3.36 \(\int \frac{1}{\sqrt{a \tanh ^3(x)}} \, dx\)

Optimal. Leaf size=64 \[ \frac{\tanh ^{-1}\left (\sqrt{\tanh (x)}\right ) \tanh ^{\frac{3}{2}}(x)}{\sqrt{a \tanh ^3(x)}}-\frac{2 \tanh (x)}{\sqrt{a \tanh ^3(x)}}-\frac{\tanh ^{\frac{3}{2}}(x) \tan ^{-1}\left (\sqrt{\tanh (x)}\right )}{\sqrt{a \tanh ^3(x)}} \]

[Out]

(-2*Tanh[x])/Sqrt[a*Tanh[x]^3] - (ArcTan[Sqrt[Tanh[x]]]*Tanh[x]^(3/2))/Sqrt[a*Tanh[x]^3] + (ArcTanh[Sqrt[Tanh[
x]]]*Tanh[x]^(3/2))/Sqrt[a*Tanh[x]^3]

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Rubi [A]  time = 0.030095, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {3658, 3474, 3476, 329, 298, 203, 206} \[ \frac{\tanh ^{-1}\left (\sqrt{\tanh (x)}\right ) \tanh ^{\frac{3}{2}}(x)}{\sqrt{a \tanh ^3(x)}}-\frac{2 \tanh (x)}{\sqrt{a \tanh ^3(x)}}-\frac{\tanh ^{\frac{3}{2}}(x) \tan ^{-1}\left (\sqrt{\tanh (x)}\right )}{\sqrt{a \tanh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Tanh[x]^3],x]

[Out]

(-2*Tanh[x])/Sqrt[a*Tanh[x]^3] - (ArcTan[Sqrt[Tanh[x]]]*Tanh[x]^(3/2))/Sqrt[a*Tanh[x]^3] + (ArcTanh[Sqrt[Tanh[
x]]]*Tanh[x]^(3/2))/Sqrt[a*Tanh[x]^3]

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
]])

Rule 3474

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a \tanh ^3(x)}} \, dx &=\frac{\tanh ^{\frac{3}{2}}(x) \int \frac{1}{\tanh ^{\frac{3}{2}}(x)} \, dx}{\sqrt{a \tanh ^3(x)}}\\ &=-\frac{2 \tanh (x)}{\sqrt{a \tanh ^3(x)}}+\frac{\tanh ^{\frac{3}{2}}(x) \int \sqrt{\tanh (x)} \, dx}{\sqrt{a \tanh ^3(x)}}\\ &=-\frac{2 \tanh (x)}{\sqrt{a \tanh ^3(x)}}-\frac{\tanh ^{\frac{3}{2}}(x) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+x^2} \, dx,x,\tanh (x)\right )}{\sqrt{a \tanh ^3(x)}}\\ &=-\frac{2 \tanh (x)}{\sqrt{a \tanh ^3(x)}}-\frac{\left (2 \tanh ^{\frac{3}{2}}(x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^4} \, dx,x,\sqrt{\tanh (x)}\right )}{\sqrt{a \tanh ^3(x)}}\\ &=-\frac{2 \tanh (x)}{\sqrt{a \tanh ^3(x)}}+\frac{\tanh ^{\frac{3}{2}}(x) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\tanh (x)}\right )}{\sqrt{a \tanh ^3(x)}}-\frac{\tanh ^{\frac{3}{2}}(x) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\tanh (x)}\right )}{\sqrt{a \tanh ^3(x)}}\\ &=-\frac{2 \tanh (x)}{\sqrt{a \tanh ^3(x)}}-\frac{\tan ^{-1}\left (\sqrt{\tanh (x)}\right ) \tanh ^{\frac{3}{2}}(x)}{\sqrt{a \tanh ^3(x)}}+\frac{\tanh ^{-1}\left (\sqrt{\tanh (x)}\right ) \tanh ^{\frac{3}{2}}(x)}{\sqrt{a \tanh ^3(x)}}\\ \end{align*}

Mathematica [C]  time = 0.0169666, size = 26, normalized size = 0.41 \[ -\frac{2 \tanh (x) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\tanh ^2(x)\right )}{\sqrt{a \tanh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Tanh[x]^3],x]

[Out]

(-2*Hypergeometric2F1[-1/4, 1, 3/4, Tanh[x]^2]*Tanh[x])/Sqrt[a*Tanh[x]^3]

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Maple [A]  time = 0.04, size = 65, normalized size = 1. \begin{align*} -{\tanh \left ( x \right ) \left ( 2\,{a}^{5/2}-{\it Artanh} \left ({\sqrt{a\tanh \left ( x \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2}\sqrt{a\tanh \left ( x \right ) }+\arctan \left ({\sqrt{a\tanh \left ( x \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2}\sqrt{a\tanh \left ( x \right ) } \right ){\frac{1}{\sqrt{a \left ( \tanh \left ( x \right ) \right ) ^{3}}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*tanh(x)^3)^(1/2),x)

[Out]

-tanh(x)*(2*a^(5/2)-arctanh((a*tanh(x))^(1/2)/a^(1/2))*a^2*(a*tanh(x))^(1/2)+arctan((a*tanh(x))^(1/2)/a^(1/2))
*a^2*(a*tanh(x))^(1/2))/(a*tanh(x)^3)^(1/2)/a^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \tanh \left (x\right )^{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(a*tanh(x)^3), x)

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Fricas [B]  time = 2.45095, size = 1763, normalized size = 27.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)^3)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*(2*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-a)*arctan((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh
(x)^2)*sqrt(-a)*sqrt(a*sinh(x)/cosh(x))/(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 - a)) + (cosh(x)^2 +
2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)^3*sinh(x) + 6*a*cosh(x)^2*sinh(x)^
2 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(-a)*sqrt(a*si
nh(x)/cosh(x)) - 2*a)/(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)
^4)) + 8*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a*sinh(x)/cosh(x)))/(a*cosh(x)^2 + 2*a*cosh(x)*s
inh(x) + a*sinh(x)^2 - a), -1/4*(2*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(a)*arctan((cosh(x)^2 +
2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a*sinh(x)/cosh(x))/sqrt(a)) - (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x
)^2 - 1)*sqrt(a)*log(2*a*cosh(x)^4 + 8*a*cosh(x)^3*sinh(x) + 12*a*cosh(x)^2*sinh(x)^2 + 8*a*cosh(x)*sinh(x)^3
+ 2*a*sinh(x)^4 + 2*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2
*(2*cosh(x)^3 + cosh(x))*sinh(x))*sqrt(a)*sqrt(a*sinh(x)/cosh(x)) - a) + 8*(cosh(x)^2 + 2*cosh(x)*sinh(x) + si
nh(x)^2 + 1)*sqrt(a*sinh(x)/cosh(x)))/(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 - a)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \tanh ^{3}{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)**3)**(1/2),x)

[Out]

Integral(1/sqrt(a*tanh(x)**3), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*tanh(x)^3)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError