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

Optimal. Leaf size=74 \[ -\frac {\left (d+e x^2\right )^{3/2}}{9 (-e)^{3/2}}+\frac {d \sqrt {d+e x^2}}{3 (-e)^{3/2}}+\frac {1}{3} x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

-1/9*(e*x^2+d)^(3/2)/(-e)^(3/2)+1/3*x^3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+1/3*d*(e*x^2+d)^(1/2)/(-e)^(3/2)

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5151, 266, 43} \[ -\frac {\left (d+e x^2\right )^{3/2}}{9 (-e)^{3/2}}+\frac {d \sqrt {d+e x^2}}{3 (-e)^{3/2}}+\frac {1}{3} x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d + e*x^2])/(3*(-e)^(3/2)) - (d + e*x^2)^(3/2)/(9*(-e)^(3/2)) + (x^3*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^
2]])/3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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^2 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {1}{3} x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{3} \sqrt {-e} \int \frac {x^3}{\sqrt {d+e x^2}} \, dx\\ &=\frac {1}{3} x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{6} \sqrt {-e} \operatorname {Subst}\left (\int \frac {x}{\sqrt {d+e x}} \, dx,x,x^2\right )\\ &=\frac {1}{3} x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{6} \sqrt {-e} \operatorname {Subst}\left (\int \left (-\frac {d}{e \sqrt {d+e x}}+\frac {\sqrt {d+e x}}{e}\right ) \, dx,x,x^2\right )\\ &=\frac {d \sqrt {d+e x^2}}{3 (-e)^{3/2}}-\frac {\left (d+e x^2\right )^{3/2}}{9 (-e)^{3/2}}+\frac {1}{3} x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 60, normalized size = 0.81 \[ \frac {1}{9} \left (\frac {\left (2 d-e x^2\right ) \sqrt {d+e x^2}}{(-e)^{3/2}}+3 x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*d - e*x^2)*Sqrt[d + e*x^2])/(-e)^(3/2) + 3*x^3*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/9

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fricas [A]  time = 0.44, size = 56, normalized size = 0.76 \[ \frac {3 \, e^{2} x^{3} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - \sqrt {e x^{2} + d} {\left (e x^{2} - 2 \, d\right )} \sqrt {-e}}{9 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*e^2*x^3*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - sqrt(e*x^2 + d)*(e*x^2 - 2*d)*sqrt(-e))/e^2

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giac [A]  time = 0.23, size = 64, normalized size = 0.86 \[ \frac {1}{3} \, x^{3} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {x^{2} e + d}}\right ) + \frac {1}{3} \, \sqrt {-x^{2} e^{2} - d e} d e^{\left (-2\right )} + \frac {1}{9} \, {\left (-x^{2} e^{2} - d e\right )}^{\frac {3}{2}} e^{\left (-3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(x*sqrt(-e)/sqrt(x^2*e + d)) + 1/3*sqrt(-x^2*e^2 - d*e)*d*e^(-2) + 1/9*(-x^2*e^2 - d*e)^(3/2)*e^
(-3)

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maple [B]  time = 0.04, size = 132, normalized size = 1.78 \[ \frac {x^{3} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{3}+\frac {\sqrt {-e}\, x^{4} \sqrt {e \,x^{2}+d}}{15 d}-\frac {4 \sqrt {-e}\, x^{2} \sqrt {e \,x^{2}+d}}{45 e}+\frac {8 \sqrt {-e}\, d \sqrt {e \,x^{2}+d}}{45 e^{2}}-\frac {\sqrt {-e}\, x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d e}+\frac {2 \sqrt {-e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{45 e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+1/15*(-e)^(1/2)/d*x^4*(e*x^2+d)^(1/2)-4/45*(-e)^(1/2)/e*x^2*(e*x^
2+d)^(1/2)+8/45*(-e)^(1/2)/e^2*d*(e*x^2+d)^(1/2)-1/15*(-e)^(1/2)/d*x^2*(e*x^2+d)^(3/2)/e+2/45*(-e)^(1/2)/e^2*(
e*x^2+d)^(3/2)

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maxima [A]  time = 0.34, size = 111, normalized size = 1.50 \[ \frac {1}{3} \, x^{3} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {{\left (3 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} - 5 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d\right )} \sqrt {-e}}{45 \, d e^{2}} + \frac {{\left (3 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x^{2} + d} d^{2}\right )} \sqrt {-e}}{45 \, d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - 1/45*(3*(e*x^2 + d)^(5/2) - 5*(e*x^2 + d)^(3/2)*d)*sqrt(-e)/(d*e^
2) + 1/45*(3*(e*x^2 + d)^(5/2) - 10*(e*x^2 + d)^(3/2)*d + 15*sqrt(e*x^2 + d)*d^2)*sqrt(-e)/(d*e^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.24, size = 70, normalized size = 0.95 \[ \begin {cases} \frac {2 i d \sqrt {d + e x^{2}}}{9 e^{\frac {3}{2}}} + \frac {i x^{3} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{3} - \frac {i x^{2} \sqrt {d + e x^{2}}}{9 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*I*d*sqrt(d + e*x**2)/(9*e**(3/2)) + I*x**3*atanh(sqrt(e)*x/sqrt(d + e*x**2))/3 - I*x**2*sqrt(d +
e*x**2)/(9*sqrt(e)), Ne(e, 0)), (0, True))

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