3.3.37 \(\int \frac {\text {ArcTan}(a x)}{x^2 (c+a^2 c x^2)^{3/2}} \, dx\) [237]

Optimal. Leaf size=103 \[ -\frac {a}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \text {ArcTan}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}{c^2 x}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{c^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.15, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5086, 5064, 272, 65, 214, 5014} \begin {gather*} -\frac {\text {ArcTan}(a x) \sqrt {a^2 c x^2+c}}{c^2 x}-\frac {a^2 x \text {ArcTan}(a x)}{c \sqrt {a^2 c x^2+c}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {a}{c \sqrt {a^2 c x^2+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a/(c*Sqrt[c + a^2*c*x^2])) - (a^2*x*ArcTan[a*x])/(c*Sqrt[c + a^2*c*x^2]) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])
/(c^2*x) - (a*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/c^(3/2)

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 5014

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[b/(c*d*Sqrt[d + e*x^2]),
x] + Simp[x*((a + b*ArcTan[c*x])/(d*Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

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 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*} \int \frac {\tan ^{-1}(a x)}{x^2 \left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=-\frac {a}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c^2 x}+\frac {a \int \frac {1}{x \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=-\frac {a}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c^2 x}+\frac {a \text {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )}{2 c}\\ &=-\frac {a}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c^2 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^2}\\ &=-\frac {a}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{c^2 x}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{c^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 122, normalized size = 1.18 \begin {gather*} -\frac {a \sqrt {c \left (1+a^2 x^2\right )}}{c^2 \left (1+a^2 x^2\right )}-\frac {\sqrt {c \left (1+a^2 x^2\right )} \left (1+2 a^2 x^2\right ) \text {ArcTan}(a x)}{c^2 x \left (1+a^2 x^2\right )}+\frac {a \log (x)}{c^{3/2}}-\frac {a \log \left (c+\sqrt {c} \sqrt {c \left (1+a^2 x^2\right )}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Sqrt[c*(1 + a^2*x^2)])/(c^2*(1 + a^2*x^2))) - (Sqrt[c*(1 + a^2*x^2)]*(1 + 2*a^2*x^2)*ArcTan[a*x])/(c^2*x*
(1 + a^2*x^2)) + (a*Log[x])/c^(3/2) - (a*Log[c + Sqrt[c]*Sqrt[c*(1 + a^2*x^2)]])/c^(3/2)

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.29, size = 231, normalized size = 2.24

method	result	size
default	\(-\frac {a \left (\arctan \left (a x \right )+i\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )-i\right ) a}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\arctan \left (a x \right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{c^{2} x}-\frac {a \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{2}}+\frac {a \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{2}}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^2/(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^2), x)

________________________________________________________________________________________

Fricas [A]
time = 2.18, size = 104, normalized size = 1.01 \begin {gather*} \frac {{\left (a^{3} x^{3} + a x\right )} \sqrt {c} \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} {\left (a x + {\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )}}{2 \, {\left (a^{2} c^{2} x^{3} + c^{2} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atan}\left (a\,x\right )}{x^2\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________