3.20 \(\int \frac {\tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^{7/2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac {2 e^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{15 d^{5/4} \sqrt {d+e x^2}}-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}} \]

[Out]

-2/5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(5/2)-4/15*e^(1/2)*(e*x^2+d)^(1/2)/d/x^(3/2)-2/15*e^(5/4)*(cos(2*arc
tan(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)/d^(5/4)/(e*x^2+d)^(1
/2)

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Rubi [A]  time = 0.07, antiderivative size = 145, 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 = {6221, 325, 329, 220} \[ -\frac {2 e^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{15 d^{5/4} \sqrt {d+e x^2}}-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]

[Out]

(-4*Sqrt[e]*Sqrt[d + e*x^2])/(15*d*x^(3/2)) - (2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(5*x^(5/2)) - (2*e^(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])/(15*d^(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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 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 )}{x^{7/2}} \, dx &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}+\frac {1}{5} \left (2 \sqrt {e}\right ) \int \frac {1}{x^{5/2} \sqrt {d+e x^2}} \, dx\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}-\frac {\left (2 e^{3/2}\right ) \int \frac {1}{\sqrt {x} \sqrt {d+e x^2}} \, dx}{15 d}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}-\frac {\left (4 e^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{15 d}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}-\frac {2 e^{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 )}{15 d^{5/4} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 142, normalized size = 0.98 \[ -\frac {2 \left (2 \sqrt {e} x \sqrt {d+e x^2}+3 d \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right )}{15 d x^{5/2}}-\frac {4 e^2 x \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {\frac {d}{e x^2}+1} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{15 d^{3/2} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]

[Out]

(-2*(2*Sqrt[e]*x*Sqrt[d + e*x^2] + 3*d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]))/(15*d*x^(5/2)) - (4*Sqrt[(I*Sqrt
[d])/Sqrt[e]]*e^2*Sqrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1])/(15*d^(3/
2)*Sqrt[d + e*x^2])

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="fricas")

[Out]

integral(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="giac")

[Out]

integrate(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(7/2), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x)

[Out]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, d \sqrt {e} \int -\frac {\sqrt {e x^{2} + d} x}{5 \, {\left ({\left (e^{2} x^{4} + d e x^{2}\right )} x^{\frac {7}{2}} - {\left (e x^{2} + d\right )} e^{\left (\log \left (e x^{2} + d\right ) + \frac {7}{2} \, \log \relax (x)\right )}\right )}}\,{d x} - \frac {\log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{5 \, x^{\frac {5}{2}}} + \frac {\log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="maxima")

[Out]

2*d*sqrt(e)*integrate(-1/5*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(7/2) - (e*x^2 + d)*e^(log(e*x^2 + d) + 7/
2*log(x))), x) - 1/5*log(sqrt(e)*x + sqrt(e*x^2 + d))/x^(5/2) + 1/5*log(-sqrt(e)*x + sqrt(e*x^2 + d))/x^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(7/2),x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**(7/2),x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**(7/2), x)

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