\(\int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx\) [126]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}-\frac {2 a (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d}+\frac {2 (a+b \text {sech}(c+d x))^{5/2}}{5 b^2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\frac {2 \left (15 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )+\frac {\sqrt {a+b \text {sech}(c+d x)} \left (-2 a^2-15 b^2+a b \text {sech}(c+d x)+3 b^2 \text {sech}^2(c+d x)\right )}{b^2}\right )}{15 d} \] Input:

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

Output:

(2*(15*Sqrt[a]*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]] + (Sqrt[a + b*Se 
ch[c + d*x]]*(-2*a^2 - 15*b^2 + a*b*Sech[c + d*x] + 3*b^2*Sech[c + d*x]^2) 
)/b^2))/(15*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.81, 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, 1584, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1584

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-2*(-(Sqrt[a]*b^2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]]) + b^3*Sech[ 
c + d*x] + (a*b^3*Sech[c + d*x]^3)/3 - (b^5*Sech[c + d*x]^5)/5))/(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 1584

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

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 665 vs. \(2 (84) = 168\).

Time = 0.51 (sec) , antiderivative size = 1589, normalized size of antiderivative = 15.89 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

[1/30*(15*(b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2 
*sinh(d*x + c)^4 + 2*b^2*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + b^2) 
*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + b^2*cosh(d*x + c))*sinh( 
d*x + c))*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)) + 4*(2 
*a*b*cosh(d*x + c)^3 - (2*a^2 + 15*b^2)*cosh(d*x + c)^4 - (2*a^2 + 15*b^2) 
*sinh(d*x + c)^4 + 2*(a*b - 2*(2*a^2 + 15*b^2)*cosh(d*x + c))*sinh(d*x + c 
)^3 + 2*a*b*cosh(d*x + c) - 2*(2*a^2 + 9*b^2)*cosh(d*x + c)^2 + 2*(3*a*b*c 
osh(d*x + c) - 3*(2*a^2 + 15*b^2)*cosh(d*x + c)^2 - 2*a^2 - 9*b^2)*sinh(d* 
x + c)^2 - 2*a^2 - 15*b^2 + 2*(3*a*b*cosh(d*x + c)^2 - 2*(2*a^2 + 15*b^2)* 
cosh(d*x + c)^3 + a*b - 2*(2*a^2 + 9*b^2)*cosh(d*x + c))*sinh(d*x + c))...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\frac {-2 \sqrt {\mathrm {sech}\left (d x +c \right ) b +a}\, \tanh \left (d x +c \right )^{2}-8 \sqrt {\mathrm {sech}\left (d x +c \right ) b +a}+\left (\int \frac {\sqrt {\mathrm {sech}\left (d x +c \right ) b +a}\, \tanh \left (d x +c \right )^{3}}{\mathrm {sech}\left (d x +c \right ) b +a}d x \right ) a d +4 \left (\int \frac {\sqrt {\mathrm {sech}\left (d x +c \right ) b +a}\, \tanh \left (d x +c \right )}{\mathrm {sech}\left (d x +c \right ) b +a}d x \right ) a d}{5 d} \] Input:

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

Output:

( - 2*sqrt(sech(c + d*x)*b + a)*tanh(c + d*x)**2 - 8*sqrt(sech(c + d*x)*b 
+ a) + int((sqrt(sech(c + d*x)*b + a)*tanh(c + d*x)**3)/(sech(c + d*x)*b + 
 a),x)*a*d + 4*int((sqrt(sech(c + d*x)*b + a)*tanh(c + d*x))/(sech(c + d*x 
)*b + a),x)*a*d)/(5*d)