\(\int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [138]

Optimal result

Integrand size = 23, antiderivative size = 610 \[ \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=-\frac {4 (a-b) \sqrt {a+b} \coth (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{b^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (8 a^2+9 b^2\right ) \coth (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{15 b^4 d}-\frac {4 \sqrt {a+b} \coth (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{b d}+\frac {2 \sqrt {a+b} \left (8 a^2-2 a b+9 b^2\right ) \coth (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{15 b^3 d}+\frac {2 \sqrt {a+b} \coth (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{a d}-\frac {8 a \sqrt {a+b \text {sech}(c+d x)} \tanh (c+d x)}{15 b^2 d}+\frac {2 \text {sech}(c+d x) \sqrt {a+b \text {sech}(c+d x)} \tanh (c+d x)}{5 b d} \] Output:

-4*(a-b)*(a+b)^(1/2)*coth(d*x+c)*EllipticE((a+b*sech(d*x+c))^(1/2)/(a+b)^( 
1/2),((a+b)/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+ 
c))/(a-b))^(1/2)/b^2/d+2/15*(a-b)*(a+b)^(1/2)*(8*a^2+9*b^2)*coth(d*x+c)*El 
lipticE((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec 
h(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/b^4/d-4*(a+b)^(1/2 
)*coth(d*x+c)*EllipticF((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^ 
(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/b/ 
d+2/15*(a+b)^(1/2)*(8*a^2-2*a*b+9*b^2)*coth(d*x+c)*EllipticF((a+b*sech(d*x 
+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2 
)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/b^3/d+2*(a+b)^(1/2)*coth(d*x+c)*Ellipti 
cPi((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(b*(1 
-sech(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/a/d-8/15*a*(a+ 
b*sech(d*x+c))^(1/2)*tanh(d*x+c)/b^2/d+2/5*sech(d*x+c)*(a+b*sech(d*x+c))^( 
1/2)*tanh(d*x+c)/b/d
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx \] Input:

Integrate[Tanh[c + d*x]^4/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

Integrate[Tanh[c + d*x]^4/Sqrt[a + b*Sech[c + d*x]], x]
 

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 610, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4383, 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.

\(\displaystyle \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cot \left (i c+i d x+\frac {\pi }{2}\right )^4}{\sqrt {a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4383

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 (a-b) \sqrt {a+b} \left (8 a^2+9 b^2\right ) \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^4 d}+\frac {2 \sqrt {a+b} \left (8 a^2-2 a b+9 b^2\right ) \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{15 b^3 d}-\frac {4 (a-b) \sqrt {a+b} \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {4 \sqrt {a+b} \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {2 \sqrt {a+b} \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {8 a \tanh (c+d x) \sqrt {a+b \text {sech}(c+d x)}}{15 b^2 d}+\frac {2 \tanh (c+d x) \text {sech}(c+d x) \sqrt {a+b \text {sech}(c+d x)}}{5 b d}\)

Input:

Int[Tanh[c + d*x]^4/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

(-4*(a - b)*Sqrt[a + b]*Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + 
 d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b) 
]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a + 
b]*(8*a^2 + 9*b^2)*Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x] 
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqr 
t[-((b*(1 + Sech[c + d*x]))/(a - b))])/(15*b^4*d) - (4*Sqrt[a + b]*Coth[c 
+ d*x]*EllipticF[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a 
 - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]) 
)/(a - b))])/(b*d) + (2*Sqrt[a + b]*(8*a^2 - 2*a*b + 9*b^2)*Coth[c + d*x]* 
EllipticF[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]* 
Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - 
b))])/(15*b^3*d) + (2*Sqrt[a + b]*Coth[c + d*x]*EllipticPi[(a + b)/a, ArcS 
in[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - S 
ech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(a*d) - 
(8*a*Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x])/(15*b^2*d) + (2*Sech[c + d*x 
]*Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x])/(5*b*d)
 

Defintions of rubi rules used

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 4383

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

Maple [F]

\[\int \frac {\tanh \left (d x +c \right )^{4}}{\sqrt {a +b \,\operatorname {sech}\left (d x +c \right )}}d x\]

Input:

int(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x)
 

Fricas [F]

\[ \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{4}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \] Input:

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

integral(tanh(d*x + c)^4/sqrt(b*sech(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\tanh ^{4}{\left (c + d x \right )}}{\sqrt {a + b \operatorname {sech}{\left (c + d x \right )}}}\, dx \] Input:

integrate(tanh(d*x+c)**4/(a+b*sech(d*x+c))**(1/2),x)
 

Output:

Integral(tanh(c + d*x)**4/sqrt(a + b*sech(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{4}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \] Input:

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(tanh(d*x + c)^4/sqrt(b*sech(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{4}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \] Input:

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(tanh(d*x + c)^4/sqrt(b*sech(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^4(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {{\mathrm {tanh}\left (c+d\,x\right )}^4}{\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}} \,d x \] Input:

int(tanh(c + d*x)^4/(a + b/cosh(c + d*x))^(1/2),x)
 

Output:

int(tanh(c + d*x)^4/(a + b/cosh(c + d*x))^(1/2), x)
 

Reduce [F]

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

int(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int((sqrt(sech(c + d*x)*b + a)*tanh(c + d*x)**4)/(sech(c + d*x)*b + a),x)