3.16 \(\int \frac{\tan ^{-1}(\frac{\sqrt{-e} x}{\sqrt{d+e x^2}})}{x^4} \, dx\)

Optimal. Leaf size=91 \[ -\frac{(-e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 d^{3/2}}-\frac{\sqrt{-e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3} \]

[Out]

-(Sqrt[-e]*Sqrt[d + e*x^2])/(6*d*x^2) - ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/(3*x^3) - ((-e)^(3/2)*ArcTanh[Sqr
t[d + e*x^2]/Sqrt[d]])/(6*d^(3/2))

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Rubi [A]  time = 0.0460897, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5151, 266, 51, 63, 208} \[ -\frac{(-e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 d^{3/2}}-\frac{\sqrt{-e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^4,x]

[Out]

-(Sqrt[-e]*Sqrt[d + e*x^2])/(6*d*x^2) - ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/(3*x^3) - ((-e)^(3/2)*ArcTanh[Sqr
t[d + e*x^2]/Sqrt[d]])/(6*d^(3/2))

Rule 5151

Int[ArcTan[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*ArcTa
n[(c*x)/Sqrt[a + b*x^2]])/(d*(m + 1)), x] - Dist[c/(d*(m + 1)), Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; F
reeQ[{a, b, c, d, m}, x] && EqQ[b + c^2, 0] && NeQ[m, -1]

Rule 266

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 208

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

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{x^4} \, dx &=-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3}+\frac{1}{3} \sqrt{-e} \int \frac{1}{x^3 \sqrt{d+e x^2}} \, dx\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3}+\frac{1}{6} \sqrt{-e} \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{d+e x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{-e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3}+\frac{(-e)^{3/2} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{12 d}\\ &=-\frac{\sqrt{-e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3}-\frac{\sqrt{-e} \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{6 d}\\ &=-\frac{\sqrt{-e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3}-\frac{(-e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.111211, size = 101, normalized size = 1.11 \[ \frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{-e}}{\sqrt{e} \sqrt{d+e x^2}}\right )}{6 d^{3/2}}-\frac{\sqrt{-e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^4,x]

[Out]

-(Sqrt[-e]*Sqrt[d + e*x^2])/(6*d*x^2) + (e^(3/2)*ArcTan[(Sqrt[d]*Sqrt[-e])/(Sqrt[e]*Sqrt[d + e*x^2])])/(6*d^(3
/2)) - ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/(3*x^3)

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Maple [A]  time = 0.039, size = 100, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,{x}^{3}}\arctan \left ({x\sqrt{-e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{e}{6}\sqrt{-e}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ){d}^{-{\frac{3}{2}}}}-{\frac{1}{6\,{d}^{2}{x}^{2}}\sqrt{-e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{e}{6\,{d}^{2}}\sqrt{-e}\sqrt{e{x}^{2}+d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^4,x)

[Out]

-1/3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^3+1/6*(-e)^(1/2)*e/d^(3/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)-1
/6*(-e)^(1/2)/d^2/x^2*(e*x^2+d)^(3/2)+1/6*(-e)^(1/2)/d^2*e*(e*x^2+d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{-d \sqrt{-e} x^{3} \int \frac{\sqrt{e x^{2} + d}}{e^{2} x^{7} + d e x^{5} -{\left (e x^{5} + d x^{3}\right )}{\left (e x^{2} + d\right )}}\,{d x} - \arctan \left (\sqrt{-e} x, \sqrt{e x^{2} + d}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^4,x, algorithm="maxima")

[Out]

1/3*(3*d*sqrt(-e)*x^3*integrate(-1/3*sqrt(e*x^2 + d)/(e^2*x^7 + d*e*x^5 - (e*x^5 + d*x^3)*(e*x^2 + d)), x) - a
rctan2(sqrt(-e)*x, sqrt(e*x^2 + d)))/x^3

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Fricas [A]  time = 2.71131, size = 460, normalized size = 5.05 \begin{align*} \left [\frac{e x^{3} \sqrt{-\frac{e}{d}} \log \left (-\frac{e^{2} x^{2} - 2 \, \sqrt{e x^{2} + d} d \sqrt{-e} \sqrt{-\frac{e}{d}} + 2 \, d e}{x^{2}}\right ) - 2 \, \sqrt{e x^{2} + d} \sqrt{-e} x - 4 \, d \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right )}{12 \, d x^{3}}, \frac{e x^{3} \sqrt{\frac{e}{d}} \arctan \left (\frac{\sqrt{e x^{2} + d} d \sqrt{-e} \sqrt{\frac{e}{d}}}{e^{2} x^{2} + d e}\right ) - \sqrt{e x^{2} + d} \sqrt{-e} x - 2 \, d \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right )}{6 \, d x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^4,x, algorithm="fricas")

[Out]

[1/12*(e*x^3*sqrt(-e/d)*log(-(e^2*x^2 - 2*sqrt(e*x^2 + d)*d*sqrt(-e)*sqrt(-e/d) + 2*d*e)/x^2) - 2*sqrt(e*x^2 +
d)*sqrt(-e)*x - 4*d*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)))/(d*x^3), 1/6*(e*x^3*sqrt(e/d)*arctan(sqrt(e*x^2 + d)*
d*sqrt(-e)*sqrt(e/d)/(e^2*x^2 + d*e)) - sqrt(e*x^2 + d)*sqrt(-e)*x - 2*d*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)))/(
d*x^3)]

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Sympy [A]  time = 8.6699, size = 82, normalized size = 0.9 \begin{align*} - \frac{\operatorname{atan}{\left (\frac{x \sqrt{- e}}{\sqrt{d + e x^{2}}} \right )}}{3 x^{3}} - \frac{\sqrt{e} \sqrt{- e} \sqrt{\frac{d}{e x^{2}} + 1}}{6 d x} + \frac{e \sqrt{- e} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{e} x} \right )}}{6 d^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x*(-e)**(1/2)/(e*x**2+d)**(1/2))/x**4,x)

[Out]

-atan(x*sqrt(-e)/sqrt(d + e*x**2))/(3*x**3) - sqrt(e)*sqrt(-e)*sqrt(d/(e*x**2) + 1)/(6*d*x) + e*sqrt(-e)*asinh
(sqrt(d)/(sqrt(e)*x))/(6*d**(3/2))

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Giac [A]  time = 1.23854, size = 112, normalized size = 1.23 \begin{align*} \frac{1}{6} \,{\left (\frac{\arctan \left (\frac{\sqrt{-x^{2} e^{2} - d e} e^{\left (-\frac{1}{2}\right )}}{\sqrt{d}}\right ) e^{\left (-\frac{3}{2}\right )}}{d^{\frac{3}{2}}} - \frac{\sqrt{-x^{2} e^{2} - d e} e^{\left (-3\right )}}{d x^{2}}\right )} e^{3} - \frac{\arctan \left (\frac{x \sqrt{-e}}{\sqrt{x^{2} e + d}}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^4,x, algorithm="giac")

[Out]

1/6*(arctan(sqrt(-x^2*e^2 - d*e)*e^(-1/2)/sqrt(d))*e^(-3/2)/d^(3/2) - sqrt(-x^2*e^2 - d*e)*e^(-3)/(d*x^2))*e^3
- 1/3*arctan(x*sqrt(-e)/sqrt(x^2*e + d))/x^3