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

Optimal. Leaf size=232 \[ -\frac{2 \sqrt [4]{d} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{\sqrt [4]{e} \sqrt{d+e x^2}}-\frac{4 \sqrt{x} \sqrt{d+e x^2}}{\sqrt{d}+\sqrt{e} x}+2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{4 \sqrt [4]{d} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{\sqrt [4]{e} \sqrt{d+e x^2}} \]

[Out]

(-4*Sqrt[x]*Sqrt[d + e*x^2])/(Sqrt[d] + Sqrt[e]*x) + 2*Sqrt[x]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] + (4*d^(1/
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])/(e^(1/4)*Sqrt[d + e*x^2]) - (2*d^(1/4)*(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])/(e^(1/4)*Sqrt[d + e*x^2])

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Rubi [A]  time = 0.135636, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6221, 329, 305, 220, 1196} \[ -\frac{4 \sqrt{x} \sqrt{d+e x^2}}{\sqrt{d}+\sqrt{e} x}+2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{2 \sqrt [4]{d} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{\sqrt [4]{e} \sqrt{d+e x^2}}+\frac{4 \sqrt [4]{d} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{\sqrt [4]{e} \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[x]*Sqrt[d + e*x^2])/(Sqrt[d] + Sqrt[e]*x) + 2*Sqrt[x]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] + (4*d^(1/
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])/(e^(1/4)*Sqrt[d + e*x^2]) - (2*d^(1/4)*(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])/(e^(1/4)*Sqrt[d + e*x^2])

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]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{\sqrt{x}} \, dx &=2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\left (2 \sqrt{e}\right ) \int \frac{\sqrt{x}}{\sqrt{d+e x^2}} \, dx\\ &=2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\left (4 \sqrt{e}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )\\ &=2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\left (4 \sqrt{d}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )+\left (4 \sqrt{d}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{e} x^2}{\sqrt{d}}}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{4 \sqrt{x} \sqrt{d+e x^2}}{\sqrt{d}+\sqrt{e} x}+2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{4 \sqrt [4]{d} \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 )}{\sqrt [4]{e} \sqrt{d+e x^2}}-\frac{2 \sqrt [4]{d} \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 )}{\sqrt [4]{e} \sqrt{d+e x^2}}\\ \end{align*}

Mathematica [C]  time = 0.10397, size = 85, normalized size = 0.37 \[ 2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{4 \sqrt{e} x^{3/2} \sqrt{\frac{e x^2}{d}+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-\frac{e x^2}{d}\right )}{3 \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.944, size = 0, normalized size = 0. \begin{align*} \int{{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ){\frac{1}{\sqrt{x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -2 \, d \sqrt{e} \int \frac{\sqrt{e x^{2} + d} x}{{\left (e x^{2} + d\right )} e^{\left (\log \left (e x^{2} + d\right ) + \frac{1}{2} \, \log \left (x\right )\right )} -{\left (e^{2} x^{4} + d e x^{2}\right )} \sqrt{x}}\,{d x} + \sqrt{x} \log \left (\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - \sqrt{x} \log \left (-\sqrt{e} x + \sqrt{e x^{2} + d}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*d*sqrt(e)*integrate(sqrt(e*x^2 + d)*x/((e*x^2 + d)*e^(log(e*x^2 + d) + 1/2*log(x)) - (e^2*x^4 + d*e*x^2)*sq
rt(x)), x) + sqrt(x)*log(sqrt(e)*x + sqrt(e*x^2 + d)) - sqrt(x)*log(-sqrt(e)*x + sqrt(e*x^2 + d))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right )}{\sqrt{x}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{\sqrt{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right )}{\sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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