3.233 $$\int \frac{\tanh (x)}{\sqrt{a+b \tanh ^2(x)}} \, dx$$

Optimal. Leaf size=29 $\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tanh ^2(x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b}}$

[Out]

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

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Rubi [A]  time = 0.0643284, antiderivative size = 29, normalized size of antiderivative = 1., 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 = {3670, 444, 63, 208} $\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tanh ^2(x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Tanh[x]/Sqrt[a + b*Tanh[x]^2],x]

[Out]

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

Rule 3670

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [A]  time = 0.0142, size = 29, normalized size = 1. $\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tanh ^2(x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.05, size = 114, normalized size = 3.9 \begin{align*}{\frac{1}{2}\ln \left ({\frac{1}{1+\tanh \left ( x \right ) } \left ( 2\,a+2\,b-2\, \left ( 1+\tanh \left ( x \right ) \right ) b+2\,\sqrt{a+b}\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}b-2\, \left ( 1+\tanh \left ( x \right ) \right ) b+a+b} \right ) } \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{1}{2}\ln \left ({\frac{1}{\tanh \left ( x \right ) -1} \left ( 2\,a+2\,b+2\, \left ( \tanh \left ( x \right ) -1 \right ) b+2\,\sqrt{a+b}\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}b+2\, \left ( \tanh \left ( x \right ) -1 \right ) b+a+b} \right ) } \right ){\frac{1}{\sqrt{a+b}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.14557, size = 3792, normalized size = 130.76 \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*tanh(x)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.03482, size = 31, normalized size = 1.07 \begin{align*} - \frac{\operatorname{atan}{\left (\frac{\sqrt{a + b \tanh ^{2}{\left (x \right )}}}{\sqrt{- a - b}} \right )}}{\sqrt{- a - b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError