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

Optimal. Leaf size=88 \[ \frac {d \sqrt {-e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 e^{3/2}}+\frac {x \sqrt {d+e x^2}}{4 \sqrt {-e}}+\frac {1}{2} x^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5151, 321, 217, 206} \[ \frac {d \sqrt {-e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 e^{3/2}}+\frac {x \sqrt {d+e x^2}}{4 \sqrt {-e}}+\frac {1}{2} x^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[d + e*x^2])/(4*Sqrt[-e]) + (x^2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/2 + (d*Sqrt[-e]*ArcTanh[(Sqrt[e]
*x)/Sqrt[d + e*x^2]])/(4*e^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 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 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {1}{2} x^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{2} \sqrt {-e} \int \frac {x^2}{\sqrt {d+e x^2}} \, dx\\ &=\frac {x \sqrt {d+e x^2}}{4 \sqrt {-e}}+\frac {1}{2} x^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {d \int \frac {1}{\sqrt {d+e x^2}} \, dx}{4 \sqrt {-e}}\\ &=\frac {x \sqrt {d+e x^2}}{4 \sqrt {-e}}+\frac {1}{2} x^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {d \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{4 \sqrt {-e}}\\ &=\frac {x \sqrt {d+e x^2}}{4 \sqrt {-e}}+\frac {1}{2} x^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {d \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 \sqrt {-e^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 59, normalized size = 0.67 \[ \frac {\left (d+2 e x^2\right ) \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\sqrt {-e} x \sqrt {d+e x^2}}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[-e]*x*Sqrt[d + e*x^2]) + (d + 2*e*x^2)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(4*e)

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fricas [A]  time = 0.69, size = 49, normalized size = 0.56 \[ -\frac {\sqrt {e x^{2} + d} \sqrt {-e} x - {\left (2 \, e x^{2} + d\right )} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{4 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(sqrt(e*x^2 + d)*sqrt(-e)*x - (2*e*x^2 + d)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)))/e

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giac [A]  time = 0.25, size = 62, normalized size = 0.70 \[ \frac {1}{2} \, x^{2} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {x^{2} e + d}}\right ) - \frac {1}{4} \, d \arcsin \left (\frac {x e}{\sqrt {-d e}}\right ) e^{\left (-1\right )} - \frac {1}{4} \, \sqrt {-x^{2} e^{2} - d e} x e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(x*sqrt(-e)/sqrt(x^2*e + d)) - 1/4*d*arcsin(x*e/sqrt(-d*e))*e^(-1) - 1/4*sqrt(-x^2*e^2 - d*e)*x*
e^(-1)

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maple [A]  time = 0.04, size = 116, normalized size = 1.32 \[ \frac {x^{2} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{2}+\frac {\sqrt {-e}\, x^{3} \sqrt {e \,x^{2}+d}}{8 d}-\frac {\sqrt {-e}\, x \sqrt {e \,x^{2}+d}}{8 e}+\frac {\sqrt {-e}\, d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{4 e^{\frac {3}{2}}}-\frac {\sqrt {-e}\, x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{8 d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+1/8*(-e)^(1/2)/d*x^3*(e*x^2+d)^(1/2)-1/8*(-e)^(1/2)/e*x*(e*x^2+d)
^(1/2)+1/4*(-e)^(1/2)/e^(3/2)*d*ln(x*e^(1/2)+(e*x^2+d)^(1/2))-1/8*(-e)^(1/2)/d*x*(e*x^2+d)^(3/2)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.99, size = 71, normalized size = 0.81 \[ \begin {cases} \frac {i d \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{4 e} + \frac {i x^{2} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{2} - \frac {i x \sqrt {d + e x^{2}}}{4 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*d*atanh(sqrt(e)*x/sqrt(d + e*x**2))/(4*e) + I*x**2*atanh(sqrt(e)*x/sqrt(d + e*x**2))/2 - I*x*sqrt
(d + e*x**2)/(4*sqrt(e)), Ne(e, 0)), (0, True))

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