3.23 \(\int \frac {\tan ^{-1}(\frac {\sqrt {-e} x}{\sqrt {d+e x^2}})}{x^{11/2}} \, dx\)

Optimal. Leaf size=186 \[ \frac {10 \sqrt {-e} e^{7/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{189 d^{9/4} \sqrt {d+e x^2}}-\frac {20 (-e)^{3/2} \sqrt {d+e x^2}}{189 d^2 x^{3/2}}-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{63 d x^{7/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}} \]

[Out]

-2/9*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(9/2)-20/189*(-e)^(3/2)*(e*x^2+d)^(1/2)/d^2/x^(3/2)-4/63*(-e)^(1/2
)*(e*x^2+d)^(1/2)/d/x^(7/2)+10/189*e^(7/4)*(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)))*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)/d^(9/4)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5151, 325, 329, 220} \[ \frac {10 \sqrt {-e} e^{7/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \text {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{189 d^{9/4} \sqrt {d+e x^2}}-\frac {20 (-e)^{3/2} \sqrt {d+e x^2}}{189 d^2 x^{3/2}}-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{63 d x^{7/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(11/2),x]

[Out]

(-4*Sqrt[-e]*Sqrt[d + e*x^2])/(63*d*x^(7/2)) - (20*(-e)^(3/2)*Sqrt[d + e*x^2])/(189*d^2*x^(3/2)) - (2*ArcTan[(
Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(9*x^(9/2)) + (10*Sqrt[-e]*e^(7/4)*(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])/(189*d^(9/4)*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 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 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 \frac {\tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}+\frac {1}{9} \left (2 \sqrt {-e}\right ) \int \frac {1}{x^{9/2} \sqrt {d+e x^2}} \, dx\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{63 d x^{7/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}+\frac {\left (10 (-e)^{3/2}\right ) \int \frac {1}{x^{5/2} \sqrt {d+e x^2}} \, dx}{63 d}\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{63 d x^{7/2}}-\frac {20 (-e)^{3/2} \sqrt {d+e x^2}}{189 d^2 x^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}+\frac {\left (10 (-e)^{5/2}\right ) \int \frac {1}{\sqrt {x} \sqrt {d+e x^2}} \, dx}{189 d^2}\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{63 d x^{7/2}}-\frac {20 (-e)^{3/2} \sqrt {d+e x^2}}{189 d^2 x^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}+\frac {\left (20 (-e)^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{189 d^2}\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{63 d x^{7/2}}-\frac {20 (-e)^{3/2} \sqrt {d+e x^2}}{189 d^2 x^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}+\frac {10 \sqrt {-e} e^{7/4} \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 )}{189 d^{9/4} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.34, size = 162, normalized size = 0.87 \[ \frac {4 \sqrt {-e} x \sqrt {d+e x^2} \left (5 e x^2-3 d\right )-42 d^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{189 d^2 x^{9/2}}+\frac {20 i (-e)^{5/2} x \sqrt {\frac {d}{e x^2}+1} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{189 d^2 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(11/2),x]

[Out]

(4*Sqrt[-e]*x*Sqrt[d + e*x^2]*(-3*d + 5*e*x^2) - 42*d^2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(189*d^2*x^(9/2)
) + (((20*I)/189)*(-e)^(5/2)*Sqrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1]
)/(d^2*Sqrt[(I*Sqrt[d])/Sqrt[e]]*Sqrt[d + e*x^2])

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {11}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(11/2),x, algorithm="fricas")

[Out]

integral(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(11/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(11/2),x, algorithm="giac")

[Out]

integrate(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(11/2), x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {11}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(11/2),x)

[Out]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(11/2),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(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(11/2),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 = -0.01 \[ \int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(11/2),x)

[Out]

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(11/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x*(-e)**(1/2)/(e*x**2+d)**(1/2))/x**(11/2),x)

[Out]

Timed out

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