]))/Sqrt[c]

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.23

method	result	size
default	\(-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right ) \sqrt {a^{2} x^{2}+1}\, a x -\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) \sqrt {a^{2} x^{2}+1}\, a x +a^{2} \arctan \left (a x \right ) x^{2}+\arctan \left (a x \right )\right )}{c x \left (a^{2} x^{2}+1\right )}\)	\(125\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.21 \[ \int \frac {\arctan (a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx=\frac {a \sqrt {c} x \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )}{2 \, c x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.64 \[ \int \frac {\arctan (a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx=-\frac {a \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) + \frac {\sqrt {a^{2} x^{2} + 1} \arctan \left (a x\right )}{x}}{\sqrt {c}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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