3.243 \(\int \frac{1}{(a+b \tanh ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=56 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}+\frac{b \tanh (x)}{a (a+b) \sqrt{a+b \tanh ^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.042802, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3661, 382, 377, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}+\frac{b \tanh (x)}{a (a+b) \sqrt{a+b \tanh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tanh[x]^2)^(-3/2),x]

[Out]

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

Rule 3661

Int[((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[(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, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 382

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

Rule 377

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

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])

Rubi steps

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

Mathematica [C]  time = 4.21493, size = 223, normalized size = 3.98 \[ -\frac{\sinh ^2(x) \left (\sqrt{2} a^2 (a+b) \tanh (x) \, _2F_1\left (2,2;\frac{7}{2};-\frac{(a+b) \sinh ^2(x)}{a}\right ) \left (-\frac{(a+b) \sinh ^2(x) ((a+b) \cosh (2 x)+a-b)}{a^2}\right )^{3/2}+\frac{15}{4} a \text{csch}(x) \text{sech}(x) ((3 a+2 b) \cosh (2 x)+3 a-2 b) \left (-2 a \sqrt{-\frac{(a+b) \sinh ^4(x) \left (a \coth ^2(x)+b\right )}{a^2}}+(a-b) \sin ^{-1}\left (\sqrt{-\frac{(a+b) \sinh ^2(x)}{a}}\right )+(a+b) \cosh (2 x) \sin ^{-1}\left (\sqrt{-\frac{(a+b) \sinh ^2(x)}{a}}\right )\right )\right )}{15 a^4 \left (-\frac{(a+b) \sinh ^2(x)}{a}\right )^{3/2} \sqrt{a+b \tanh ^2(x)} \sqrt{\frac{b \sinh ^2(x)}{a}+\cosh ^2(x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Tanh[x]^2)^(-3/2),x]

[Out]

-(Sinh[x]^2*((15*a*(3*a - 2*b + (3*a + 2*b)*Cosh[2*x])*Csch[x]*Sech[x]*((a - b)*ArcSin[Sqrt[-(((a + b)*Sinh[x]
^2)/a)]] + (a + b)*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]]*Cosh[2*x] - 2*a*Sqrt[-(((a + b)*(b + a*Coth[x]^2)*Si
nh[x]^4)/a^2)]))/4 + Sqrt[2]*a^2*(a + b)*Hypergeometric2F1[2, 2, 7/2, -(((a + b)*Sinh[x]^2)/a)]*(-(((a + b)*(a
- b + (a + b)*Cosh[2*x])*Sinh[x]^2)/a^2))^(3/2)*Tanh[x]))/(15*a^4*(-(((a + b)*Sinh[x]^2)/a))^(3/2)*Sqrt[Cosh[
x]^2 + (b*Sinh[x]^2)/a]*Sqrt[a + b*Tanh[x]^2])

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Maple [B]  time = 0.026, size = 272, normalized size = 4.9 \begin{align*}{\frac{1}{2\,b+2\,a}{\frac{1}{\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}b-2\, \left ( 1+\tanh \left ( x \right ) \right ) b+a+b}}}}+{\frac{b \left ( 2\, \left ( 1+\tanh \left ( x \right ) \right ) b-2\,b \right ) }{ \left ( a+b \right ) \left ( 4\,b \left ( a+b \right ) -4\,{b}^{2} \right ) }{\frac{1}{\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}b-2\, \left ( 1+\tanh \left ( x \right ) \right ) b+a+b}}}}-{\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 ) \left ( a+b \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{2\,b+2\,a}{\frac{1}{\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}b+2\, \left ( \tanh \left ( x \right ) -1 \right ) b+a+b}}}}+{\frac{b \left ( 2\, \left ( \tanh \left ( x \right ) -1 \right ) b+2\,b \right ) }{ \left ( a+b \right ) \left ( 4\,b \left ( a+b \right ) -4\,{b}^{2} \right ) }{\frac{1}{\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}b+2\, \left ( \tanh \left ( x \right ) -1 \right ) b+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 ) \left ( a+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(a+b)/((1+tanh(x))^2*b-2*(1+tanh(x))*b+a+b)^(1/2)+b/(a+b)*(2*(1+tanh(x))*b-2*b)/(4*b*(a+b)-4*b^2)/((1+tanh
(x))^2*b-2*(1+tanh(x))*b+a+b)^(1/2)-1/2/(a+b)^(3/2)*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)/((tanh(x)-1)^2*b+2*(tanh(x)-1)*b+a+b)^(1/2)+b/(a+b)*(2*(ta
nh(x)-1)*b+2*b)/(4*b*(a+b)-4*b^2)/((tanh(x)-1)^2*b+2*(tanh(x)-1)*b+a+b)^(1/2)+1/2/(a+b)^(3/2)*ln((2*a+2*b+2*(t
anh(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{1}{{\left (b \tanh \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(1/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.09694, size = 6730, normalized size = 120.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((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 - a*b)*cosh(x)
^2 + 2*(3*(a^2 + a*b)*cosh(x)^2 + a^2 - a*b)*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x)^3 + (a^2 - a*b)*co
sh(x))*sinh(x))*sqrt(a + b)*log(-((a*b^2 + b^3)*cosh(x)^8 + 8*(a*b^2 + b^3)*cosh(x)*sinh(x)^7 + (a*b^2 + b^3)*
sinh(x)^8 - 2*(a*b^2 + 2*b^3)*cosh(x)^6 - 2*(a*b^2 + 2*b^3 - 14*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^6 + 4*(14*(a*
b^2 + b^3)*cosh(x)^3 - 3*(a*b^2 + 2*b^3)*cosh(x))*sinh(x)^5 + (a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^4 + (70*
(a*b^2 + b^3)*cosh(x)^4 + a^3 - a^2*b + 4*a*b^2 + 6*b^3 - 30*(a*b^2 + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a*b
^2 + b^3)*cosh(x)^5 - 10*(a*b^2 + 2*b^3)*cosh(x)^3 + (a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x))*sinh(x)^3 + a^3
+ 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2 + 2*(14*(a*b^2 + b^3)*cosh(x)^6 - 15*(a*b^2 +
2*b^3)*cosh(x)^4 + a^3 - 3*a*b^2 - 2*b^3 + 3*(a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^2)*sinh(x)^2 + sqrt(2)*(b
^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 - 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4
+ 4*(5*b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 - 2*a*b - 3*b^2)*cosh(x)^2 + (15*b^2*cosh(x)^4 - 18*b^
2*cosh(x)^2 - a^2 + 2*a*b + 3*b^2)*sinh(x)^2 - a^2 - 2*a*b - b^2 + 2*(3*b^2*cosh(x)^5 - 6*b^2*cosh(x)^3 - (a^2
- 2*a*b - 3*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*b^2 + b^3)*cosh(x)^7 - 3*(a*b^2 + 2*b^3)*cosh(x)^5 + (a^3 - a^2*
b + 4*a*b^2 + 6*b^3)*cosh(x)^3 + (a^3 - 3*a*b^2 - 2*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)) +
((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 - a*b)*cosh(x)^2 +
2*(3*(a^2 + a*b)*cosh(x)^2 + a^2 - a*b)*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x)^3 + (a^2 - a*b)*cosh(x)
)*sinh(x))*sqrt(a + b)*log(((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 + 2*a*cosh(x)^
2 + 2*(3*(a + b)*cosh(x)^2 + a)*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 + a*cosh(x))*sinh(x) + a + b)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*sqrt(2)*((a*b + b^
2)*cosh(x)^2 + 2*(a*b + b^2)*cosh(x)*sinh(x) + (a*b + b^2)*sinh(x)^2 - a*b - b^2)*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^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*co
sh(x)^4 + 4*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)*sinh(x)^3 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sinh(x
)^4 + a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 + 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*cosh(x)^2 + 2*(a^4 + a^3*b - a^2*b
^2 - a*b^3 + 3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^
3)*cosh(x)^3 + (a^4 + a^3*b - a^2*b^2 - a*b^3)*cosh(x))*sinh(x)), -1/2*(((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 - a*b)*cosh(x)^2 + 2*(3*(a^2 + a*b)*cosh(x)^2 + a^2 - a*b)
*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x)^3 + (a^2 - a*b)*cosh(x))*sinh(x))*sqrt(-a - b)*arctan(sqrt(2)*
(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 - a - b)*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 + b^2)*cosh(x)^4 + 4*(a*b + b^2)*cosh(x)*sinh
(x)^3 + (a*b + b^2)*sinh(x)^4 + (a^2 - a*b - 2*b^2)*cosh(x)^2 + (6*(a*b + b^2)*cosh(x)^2 + a^2 - a*b - 2*b^2)*
sinh(x)^2 + a^2 + 2*a*b + b^2 + 2*(2*(a*b + b^2)*cosh(x)^3 + (a^2 - a*b - 2*b^2)*cosh(x))*sinh(x))) + ((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 - a*b)*cosh(x)^2 + 2*(3*(a^2
+ a*b)*cosh(x)^2 + a^2 - a*b)*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x)^3 + (a^2 - a*b)*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)*cosh(x)^3 + (a - b)*cosh(x))*sinh(x) + a + b)) - 2*sqrt(2)*((a*b + b^2)*cosh(x)^2 + 2*(a*b + b^2)*cosh(
x)*sinh(x) + (a*b + b^2)*sinh(x)^2 - a*b - b^2)*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^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^4 + 4*(a^4 + 3*a^3*b + 3*a^2
*b^2 + a*b^3)*cosh(x)*sinh(x)^3 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sinh(x)^4 + a^4 + 3*a^3*b + 3*a^2*b^2 +
a*b^3 + 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*cosh(x)^2 + 2*(a^4 + a^3*b - a^2*b^2 - a*b^3 + 3*(a^4 + 3*a^3*b + 3*
a^2*b^2 + a*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^3 + (a^4 + a^3*b - a^2*
b^2 - a*b^3)*cosh(x))*sinh(x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError