3.26 \(\int x^{3/2} \tan ^{-1}(\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}) \, dx\)

Optimal. Leaf size=296 \[ \frac{6 d^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{25 e^{7/4} \sqrt{d+e x^2}}-\frac{12 d^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{7/4} \sqrt{d+e x^2}}+\frac{12 d \sqrt{-e} \sqrt{x} \sqrt{d+e x^2}}{25 e^{3/2} \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{-e}}+\frac{2}{5} x^{5/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right ) \]

[Out]

(4*x^(3/2)*Sqrt[d + e*x^2])/(25*Sqrt[-e]) + (12*d*Sqrt[-e]*Sqrt[x]*Sqrt[d + e*x^2])/(25*e^(3/2)*(Sqrt[d] + Sqr
t[e]*x)) + (2*x^(5/2)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/5 - (12*d^(5/4)*Sqrt[-e]*(Sqrt[d] + Sqrt[e]*x)*Sqr
t[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(25*e^(7/4)*Sqrt[d
+ e*x^2]) + (6*d^(5/4)*Sqrt[-e]*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*A
rcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(25*e^(7/4)*Sqrt[d + e*x^2])

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Rubi [A]  time = 0.161298, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5151, 321, 329, 305, 220, 1196} \[ \frac{6 d^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{7/4} \sqrt{d+e x^2}}-\frac{12 d^{5/4} \sqrt{-e} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{7/4} \sqrt{d+e x^2}}+\frac{12 d \sqrt{-e} \sqrt{x} \sqrt{d+e x^2}}{25 e^{3/2} \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{-e}}+\frac{2}{5} x^{5/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^(3/2)*Sqrt[d + e*x^2])/(25*Sqrt[-e]) + (12*d*Sqrt[-e]*Sqrt[x]*Sqrt[d + e*x^2])/(25*e^(3/2)*(Sqrt[d] + Sqr
t[e]*x)) + (2*x^(5/2)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/5 - (12*d^(5/4)*Sqrt[-e]*(Sqrt[d] + Sqrt[e]*x)*Sqr
t[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(25*e^(7/4)*Sqrt[d
+ e*x^2]) + (6*d^(5/4)*Sqrt[-e]*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*A
rcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(25*e^(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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 x^{3/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right ) \, dx &=\frac{2}{5} x^{5/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{5} \left (2 \sqrt{-e}\right ) \int \frac{x^{5/2}}{\sqrt{d+e x^2}} \, dx\\ &=\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{-e}}+\frac{2}{5} x^{5/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{(6 d) \int \frac{\sqrt{x}}{\sqrt{d+e x^2}} \, dx}{25 \sqrt{-e}}\\ &=\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{-e}}+\frac{2}{5} x^{5/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{(12 d) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{25 \sqrt{-e}}\\ &=\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{-e}}+\frac{2}{5} x^{5/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (12 d^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{25 \sqrt{-e^2}}+\frac{\left (12 d^{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 )}{25 \sqrt{-e^2}}\\ &=\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{-e}}-\frac{12 d \sqrt{x} \sqrt{d+e x^2}}{25 \sqrt{-e^2} \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{2}{5} x^{5/2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )+\frac{12 d^{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 )}{25 \sqrt [4]{e} \sqrt{-e^2} \sqrt{d+e x^2}}-\frac{6 d^{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 )}{25 \sqrt [4]{e} \sqrt{-e^2} \sqrt{d+e x^2}}\\ \end{align*}

Mathematica [C]  time = 0.109487, size = 119, normalized size = 0.4 \[ \frac{2 x^{3/2} \left (2 d \sqrt{-e} \sqrt{\frac{e x^2}{d}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{e x^2}{d}\right )-2 \sqrt{-e} \left (d+e x^2\right )+5 e x \sqrt{d+e x^2} \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )\right )}{25 e \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(-2*Sqrt[-e]*(d + e*x^2) + 5*e*x*Sqrt[d + e*x^2]*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]] + 2*d*Sqrt[-e
]*Sqrt[1 + (e*x^2)/d]*Hypergeometric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)]))/(25*e*Sqrt[d + e*x^2])

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Maple [F]  time = 0.296, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{3}{2}}}\arctan \left ({x\sqrt{-e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x)

[Out]

int(x^(3/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2}{5} \, x^{\frac{5}{2}} \arctan \left (\sqrt{-e} x, \sqrt{e x^{2} + d}\right ) - 2 \, d \sqrt{-e} \int -\frac{x e^{\left (\frac{1}{2} \, \log \left (e x^{2} + d\right ) + \frac{3}{2} \, \log \left (x\right )\right )}}{5 \,{\left (e^{2} x^{4} + d e x^{2} -{\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

2/5*x^(5/2)*arctan2(sqrt(-e)*x, sqrt(e*x^2 + d)) - 2*d*sqrt(-e)*integrate(-1/5*x*e^(1/2*log(e*x^2 + d) + 3/2*l
og(x))/(e^2*x^4 + d*e*x^2 - (e*x^2 + d)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

integral(x^(3/2)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)), x)

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Sympy [C]  time = 40.5369, size = 75, normalized size = 0.25 \begin{align*} \frac{2 x^{\frac{5}{2}} \operatorname{atan}{\left (\frac{x \sqrt{- e}}{\sqrt{d + e x^{2}}} \right )}}{5} - \frac{x^{\frac{7}{2}} \sqrt{- e} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{e x^{2} e^{i \pi }}{d}} \right )}}{5 \sqrt{d} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*atan(x*(-e)**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

2*x**(5/2)*atan(x*sqrt(-e)/sqrt(d + e*x**2))/5 - x**(7/2)*sqrt(-e)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), e*x**
2*exp_polar(I*pi)/d)/(5*sqrt(d)*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

integrate(x^(3/2)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)), x)