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

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

[Out]

-2/5*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(5/2)-4/15*(-e)^(1/2)*(e*x^2+d)^(1/2)/d/x^(3/2)-2/15*e^(3/4)*(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^(5
/4)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 156, normalized size of antiderivative = 1.00, 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 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^{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*}

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Mathematica [C]  time = 0.29, size = 150, normalized size = 0.96 \[ -\frac {2 \left (2 \sqrt {-e} x \sqrt {d+e x^2}+3 d \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\right )}{15 d x^{5/2}}+\frac {4 i (-e)^{3/2} x \sqrt {\frac {d}{e x^2}+1} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{15 d \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}} \]

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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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(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(-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^(7/2),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 \[ \int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 146.13, size = 78, normalized size = 0.50 \[ - \frac {2 \operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{5 x^{\frac {5}{2}}} + \frac {\sqrt {- e} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5 \sqrt {d} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]

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]

-2*atan(x*sqrt(-e)/sqrt(d + e*x**2))/(5*x**(5/2)) + sqrt(-e)*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), e*x**2*exp
_polar(I*pi)/d)/(5*sqrt(d)*x**(3/2)*gamma(1/4))

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