\(\int \frac {\tanh ^5(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [133]

Optimal result

Integrand size = 23, antiderivative size = 148 \[ \int \frac {\tanh ^5(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}{b^4 d}-\frac {2 \left (3 a^2-2 b^2\right ) (a+b \text {sech}(c+d x))^{3/2}}{3 b^4 d}+\frac {6 a (a+b \text {sech}(c+d x))^{5/2}}{5 b^4 d}-\frac {2 (a+b \text {sech}(c+d x))^{7/2}}{7 b^4 d} \] Output:

2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))/a^(1/2)/d+2*a*(a^2-2*b^2)*(a+b* 
sech(d*x+c))^(1/2)/b^4/d-2/3*(3*a^2-2*b^2)*(a+b*sech(d*x+c))^(3/2)/b^4/d+6 
/5*a*(a+b*sech(d*x+c))^(5/2)/b^4/d-2/7*(a+b*sech(d*x+c))^(7/2)/b^4/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int \frac {\tanh ^5(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \left (\frac {105 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\sqrt {a+b \text {sech}(c+d x)} \left (48 a^3-140 a b^2+\left (-24 a^2 b+70 b^3\right ) \text {sech}(c+d x)+18 a b^2 \text {sech}^2(c+d x)-15 b^3 \text {sech}^3(c+d x)\right )}{b^4}\right )}{105 d} \] Input:

Integrate[Tanh[c + d*x]^5/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

(2*((105*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/Sqrt[a] + (Sqrt[a + b 
*Sech[c + d*x]]*(48*a^3 - 140*a*b^2 + (-24*a^2*b + 70*b^3)*Sech[c + d*x] + 
 18*a*b^2*Sech[c + d*x]^2 - 15*b^3*Sech[c + d*x]^3))/b^4))/(105*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4373, 517, 25, 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.

\(\displaystyle \int \frac {\tanh ^5(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \cot \left (i c+i d x+\frac {\pi }{2}\right )^5}{\sqrt {a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^5}{\sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}}dx\)

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1467

\(\displaystyle \frac {2 \int \left (-b^6 \text {sech}^6(c+d x)+3 a b^4 \text {sech}^4(c+d x)-b^2 \left (3 a^2-2 b^2\right ) \text {sech}^2(c+d x)+a^3-2 a b^2+\frac {b^4}{a-b^2 \text {sech}^2(c+d x)}\right )d\sqrt {a+b \text {sech}(c+d x)}}{b^4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (-a \left (a^2-2 b^2\right ) \sqrt {a+b \text {sech}(c+d x)}+\frac {1}{3} b^3 \left (3 a^2-2 b^2\right ) \text {sech}^3(c+d x)-\frac {b^4 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {3}{5} a b^5 \text {sech}^5(c+d x)+\frac {1}{7} b^7 \text {sech}^7(c+d x)\right )}{b^4 d}\)

Input:

Int[Tanh[c + d*x]^5/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

(-2*(-((b^4*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/Sqrt[a]) + (b^3*(3 
*a^2 - 2*b^2)*Sech[c + d*x]^3)/3 - (3*a*b^5*Sech[c + d*x]^5)/5 + (b^7*Sech 
[c + d*x]^7)/7 - a*(a^2 - 2*b^2)*Sqrt[a + b*Sech[c + d*x]]))/(b^4*d)
 

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 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]

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

Input:

int(tanh(d*x+c)^5/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int(tanh(d*x+c)^5/(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. 1277 vs. \(2 (128) = 256\).

Time = 0.54 (sec) , antiderivative size = 2813, normalized size of antiderivative = 19.01 \[ \int \frac {\tanh ^5(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

[1/210*(105*(b^4*cosh(d*x + c)^6 + 6*b^4*cosh(d*x + c)*sinh(d*x + c)^5 + b 
^4*sinh(d*x + c)^6 + 3*b^4*cosh(d*x + c)^4 + 3*b^4*cosh(d*x + c)^2 + 3*(5* 
b^4*cosh(d*x + c)^2 + b^4)*sinh(d*x + c)^4 + b^4 + 4*(5*b^4*cosh(d*x + c)^ 
3 + 3*b^4*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^4*cosh(d*x + c)^4 + 6*b^ 
4*cosh(d*x + c)^2 + b^4)*sinh(d*x + c)^2 + 6*(b^4*cosh(d*x + c)^5 + 2*b^4* 
cosh(d*x + c)^3 + b^4*cosh(d*x + c))*sinh(d*x + c))*sqrt(a)*log(-(2*a^2*co 
sh(d*x + c)^4 + 2*a^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^3 + 4*(2*a^2*c 
osh(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)) + 16*((12*a^4 - 35*a^2*b^2)*cosh(d*x + 
 c)^6 + (12*a^4 - 35*a^2*b^2)*sinh(d*x + c)^6 - (12*a^3*b - 35*a*b^3)*cosh 
(d*x + c)^5 - (12*a^3*b - 35*a*b^3 - 6*(12*a^4 - 35*a^2*b^2)*cosh(d*x + c) 
)*sinh(d*x + c)^5 + 3*(12*a^4 - 29*a^2*b^2)*cosh(d*x + c)^4 + (36*a^4 -...
 

Sympy [F]

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

integrate(tanh(d*x+c)**5/(a+b*sech(d*x+c))**(1/2),x)
 

Output:

Integral(tanh(c + d*x)**5/sqrt(a + b*sech(c + d*x)), x)
 

Maxima [F]

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

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(tanh(d*x + c)^5/sqrt(b*sech(d*x + c) + a), x)
 

Giac [F]

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

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(tanh(d*x + c)^5/sqrt(b*sech(d*x + c) + a), x)
 

Mupad [F(-1)]

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

int(tanh(c + d*x)^5/(a + b/cosh(c + d*x))^(1/2),x)
 

Output:

int(tanh(c + d*x)^5/(a + b/cosh(c + d*x))^(1/2), x)
 

Reduce [F]

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

int(tanh(d*x+c)^5/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int((sqrt(sech(c + d*x)*b + a)*tanh(c + d*x)**5)/(sech(c + d*x)*b + a),x)