\(\int \frac {\tanh ^3(x)}{(a+b \text {sech}^2(x))^{5/2}} \, dx\) [214]

Optimal result

Integrand size = 17, antiderivative size = 68 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {a+b}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}-\frac {1}{a^2 \sqrt {a+b \text {sech}^2(x)}} \] Output:

arctanh((a+b*sech(x)^2)^(1/2)/a^(1/2))/a^(5/2)-1/3*(a+b)/a/b/(a+b*sech(x)^ 
2)^(3/2)-1/a^2/(a+b*sech(x)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.82 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cosh (2 x)) \left (3 \sqrt {a} (a+2 b)^2 \cosh (x)+a^{3/2} (a+4 b) \cosh (3 x)-3 \sqrt {2} b (a+2 b+a \cosh (2 x))^{3/2} \log \left (\sqrt {2} \sqrt {a} \cosh (x)+\sqrt {a+2 b+a \cosh (2 x)}\right )\right ) \text {sech}^5(x)}{24 a^{5/2} b \left (a+b \text {sech}^2(x)\right )^{5/2}} \] Input:

Integrate[Tanh[x]^3/(a + b*Sech[x]^2)^(5/2),x]
 

Output:

-1/24*((a + 2*b + a*Cosh[2*x])*(3*Sqrt[a]*(a + 2*b)^2*Cosh[x] + a^(3/2)*(a 
 + 4*b)*Cosh[3*x] - 3*Sqrt[2]*b*(a + 2*b + a*Cosh[2*x])^(3/2)*Log[Sqrt[2]* 
Sqrt[a]*Cosh[x] + Sqrt[a + 2*b + a*Cosh[2*x]]])*Sech[x]^5)/(a^(5/2)*b*(a + 
 b*Sech[x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {3042, 26, 4627, 25, 354, 87, 61, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \tan (i x)^3}{\left (a+b \sec (i x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\tan (i x)^3}{\left (b \sec (i x)^2+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4627

\(\displaystyle \int -\frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (a+b \text {sech}^2(x)\right )^{5/2}}d\text {sech}(x)\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (b \text {sech}^2(x)+a\right )^{5/2}}d\text {sech}(x)\)

\(\Big \downarrow \) 354

\(\displaystyle -\frac {1}{2} \int \frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (b \text {sech}^2(x)+a\right )^{5/2}}d\text {sech}^2(x)\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\cosh (x)}{\left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}^2(x)}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\int \frac {\cosh (x)}{\sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a}+\frac {2}{a \sqrt {a+b \text {sech}^2(x)}}}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 \int \frac {1}{\frac {\text {sech}^4(x)}{b}-\frac {a}{b}}d\sqrt {b \text {sech}^2(x)+a}}{a b}+\frac {2}{a \sqrt {a+b \text {sech}^2(x)}}}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2}{a \sqrt {a+b \text {sech}^2(x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{3/2}}}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\)

Input:

Int[Tanh[x]^3/(a + b*Sech[x]^2)^(5/2),x]
 

Output:

((-2*(a + b))/(3*a*b*(a + b*Sech[x]^2)^(3/2)) - ((-2*ArcTanh[Sqrt[a + b*Se 
ch[x]^2]/Sqrt[a]])/a^(3/2) + 2/(a*Sqrt[a + b*Sech[x]^2]))/a)/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 61

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

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(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] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

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

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

Maple [F]

\[\int \frac {\tanh \left (x \right )^{3}}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {5}{2}}}d x\]

Input:

int(tanh(x)^3/(a+sech(x)^2*b)^(5/2),x)
 

Output:

int(tanh(x)^3/(a+sech(x)^2*b)^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1954 vs. \(2 (56) = 112\).

Time = 0.32 (sec) , antiderivative size = 4644, normalized size of antiderivative = 68.29 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(tanh(x)^3/(a+b*sech(x)^2)^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(tanh(x)**3/(a+b*sech(x)**2)**(5/2),x)
 

Output:

Integral(tanh(x)**3/(a + b*sech(x)**2)**(5/2), x)
 

Maxima [F]

\[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(tanh(x)^3/(a+b*sech(x)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate(tanh(x)^3/(b*sech(x)^2 + a)^(5/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (56) = 112\).

Time = 0.22 (sec) , antiderivative size = 285, normalized size of antiderivative = 4.19 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=-\frac {{\left ({\left (\frac {{\left (a^{8} b + 6 \, a^{7} b^{2} + 9 \, a^{6} b^{3} + 4 \, a^{5} b^{4}\right )} e^{\left (2 \, x\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}} + \frac {3 \, {\left (a^{8} b + 6 \, a^{7} b^{2} + 13 \, a^{6} b^{3} + 12 \, a^{5} b^{4} + 4 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} + \frac {3 \, {\left (a^{8} b + 6 \, a^{7} b^{2} + 13 \, a^{6} b^{3} + 12 \, a^{5} b^{4} + 4 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} + \frac {a^{8} b + 6 \, a^{7} b^{2} + 9 \, a^{6} b^{3} + 4 \, a^{5} b^{4}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}}{3 \, {\left (a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + 4 \, b e^{\left (2 \, x\right )} + a\right )}^{\frac {3}{2}}} \] Input:

integrate(tanh(x)^3/(a+b*sech(x)^2)^(5/2),x, algorithm="giac")
 

Output:

-1/3*((((a^8*b + 6*a^7*b^2 + 9*a^6*b^3 + 4*a^5*b^4)*e^(2*x)/(a^8*b^2 + 2*a 
^7*b^3 + a^6*b^4) + 3*(a^8*b + 6*a^7*b^2 + 13*a^6*b^3 + 12*a^5*b^4 + 4*a^4 
*b^5)/(a^8*b^2 + 2*a^7*b^3 + a^6*b^4))*e^(2*x) + 3*(a^8*b + 6*a^7*b^2 + 13 
*a^6*b^3 + 12*a^5*b^4 + 4*a^4*b^5)/(a^8*b^2 + 2*a^7*b^3 + a^6*b^4))*e^(2*x 
) + (a^8*b + 6*a^7*b^2 + 9*a^6*b^3 + 4*a^5*b^4)/(a^8*b^2 + 2*a^7*b^3 + a^6 
*b^4))/(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a)^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^3}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2}} \,d x \] Input:

int(tanh(x)^3/(a + b/cosh(x)^2)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int(tanh(x)^3/(a + b/cosh(x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (x \right )^{2} b +a}\, \tanh \left (x \right )^{3}}{\mathrm {sech}\left (x \right )^{6} b^{3}+3 \mathrm {sech}\left (x \right )^{4} a \,b^{2}+3 \mathrm {sech}\left (x \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(tanh(x)^3/(a+b*sech(x)^2)^(5/2),x)
 

Output:

int((sqrt(sech(x)**2*b + a)*tanh(x)**3)/(sech(x)**6*b**3 + 3*sech(x)**4*a* 
b**2 + 3*sech(x)**2*a**2*b + a**3),x)