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

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

[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 = 0.04, 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 = {6221, 271, 264} \[ -\frac {4 e^{5/2} \sqrt {d+e x^2}}{45 d^3 x}+\frac {2 e^{3/2} \sqrt {d+e x^2}}{45 d^2 x^3}-\frac {\sqrt {e} \sqrt {d+e x^2}}{30 d x^5}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(30*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 264

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 271

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 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} \, 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 = 0.05, size = 74, normalized size = 0.70 \[ \frac {\sqrt {e} x \sqrt {d+e x^2} \left (-3 d^2+4 d e x^2-8 e^2 x^4\right )-15 d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{90 d^3 x^6} \]

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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fricas [A]  time = 0.57, size = 78, normalized size = 0.74 \[ -\frac {15 \, d^{3} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) + 2 \, {\left (8 \, e^{2} x^{5} - 4 \, d e x^{3} + 3 \, d^{2} x\right )} \sqrt {e x^{2} + d} \sqrt {e}}{180 \, d^{3} x^{6}} \]

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*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) + 2*(8*e^2*x^5 - 4*d*e*x^3 + 3*d^2*x)*sqrt(e
*x^2 + d)*sqrt(e))/(d^3*x^6)

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giac [A]  time = 0.45, size = 134, normalized size = 1.28 \[ \frac {8 \, {\left (10 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{4} d^{2} e^{2} - 5 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} d^{3} e^{2} + d^{4} e^{2}\right )} e}{45 \, {\left ({\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} - d\right )}^{5} d^{2}} - \frac {\log \left (-\frac {\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} + 1}{\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} - 1}\right )}{12 \, x^{6}} \]

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]

8/45*(10*(x*e^(1/2) - sqrt(x^2*e + d))^4*d^2*e^2 - 5*(x*e^(1/2) - sqrt(x^2*e + d))^2*d^3*e^2 + d^4*e^2)*e/(((x
*e^(1/2) - sqrt(x^2*e + d))^2 - d)^5*d^2) - 1/12*log(-(x*e^(1/2)/sqrt(x^2*e + d) + 1)/(x*e^(1/2)/sqrt(x^2*e +
d) - 1))/x^6

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maple [A]  time = 0.03, size = 110, normalized size = 1.05 \[ -\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{6 x^{6}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{6 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{6 d} \]

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)

[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 = 0.34, size = 102, normalized size = 0.97 \[ -\frac {{\left (2 \, e^{2} x^{4} + d e x^{2} - d^{2}\right )} e^{\frac {3}{2}}}{18 \, \sqrt {e x^{2} + d} d^{3} x^{3}} - \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{6 \, x^{6}} + \frac {{\left (2 \, e^{2} x^{4} - d e x^{2} - 3 \, d^{2}\right )} \sqrt {e x^{2} + d} \sqrt {e}}{90 \, d^{3} x^{5}} \]

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

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

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,x)

[Out]

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

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

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,x)

[Out]

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

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