\(\int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\) [142]

Optimal result

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.59 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=-\frac {2 \left (5 b^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \text {sech}(c+d x)}{a}\right )+a \left (4 a \left (4 a^2-5 b^2\right )+2 b \left (4 a^2-5 b^2\right ) \text {sech}(c+d x)-2 a b^2 \text {sech}^2(c+d x)+b^3 \text {sech}^3(c+d x)\right )\right )}{5 a b^4 d \sqrt {a+b \text {sech}(c+d x)}} \] Input:

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

Output:

(-2*(5*b^4*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sech[c + d*x])/a] + a*(4 
*a*(4*a^2 - 5*b^2) + 2*b*(4*a^2 - 5*b^2)*Sech[c + d*x] - 2*a*b^2*Sech[c + 
d*x]^2 + b^3*Sech[c + d*x]^3)))/(5*a*b^4*d*Sqrt[a + b*Sech[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.80, 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, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \cot \left (i c+i d x+\frac {\pi }{2}\right )^5}{\left (a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^5}{\left (a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1584

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1743 vs. \(2 (132) = 264\).

Time = 0.52 (sec) , antiderivative size = 3745, normalized size of antiderivative = 25.30 \[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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