3.211 \(\int \frac{\tanh ^6(x)}{(a+b \text{sech}^2(x))^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.338873, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {4141, 1975, 470, 578, 523, 217, 203, 377, 206} \[ -\frac{\left (\frac{1}{a^2}-\frac{1}{b^2}\right ) \tanh (x)}{\sqrt{a-b \tanh ^2(x)+b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a-b \tanh ^2(x)+b}}\right )}{a^{5/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} \tanh (x)}{\sqrt{a-b \tanh ^2(x)+b}}\right )}{b^{5/2}}-\frac{(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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]

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

Mathematica [A]  time = 0.692483, size = 178, normalized size = 1.51 \[ \frac{\text{sech}^5(x) \left (\frac{2 (a+b) \sinh (x) \left (3 a^2+a (3 a-4 b) \cosh (2 x)+4 a b-6 b^2\right ) (a \cosh (2 x)+a+2 b)}{3 a^2 b^2}+\frac{\sqrt{2} (a \cosh (2 x)+a+2 b)^{5/2} \left (b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sinh (x)}{\sqrt{a \cosh (2 x)+a+2 b}}\right )-a^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sinh (x)}{\sqrt{a \cosh (2 x)+a+2 b}}\right )\right )}{a^{5/2} b^{5/2}}\right )}{8 \left (a+b \text{sech}^2(x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[x]^5*((Sqrt[2]*(-(a^(5/2)*ArcTan[(Sqrt[2]*Sqrt[b]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]]) + b^(5/2)*ArcTa
nh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]])*(a + 2*b + a*Cosh[2*x])^(5/2))/(a^(5/2)*b^(5/2)) +
(2*(a + b)*(a + 2*b + a*Cosh[2*x])*(3*a^2 + 4*a*b - 6*b^2 + a*(3*a - 4*b)*Cosh[2*x])*Sinh[x])/(3*a^2*b^2)))/(8
*(a + b*Sech[x]^2)^(5/2))

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Maple [F]  time = 0.106, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tanh \left ( x \right ) \right ) ^{6} \left ( a+b \left ({\rm sech} \left (x\right ) \right ) ^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (x\right )^{6}}{{\left (b \operatorname{sech}\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*sech(x)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)^6/(b*sech(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*sech(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 \operatorname{sech}^{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*sech(x)**2)**(5/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (x\right )^{6}}{{\left (b \operatorname{sech}\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*sech(x)^2)^(5/2),x, algorithm="giac")

[Out]

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