3.19 \(\int x^{5/2} \tan ^{-1}(\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}) \, dx\)

Optimal. Leaf size=181 \[ -\frac {10 d^{7/4} \sqrt {-e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{147 e^{9/4} \sqrt {d+e x^2}}+\frac {20 d \sqrt {x} \sqrt {d+e x^2}}{147 (-e)^{3/2}}+\frac {4 x^{5/2} \sqrt {d+e x^2}}{49 \sqrt {-e}}+\frac {2}{7} x^{7/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

2/7*x^(7/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+4/49*x^(5/2)*(e*x^2+d)^(1/2)/(-e)^(1/2)+20/147*d*x^(1/2)*(e*x
^2+d)^(1/2)/(-e)^(3/2)-10/147*d^(7/4)*(cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))^2)^(1/2)/cos(2*arctan(e^(1/4)*x^
(1/2)/d^(1/4)))*EllipticF(sin(2*arctan(e^(1/4)*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(-e)^(1/2)*(d^(1/2)+x*e^(1/2))*(
(e*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(9/4)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5151, 321, 329, 220} \[ -\frac {10 d^{7/4} \sqrt {-e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \text {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{147 e^{9/4} \sqrt {d+e x^2}}+\frac {4 x^{5/2} \sqrt {d+e x^2}}{49 \sqrt {-e}}+\frac {20 d \sqrt {x} \sqrt {d+e x^2}}{147 (-e)^{3/2}}+\frac {2}{7} x^{7/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(20*d*Sqrt[x]*Sqrt[d + e*x^2])/(147*(-e)^(3/2)) + (4*x^(5/2)*Sqrt[d + e*x^2])/(49*Sqrt[-e]) + (2*x^(7/2)*ArcTa
n[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/7 - (10*d^(7/4)*Sqrt[-e]*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sq
rt[e]*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(147*e^(9/4)*Sqrt[d + e*x^2])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && 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^{5/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {2}{7} x^{7/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{7} \left (2 \sqrt {-e}\right ) \int \frac {x^{7/2}}{\sqrt {d+e x^2}} \, dx\\ &=\frac {4 x^{5/2} \sqrt {d+e x^2}}{49 \sqrt {-e}}+\frac {2}{7} x^{7/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {(10 d) \int \frac {x^{3/2}}{\sqrt {d+e x^2}} \, dx}{49 \sqrt {-e}}\\ &=\frac {20 d \sqrt {x} \sqrt {d+e x^2}}{147 (-e)^{3/2}}+\frac {4 x^{5/2} \sqrt {d+e x^2}}{49 \sqrt {-e}}+\frac {2}{7} x^{7/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {\left (10 d^2\right ) \int \frac {1}{\sqrt {x} \sqrt {d+e x^2}} \, dx}{147 (-e)^{3/2}}\\ &=\frac {20 d \sqrt {x} \sqrt {d+e x^2}}{147 (-e)^{3/2}}+\frac {4 x^{5/2} \sqrt {d+e x^2}}{49 \sqrt {-e}}+\frac {2}{7} x^{7/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {\left (20 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{147 (-e)^{3/2}}\\ &=\frac {20 d \sqrt {x} \sqrt {d+e x^2}}{147 (-e)^{3/2}}+\frac {4 x^{5/2} \sqrt {d+e x^2}}{49 \sqrt {-e}}+\frac {2}{7} x^{7/2} \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {10 d^{7/4} \sqrt {-e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{147 e^{9/4} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.46, size = 158, normalized size = 0.87 \[ \frac {2}{147} \sqrt {x} \left (\frac {2 \left (5 d-3 e x^2\right ) \sqrt {d+e x^2}}{(-e)^{3/2}}+21 x^3 \tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\right )-\frac {20 i d^2 x \sqrt {\frac {d}{e x^2}+1} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{147 (-e)^{3/2} \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(2*Sqrt[x]*((2*(5*d - 3*e*x^2)*Sqrt[d + e*x^2])/(-e)^(3/2) + 21*x^3*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]))/147
- (((20*I)/147)*d^2*Sqrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1])/(Sqrt[
(I*Sqrt[d])/Sqrt[e]]*(-e)^(3/2)*Sqrt[d + e*x^2])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

integral(x^(5/2)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:exp(
1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=e
xp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp
(1/2)^2=exp(1)exp(1/2)^2=exp(1)Unable to transpose Error: Bad Argument Value

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int x^{\frac {5}{2}} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x)

[Out]

int(x^(5/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2)),x)

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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^(5/2)*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 \[ \int x^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2)),x)

[Out]

int(x^(5/2)*atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2)), x)

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sympy [C]  time = 150.30, size = 75, normalized size = 0.41 \[ \frac {2 x^{\frac {7}{2}} \operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{7} - \frac {x^{\frac {9}{2}} \sqrt {- e} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{7 \sqrt {d} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*atan(x*(-e)**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

2*x**(7/2)*atan(x*sqrt(-e)/sqrt(d + e*x**2))/7 - x**(9/2)*sqrt(-e)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), e*x**
2*exp_polar(I*pi)/d)/(7*sqrt(d)*gamma(13/4))

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