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

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

[Out]

2/5*x^(5/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+4/25*x^(3/2)*(e*x^2+d)^(1/2)/(-e)^(1/2)+12/25*d*(-e)^(1/2)*x^
(1/2)*(e*x^2+d)^(1/2)/e^(3/2)/(d^(1/2)+x*e^(1/2))-12/25*d^(5/4)*(cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))^2)^(1/
2)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticE(sin(2*arctan(e^(1/4)*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(-e)^(
1/2)*(d^(1/2)+x*e^(1/2))*((e*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(7/4)/(e*x^2+d)^(1/2)+6/25*d^(5/4)*(cos(2*a
rctan(e^(1/4)*x^(1/2)/d^(1/4)))^2)^(1/2)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticF(sin(2*arctan(e^(1/4)
*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(-e)^(1/2)*(d^(1/2)+x*e^(1/2))*((e*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(7/4)
/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 296, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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]

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*}

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Mathematica [C]  time = 0.12, size = 119, normalized size = 0.40 \[ \frac {2 x^{3/2} \left (2 d \sqrt {-e} \sqrt {\frac {e x^2}{d}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {e x^2}{d}\right )-2 \sqrt {-e} \left (d+e x^2\right )+5 e x \sqrt {d+e x^2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\right )}{25 e \sqrt {d+e x^2}} \]

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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fricas [F]  time = 1.27, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ), x\right ) \]

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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giac [A]  time = 0.26, size = 1, normalized size = 0.00 \[ +\infty \]

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]

+Infinity

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )\, dx \]

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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found sqrt(-_SAGE_VAR_e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 13.07, size = 75, normalized size = 0.25 \[ \frac {2 x^{\frac {5}{2}} \operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{5} - \frac {x^{\frac {7}{2}} \sqrt {- e} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5 \sqrt {d} \Gamma \left (\frac {11}{4}\right )} \]

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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