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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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^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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.45, size = 65, normalized size = 0.56 \[ -\frac {{\left (2 \, e x^{3} - 3 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {-e} - {\left (8 \, e^{2} x^{4} - 3 \, d^{2}\right )} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{32 \, e^{2}} \]

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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giac [A]  time = 0.25, size = 75, normalized size = 0.65 \[ \frac {1}{4} \, x^{4} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {x^{2} e + d}}\right ) + \frac {3}{32} \, d^{2} \arcsin \left (\frac {x e}{\sqrt {-d e}}\right ) e^{\left (-2\right )} - \frac {1}{32} \, \sqrt {-x^{2} e^{2} - d e} {\left (2 \, x^{2} e^{\left (-1\right )} - 3 \, d e^{\left (-2\right )}\right )} x \]

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

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maple [A]  time = 0.05, size = 163, normalized size = 1.41 \[ \frac {x^{4} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{4}+\frac {\sqrt {-e}\, x^{5} \sqrt {e \,x^{2}+d}}{24 d}-\frac {5 \sqrt {-e}\, x^{3} \sqrt {e \,x^{2}+d}}{96 e}+\frac {\sqrt {-e}\, d x \sqrt {e \,x^{2}+d}}{16 e^{2}}-\frac {3 \sqrt {-e}\, d^{2} \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{32 e^{\frac {5}{2}}}-\frac {\sqrt {-e}\, x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{24 d e}+\frac {\sqrt {-e}\, x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{32 e^{2}} \]

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(-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^3*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 \[ \int x^3\,\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.46, size = 102, normalized size = 0.88 \[ \begin {cases} - \frac {3 i d^{2} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{32 e^{2}} + \frac {3 i d x \sqrt {d + e x^{2}}}{32 e^{\frac {3}{2}}} + \frac {i x^{4} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{4} - \frac {i x^{3} \sqrt {d + e x^{2}}}{16 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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