3.22 \(\int \frac{\tan ^{-1}(\frac{\sqrt{-e} x}{\sqrt{d+e x^2}})}{x^{7/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac{2 \sqrt{-e} e^{3/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{15 d^{5/4} \sqrt{d+e x^2}}-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{15 d x^{3/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^{5/2}} \]

[Out]

(-4*Sqrt[-e]*Sqrt[d + e*x^2])/(15*d*x^(3/2)) - (2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(5*x^(5/2)) - (2*Sqrt[
-e]*e^(3/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x
])/d^(1/4)], 1/2])/(15*d^(5/4)*Sqrt[d + e*x^2])

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Rubi [A]  time = 0.0749551, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {5151, 325, 329, 220} \[ -\frac{2 \sqrt{-e} e^{3/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{15 d^{5/4} \sqrt{d+e x^2}}-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{15 d x^{3/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]

[Out]

(-4*Sqrt[-e]*Sqrt[d + e*x^2])/(15*d*x^(3/2)) - (2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(5*x^(5/2)) - (2*Sqrt[
-e]*e^(3/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x
])/d^(1/4)], 1/2])/(15*d^(5/4)*Sqrt[d + e*x^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 325

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] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{x^{7/2}} \, dx &=-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^{5/2}}+\frac{1}{5} \left (2 \sqrt{-e}\right ) \int \frac{1}{x^{5/2} \sqrt{d+e x^2}} \, dx\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{15 d x^{3/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^{5/2}}+\frac{\left (2 (-e)^{3/2}\right ) \int \frac{1}{\sqrt{x} \sqrt{d+e x^2}} \, dx}{15 d}\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{15 d x^{3/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^{5/2}}+\frac{\left (4 (-e)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{15 d}\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{15 d x^{3/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^{5/2}}+\frac{2 (-e)^{3/2} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{15 d^{5/4} \sqrt [4]{e} \sqrt{d+e x^2}}\\ \end{align*}

Mathematica [C]  time = 0.266727, size = 150, normalized size = 0.96 \[ -\frac{2 \left (2 \sqrt{-e} x \sqrt{d+e x^2}+3 d \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )\right )}{15 d x^{5/2}}+\frac{4 i (-e)^{3/2} x \sqrt{\frac{d}{e x^2}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{d}}{\sqrt{e}}}}{\sqrt{x}}\right ),-1\right )}{15 d \sqrt{\frac{i \sqrt{d}}{\sqrt{e}}} \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]

[Out]

(-2*(2*Sqrt[-e]*x*Sqrt[d + e*x^2] + 3*d*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]))/(15*d*x^(5/2)) + (((4*I)/15)*(-
e)^(3/2)*Sqrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1])/(d*Sqrt[(I*Sqrt[d]
)/Sqrt[e]]*Sqrt[d + e*x^2])

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Maple [F]  time = 0.29, size = 0, normalized size = 0. \begin{align*} \int{\arctan \left ({x\sqrt{-e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ){x}^{-{\frac{7}{2}}}}\, dx \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^(7/2),x)

[Out]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (-d \sqrt{-e} x^{\frac{5}{2}} \int -\frac{\sqrt{e x^{2} + d} x}{{\left (e x^{2} + d\right )}^{2} x^{\frac{7}{2}} -{\left (e^{2} x^{4} + d e x^{2}\right )} x^{\frac{7}{2}}}\,{d x} - \arctan \left (\sqrt{-e} x, \sqrt{e x^{2} + d}\right )\right )}}{5 \, x^{\frac{5}{2}}} \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^(7/2),x, algorithm="maxima")

[Out]

2/5*(5*d*sqrt(-e)*x^(5/2)*integrate(-1/5*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(7/2) - (e*x^2 + d)*e^(log(e
*x^2 + d) + 7/2*log(x))), x) - arctan2(sqrt(-e)*x, sqrt(e*x^2 + d)))/x^(5/2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right )}{x^{\frac{7}{2}}}, x\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^(7/2),x, algorithm="fricas")

[Out]

integral(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(7/2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right )}{x^{\frac{7}{2}}}\,{d x} \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^(7/2),x, algorithm="giac")

[Out]

integrate(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(7/2), x)