3.1.6 \(\int \frac {\tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^5} \, dx\) [6]

Optimal. Leaf size=79 \[ -\frac {\sqrt {e} \sqrt {d+e x^2}}{12 d x^3}+\frac {e^{3/2} \sqrt {d+e x^2}}{6 d^2 x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 x^4} \]

[Out]

-1/4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^4+1/6*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x-1/12*e^(1/2)*(e*x^2+d)^(1/2)/d/x
^3

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Rubi [A]
time = 0.02, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6356, 277, 270} \begin {gather*} \frac {e^{3/2} \sqrt {d+e x^2}}{6 d^2 x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 x^4}-\frac {\sqrt {e} \sqrt {d+e x^2}}{12 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^5,x]

[Out]

-1/12*(Sqrt[e]*Sqrt[d + e*x^2])/(d*x^3) + (e^(3/2)*Sqrt[d + e*x^2])/(6*d^2*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e
*x^2]]/(4*x^4)

Rule 270

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

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 \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^5} \, dx &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 x^4}+\frac {1}{4} \sqrt {e} \int \frac {1}{x^4 \sqrt {d+e x^2}} \, dx\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{12 d x^3}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 x^4}-\frac {e^{3/2} \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx}{6 d}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{12 d x^3}+\frac {e^{3/2} \sqrt {d+e x^2}}{6 d^2 x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 63, normalized size = 0.80 \begin {gather*} \frac {\sqrt {e} x \sqrt {d+e x^2} \left (-d+2 e x^2\right )-3 d^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{12 d^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^5,x]

[Out]

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-d + 2*e*x^2) - 3*d^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(12*d^2*x^4)

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Maple [A]
time = 0.01, size = 62, normalized size = 0.78

method	result	size
default	\(-\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{4 x^{4}}+\frac {e^{\frac {3}{2}} \sqrt {e \,x^{2}+d}}{4 d^{2} x}-\frac {\sqrt {e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{12 d^{2} x^{3}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^4+1/4*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x-1/12*e^(1/2)/d^2/x^3*(e*x^2+d)^(
3/2)

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Maxima [A]
time = 0.28, size = 61, normalized size = 0.77 \begin {gather*} \frac {\sqrt {x^{2} e + d} e^{\frac {3}{2}}}{4 \, d^{2} x} - \frac {{\left (x^{2} e + d\right )}^{\frac {3}{2}} e^{\frac {1}{2}}}{12 \, d^{2} x^{3}} - \frac {\operatorname {artanh}\left (\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}}\right )}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^5,x, algorithm="maxima")

[Out]

1/4*sqrt(x^2*e + d)*e^(3/2)/(d^2*x) - 1/12*(x^2*e + d)^(3/2)*e^(1/2)/(d^2*x^3) - 1/4*arctanh(x*e^(1/2)/sqrt(x^
2*e + d))/x^4

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (61) = 122\).
time = 0.35, size = 176, normalized size = 2.23 \begin {gather*} -\frac {3 \, d^{2} \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 ) - 2 \, {\left (2 \, x^{3} \cosh \left (\frac {1}{2}\right )^{3} + 6 \, x^{3} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{2} + 2 \, x^{3} \sinh \left (\frac {1}{2}\right )^{3} - d x \cosh \left (\frac {1}{2}\right ) + {\left (6 \, x^{3} \cosh \left (\frac {1}{2}\right )^{2} - d x\right )} \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 )}}}{24 \, d^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^5,x, algorithm="fricas")

[Out]

-1/24*(3*d^2*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) - 2*(2*x^3*cosh(1/2)^3
+ 6*x^3*cosh(1/2)*sinh(1/2)^2 + 2*x^3*sinh(1/2)^3 - d*x*cosh(1/2) + (6*x^3*cosh(1/2)^2 - d*x)*sinh(1/2))*sqrt
(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2) - sinh(1/2))))/(d^2*x^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**5,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**5, x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^5,x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^5,x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^5, x)

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