3.35 \(\int \sqrt{a \tanh ^3(x)} \, dx\)

Optimal. Leaf size=63 \[ \frac{\tanh ^{-1}\left (\sqrt{\tanh (x)}\right ) \sqrt{a \tanh ^3(x)}}{\tanh ^{\frac{3}{2}}(x)}+\frac{\sqrt{a \tanh ^3(x)} \tan ^{-1}\left (\sqrt{\tanh (x)}\right )}{\tanh ^{\frac{3}{2}}(x)}-2 \coth (x) \sqrt{a \tanh ^3(x)} \]

[Out]

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

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Rubi [A]  time = 0.0296038, antiderivative size = 63, 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, 3473, 3476, 329, 212, 206, 203} \[ \frac{\tanh ^{-1}\left (\sqrt{\tanh (x)}\right ) \sqrt{a \tanh ^3(x)}}{\tanh ^{\frac{3}{2}}(x)}+\frac{\sqrt{a \tanh ^3(x)} \tan ^{-1}\left (\sqrt{\tanh (x)}\right )}{\tanh ^{\frac{3}{2}}(x)}-2 \coth (x) \sqrt{a \tanh ^3(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 212

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0330763, size = 40, normalized size = 0.63 \[ \frac{\sqrt{a \tanh ^3(x)} \left (\tanh ^{-1}\left (\sqrt{\tanh (x)}\right )-2 \sqrt{\tanh (x)}+\tan ^{-1}\left (\sqrt{\tanh (x)}\right )\right )}{\tanh ^{\frac{3}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 62, normalized size = 1. \begin{align*} -{\frac{1}{\tanh \left ( x \right ) }\sqrt{a \left ( \tanh \left ( x \right ) \right ) ^{3}} \left ( 2\,\sqrt{a\tanh \left ( x \right ) }-\sqrt{a}{\it Artanh} \left ({\sqrt{a\tanh \left ( x \right ) }{\frac{1}{\sqrt{a}}}} \right ) -\sqrt{a}\arctan \left ({\sqrt{a\tanh \left ( x \right ) }{\frac{1}{\sqrt{a}}}} \right ) \right ){\frac{1}{\sqrt{a\tanh \left ( x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*tanh(x)^3), x)

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Fricas [B]  time = 2.31753, size = 1237, normalized size = 19.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*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)) + 1/4*sqrt(-a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)^3*sinh(x) + 6*a*cos
h(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)*sqr
t(-a)*sqrt(a*sinh(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)) - 2*sqrt(a*sinh(x)/cosh(x)), -1/2*sqrt(a)*arctan(sqrt(a)*sqrt(a*sinh(x)/cosh(x))/(a*cosh(x
)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 - a)) + 1/4*sqrt(a)*log(2*a*cosh(x)^4 + 8*a*cosh(x)^3*sinh(x) + 12*a*c
osh(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) - 2*sqrt(a*sinh(x)/cosh(x))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*tanh(x)**3), x)

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Giac [B]  time = 1.24178, size = 155, normalized size = 2.46 \begin{align*} \sqrt{a} \arctan \left (-\frac{\sqrt{a} e^{\left (2 \, x\right )} - \sqrt{a e^{\left (4 \, x\right )} - a}}{\sqrt{a}}\right ) \mathrm{sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \frac{1}{2} \, \sqrt{a} \log \left ({\left | -\sqrt{a} e^{\left (2 \, x\right )} + \sqrt{a e^{\left (4 \, x\right )} - a} \right |}\right ) \mathrm{sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \frac{4 \, a \mathrm{sgn}\left (e^{\left (4 \, x\right )} - 1\right )}{\sqrt{a} e^{\left (2 \, x\right )} - \sqrt{a e^{\left (4 \, x\right )} - a} + \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*arctan(-(sqrt(a)*e^(2*x) - sqrt(a*e^(4*x) - a))/sqrt(a))*sgn(e^(4*x) - 1) - 1/2*sqrt(a)*log(abs(-sqrt(
a)*e^(2*x) + sqrt(a*e^(4*x) - a)))*sgn(e^(4*x) - 1) - 4*a*sgn(e^(4*x) - 1)/(sqrt(a)*e^(2*x) - sqrt(a*e^(4*x) -
a) + sqrt(a))