3.24 \(\int x^{3/2} \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\)

Optimal. Leaf size=269 \[ \frac {6 d^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{25 e^{5/4} \sqrt {d+e x^2}}-\frac {12 d^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{25 e^{5/4} \sqrt {d+e x^2}}+\frac {12 d \sqrt {x} \sqrt {d+e x^2}}{25 e \left (\sqrt {d}+\sqrt {e} x\right )}-\frac {4 x^{3/2} \sqrt {d+e x^2}}{25 \sqrt {e}}+\frac {2}{5} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

2/5*x^(5/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))-4/25*x^(3/2)*(e*x^2+d)^(1/2)/e^(1/2)+12/25*d*x^(1/2)*(e*x^2+d)^
(1/2)/e/(d^(1/2)+x*e^(1/2))-12/25*d^(5/4)*(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)))*EllipticE(sin(2*arctan(e^(1/4)*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(d^(1/2)+x*e^(1/2))*((e*x^2+
d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(5/4)/(e*x^2+d)^(1/2)+6/25*d^(5/4)*(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))*(
d^(1/2)+x*e^(1/2))*((e*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(5/4)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6221, 321, 329, 305, 220, 1196} \[ \frac {6 d^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{25 e^{5/4} \sqrt {d+e x^2}}-\frac {12 d^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{25 e^{5/4} \sqrt {d+e x^2}}-\frac {4 x^{3/2} \sqrt {d+e x^2}}{25 \sqrt {e}}+\frac {12 d \sqrt {x} \sqrt {d+e x^2}}{25 e \left (\sqrt {d}+\sqrt {e} x\right )}+\frac {2}{5} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*x^(3/2)*Sqrt[d + e*x^2])/(25*Sqrt[e]) + (12*d*Sqrt[x]*Sqrt[d + e*x^2])/(25*e*(Sqrt[d] + Sqrt[e]*x)) + (2*x
^(5/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5 - (12*d^(5/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] +
Sqrt[e]*x)^2]*EllipticE[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(25*e^(5/4)*Sqrt[d + e*x^2]) + (6*d^(5/4)*
(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)],
1/2])/(25*e^(5/4)*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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 6221

Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*ArcT
anh[(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] /;
FreeQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {2}{5} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{5} \left (2 \sqrt {e}\right ) \int \frac {x^{5/2}}{\sqrt {d+e x^2}} \, dx\\ &=-\frac {4 x^{3/2} \sqrt {d+e x^2}}{25 \sqrt {e}}+\frac {2}{5} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {(6 d) \int \frac {\sqrt {x}}{\sqrt {d+e x^2}} \, dx}{25 \sqrt {e}}\\ &=-\frac {4 x^{3/2} \sqrt {d+e x^2}}{25 \sqrt {e}}+\frac {2}{5} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {(12 d) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{25 \sqrt {e}}\\ &=-\frac {4 x^{3/2} \sqrt {d+e x^2}}{25 \sqrt {e}}+\frac {2}{5} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {\left (12 d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{25 e}-\frac {\left (12 d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {e} x^2}{\sqrt {d}}}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{25 e}\\ &=-\frac {4 x^{3/2} \sqrt {d+e x^2}}{25 \sqrt {e}}+\frac {12 d \sqrt {x} \sqrt {d+e x^2}}{25 e \left (\sqrt {d}+\sqrt {e} x\right )}+\frac {2}{5} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {12 d^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{25 e^{5/4} \sqrt {d+e x^2}}+\frac {6 d^{5/4} \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 )}{25 e^{5/4} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 109, normalized size = 0.41 \[ \frac {2 x^{3/2} \left (2 d \sqrt {\frac {e x^2}{d}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {e x^2}{d}\right )-2 \left (d+e x^2\right )+5 \sqrt {e} x \sqrt {d+e x^2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right )}{25 \sqrt {e} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(-2*(d + e*x^2) + 5*Sqrt[e]*x*Sqrt[d + e*x^2]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] + 2*d*Sqrt[1 + (
e*x^2)/d]*Hypergeometric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)]))/(25*Sqrt[e]*Sqrt[d + e*x^2])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{\frac {3}{2}} \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^(3/2)*arctanh(sqrt(e)*x/sqrt(e*x^2 + d)), x)

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giac [A]  time = 0.32, size = 1, normalized size = 0.00 \[ +\infty \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

+Infinity

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{5} \, x^{\frac {5}{2}} \log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \frac {1}{5} \, x^{\frac {5}{2}} \log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - 2 \, d \sqrt {e} \int -\frac {x e^{\left (\frac {1}{2} \, \log \left (e x^{2} + d\right ) + \frac {3}{2} \, \log \relax (x)\right )}}{5 \, {\left (e^{2} x^{4} + d e x^{2} - {\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^(5/2)*log(sqrt(e)*x + sqrt(e*x^2 + d)) - 1/5*x^(5/2)*log(-sqrt(e)*x + sqrt(e*x^2 + d)) - 2*d*sqrt(e)*int
egrate(-1/5*x*e^(1/2*log(e*x^2 + d) + 3/2*log(x))/(e^2*x^4 + d*e*x^2 - (e*x^2 + d)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**(3/2)*atanh(sqrt(e)*x/sqrt(d + e*x**2)), x)

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