3.241 \(\int \frac{\tanh ^2(x)}{(a+b \tanh ^2(x))^{3/2}} \, dx\)
Optimal. Leaf size=53 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}-\frac{\tanh (x)}{(a+b) \sqrt{a+b \tanh ^2(x)}} \]
[Out]
ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]]/(a + b)^(3/2) - Tanh[x]/((a + b)*Sqrt[a + b*Tanh[x]^2])
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Rubi [A] time = 0.0992793, antiderivative size = 53, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used =
{3670, 471, 377, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}-\frac{\tanh (x)}{(a+b) \sqrt{a+b \tanh ^2(x)}} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[x]^2/(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) - Tanh[x]/((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 471
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
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{\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx &=\operatorname{Subst}\left (\int \frac{x^2}{\left (1-x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )\\ &=-\frac{\tanh (x)}{(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{\tanh (x)}{(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{\tanh (x)}{(a+b) \sqrt{a+b \tanh ^2(x)}}\\ \end{align*}
Mathematica [B] time = 1.45602, size = 112, normalized size = 2.11 \[ \frac{\tanh (x) \left (\tanh ^{-1}\left (\frac{\sqrt{\frac{(a+b) \tanh ^2(x)}{a}}}{\sqrt{\frac{b \tanh ^2(x)}{a}+1}}\right ) \sqrt{\frac{(a+b) \tanh ^2(x)}{a}} \left (a \coth ^2(x)+b\right )-(a+b) \sqrt{\frac{b \tanh ^2(x)}{a}+1}\right )}{(a+b)^2 \sqrt{a+b \tanh ^2(x)} \sqrt{\frac{b \tanh ^2(x)}{a}+1}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[x]^2/(a + b*Tanh[x]^2)^(3/2),x]
[Out]
(Tanh[x]*(ArcTanh[Sqrt[((a + b)*Tanh[x]^2)/a]/Sqrt[1 + (b*Tanh[x]^2)/a]]*(b + a*Coth[x]^2)*Sqrt[((a + b)*Tanh[
x]^2)/a] - (a + b)*Sqrt[1 + (b*Tanh[x]^2)/a]))/((a + b)^2*Sqrt[a + b*Tanh[x]^2]*Sqrt[1 + (b*Tanh[x]^2)/a])
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Maple [B] time = 0.021, size = 289, normalized size = 5.5 \begin{align*} -{\frac{\tanh \left ( x \right ) }{a}{\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)^2/(a+b*tanh(x)^2)^(3/2),x)
[Out]
-tanh(x)/a/(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 )^{2}}{{\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)^2/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate(tanh(x)^2/(b*tanh(x)^2 + a)^(3/2), x)
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Fricas [B] time = 3.09922, size = 6395, normalized size = 120.66 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^2/(a+b*tanh(x)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/4*(((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)*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)*s
inh(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 + 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)*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*co
sh(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)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^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 + 3*a^2*b +
3*a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)*sinh(x)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b
^3)*sinh(x)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2 + 2*(a^3 + a^2*b - a*b
^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)
^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(x))*sinh(x)), -1/2*(((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)*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) + sin
h(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)*c
osh(x)^3 + (a^2 - a*b - 2*b^2)*cosh(x))*sinh(x))) + ((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)*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)*cosh(x)^2 +
2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^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 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 + 3*a^2*b + 3
*a*b^2 + b^3)*cosh(x)*sinh(x)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(x)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 +
2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2 + 2*(a^3 + a^2*b - a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos
h(x)^2)*sinh(x)^2 + 4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(x))*sinh(x
))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{2}{\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)**2/(a+b*tanh(x)**2)**(3/2),x)
[Out]
Integral(tanh(x)**2/(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)^2/(a+b*tanh(x)^2)^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError