3.39 \(\int \frac {1}{\sqrt {a \tanh ^4(x)}} \, dx\)

Optimal. Leaf size=31 \[ \frac {x \tanh ^2(x)}{\sqrt {a \tanh ^4(x)}}-\frac {\tanh (x)}{\sqrt {a \tanh ^4(x)}} \]

[Out]

-tanh(x)/(a*tanh(x)^4)^(1/2)+x*tanh(x)^2/(a*tanh(x)^4)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3658, 3473, 8} \[ \frac {x \tanh ^2(x)}{\sqrt {a \tanh ^4(x)}}-\frac {\tanh (x)}{\sqrt {a \tanh ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Tanh[x]^4],x]

[Out]

-(Tanh[x]/Sqrt[a*Tanh[x]^4]) + (x*Tanh[x]^2)/Sqrt[a*Tanh[x]^4]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a \tanh ^4(x)}} \, dx &=\frac {\tanh ^2(x) \int \coth ^2(x) \, dx}{\sqrt {a \tanh ^4(x)}}\\ &=-\frac {\tanh (x)}{\sqrt {a \tanh ^4(x)}}+\frac {\tanh ^2(x) \int 1 \, dx}{\sqrt {a \tanh ^4(x)}}\\ &=-\frac {\tanh (x)}{\sqrt {a \tanh ^4(x)}}+\frac {x \tanh ^2(x)}{\sqrt {a \tanh ^4(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 19, normalized size = 0.61 \[ \frac {\tanh (x) (x \tanh (x)-1)}{\sqrt {a \tanh ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Tanh[x]^4],x]

[Out]

(Tanh[x]*(-1 + x*Tanh[x]))/Sqrt[a*Tanh[x]^4]

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fricas [B]  time = 0.56, size = 238, normalized size = 7.68 \[ \frac {{\left (x \cosh \relax (x)^{2} + {\left (x e^{\left (4 \, x\right )} + 2 \, x e^{\left (2 \, x\right )} + x\right )} \sinh \relax (x)^{2} + {\left (x \cosh \relax (x)^{2} - x - 2\right )} e^{\left (4 \, x\right )} + 2 \, {\left (x \cosh \relax (x)^{2} - x - 2\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x \cosh \relax (x) e^{\left (4 \, x\right )} + 2 \, x \cosh \relax (x) e^{\left (2 \, x\right )} + x \cosh \relax (x)\right )} \sinh \relax (x) - x - 2\right )} \sqrt {\frac {a e^{\left (8 \, x\right )} - 4 \, a e^{\left (6 \, x\right )} + 6 \, a e^{\left (4 \, x\right )} - 4 \, a e^{\left (2 \, x\right )} + a}{e^{\left (8 \, x\right )} + 4 \, e^{\left (6 \, x\right )} + 6 \, e^{\left (4 \, x\right )} + 4 \, e^{\left (2 \, x\right )} + 1}}}{a \cosh \relax (x)^{2} + {\left (a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a\right )} \sinh \relax (x)^{2} + {\left (a \cosh \relax (x)^{2} - a\right )} e^{\left (4 \, x\right )} - 2 \, {\left (a \cosh \relax (x)^{2} - a\right )} e^{\left (2 \, x\right )} + 2 \, {\left (a \cosh \relax (x) e^{\left (4 \, x\right )} - 2 \, a \cosh \relax (x) e^{\left (2 \, x\right )} + a \cosh \relax (x)\right )} \sinh \relax (x) - a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)^4)^(1/2),x, algorithm="fricas")

[Out]

(x*cosh(x)^2 + (x*e^(4*x) + 2*x*e^(2*x) + x)*sinh(x)^2 + (x*cosh(x)^2 - x - 2)*e^(4*x) + 2*(x*cosh(x)^2 - x -
2)*e^(2*x) + 2*(x*cosh(x)*e^(4*x) + 2*x*cosh(x)*e^(2*x) + x*cosh(x))*sinh(x) - x - 2)*sqrt((a*e^(8*x) - 4*a*e^
(6*x) + 6*a*e^(4*x) - 4*a*e^(2*x) + a)/(e^(8*x) + 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) + 1))/(a*cosh(x)^2 + (a*e^
(4*x) - 2*a*e^(2*x) + a)*sinh(x)^2 + (a*cosh(x)^2 - a)*e^(4*x) - 2*(a*cosh(x)^2 - a)*e^(2*x) + 2*(a*cosh(x)*e^
(4*x) - 2*a*cosh(x)*e^(2*x) + a*cosh(x))*sinh(x) - a)

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giac [A]  time = 0.13, size = 19, normalized size = 0.61 \[ \frac {x}{\sqrt {a}} - \frac {2}{\sqrt {a} {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)^4)^(1/2),x, algorithm="giac")

[Out]

x/sqrt(a) - 2/(sqrt(a)*(e^(2*x) - 1))

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maple [A]  time = 0.10, size = 32, normalized size = 1.03 \[ -\frac {\tanh \relax (x ) \left (\ln \left (\tanh \relax (x )-1\right ) \tanh \relax (x )-\ln \left (1+\tanh \relax (x )\right ) \tanh \relax (x )+2\right )}{2 \sqrt {a \left (\tanh ^{4}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*tanh(x)^4)^(1/2),x)

[Out]

-1/2*tanh(x)*(ln(tanh(x)-1)*tanh(x)-ln(1+tanh(x))*tanh(x)+2)/(a*tanh(x)^4)^(1/2)

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maxima [A]  time = 0.44, size = 23, normalized size = 0.74 \[ \frac {x}{\sqrt {a}} + \frac {2 \, \sqrt {a}}{a e^{\left (-2 \, x\right )} - a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)^4)^(1/2),x, algorithm="maxima")

[Out]

x/sqrt(a) + 2*sqrt(a)/(a*e^(-2*x) - a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{\sqrt {a\,{\mathrm {tanh}\relax (x)}^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*tanh(x)^4)^(1/2),x)

[Out]

int(1/(a*tanh(x)^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a \tanh ^{4}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)**4)**(1/2),x)

[Out]

Integral(1/sqrt(a*tanh(x)**4), x)

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