\(\int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\) [146]

Optimal result

Integrand size = 23, antiderivative size = 217 \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {(4 a-7 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}-\frac {(4 a+7 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}-\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}-\frac {\coth ^2(c+d x) \sqrt {a+b \text {sech}(c+d x)} \left (a^2+b^2-2 a b \text {sech}(c+d x)\right )}{2 \left (a^2-b^2\right )^2 d} \] Output:

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

Mathematica [C] (verified)

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

Time = 1.11 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.45 \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=-\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a-b}\right )}{(a-b) \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a+b}\right )}{(a+b) \sqrt {a+b \text {sech}(c+d x)}}+\frac {4 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \text {sech}(c+d x)}{a}\right )}{a \sqrt {a+b \text {sech}(c+d x)}}+\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a-b}\right )}{(a-b)^2 \sqrt {a+b \text {sech}(c+d x)}}-\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a+b}\right )}{(a+b)^2 \sqrt {a+b \text {sech}(c+d x)}}}{2 b d} \] Input:

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

Output:

-1/2*((-2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] + (2 
*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] - (2*a*Hyperg 
eometric2F1[-1/2, 1, 1/2, (a + b*Sech[c + d*x])/(a - b)])/((a - b)*Sqrt[a 
+ b*Sech[c + d*x]]) + (2*a*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sech[c + 
 d*x])/(a + b)])/((a + b)*Sqrt[a + b*Sech[c + d*x]]) + (4*b*Hypergeometric 
2F1[-1/2, 1, 1/2, 1 + (b*Sech[c + d*x])/a])/(a*Sqrt[a + b*Sech[c + d*x]]) 
+ (b^2*Hypergeometric2F1[-1/2, 2, 1/2, (a + b*Sech[c + d*x])/(a - b)])/((a 
 - b)^2*Sqrt[a + b*Sech[c + d*x]]) - (b^2*Hypergeometric2F1[-1/2, 2, 1/2, 
(a + b*Sech[c + d*x])/(a + b)])/((a + b)^2*Sqrt[a + b*Sech[c + d*x]]))/(b* 
d)
 

Rubi [A] (warning: unable to verify)

Time = 0.87 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.43, 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, 561, 25, 1674, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1674

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b^4 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} b^4}+\frac {\cosh (c+d x)}{a b \left (a^2-b^2\right )^2}+\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 b^4 (a-b)^{5/2}}+\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 b^4 (a+b)^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{8 b^3 (a-b)^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{8 b^3 (a+b)^{5/2}}-\frac {\text {sech}(c+d x)}{8 b^2 (a-b)^2 \left (a-b^2 \text {sech}^2(c+d x)-b\right )}+\frac {\text {sech}(c+d x)}{8 b^2 (a+b)^2 \left (a-b^2 \text {sech}^2(c+d x)+b\right )}\right )}{d}\)

Input:

Int[Coth[c + d*x]^3/(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)*b^4)) + ((2 
*a - 3*b)*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(5/2) 
*b^4) - ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(8*(a - b)^(5/2)*b^ 
3) + ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]]/(8*b^3*(a + b)^(5/2)) 
+ ((2*a + 3*b)*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]])/(4*b^4*(a + 
 b)^(5/2)) + Cosh[c + d*x]/(a*b*(a^2 - b^2)^2) - Sech[c + d*x]/(8*(a - b)^ 
2*b^2*(a - b - b^2*Sech[c + d*x]^2)) + Sech[c + d*x]/(8*b^2*(a + b)^2*(a + 
 b - b^2*Sech[c + d*x]^2))))/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 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1674

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, p, q}, x] && N 
eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
 

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]

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

Input:

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

Output:

int(coth(d*x+c)^3/(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. 6123 vs. \(2 (193) = 386\).

Time = 6.47 (sec) , antiderivative size = 53763, normalized size of antiderivative = 247.76 \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sech(c + d*x)*b + a)*coth(c + d*x)**3)/(sech(c + d*x)**2*b**2 + 
2*sech(c + d*x)*a*b + a**2),x)