\(\int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [134]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {2 a \sqrt {a+b \text {sech}(c+d x)}}{b^2 d}+\frac {2 (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {(-2 a+b \text {sech}(c+d x)) \sqrt {a+b \text {sech}(c+d x)}}{b^2}\right )}{3 d} \] Input:

Integrate[Tanh[c + d*x]^3/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 26, 4373, 517, 1467, 2009}

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.

\(\displaystyle \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \cot \left (i c+i d x+\frac {\pi }{2}\right )^3}{\sqrt {a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^3}{\sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}}dx\)

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 517

\(\displaystyle -\frac {2 \int \frac {b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2}{a-b^2 \text {sech}^2(c+d x)}d\sqrt {a+b \text {sech}(c+d x)}}{b^2 d}\)

\(\Big \downarrow \) 1467

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tanh[c + d*x]^3/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

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

Defintions of rubi rules used

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 517

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1))   Subst[Int[x^(2*n + 1)*(-c + x^ 
2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
 

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4373

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1))   Subst[Int[(b^2 - x^ 
2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, 
 d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
 

Maple [F]

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

Input:

int(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (67) = 134\).

Time = 0.69 (sec) , antiderivative size = 925, normalized size of antiderivative = 11.71 \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sin 
h(d*x + c)^2 + b^2)*sqrt(a)*log(-(2*a^2*cosh(d*x + c)^4 + 2*a^2*sinh(d*x + 
 c)^4 + 4*a*b*cosh(d*x + c)^3 + 4*(2*a^2*cosh(d*x + c) + a*b)*sinh(d*x + c 
)^3 + 4*a*b*cosh(d*x + c) + (4*a^2 + b^2)*cosh(d*x + c)^2 + (12*a^2*cosh(d 
*x + c)^2 + 12*a*b*cosh(d*x + c) + 4*a^2 + b^2)*sinh(d*x + c)^2 + 2*a^2 + 
2*(a*cosh(d*x + c)^4 + a*sinh(d*x + c)^4 + b*cosh(d*x + c)^3 + (4*a*cosh(d 
*x + c) + b)*sinh(d*x + c)^3 + 2*a*cosh(d*x + c)^2 + (6*a*cosh(d*x + c)^2 
+ 3*b*cosh(d*x + c) + 2*a)*sinh(d*x + c)^2 + b*cosh(d*x + c) + (4*a*cosh(d 
*x + c)^3 + 3*b*cosh(d*x + c)^2 + 4*a*cosh(d*x + c) + b)*sinh(d*x + c) + a 
)*sqrt(a)*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)) + 2*(4*a^2*cosh(d*x + 
c)^3 + 6*a*b*cosh(d*x + c)^2 + 2*a*b + (4*a^2 + b^2)*cosh(d*x + c))*sinh(d 
*x + c))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^ 
2)) - 8*(a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 - a*b*cosh(d*x + c) + a 
^2 + (2*a^2*cosh(d*x + c) - a*b)*sinh(d*x + c))*sqrt((a*cosh(d*x + c) + b) 
/cosh(d*x + c)))/(a*b^2*d*cosh(d*x + c)^2 + 2*a*b^2*d*cosh(d*x + c)*sinh(d 
*x + c) + a*b^2*d*sinh(d*x + c)^2 + a*b^2*d), -1/3*(3*(b^2*cosh(d*x + c)^2 
 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + b^2)*sqrt(-a) 
*arctan((a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + b*cosh(d*x + c) + (2*a*co 
sh(d*x + c) + b)*sinh(d*x + c) + a)*sqrt(-a)*sqrt((a*cosh(d*x + c) + b)/co 
sh(d*x + c))/(a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*...
 

Sympy [F]

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

integrate(tanh(d*x+c)**3/(a+b*sech(d*x+c))**(1/2),x)
 

Output:

Integral(tanh(c + d*x)**3/sqrt(a + b*sech(c + d*x)), x)
 

Maxima [F]

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

integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(tanh(d*x + c)^3/sqrt(b*sech(d*x + c) + a), x)
 

Giac [F]

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

integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(tanh(d*x + c)^3/sqrt(b*sech(d*x + c) + a), x)
 

Mupad [F(-1)]

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

int(tanh(c + d*x)^3/(a + b/cosh(c + d*x))^(1/2),x)
 

Output:

int(tanh(c + d*x)^3/(a + b/cosh(c + d*x))^(1/2), x)
 

Reduce [F]

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

int(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int((sqrt(sech(c + d*x)*b + a)*tanh(c + d*x)**3)/(sech(c + d*x)*b + a),x)