3.205 \(\int \frac{\tanh ^3(x)}{(a+b \text{sech}^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=49 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}^2(x)}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{a+b}{a b \sqrt{a+b \text{sech}^2(x)}} \]

[Out]

ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]]/a^(3/2) - (a + b)/(a*b*Sqrt[a + b*Sech[x]^2])

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Rubi [A]  time = 0.114608, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {4139, 446, 78, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}^2(x)}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{a+b}{a b \sqrt{a+b \text{sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Sech[x]^2)^(3/2),x]

[Out]

ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]]/a^(3/2) - (a + b)/(a*b*Sqrt[a + b*Sech[x]^2])

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [B]  time = 0.229031, size = 103, normalized size = 2.1 \[ \frac{\text{sech}^3(x) \left (\sqrt{2} (a \cosh (2 x)+a+2 b)^{3/2} \log \left (\sqrt{a \cosh (2 x)+a+2 b}+\sqrt{2} \sqrt{a} \cosh (x)\right )-\frac{2 \sqrt{a} (a+b) \cosh (x) (a \cosh (2 x)+a+2 b)}{b}\right )}{4 a^{3/2} \left (a+b \text{sech}^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^3/(a + b*Sech[x]^2)^(3/2),x]

[Out]

(((-2*Sqrt[a]*(a + b)*Cosh[x]*(a + 2*b + a*Cosh[2*x]))/b + Sqrt[2]*(a + 2*b + a*Cosh[2*x])^(3/2)*Log[Sqrt[2]*S
qrt[a]*Cosh[x] + Sqrt[a + 2*b + a*Cosh[2*x]]])*Sech[x]^3)/(4*a^(3/2)*(a + b*Sech[x]^2)^(3/2))

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Maple [F]  time = 0.098, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tanh \left ( x \right ) \right ) ^{3} \left ( a+b \left ({\rm sech} \left (x\right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a+b*sech(x)^2)^(3/2),x)

[Out]

int(tanh(x)^3/(a+b*sech(x)^2)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (x\right )^{3}}{{\left (b \operatorname{sech}\left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tanh(x)^3/(b*sech(x)^2 + a)^(3/2), x)

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Fricas [B]  time = 2.74828, size = 6137, normalized size = 125.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((a*b*cosh(x)^4 + 4*a*b*cosh(x)*sinh(x)^3 + a*b*sinh(x)^4 + 2*(a*b + 2*b^2)*cosh(x)^2 + 2*(3*a*b*cosh(x)^
2 + a*b + 2*b^2)*sinh(x)^2 + a*b + 4*(a*b*cosh(x)^3 + (a*b + 2*b^2)*cosh(x))*sinh(x))*sqrt(a)*log(((a^3 + 2*a^
2*b + a*b^2)*cosh(x)^8 + 8*(a^3 + 2*a^2*b + a*b^2)*cosh(x)*sinh(x)^7 + (a^3 + 2*a^2*b + a*b^2)*sinh(x)^8 + 2*(
2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x)^6 + 2*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3 + 14*(a^3 + 2*a^2*b + a*b^2)*c
osh(x)^2)*sinh(x)^6 + 4*(14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^3 + 3*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x))*s
inh(x)^5 + (6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x)^4 + (70*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^4 + 6*a^3 + 14*a^2*b +
9*a*b^2 + 30*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^5
+ 10*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x)^3 + (6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x))*sinh(x)^3 + a^3 + 2*
(2*a^3 + 3*a^2*b)*cosh(x)^2 + 2*(14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^6 + 15*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*c
osh(x)^4 + 2*a^3 + 3*a^2*b + 3*(6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x)^2)*sinh(x)^2 + sqrt(2)*((a^2 + 2*a*b + b^2
)*cosh(x)^6 + 6*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^5 + (a^2 + 2*a*b + b^2)*sinh(x)^6 + 3*(a^2 + 2*a*b + b^2)*
cosh(x)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(x)^2 + a^2 + 2*a*b + b^2)*sinh(x)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(
x)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(x))*sinh(x)^3 + (3*a^2 + 4*a*b)*cosh(x)^2 + (15*(a^2 + 2*a*b + b^2)*cosh(x)^
4 + 18*(a^2 + 2*a*b + b^2)*cosh(x)^2 + 3*a^2 + 4*a*b)*sinh(x)^2 + a^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(x)^5 + 6
*(a^2 + 2*a*b + b^2)*cosh(x)^3 + (3*a^2 + 4*a*b)*cosh(x))*sinh(x))*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a
+ 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^7 + 3*(2*a^3 + 5*a
^2*b + 4*a*b^2 + b^3)*cosh(x)^5 + (6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x)^3 + (2*a^3 + 3*a^2*b)*cosh(x))*sinh(x))
/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 +
6*cosh(x)*sinh(x)^5 + sinh(x)^6)) + (a*b*cosh(x)^4 + 4*a*b*cosh(x)*sinh(x)^3 + a*b*sinh(x)^4 + 2*(a*b + 2*b^2
)*cosh(x)^2 + 2*(3*a*b*cosh(x)^2 + a*b + 2*b^2)*sinh(x)^2 + a*b + 4*(a*b*cosh(x)^3 + (a*b + 2*b^2)*cosh(x))*si
nh(x))*sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*b*cosh(x)^2 + 2*(3*a*cosh(x)^2 + b)
*sinh(x)^2 + sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 +
a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(a*cosh(x)^3 + b*cosh(x))*sinh(x) + a)/(cosh(x)^2 +
2*cosh(x)*sinh(x) + sinh(x)^2)) - 4*sqrt(2)*((a^2 + a*b)*cosh(x)^2 + 2*(a^2 + a*b)*cosh(x)*sinh(x) + (a^2 + a
*b)*sinh(x)^2 + a^2 + a*b)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)
^2)))/(a^3*b*cosh(x)^4 + 4*a^3*b*cosh(x)*sinh(x)^3 + a^3*b*sinh(x)^4 + a^3*b + 2*(a^3*b + 2*a^2*b^2)*cosh(x)^2
+ 2*(3*a^3*b*cosh(x)^2 + a^3*b + 2*a^2*b^2)*sinh(x)^2 + 4*(a^3*b*cosh(x)^3 + (a^3*b + 2*a^2*b^2)*cosh(x))*sin
h(x)), -1/2*((a*b*cosh(x)^4 + 4*a*b*cosh(x)*sinh(x)^3 + a*b*sinh(x)^4 + 2*(a*b + 2*b^2)*cosh(x)^2 + 2*(3*a*b*c
osh(x)^2 + a*b + 2*b^2)*sinh(x)^2 + a*b + 4*(a*b*cosh(x)^3 + (a*b + 2*b^2)*cosh(x))*sinh(x))*sqrt(-a)*arctan(s
qrt(2)*((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a)*sqrt(-a)*sqrt((a*cosh(x)^2 + a*
sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh(
x)*sinh(x)^3 + (a^2 + a*b)*sinh(x)^4 + (2*a^2 + 3*a*b)*cosh(x)^2 + (6*(a^2 + a*b)*cosh(x)^2 + 2*a^2 + 3*a*b)*s
inh(x)^2 + a^2 + 2*(2*(a^2 + a*b)*cosh(x)^3 + (2*a^2 + 3*a*b)*cosh(x))*sinh(x))) + (a*b*cosh(x)^4 + 4*a*b*cosh
(x)*sinh(x)^3 + a*b*sinh(x)^4 + 2*(a*b + 2*b^2)*cosh(x)^2 + 2*(3*a*b*cosh(x)^2 + a*b + 2*b^2)*sinh(x)^2 + a*b
+ 4*(a*b*cosh(x)^3 + (a*b + 2*b^2)*cosh(x))*sinh(x))*sqrt(-a)*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) +
sinh(x)^2 - 1)*sqrt(-a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)
)/(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*(a + 2*b)*cosh(x)^2 + 2*(3*a*cosh(x)^2 + a + 2*b)*sin
h(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sinh(x) + a)) + 2*sqrt(2)*((a^2 + a*b)*cosh(x)^2 + 2*(a^2 + a*b)*
cosh(x)*sinh(x) + (a^2 + a*b)*sinh(x)^2 + a^2 + a*b)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2
*cosh(x)*sinh(x) + sinh(x)^2)))/(a^3*b*cosh(x)^4 + 4*a^3*b*cosh(x)*sinh(x)^3 + a^3*b*sinh(x)^4 + a^3*b + 2*(a^
3*b + 2*a^2*b^2)*cosh(x)^2 + 2*(3*a^3*b*cosh(x)^2 + a^3*b + 2*a^2*b^2)*sinh(x)^2 + 4*(a^3*b*cosh(x)^3 + (a^3*b
+ 2*a^2*b^2)*cosh(x))*sinh(x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\left (x \right )}}{\left (a + b \operatorname{sech}^{2}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**3/(a+b*sech(x)**2)**(3/2),x)

[Out]

Integral(tanh(x)**3/(a + b*sech(x)**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (x\right )^{3}}{{\left (b \operatorname{sech}\left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tanh(x)^3/(b*sech(x)^2 + a)^(3/2), x)