3.1.18 \(\int \sqrt {x} \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\) [18]

Optimal. Leaf size=142 \[ -\frac {4 \sqrt {x} \sqrt {d+e x^2}}{9 \sqrt {e}}+\frac {2}{3} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {2 d^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{9 e^{3/4} \sqrt {d+e x^2}} \]

[Out]

2/3*x^(3/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))-4/9*x^(1/2)*(e*x^2+d)^(1/2)/e^(1/2)+2/9*d^(3/4)*(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^(3/4)/(e*x^2+d)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6356, 327, 335, 226} \begin {gather*} \frac {2 d^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{9 e^{3/4} \sqrt {d+e x^2}}-\frac {4 \sqrt {x} \sqrt {d+e x^2}}{9 \sqrt {e}}+\frac {2}{3} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[x]*Sqrt[d + e*x^2])/(9*Sqrt[e]) + (2*x^(3/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/3 + (2*d^(3/4)*(Sq
rt[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])/(9*e^(3/4)*Sqrt[d + e*x^2])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 327

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 335

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 6356

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 \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {2}{3} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{3} \left (2 \sqrt {e}\right ) \int \frac {x^{3/2}}{\sqrt {d+e x^2}} \, dx\\ &=-\frac {4 \sqrt {x} \sqrt {d+e x^2}}{9 \sqrt {e}}+\frac {2}{3} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {(2 d) \int \frac {1}{\sqrt {x} \sqrt {d+e x^2}} \, dx}{9 \sqrt {e}}\\ &=-\frac {4 \sqrt {x} \sqrt {d+e x^2}}{9 \sqrt {e}}+\frac {2}{3} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {(4 d) \text {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{9 \sqrt {e}}\\ &=-\frac {4 \sqrt {x} \sqrt {d+e x^2}}{9 \sqrt {e}}+\frac {2}{3} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {2 d^{3/4} \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 )}{9 e^{3/4} \sqrt {d+e x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.20, size = 135, normalized size = 0.95 \begin {gather*} \frac {2}{9} \sqrt {x} \left (-\frac {2 \sqrt {d+e x^2}}{\sqrt {e}}+3 x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right )+\frac {4 \sqrt {d} \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {1+\frac {d}{e x^2}} x F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{9 \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*((-2*Sqrt[d + e*x^2])/Sqrt[e] + 3*x*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]))/9 + (4*Sqrt[d]*Sqrt[(I*S
qrt[d])/Sqrt[e]]*Sqrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1])/(9*Sqrt[d
+ e*x^2])

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 188, normalized size = 1.32 \begin {gather*} \frac {3 \, {\left (x \cosh \left (\frac {1}{2}\right )^{2} + 2 \, x \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + x \sinh \left (\frac {1}{2}\right )^{2}\right )} \sqrt {x} \log \left (\frac {2 \, x^{2} \cosh \left (\frac {1}{2}\right )^{2} + 4 \, x^{2} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + 2 \, x^{2} \sinh \left (\frac {1}{2}\right )^{2} + 2 \, {\left (x \cosh \left (\frac {1}{2}\right ) + x \sinh \left (\frac {1}{2}\right )\right )} \sqrt {\frac {{\left (x^{2} + d\right )} \cosh \left (\frac {1}{2}\right ) + {\left (x^{2} - d\right )} \sinh \left (\frac {1}{2}\right )}{\cosh \left (\frac {1}{2}\right ) - \sinh \left (\frac {1}{2}\right )}} + d}{d}\right ) - 4 \, \sqrt {x} \sqrt {\frac {{\left (x^{2} + d\right )} \cosh \left (\frac {1}{2}\right ) + {\left (x^{2} - d\right )} \sinh \left (\frac {1}{2}\right )}{\cosh \left (\frac {1}{2}\right ) - \sinh \left (\frac {1}{2}\right )}} {\left (\cosh \left (\frac {1}{2}\right ) + \sinh \left (\frac {1}{2}\right )\right )} + 4 \, d {\rm weierstrassPInverse}\left (-\frac {4 \, d}{\cosh \left (\frac {1}{2}\right )^{2} + 2 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + \sinh \left (\frac {1}{2}\right )^{2}}, 0, x\right )}{9 \, {\left (\cosh \left (\frac {1}{2}\right )^{2} + 2 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + \sinh \left (\frac {1}{2}\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*(x*cosh(1/2)^2 + 2*x*cosh(1/2)*sinh(1/2) + x*sinh(1/2)^2)*sqrt(x)*log((2*x^2*cosh(1/2)^2 + 4*x^2*cosh(1
/2)*sinh(1/2) + 2*x^2*sinh(1/2)^2 + 2*(x*cosh(1/2) + x*sinh(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1
/2))/(cosh(1/2) - sinh(1/2))) + d)/d) - 4*sqrt(x)*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2)
- sinh(1/2)))*(cosh(1/2) + sinh(1/2)) + 4*d*weierstrassPInverse(-4*d/(cosh(1/2)^2 + 2*cosh(1/2)*sinh(1/2) + si
nh(1/2)^2), 0, x))/(cosh(1/2)^2 + 2*cosh(1/2)*sinh(1/2) + sinh(1/2)^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {x}\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________