3.12 \(\int x^4 \tan ^{-1}(\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}) \, dx\)

Optimal. Leaf size=99 \[ \frac {d^2 \sqrt {d+e x^2}}{5 (-e)^{5/2}}+\frac {\left (d+e x^2\right )^{5/2}}{25 (-e)^{5/2}}-\frac {2 d \left (d+e x^2\right )^{3/2}}{15 (-e)^{5/2}}+\frac {1}{5} x^5 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

-2/15*d*(e*x^2+d)^(3/2)/(-e)^(5/2)+1/25*(e*x^2+d)^(5/2)/(-e)^(5/2)+1/5*x^5*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)
)+1/5*d^2*(e*x^2+d)^(1/2)/(-e)^(5/2)

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Rubi [A]  time = 0.05, antiderivative size = 99, 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} \[ \frac {d^2 \sqrt {d+e x^2}}{5 (-e)^{5/2}}+\frac {\left (d+e x^2\right )^{5/2}}{25 (-e)^{5/2}}-\frac {2 d \left (d+e x^2\right )^{3/2}}{15 (-e)^{5/2}}+\frac {1}{5} x^5 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*Sqrt[d + e*x^2])/(5*(-e)^(5/2)) - (2*d*(d + e*x^2)^(3/2))/(15*(-e)^(5/2)) + (d + e*x^2)^(5/2)/(25*(-e)^(5
/2)) + (x^5*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/5

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^4 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {1}{5} x^5 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{5} \sqrt {-e} \int \frac {x^5}{\sqrt {d+e x^2}} \, dx\\ &=\frac {1}{5} x^5 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {-e} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {d+e x}} \, dx,x,x^2\right )\\ &=\frac {1}{5} x^5 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {-e} \operatorname {Subst}\left (\int \left (\frac {d^2}{e^2 \sqrt {d+e x}}-\frac {2 d \sqrt {d+e x}}{e^2}+\frac {(d+e x)^{3/2}}{e^2}\right ) \, dx,x,x^2\right )\\ &=\frac {d^2 \sqrt {d+e x^2}}{5 (-e)^{5/2}}-\frac {2 d \left (d+e x^2\right )^{3/2}}{15 (-e)^{5/2}}+\frac {\left (d+e x^2\right )^{5/2}}{25 (-e)^{5/2}}+\frac {1}{5} x^5 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 72, normalized size = 0.73 \[ \frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{75 (-e)^{5/2}}+\frac {1}{5} x^5 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4))/(75*(-e)^(5/2)) + (x^5*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])
/5

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fricas [A]  time = 0.49, size = 68, normalized size = 0.69 \[ \frac {15 \, e^{3} x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (3 \, e^{2} x^{4} - 4 \, d e x^{2} + 8 \, d^{2}\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{75 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

1/75*(15*e^3*x^5*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (3*e^2*x^4 - 4*d*e*x^2 + 8*d^2)*sqrt(e*x^2 + d)*sqrt(-e)
)/e^3

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giac [A]  time = 0.25, size = 102, normalized size = 1.03 \[ \frac {1}{5} \, x^{5} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {x^{2} e + d}}\right ) - \frac {1}{5} \, \sqrt {-x^{2} e^{2} - d e} d^{2} e^{\left (-3\right )} - \frac {1}{75} \, {\left (10 \, {\left (-x^{2} e^{2} - d e\right )}^{\frac {3}{2}} d e + 3 \, {\left (x^{2} e^{2} + d e\right )}^{2} \sqrt {-x^{2} e^{2} - d e}\right )} e^{\left (-5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

1/5*x^5*arctan(x*sqrt(-e)/sqrt(x^2*e + d)) - 1/5*sqrt(-x^2*e^2 - d*e)*d^2*e^(-3) - 1/75*(10*(-x^2*e^2 - d*e)^(
3/2)*d*e + 3*(x^2*e^2 + d*e)^2*sqrt(-x^2*e^2 - d*e))*e^(-5)

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maple [B]  time = 0.04, size = 180, normalized size = 1.82 \[ \frac {x^{5} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{5}+\frac {\sqrt {-e}\, x^{6} \sqrt {e \,x^{2}+d}}{35 d}-\frac {6 \sqrt {-e}\, x^{4} \sqrt {e \,x^{2}+d}}{175 e}+\frac {8 \sqrt {-e}\, d \,x^{2} \sqrt {e \,x^{2}+d}}{175 e^{2}}-\frac {16 \sqrt {-e}\, d^{2} \sqrt {e \,x^{2}+d}}{175 e^{3}}-\frac {\sqrt {-e}\, x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{35 d e}+\frac {4 \sqrt {-e}\, x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{175 e^{2}}-\frac {8 \sqrt {-e}\, d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{525 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+1/35*(-e)^(1/2)/d*x^6*(e*x^2+d)^(1/2)-6/175*(-e)^(1/2)/e*x^4*(e*x
^2+d)^(1/2)+8/175*(-e)^(1/2)/e^2*d*x^2*(e*x^2+d)^(1/2)-16/175*(-e)^(1/2)/e^3*d^2*(e*x^2+d)^(1/2)-1/35*(-e)^(1/
2)/d*x^4*(e*x^2+d)^(3/2)/e+4/175*(-e)^(1/2)/e^2*x^2*(e*x^2+d)^(3/2)-8/525*(-e)^(1/2)*d/e^3*(e*x^2+d)^(3/2)

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maxima [A]  time = 0.34, size = 139, normalized size = 1.40 \[ \frac {1}{5} \, x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {{\left (15 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 42 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}\right )} \sqrt {-e}}{525 \, d e^{3}} + \frac {{\left (5 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x^{2} + d} d^{3}\right )} \sqrt {-e}}{175 \, d e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/5*x^5*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - 1/525*(15*(e*x^2 + d)^(7/2) - 42*(e*x^2 + d)^(5/2)*d + 35*(e*x^2
+ d)^(3/2)*d^2)*sqrt(-e)/(d*e^3) + 1/175*(5*(e*x^2 + d)^(7/2) - 21*(e*x^2 + d)^(5/2)*d + 35*(e*x^2 + d)^(3/2)*
d^2 - 35*sqrt(e*x^2 + d)*d^3)*sqrt(-e)/(d*e^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2)),x)

[Out]

int(x^4*atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2)), x)

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sympy [A]  time = 3.56, size = 97, normalized size = 0.98 \[ \begin {cases} - \frac {8 i d^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {5}{2}}} + \frac {4 i d x^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {3}{2}}} + \frac {i x^{5} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{5} - \frac {i x^{4} \sqrt {d + e x^{2}}}{25 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*atan(x*(-e)**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((-8*I*d**2*sqrt(d + e*x**2)/(75*e**(5/2)) + 4*I*d*x**2*sqrt(d + e*x**2)/(75*e**(3/2)) + I*x**5*atanh
(sqrt(e)*x/sqrt(d + e*x**2))/5 - I*x**4*sqrt(d + e*x**2)/(25*sqrt(e)), Ne(e, 0)), (0, True))

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