### 3.240 $$\int \frac{\tanh ^3(x)}{(a+b \tanh ^2(x))^{3/2}} \, dx$$

Optimal. Leaf size=52 $\frac{a}{b (a+b) \sqrt{a+b \tanh ^2(x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tanh ^2(x)}}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.123291, antiderivative size = 52, 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 = {3670, 446, 78, 63, 208} $\frac{a}{b (a+b) \sqrt{a+b \tanh ^2(x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tanh ^2(x)}}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.118817, size = 52, normalized size = 1. $\frac{a}{b (a+b) \sqrt{a+b \tanh ^2(x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tanh ^2(x)}}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.023, size = 287, normalized size = 5.5 \begin{align*}{\frac{1}{b}{\frac{1}{\sqrt{a+b \left ( \tanh \left ( x \right ) \right ) ^{2}}}}}-{\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(tanh(x)^3/(a+b*tanh(x)^2)^(3/2),x)

[Out]

1/b/(a+b*tanh(x)^2)^(1/2)-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*(tanh(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*(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 )^{3}}{{\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(tanh(x)^3/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.07382, size = 6730, normalized size = 129.42 \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*tanh(x)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(((a*b + b^2)*cosh(x)^4 + 4*(a*b + b^2)*cosh(x)*sinh(x)^3 + (a*b + b^2)*sinh(x)^4 + 2*(a*b - b^2)*cosh(x)
^2 + 2*(3*(a*b + b^2)*cosh(x)^2 + a*b - b^2)*sinh(x)^2 + a*b + b^2 + 4*((a*b + b^2)*cosh(x)^3 + (a*b - b^2)*co
sh(x))*sinh(x))*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)*s
inh(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)*cosh(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) + 1
5*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 + b^2)*cosh(x)^4 + 4*(a*b + b^2)*cosh(x)*sinh(x)^3 + (a*b + b^2)*sinh(x)^4 + 2*(a*b - b^2)*cosh(x)^2 + 2
*(3*(a*b + b^2)*cosh(x)^2 + a*b - b^2)*sinh(x)^2 + a*b + b^2 + 4*((a*b + b^2)*cosh(x)^3 + (a*b - b^2)*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*b*cosh(x)^
2 + 2*(3*(a + b)*cosh(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*cosh(x))*sinh(x) + a + b)/(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 + b)*cosh(x)^2 + (a
+ b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*co
sh(x)^4 + 4*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cosh(x)*sinh(x)^3 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*sinh(x
)^4 + a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4 + 2*(a^3*b + a^2*b^2 - a*b^3 - b^4)*cosh(x)^2 + 2*(a^3*b + a^2*b^2 - a
*b^3 - b^4 + 3*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^
4)*cosh(x)^3 + (a^3*b + a^2*b^2 - a*b^3 - b^4)*cosh(x))*sinh(x)), -1/2*(((a*b + b^2)*cosh(x)^4 + 4*(a*b + b^2)
*cosh(x)*sinh(x)^3 + (a*b + b^2)*sinh(x)^4 + 2*(a*b - b^2)*cosh(x)^2 + 2*(3*(a*b + b^2)*cosh(x)^2 + a*b - b^2)
*sinh(x)^2 + a*b + b^2 + 4*((a*b + b^2)*cosh(x)^3 + (a*b - b^2)*cosh(x))*sinh(x))*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)*sinh(
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)*sinh
(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))) + ((a*b +
b^2)*cosh(x)^4 + 4*(a*b + b^2)*cosh(x)*sinh(x)^3 + (a*b + b^2)*sinh(x)^4 + 2*(a*b - b^2)*cosh(x)^2 + 2*(3*(a*b
+ b^2)*cosh(x)^2 + a*b - b^2)*sinh(x)^2 + a*b + b^2 + 4*((a*b + b^2)*cosh(x)^3 + (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)*cosh(x)^3 + (a - b)*cosh(x))*sinh(x) + a + b)) - 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 + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^
2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cosh(x)^4 + 4*(a^3*b + 3*a^2*b^2 + 3
*a*b^3 + b^4)*cosh(x)*sinh(x)^3 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*sinh(x)^4 + a^3*b + 3*a^2*b^2 + 3*a*b^3
+ b^4 + 2*(a^3*b + a^2*b^2 - a*b^3 - b^4)*cosh(x)^2 + 2*(a^3*b + a^2*b^2 - a*b^3 - b^4 + 3*(a^3*b + 3*a^2*b^2
+ 3*a*b^3 + b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cosh(x)^3 + (a^3*b + a^2*b^2 -
a*b^3 - b^4)*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 \tanh ^{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*tanh(x)**2)**(3/2),x)

[Out]

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

[Out]

Exception raised: TypeError