### 3.246 $$\int \frac{\tanh ^6(x)}{(a+b \tanh ^2(x))^{5/2}} \, dx$$

Optimal. Leaf size=118 $\frac{a (a+2 b) \tanh (x)}{b^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{b^{5/2}}+\frac{a \tanh ^3(x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}$

[Out]

-(ArcTanh[(Sqrt[b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]]/b^(5/2)) + ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]
^2]]/(a + b)^(5/2) + (a*Tanh[x]^3)/(3*b*(a + b)*(a + b*Tanh[x]^2)^(3/2)) + (a*(a + 2*b)*Tanh[x])/(b^2*(a + b)^
2*Sqrt[a + b*Tanh[x]^2])

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Rubi [A]  time = 0.220451, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.412, Rules used = {3670, 470, 578, 523, 217, 206, 377} $\frac{a (a+2 b) \tanh (x)}{b^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{b^{5/2}}+\frac{a \tanh ^3(x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Tanh[x]^6/(a + b*Tanh[x]^2)^(5/2),x]

[Out]

-(ArcTanh[(Sqrt[b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]]/b^(5/2)) + ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]
^2]]/(a + b)^(5/2) + (a*Tanh[x]^3)/(3*b*(a + b)*(a + b*Tanh[x]^2)^(3/2)) + (a*(a + 2*b)*Tanh[x])/(b^2*(a + b)^
2*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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 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] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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])

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]

Rubi steps

\begin{align*} \int \frac{\tanh ^6(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx &=\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=\frac{a \tanh ^3(x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a-3 (a+b) x^2\right )}{\left (1-x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )}{3 b (a+b)}\\ &=\frac{a \tanh ^3(x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{a (a+2 b) \tanh (x)}{b^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{\operatorname{Subst}\left (\int \frac{3 a (a+2 b)-3 (a+b)^2 x^2}{\left (1-x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tanh (x)\right )}{3 b^2 (a+b)^2}\\ &=\frac{a \tanh ^3(x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{a (a+2 b) \tanh (x)}{b^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tanh (x)\right )}{b^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tanh (x)\right )}{(a+b)^2}\\ &=\frac{a \tanh ^3(x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{a (a+2 b) \tanh (x)}{b^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{b^2}+\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)^2}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{b^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}+\frac{a \tanh ^3(x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{a (a+2 b) \tanh (x)}{b^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}\\ \end{align*}

Mathematica [C]  time = 1.86165, size = 231, normalized size = 1.96 $\frac{\sqrt{\text{sech}^2(x) ((a+b) \cosh (2 x)+a-b)} \left (\frac{a (a+b) \sinh (2 x) \left (\left (3 a^2+10 a b+7 b^2\right ) \cosh (2 x)+3 a^2+2 a b-7 b^2\right )}{((a+b) \cosh (2 x)+a-b)^2}-\frac{3 \sqrt{2} a \coth (x) \left (\left (a^2+3 a b+2 b^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\text{csch}^2(x) ((a+b) \cosh (2 x)+a-b)}{b}}}{\sqrt{2}}\right ),1\right )+b^2 \Pi \left (\frac{b}{a+b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a-b+(a+b) \cosh (2 x)) \text{csch}^2(x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{b \sqrt{\frac{\text{csch}^2(x) ((a+b) \cosh (2 x)+a-b)}{b}}}\right )}{3 \sqrt{2} b^2 (a+b)^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Tanh[x]^6/(a + b*Tanh[x]^2)^(5/2),x]

[Out]

(Sqrt[(a - b + (a + b)*Cosh[2*x])*Sech[x]^2]*((-3*Sqrt[2]*a*Coth[x]*((a^2 + 3*a*b + 2*b^2)*EllipticF[ArcSin[Sq
rt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]/Sqrt[2]], 1] + b^2*EllipticPi[b/(a + b), ArcSin[Sqrt[((a - b + (
a + b)*Cosh[2*x])*Csch[x]^2)/b]/Sqrt[2]], 1]))/(b*Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]) + (a*(a + b
)*(3*a^2 + 2*a*b - 7*b^2 + (3*a^2 + 10*a*b + 7*b^2)*Cosh[2*x])*Sinh[2*x])/(a - b + (a + b)*Cosh[2*x])^2))/(3*S
qrt[2]*b^2*(a + b)^3)

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Maple [B]  time = 0.032, size = 549, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^6/(a+b*tanh(x)^2)^(5/2),x)

[Out]

1/3*tanh(x)^3/b/(a+b*tanh(x)^2)^(3/2)+1/b^2*tanh(x)/(a+b*tanh(x)^2)^(1/2)-1/b^(5/2)*ln(tanh(x)*b^(1/2)+(a+b*ta
nh(x)^2)^(1/2))+1/3*tanh(x)/b/(a+b*tanh(x)^2)^(3/2)-1/3/a/b*tanh(x)/(a+b*tanh(x)^2)^(1/2)-1/3*tanh(x)/a/(a+b*t
anh(x)^2)^(3/2)-2/3/a^2*tanh(x)/(a+b*tanh(x)^2)^(1/2)+1/6/(a+b)/((1+tanh(x))^2*b-2*(1+tanh(x))*b+a+b)^(3/2)+1/
6*b/(a+b)/a/((1+tanh(x))^2*b-2*(1+tanh(x))*b+a+b)^(3/2)*tanh(x)+1/3*b/(a+b)/a^2/((1+tanh(x))^2*b-2*(1+tanh(x))
*b+a+b)^(1/2)*tanh(x)+1/2/(a+b)^2/((1+tanh(x))^2*b-2*(1+tanh(x))*b+a+b)^(1/2)+1/2/(a+b)^2/a/((1+tanh(x))^2*b-2
*(1+tanh(x))*b+a+b)^(1/2)*b*tanh(x)-1/2/(a+b)^(5/2)*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)))-1/6/(a+b)/((tanh(x)-1)^2*b+2*(tanh(x)-1)*b+a+b)^(3/2)+1/6*b/(a+b)/a/
((tanh(x)-1)^2*b+2*(tanh(x)-1)*b+a+b)^(3/2)*tanh(x)+1/3*b/(a+b)/a^2/((tanh(x)-1)^2*b+2*(tanh(x)-1)*b+a+b)^(1/2
)*tanh(x)-1/2/(a+b)^2/((tanh(x)-1)^2*b+2*(tanh(x)-1)*b+a+b)^(1/2)+1/2/(a+b)^2/a/((tanh(x)-1)^2*b+2*(tanh(x)-1)
*b+a+b)^(1/2)*b*tanh(x)+1/2/(a+b)^(5/2)*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 )^{6}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+b*tanh(x)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)^6/(b*tanh(x)^2 + a)^(5/2), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+b*tanh(x)^2)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{6}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**6/(a+b*tanh(x)**2)**(5/2),x)

[Out]

Integral(tanh(x)**6/(a + b*tanh(x)**2)**(5/2), 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)^6/(a+b*tanh(x)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError