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

Optimal. Leaf size=105 \[ -\frac {\sqrt {e} \sqrt {d+e x^2}}{30 d x^5}+\frac {2 e^{3/2} \sqrt {d+e x^2}}{45 d^2 x^3}-\frac {4 e^{5/2} \sqrt {d+e x^2}}{45 d^3 x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 x^6} \]

[Out]

-1/6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6+2/45*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x^3-4/45*e^(5/2)*(e*x^2+d)^(1/2)/
d^3/x-1/30*e^(1/2)*(e*x^2+d)^(1/2)/d/x^5

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Rubi [A]
time = 0.03, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {4 e^{5/2} \sqrt {d+e x^2}}{45 d^3 x}+\frac {2 e^{3/2} \sqrt {d+e x^2}}{45 d^2 x^3}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 x^6}-\frac {\sqrt {e} \sqrt {d+e x^2}}{30 d x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(Sqrt[e]*Sqrt[d + e*x^2])/(d*x^5) + (2*e^(3/2)*Sqrt[d + e*x^2])/(45*d^2*x^3) - (4*e^(5/2)*Sqrt[d + e*x^2
])/(45*d^3*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(6*x^6)

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^7} \, dx &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 x^6}+\frac {1}{6} \sqrt {e} \int \frac {1}{x^6 \sqrt {d+e x^2}} \, dx\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{30 d x^5}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 x^6}-\frac {\left (2 e^{3/2}\right ) \int \frac {1}{x^4 \sqrt {d+e x^2}} \, dx}{15 d}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{30 d x^5}+\frac {2 e^{3/2} \sqrt {d+e x^2}}{45 d^2 x^3}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 x^6}+\frac {\left (4 e^{5/2}\right ) \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx}{45 d^2}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{30 d x^5}+\frac {2 e^{3/2} \sqrt {d+e x^2}}{45 d^2 x^3}-\frac {4 e^{5/2} \sqrt {d+e x^2}}{45 d^3 x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 x^6}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 74, normalized size = 0.70 \begin {gather*} \frac {\sqrt {e} x \sqrt {d+e x^2} \left (-3 d^2+4 d e x^2-8 e^2 x^4\right )-15 d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{90 d^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-3*d^2 + 4*d*e*x^2 - 8*e^2*x^4) - 15*d^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(90
*d^3*x^6)

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Maple [A]
time = 0.01, size = 110, normalized size = 1.05

method	result	size
default	\(-\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{6 x^{6}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{6 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{6 d}\)	\(110\)

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,x,method=_RETURNVERBOSE)

[Out]

-1/6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6-1/6*e^(3/2)/d*(-1/3/d/x^3*(e*x^2+d)^(1/2)+2/3*e/d^2/x*(e*x^2+d)^(1
/2))+1/6*e^(1/2)/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 = 0.26, size = 102, normalized size = 0.97 \begin {gather*} -\frac {{\left (2 \, x^{4} e^{2} + d x^{2} e - d^{2}\right )} e^{\frac {3}{2}}}{18 \, \sqrt {x^{2} e + d} d^{3} x^{3}} - \frac {\operatorname {artanh}\left (\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}}\right )}{6 \, x^{6}} + \frac {{\left (2 \, x^{4} e^{2} - d x^{2} e - 3 \, d^{2}\right )} \sqrt {x^{2} e + d} e^{\frac {1}{2}}}{90 \, d^{3} x^{5}} \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^7,x, algorithm="maxima")

[Out]

-1/18*(2*x^4*e^2 + d*x^2*e - d^2)*e^(3/2)/(sqrt(x^2*e + d)*d^3*x^3) - 1/6*arctanh(x*e^(1/2)/sqrt(x^2*e + d))/x
^6 + 1/90*(2*x^4*e^2 - d*x^2*e - 3*d^2)*sqrt(x^2*e + d)*e^(1/2)/(d^3*x^5)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (81) = 162\).
time = 0.36, size = 246, normalized size = 2.34 \begin {gather*} -\frac {15 \, d^{3} \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 (8 \, x^{5} \cosh \left (\frac {1}{2}\right )^{5} + 40 \, x^{5} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{4} + 8 \, x^{5} \sinh \left (\frac {1}{2}\right )^{5} - 4 \, d x^{3} \cosh \left (\frac {1}{2}\right )^{3} + 3 \, d^{2} x \cosh \left (\frac {1}{2}\right ) + 4 \, {\left (20 \, x^{5} \cosh \left (\frac {1}{2}\right )^{2} - d x^{3}\right )} \sinh \left (\frac {1}{2}\right )^{3} + 4 \, {\left (20 \, x^{5} \cosh \left (\frac {1}{2}\right )^{3} - 3 \, d x^{3} \cosh \left (\frac {1}{2}\right )\right )} \sinh \left (\frac {1}{2}\right )^{2} + {\left (40 \, x^{5} \cosh \left (\frac {1}{2}\right )^{4} - 12 \, d x^{3} \cosh \left (\frac {1}{2}\right )^{2} + 3 \, d^{2} 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 )}}}{180 \, d^{3} x^{6}} \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^7,x, algorithm="fricas")

[Out]

-1/180*(15*d^3*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*(8*x^5*cosh(1/2)
^5 + 40*x^5*cosh(1/2)*sinh(1/2)^4 + 8*x^5*sinh(1/2)^5 - 4*d*x^3*cosh(1/2)^3 + 3*d^2*x*cosh(1/2) + 4*(20*x^5*co
sh(1/2)^2 - d*x^3)*sinh(1/2)^3 + 4*(20*x^5*cosh(1/2)^3 - 3*d*x^3*cosh(1/2))*sinh(1/2)^2 + (40*x^5*cosh(1/2)^4
- 12*d*x^3*cosh(1/2)^2 + 3*d^2*x)*sinh(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2) - sin
h(1/2))))/(d^3*x^6)

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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^{7}}\, 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**7,x)

[Out]

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

[Out]

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

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