3.198 \(\int \frac {\tanh (x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx\)

Optimal. Leaf size=25 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

arctanh((a+b*sech(x)^2)^(1/2)/a^(1/2))/a^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4139, 266, 63, 208} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/Sqrt[a + b*Sech[x]^2],x]

[Out]

ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]]/Sqrt[a]

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]

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 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])

Rubi steps

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

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Mathematica [B]  time = 0.08, size = 70, normalized size = 2.80 \[ \frac {\text {sech}(x) \sqrt {a \cosh (2 x)+a+2 b} \log \left (\sqrt {a \cosh (2 x)+a+2 b}+\sqrt {2} \sqrt {a} \cosh (x)\right )}{\sqrt {2} \sqrt {a} \sqrt {a+b \text {sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/Sqrt[a + b*Sech[x]^2],x]

[Out]

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

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fricas [B]  time = 0.46, size = 1430, normalized size = 57.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(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 + (1
5*(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)) + sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)*sin
h(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)))/a, -1/2*(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)/(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)*sinh(x)^2 + a^2 + 2*(2*(a^2 + a*b)*cosh(x)^3 + (2*a^2 + 3*a*b)*cosh(x))*sinh(x))) + sqrt(-a)*arctan(s
qrt(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)/(co
sh(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)*cos
h(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)))/a]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [A]  time = 0.12, size = 30, normalized size = 1.20 \[ \frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \mathrm {sech}\relax (x )^{2}}}{\mathrm {sech}\relax (x )}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a+b*sech(x)^2)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \relax (x)}{\sqrt {b \operatorname {sech}\relax (x)^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.67, size = 19, normalized size = 0.76 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{{\mathrm {cosh}\relax (x)}^2}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a + b/cosh(x)^2)^(1/2),x)

[Out]

atanh((a + b/cosh(x)^2)^(1/2)/a^(1/2))/a^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh {\relax (x )}}{\sqrt {a + b \operatorname {sech}^{2}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*sech(x)**2)**(1/2),x)

[Out]

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

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