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

Optimal. Leaf size=116 \[ -\frac{3 d^2 \sqrt{-e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{32 e^{5/2}}+\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{-e}}+\frac{3 d x \sqrt{d+e x^2}}{32 (-e)^{3/2}}+\frac{1}{4} x^4 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right ) \]

[Out]

(3*d*x*Sqrt[d + e*x^2])/(32*(-e)^(3/2)) + (x^3*Sqrt[d + e*x^2])/(16*Sqrt[-e]) + (x^4*ArcTan[(Sqrt[-e]*x)/Sqrt[
d + e*x^2]])/4 - (3*d^2*Sqrt[-e]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(32*e^(5/2))

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Rubi [A]  time = 0.0419833, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {5151, 321, 217, 206} \[ -\frac{3 d^2 \sqrt{-e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{32 e^{5/2}}+\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{-e}}+\frac{3 d x \sqrt{d+e x^2}}{32 (-e)^{3/2}}+\frac{1}{4} x^4 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*x*Sqrt[d + e*x^2])/(32*(-e)^(3/2)) + (x^3*Sqrt[d + e*x^2])/(16*Sqrt[-e]) + (x^4*ArcTan[(Sqrt[-e]*x)/Sqrt[
d + e*x^2]])/4 - (3*d^2*Sqrt[-e]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(32*e^(5/2))

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]

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 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 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])

Rubi steps

\begin{align*} \int x^3 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right ) \, dx &=\frac{1}{4} x^4 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{4} \sqrt{-e} \int \frac{x^4}{\sqrt{d+e x^2}} \, dx\\ &=\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{-e}}+\frac{1}{4} x^4 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{(3 d) \int \frac{x^2}{\sqrt{d+e x^2}} \, dx}{16 \sqrt{-e}}\\ &=\frac{3 d x \sqrt{d+e x^2}}{32 (-e)^{3/2}}+\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{-e}}+\frac{1}{4} x^4 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (3 d^2\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{32 (-e)^{3/2}}\\ &=\frac{3 d x \sqrt{d+e x^2}}{32 (-e)^{3/2}}+\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{-e}}+\frac{1}{4} x^4 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{32 (-e)^{3/2}}\\ &=\frac{3 d x \sqrt{d+e x^2}}{32 (-e)^{3/2}}+\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{-e}}+\frac{1}{4} x^4 \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )-\frac{3 d^2 \sqrt{-e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{32 e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0528404, size = 74, normalized size = 0.64 \[ \frac{\left (8 e^2 x^4-3 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )+\sqrt{-e} x \sqrt{d+e x^2} \left (3 d-2 e x^2\right )}{32 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-e]*x*(3*d - 2*e*x^2)*Sqrt[d + e*x^2] + (-3*d^2 + 8*e^2*x^4)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(32*e
^2)

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Maple [A]  time = 0.041, size = 163, normalized size = 1.4 \begin{align*}{\frac{{x}^{4}}{4}\arctan \left ({x\sqrt{-e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{{x}^{5}}{24\,d}\sqrt{-e}\sqrt{e{x}^{2}+d}}-{\frac{5\,{x}^{3}}{96\,e}\sqrt{-e}\sqrt{e{x}^{2}+d}}+{\frac{dx}{16\,{e}^{2}}\sqrt{-e}\sqrt{e{x}^{2}+d}}-{\frac{3\,{d}^{2}}{32}\sqrt{-e}\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{5}{2}}}}-{\frac{{x}^{3}}{24\,de}\sqrt{-e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{x}{32\,{e}^{2}}\sqrt{-e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+1/24*(-e)^(1/2)/d*x^5*(e*x^2+d)^(1/2)-5/96*(-e)^(1/2)/e*x^3*(e*x^
2+d)^(1/2)+1/16*(-e)^(1/2)/e^2*d*x*(e*x^2+d)^(1/2)-3/32*(-e)^(1/2)/e^(5/2)*d^2*ln(x*e^(1/2)+(e*x^2+d)^(1/2))-1
/24*(-e)^(1/2)/d*x^3*(e*x^2+d)^(3/2)/e+1/32*(-e)^(1/2)/e^2*x*(e*x^2+d)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (\sqrt{-e} x, \sqrt{e x^{2} + d}\right ) - d \sqrt{-e} \int -\frac{\sqrt{e x^{2} + d} x^{4}}{4 \,{\left (e^{2} x^{4} + d e x^{2} -{\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/4*x^4*arctan2(sqrt(-e)*x, sqrt(e*x^2 + d)) - d*sqrt(-e)*integrate(-1/4*sqrt(e*x^2 + d)*x^4/(e^2*x^4 + d*e*x^
2 - (e*x^2 + d)^2), x)

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Fricas [A]  time = 2.27933, size = 153, normalized size = 1.32 \begin{align*} -\frac{{\left (2 \, e x^{3} - 3 \, d x\right )} \sqrt{e x^{2} + d} \sqrt{-e} -{\left (8 \, e^{2} x^{4} - 3 \, d^{2}\right )} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right )}{32 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

-1/32*((2*e*x^3 - 3*d*x)*sqrt(e*x^2 + d)*sqrt(-e) - (8*e^2*x^4 - 3*d^2)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)))/e^
2

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Sympy [A]  time = 2.47892, size = 102, normalized size = 0.88 \begin{align*} \begin{cases} - \frac{3 i d^{2} \operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{32 e^{2}} + \frac{3 i d x \sqrt{d + e x^{2}}}{32 e^{\frac{3}{2}}} + \frac{i x^{4} \operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{4} - \frac{i x^{3} \sqrt{d + e x^{2}}}{16 \sqrt{e}} & \text{for}\: e \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*I*d**2*atanh(sqrt(e)*x/sqrt(d + e*x**2))/(32*e**2) + 3*I*d*x*sqrt(d + e*x**2)/(32*e**(3/2)) + I*
x**4*atanh(sqrt(e)*x/sqrt(d + e*x**2))/4 - I*x**3*sqrt(d + e*x**2)/(16*sqrt(e)), Ne(e, 0)), (0, True))

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Giac [A]  time = 1.16984, size = 101, normalized size = 0.87 \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (\frac{x \sqrt{-e}}{\sqrt{x^{2} e + d}}\right ) + \frac{3}{32} \, d^{2} \arcsin \left (\frac{x e}{\sqrt{-d e}}\right ) e^{\left (-2\right )} - \frac{1}{32} \, \sqrt{-x^{2} e^{2} - d e}{\left (2 \, x^{2} e^{\left (-1\right )} - 3 \, d e^{\left (-2\right )}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

1/4*x^4*arctan(x*sqrt(-e)/sqrt(x^2*e + d)) + 3/32*d^2*arcsin(x*e/sqrt(-d*e))*e^(-2) - 1/32*sqrt(-x^2*e^2 - d*e
)*(2*x^2*e^(-1) - 3*d*e^(-2))*x