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

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

[Out]

2*x^(1/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))-4*x^(1/2)*(e*x^2+d)^(1/2)/(d^(1/2)+x*e^(1/2))+4*d^(1/4)*(cos(2*ar
ctan(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^(1/4)/(e*x^2+d)^(
1/2)-2*d^(1/4)*(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)))*Ellipti
cF(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^(1/4)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 \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*}

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Mathematica [C]  time = 0.11, size = 85, normalized size = 0.37 \[ 2 \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {4 \sqrt {e} x^{3/2} \sqrt {\frac {e x^2}{d}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {e x^2}{d}\right )}{3 \sqrt {d+e x^2}} \]

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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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{\sqrt {x}}, x\right ) \]

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

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)

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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