3.24 \(\int \frac {\tan ^{-1}(\frac {\sqrt {-e} x}{\sqrt {d+e x^2}})}{x^{15/2}} \, dx\)

Optimal. Leaf size=216 \[ -\frac {30 \sqrt {-e} e^{11/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}}-\frac {60 (-e)^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {36 (-e)^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}} \]

[Out]

-2/13*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(13/2)-36/1001*(-e)^(3/2)*(e*x^2+d)^(1/2)/d^2/x^(7/2)-60/1001*(-e
)^(5/2)*(e*x^2+d)^(1/2)/d^3/x^(3/2)-4/143*(-e)^(1/2)*(e*x^2+d)^(1/2)/d/x^(11/2)-30/1001*e^(11/4)*(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^(13/4)/(e*
x^2+d)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5151, 325, 329, 220} \[ -\frac {30 \sqrt {-e} e^{11/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \text {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}}-\frac {60 (-e)^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {36 (-e)^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]

[Out]

(-4*Sqrt[-e]*Sqrt[d + e*x^2])/(143*d*x^(11/2)) - (36*(-e)^(3/2)*Sqrt[d + e*x^2])/(1001*d^2*x^(7/2)) - (60*(-e)
^(5/2)*Sqrt[d + e*x^2])/(1001*d^3*x^(3/2)) - (2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(13*x^(13/2)) - (30*Sqrt
[-e]*e^(11/4)*(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])/(1001*d^(13/4)*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^{15/2}} \, dx &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {1}{13} \left (2 \sqrt {-e}\right ) \int \frac {1}{x^{13/2} \sqrt {d+e x^2}} \, dx\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {\left (18 (-e)^{3/2}\right ) \int \frac {1}{x^{9/2} \sqrt {d+e x^2}} \, dx}{143 d}\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {36 (-e)^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {\left (90 (-e)^{5/2}\right ) \int \frac {1}{x^{5/2} \sqrt {d+e x^2}} \, dx}{1001 d^2}\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {36 (-e)^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 (-e)^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {\left (30 (-e)^{7/2}\right ) \int \frac {1}{\sqrt {x} \sqrt {d+e x^2}} \, dx}{1001 d^3}\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {36 (-e)^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 (-e)^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {\left (60 (-e)^{7/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{1001 d^3}\\ &=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {36 (-e)^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 (-e)^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {30 (-e)^{7/2} \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 )}{1001 d^{13/4} \sqrt [4]{e} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.61, size = 171, normalized size = 0.79 \[ \frac {2 \left (\frac {30 i (-e)^{7/2} x^{15/2} \sqrt {\frac {d}{e x^2}+1} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{d^3 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}}-\frac {2 \sqrt {-e} \sqrt {d+e x^2} \left (7 d^2 x-9 d e x^3+15 e^2 x^5\right )}{d^3}-77 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\right )}{1001 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]

[Out]

(2*((-2*Sqrt[-e]*Sqrt[d + e*x^2]*(7*d^2*x - 9*d*e*x^3 + 15*e^2*x^5))/d^3 - 77*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x
^2]] + ((30*I)*(-e)^(7/2)*Sqrt[1 + d/(e*x^2)]*x^(15/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]],
-1])/(d^3*Sqrt[(I*Sqrt[d])/Sqrt[e]]*Sqrt[d + e*x^2])))/(1001*x^(13/2))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="fricas")

[Out]

integral(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(15/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="giac")

[Out]

integrate(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(15/2), x)

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {15}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x)

[Out]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),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^(15/2),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 \[ \int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{15/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2),x)

[Out]

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2), 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**(15/2),x)

[Out]

Timed out

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