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

Optimal. Leaf size=43 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}^2(x)}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{1}{a \sqrt{a+b \text{sech}^2(x)}} \]

[Out]

ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]]/a^(3/2) - 1/(a*Sqrt[a + b*Sech[x]^2])

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Rubi [A]  time = 0.0729075, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4139, 266, 51, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}^2(x)}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{1}{a \sqrt{a+b \text{sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]]/a^(3/2) - 1/(a*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 266

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

Rule 51

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

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

Mathematica [B]  time = 0.396049, size = 98, normalized size = 2.28 \[ -\frac{\text{sech}^3(x) (a \cosh (2 x)+a+2 b) \left (2 \sqrt{a} \cosh (x)-\sqrt{2} \sqrt{a \cosh (2 x)+a+2 b} \log \left (\sqrt{a \cosh (2 x)+a+2 b}+\sqrt{2} \sqrt{a} \cosh (x)\right )\right )}{4 a^{3/2} \left (a+b \text{sech}^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cosh[2*x])*(2*Sqrt[a]*Cosh[x] - Sqrt[2]*Sqrt[a + 2*b + a*Cosh[2*x]]*Log[Sqrt[2]*Sqrt[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 [A]  time = 0.015, size = 46, normalized size = 1.1 \begin{align*} -{\frac{1}{a}{\frac{1}{\sqrt{a+b \left ({\rm sech} \left (x\right ) \right ) ^{2}}}}}+{\ln \left ({\frac{1}{{\rm sech} \left (x\right )} \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ({\rm sech} \left (x\right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/(a+b*sech(x)^2)^(1/2)+1/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sech(x)^2)^(1/2))/sech(x))

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

[Out]

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

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Fricas [B]  time = 2.61665, size = 5840, normalized size = 135.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((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)
*sinh(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sinh(x) + a)*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)*cosh(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))*sinh(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)*cosh(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)*cos
h(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*c
osh(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)*co
sh(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 + s
inh(x)^6)) + (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)*sinh(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sinh(x) + a)*sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)*s
inh(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*
cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 + a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*c
osh(x)*sinh(x) + sinh(x)^2)))/(a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^3*sinh(x)^4 + a^3 + 2*(a^3 + 2*a^2*
b)*cosh(x)^2 + 2*(3*a^3*cosh(x)^2 + a^3 + 2*a^2*b)*sinh(x)^2 + 4*(a^3*cosh(x)^3 + (a^3 + 2*a^2*b)*cosh(x))*sin
h(x)), -1/2*((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)*sinh(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sinh(x) + a)*sqrt(-a)*arctan(sqrt(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)/(co
sh(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)*sinh(x)^2 + a^2 + 2*(2*(a
^2 + a*b)*cosh(x)^3 + (2*a^2 + 3*a*b)*cosh(x))*sinh(x))) + (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)*sinh(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sinh(x)
+ a)*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*si
nh(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)*sinh(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sin
h(x) + a)) + 2*sqrt(2)*(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 + 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^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^3*sinh(x)
^4 + a^3 + 2*(a^3 + 2*a^2*b)*cosh(x)^2 + 2*(3*a^3*cosh(x)^2 + a^3 + 2*a^2*b)*sinh(x)^2 + 4*(a^3*cosh(x)^3 + (a
^3 + 2*a^2*b)*cosh(x))*sinh(x))]

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Sympy [A]  time = 6.61079, size = 44, normalized size = 1.02 \begin{align*} - \frac{1}{a \sqrt{a + b \operatorname{sech}^{2}{\left (x \right )}}} - \frac{\operatorname{atan}{\left (\frac{\sqrt{a + b \operatorname{sech}^{2}{\left (x \right )}}}{\sqrt{- a}} \right )}}{a \sqrt{- a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*sqrt(a + b*sech(x)**2)) - atan(sqrt(a + b*sech(x)**2)/sqrt(-a))/(a*sqrt(-a))

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

[Out]

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