3.29 \(\int \frac{\tan ^{-1}(\frac{\sqrt{-e} x}{\sqrt{d+e x^2}})}{x^{9/2}} \, dx\)

Optimal. Leaf size=331 \[ -\frac{6 e^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{12 e^{3/2} \sqrt{-e} \sqrt{x} \sqrt{d+e x^2}}{35 d^2 \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{12 e^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{12 (-e)^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}} \]

[Out]

(-4*Sqrt[-e]*Sqrt[d + e*x^2])/(35*d*x^(5/2)) - (12*(-e)^(3/2)*Sqrt[d + e*x^2])/(35*d^2*Sqrt[x]) - (12*Sqrt[-e]
*e^(3/2)*Sqrt[x]*Sqrt[d + e*x^2])/(35*d^2*(Sqrt[d] + Sqrt[e]*x)) - (2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(7
*x^(7/2)) + (12*Sqrt[-e]*e^(5/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*A
rcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(35*d^(7/4)*Sqrt[d + e*x^2]) - (6*Sqrt[-e]*e^(5/4)*(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])/(35*d^(7/4)*
Sqrt[d + e*x^2])

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Rubi [A]  time = 0.193637, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5151, 325, 329, 305, 220, 1196} \[ -\frac{12 e^{3/2} \sqrt{-e} \sqrt{x} \sqrt{d+e x^2}}{35 d^2 \left (\sqrt{d}+\sqrt{e} x\right )}-\frac{6 e^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}+\frac{12 e^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{12 (-e)^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(9/2),x]

[Out]

(-4*Sqrt[-e]*Sqrt[d + e*x^2])/(35*d*x^(5/2)) - (12*(-e)^(3/2)*Sqrt[d + e*x^2])/(35*d^2*Sqrt[x]) - (12*Sqrt[-e]
*e^(3/2)*Sqrt[x]*Sqrt[d + e*x^2])/(35*d^2*(Sqrt[d] + Sqrt[e]*x)) - (2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(7
*x^(7/2)) + (12*Sqrt[-e]*e^(5/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*A
rcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(35*d^(7/4)*Sqrt[d + e*x^2]) - (6*Sqrt[-e]*e^(5/4)*(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])/(35*d^(7/4)*
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 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{x^{9/2}} \, dx &=-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}+\frac{1}{7} \left (2 \sqrt{-e}\right ) \int \frac{1}{x^{7/2} \sqrt{d+e x^2}} \, dx\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}+\frac{\left (6 (-e)^{3/2}\right ) \int \frac{1}{x^{3/2} \sqrt{d+e x^2}} \, dx}{35 d}\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{12 (-e)^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}-\frac{\left (6 (-e)^{5/2}\right ) \int \frac{\sqrt{x}}{\sqrt{d+e x^2}} \, dx}{35 d^2}\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{12 (-e)^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}-\frac{\left (12 (-e)^{5/2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{35 d^2}\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{12 (-e)^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}-\frac{\left (12 \sqrt{-e} e^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{35 d^{3/2}}+\frac{\left (12 \sqrt{-e} e^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{e} x^2}{\sqrt{d}}}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{35 d^{3/2}}\\ &=-\frac{4 \sqrt{-e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{12 (-e)^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{12 \sqrt{-e} e^{3/2} \sqrt{x} \sqrt{d+e x^2}}{35 d^2 \left (\sqrt{d}+\sqrt{e} x\right )}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}+\frac{12 \sqrt{-e} e^{5/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{6 \sqrt{-e} e^{5/4} \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 )}{35 d^{7/4} \sqrt{d+e x^2}}\\ \end{align*}

Mathematica [C]  time = 0.117839, size = 137, normalized size = 0.41 \[ \frac{4 \sqrt{-e} x \left (-d^2+2 d e x^2+3 e^2 x^4\right )-10 d^2 \sqrt{d+e x^2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-4 (-e)^{5/2} x^5 \sqrt{\frac{e x^2}{d}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{e x^2}{d}\right )}{35 d^2 x^{7/2} \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(9/2),x]

[Out]

(4*Sqrt[-e]*x*(-d^2 + 2*d*e*x^2 + 3*e^2*x^4) - 10*d^2*Sqrt[d + e*x^2]*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]] - 4
*(-e)^(5/2)*x^5*Sqrt[1 + (e*x^2)/d]*Hypergeometric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)])/(35*d^2*x^(7/2)*Sqrt[d + e
*x^2])

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Maple [F]  time = 0.313, size = 0, normalized size = 0. \begin{align*} \int{\arctan \left ({x\sqrt{-e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ){x}^{-{\frac{9}{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^(9/2),x)

[Out]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(9/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (-d \sqrt{-e} x^{\frac{7}{2}} \int -\frac{\sqrt{e x^{2} + d} x}{{\left (e x^{2} + d\right )}^{2} x^{\frac{9}{2}} -{\left (e^{2} x^{4} + d e x^{2}\right )} x^{\frac{9}{2}}}\,{d x} - \arctan \left (\sqrt{-e} x, \sqrt{e x^{2} + d}\right )\right )}}{7 \, x^{\frac{7}{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^(9/2),x, algorithm="maxima")

[Out]

2/7*(7*d*sqrt(-e)*x^(7/2)*integrate(-1/7*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(9/2) - (e*x^2 + d)*e^(log(e
*x^2 + d) + 9/2*log(x))), x) - arctan2(sqrt(-e)*x, sqrt(e*x^2 + d)))/x^(7/2)

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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{9}{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^(9/2),x, algorithm="fricas")

[Out]

integral(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(9/2), 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**(9/2),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{9}{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^(9/2),x, algorithm="giac")

[Out]

integrate(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(9/2), x)