3.1.8 \(\int \frac {\tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^9} \, dx\) [8]

Optimal. Leaf size=131 \[ -\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{140 d^2 x^5}-\frac {e^{5/2} \sqrt {d+e x^2}}{35 d^3 x^3}+\frac {2 e^{7/2} \sqrt {d+e x^2}}{35 d^4 x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8} \]

[Out]

-1/8*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^8+3/140*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x^5-1/35*e^(5/2)*(e*x^2+d)^(1/2)
/d^3/x^3+2/35*e^(7/2)*(e*x^2+d)^(1/2)/d^4/x-1/56*e^(1/2)*(e*x^2+d)^(1/2)/d/x^7

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Rubi [A]
time = 0.03, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {2 e^{7/2} \sqrt {d+e x^2}}{35 d^4 x}-\frac {e^{5/2} \sqrt {d+e x^2}}{35 d^3 x^3}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{140 d^2 x^5}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}-\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^9,x]

[Out]

-1/56*(Sqrt[e]*Sqrt[d + e*x^2])/(d*x^7) + (3*e^(3/2)*Sqrt[d + e*x^2])/(140*d^2*x^5) - (e^(5/2)*Sqrt[d + e*x^2]
)/(35*d^3*x^3) + (2*e^(7/2)*Sqrt[d + e*x^2])/(35*d^4*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(8*x^8)

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^9} \, dx &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}+\frac {1}{8} \sqrt {e} \int \frac {1}{x^8 \sqrt {d+e x^2}} \, dx\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}-\frac {\left (3 e^{3/2}\right ) \int \frac {1}{x^6 \sqrt {d+e x^2}} \, dx}{28 d}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{140 d^2 x^5}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}+\frac {\left (3 e^{5/2}\right ) \int \frac {1}{x^4 \sqrt {d+e x^2}} \, dx}{35 d^2}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{140 d^2 x^5}-\frac {e^{5/2} \sqrt {d+e x^2}}{35 d^3 x^3}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}-\frac {\left (2 e^{7/2}\right ) \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx}{35 d^3}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{140 d^2 x^5}-\frac {e^{5/2} \sqrt {d+e x^2}}{35 d^3 x^3}+\frac {2 e^{7/2} \sqrt {d+e x^2}}{35 d^4 x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 85, normalized size = 0.65 \begin {gather*} \frac {\sqrt {e} x \sqrt {d+e x^2} \left (-5 d^3+6 d^2 e x^2-8 d e^2 x^4+16 e^3 x^6\right )-35 d^4 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{280 d^4 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^9,x]

[Out]

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-5*d^3 + 6*d^2*e*x^2 - 8*d*e^2*x^4 + 16*e^3*x^6) - 35*d^4*ArcTanh[(Sqrt[e]*x)/Sqrt
[d + e*x^2]])/(280*d^4*x^8)

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Maple [A]
time = 0.01, size = 158, normalized size = 1.21

method	result	size
default	\(-\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{8 x^{8}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{5 d \,x^{5}}-\frac {4 e \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{5 d}\right )}{8 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 d \,x^{7}}-\frac {4 e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{7 d}\right )}{8 d}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/8*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^8-1/8*e^(3/2)/d*(-1/5/d/x^5*(e*x^2+d)^(1/2)-4/5*e/d*(-1/3/d/x^3*(e*x
^2+d)^(1/2)+2/3*e/d^2/x*(e*x^2+d)^(1/2)))+1/8*e^(1/2)/d*(-1/7/d/x^7*(e*x^2+d)^(3/2)-4/7*e/d*(-1/5/d/x^5*(e*x^2
+d)^(3/2)+2/15*e/d^2/x^3*(e*x^2+d)^(3/2)))

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Maxima [A]
time = 0.28, size = 123, normalized size = 0.94 \begin {gather*} -\frac {\operatorname {artanh}\left (\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}}\right )}{8 \, x^{8}} + \frac {{\left (8 \, x^{6} e^{3} + 4 \, d x^{4} e^{2} - d^{2} x^{2} e + 3 \, d^{3}\right )} e^{\frac {3}{2}}}{120 \, \sqrt {x^{2} e + d} d^{4} x^{5}} - \frac {{\left (8 \, x^{6} e^{3} - 4 \, d x^{4} e^{2} + 3 \, d^{2} x^{2} e + 15 \, d^{3}\right )} \sqrt {x^{2} e + d} e^{\frac {1}{2}}}{840 \, d^{4} x^{7}} \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^9,x, algorithm="maxima")

[Out]

-1/8*arctanh(x*e^(1/2)/sqrt(x^2*e + d))/x^8 + 1/120*(8*x^6*e^3 + 4*d*x^4*e^2 - d^2*x^2*e + 3*d^3)*e^(3/2)/(sqr
t(x^2*e + d)*d^4*x^5) - 1/840*(8*x^6*e^3 - 4*d*x^4*e^2 + 3*d^2*x^2*e + 15*d^3)*sqrt(x^2*e + d)*e^(1/2)/(d^4*x^
7)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (101) = 202\).
time = 0.36, size = 340, normalized size = 2.60 \begin {gather*} -\frac {35 \, d^{4} \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 (16 \, x^{7} \cosh \left (\frac {1}{2}\right )^{7} + 112 \, x^{7} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{6} + 16 \, x^{7} \sinh \left (\frac {1}{2}\right )^{7} - 8 \, d x^{5} \cosh \left (\frac {1}{2}\right )^{5} + 6 \, d^{2} x^{3} \cosh \left (\frac {1}{2}\right )^{3} + 8 \, {\left (42 \, x^{7} \cosh \left (\frac {1}{2}\right )^{2} - d x^{5}\right )} \sinh \left (\frac {1}{2}\right )^{5} - 5 \, d^{3} x \cosh \left (\frac {1}{2}\right ) + 40 \, {\left (14 \, x^{7} \cosh \left (\frac {1}{2}\right )^{3} - d x^{5} \cosh \left (\frac {1}{2}\right )\right )} \sinh \left (\frac {1}{2}\right )^{4} + 2 \, {\left (280 \, x^{7} \cosh \left (\frac {1}{2}\right )^{4} - 40 \, d x^{5} \cosh \left (\frac {1}{2}\right )^{2} + 3 \, d^{2} x^{3}\right )} \sinh \left (\frac {1}{2}\right )^{3} + 2 \, {\left (168 \, x^{7} \cosh \left (\frac {1}{2}\right )^{5} - 40 \, d x^{5} \cosh \left (\frac {1}{2}\right )^{3} + 9 \, d^{2} x^{3} \cosh \left (\frac {1}{2}\right )\right )} \sinh \left (\frac {1}{2}\right )^{2} + {\left (112 \, x^{7} \cosh \left (\frac {1}{2}\right )^{6} - 40 \, d x^{5} \cosh \left (\frac {1}{2}\right )^{4} + 18 \, d^{2} x^{3} \cosh \left (\frac {1}{2}\right )^{2} - 5 \, d^{3} 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 )}}}{560 \, d^{4} x^{8}} \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^9,x, algorithm="fricas")

[Out]

-1/560*(35*d^4*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*sin
h(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2) - sinh(1/2))) + d)/d) - 2*(16*x^7*cosh(1/2
)^7 + 112*x^7*cosh(1/2)*sinh(1/2)^6 + 16*x^7*sinh(1/2)^7 - 8*d*x^5*cosh(1/2)^5 + 6*d^2*x^3*cosh(1/2)^3 + 8*(42
*x^7*cosh(1/2)^2 - d*x^5)*sinh(1/2)^5 - 5*d^3*x*cosh(1/2) + 40*(14*x^7*cosh(1/2)^3 - d*x^5*cosh(1/2))*sinh(1/2
)^4 + 2*(280*x^7*cosh(1/2)^4 - 40*d*x^5*cosh(1/2)^2 + 3*d^2*x^3)*sinh(1/2)^3 + 2*(168*x^7*cosh(1/2)^5 - 40*d*x
^5*cosh(1/2)^3 + 9*d^2*x^3*cosh(1/2))*sinh(1/2)^2 + (112*x^7*cosh(1/2)^6 - 40*d*x^5*cosh(1/2)^4 + 18*d^2*x^3*c
osh(1/2)^2 - 5*d^3*x)*sinh(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2) - sinh(1/2))))/(d
^4*x^8)

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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^{9}}\, 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**9,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**9, 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^9,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^9} \,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^9,x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^9, x)

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