\(\int \frac {\sinh ^3(c+d x)}{(a+b \text {sech}^2(c+d x))^3} \, dx\) [42]

Optimal result

Integrand size = 23, antiderivative size = 154 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {5 \sqrt {b} (3 a+7 b) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{8 a^{9/2} d}-\frac {(a+3 b) \cosh (c+d x)}{a^4 d}+\frac {\cosh ^3(c+d x)}{3 a^3 d}+\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 d \left (b+a \cosh ^2(c+d x)\right )^2}-\frac {b (9 a+13 b) \cosh (c+d x)}{8 a^4 d \left (b+a \cosh ^2(c+d x)\right )} \] Output:

5/8*b^(1/2)*(3*a+7*b)*arctan(a^(1/2)*cosh(d*x+c)/b^(1/2))/a^(9/2)/d-(a+3*b 
)*cosh(d*x+c)/a^4/d+1/3*cosh(d*x+c)^3/a^3/d+1/4*b^2*(a+b)*cosh(d*x+c)/a^4/ 
d/(b+a*cosh(d*x+c)^2)^2-1/8*b*(9*a+13*b)*cosh(d*x+c)/a^4/d/(b+a*cosh(d*x+c 
)^2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 8.42 (sec) , antiderivative size = 1217, normalized size of antiderivative = 7.90 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])^3*Sech[c + d*x]^6*((24*(3*a - 4*b)*(ArcTa 
n[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x 
)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[( 
d*x)/2]))/Sqrt[b]] + ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh 
[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Co 
sh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]]))/(a^(3/2)*b^(5/2)) - (54*(Ar 
cTan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]] + ArcTan[(Sqrt[a 
] + I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]]))/(Sqrt[a]*b^(5/2)) - (36*Co 
sh[c + d*x]*(3*a + 10*b + 3*a*Cosh[2*(c + d*x)]))/(b^2*(a + 2*b + a*Cosh[2 
*(c + d*x)])^2) + (48*Cosh[c + d*x]*(3*a^2 + 6*a*b + 8*b^2 + a*(3*a - 4*b) 
*Cosh[2*(c + d*x)]))/(a*b^2*(a + 2*b + a*Cosh[2*(c + d*x)])^2) + (3*(3*a^4 
 - 40*a^3*b + 720*a^2*b^2 + 6720*a*b^3 + 8960*b^4)*ArcTan[((Sqrt[a] - I*Sq 
rt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sq 
rt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] 
 + 3*(3*a^4 - 40*a^3*b + 720*a^2*b^2 + 6720*a*b^3 + 8960*b^4)*ArcTan[((Sqr 
t[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + 
Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2] 
))/Sqrt[b]] + (2*Sqrt[a]*Sqrt[b]*Cosh[c + d*x]*(9*a^5 - 90*a^4*b - 10144*a 
^3*b^2 - 48672*a^2*b^3 - 85120*a*b^4 - 53760*b^5 + a*(9*a^4 - 120*a^3*b - 
12432*a^2*b^2 - 47936*a*b^3 - 44800*b^4)*Cosh[2*(c + d*x)] - 128*a^2*b^...
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 26, 4621, 360, 25, 2345, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \sin (i c+i d x)^3}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\sin (i c+i d x)^3}{\left (b \sec (i c+i d x)^2+a\right )^3}dx\)

\(\Big \downarrow \) 4621

\(\displaystyle -\frac {\int \frac {\cosh ^6(c+d x) \left (1-\cosh ^2(c+d x)\right )}{\left (a \cosh ^2(c+d x)+b\right )^3}d\cosh (c+d x)}{d}\)

\(\Big \downarrow \) 360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2345

\(\displaystyle -\frac {\frac {\frac {b (9 a+13 b) \cosh (c+d x)}{2 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\int \frac {8 a^2 b \cosh ^4(c+d x)-8 a b (a+2 b) \cosh ^2(c+d x)+b^2 (7 a+11 b)}{a \cosh ^2(c+d x)+b}d\cosh (c+d x)}{2 b}}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 1467

\(\displaystyle -\frac {\frac {\frac {b (9 a+13 b) \cosh (c+d x)}{2 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\int \left (8 a b \cosh ^2(c+d x)-8 b (a+3 b)+\frac {5 \left (7 b^3+3 a b^2\right )}{a \cosh ^2(c+d x)+b}\right )d\cosh (c+d x)}{2 b}}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {\frac {b (9 a+13 b) \cosh (c+d x)}{2 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\frac {5 b^{3/2} (3 a+7 b) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{\sqrt {a}}+\frac {8}{3} a b \cosh ^3(c+d x)-8 b (a+3 b) \cosh (c+d x)}{2 b}}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\)

Input:

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

Output:

-((-1/4*(b^2*(a + b)*Cosh[c + d*x])/(a^4*(b + a*Cosh[c + d*x]^2)^2) + ((b* 
(9*a + 13*b)*Cosh[c + d*x])/(2*(b + a*Cosh[c + d*x]^2)) - ((5*b^(3/2)*(3*a 
 + 7*b)*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/Sqrt[a] - 8*b*(a + 3*b)*C 
osh[c + d*x] + (8*a*b*Cosh[c + d*x]^3)/3)/(2*b))/(4*a^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 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

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 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4621

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
   Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), 
x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 
2] && IntegerQ[n] && IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(347\) vs. \(2(138)=276\).

Time = 0.20 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.26

\[\frac {\frac {1}{3 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a +6 b}{2 a^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b \left (\frac {\left (-\frac {9}{8} a^{2}+\frac {1}{4} a b +\frac {11}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {\left (27 a^{3}+15 a^{2} b +5 a \,b^{2}+33 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 \left (a +b \right )}+\left (-\frac {27}{8} a^{2}-\frac {5}{4} a b +\frac {33}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {9 a^{2}}{8}-\frac {5 a b}{2}-\frac {11 b^{2}}{8}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {5 \left (3 a +7 b \right ) \arctan \left (\frac {2 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 a -2 b}{4 \sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{4}}-\frac {1}{3 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -6 b}{2 a^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\]

Input:

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

Output:

1/d*(1/3/a^3/(tanh(1/2*d*x+1/2*c)+1)^3-1/2/a^3/(tanh(1/2*d*x+1/2*c)+1)^2-1 
/2*(a+6*b)/a^4/(tanh(1/2*d*x+1/2*c)+1)+2*b/a^4*(((-9/8*a^2+1/4*a*b+11/8*b^ 
2)*tanh(1/2*d*x+1/2*c)^6-1/8*(27*a^3+15*a^2*b+5*a*b^2+33*b^3)/(a+b)*tanh(1 
/2*d*x+1/2*c)^4+(-27/8*a^2-5/4*a*b+33/8*b^2)*tanh(1/2*d*x+1/2*c)^2-9/8*a^2 
-5/2*a*b-11/8*b^2)/(tanh(1/2*d*x+1/2*c)^4*a+tanh(1/2*d*x+1/2*c)^4*b+2*tanh 
(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)^2+5/16*(3*a+7*b)/(a*b)^ 
(1/2)*arctan(1/4*(2*(a+b)*tanh(1/2*d*x+1/2*c)^2+2*a-2*b)/(a*b)^(1/2)))-1/3 
/a^3/(tanh(1/2*d*x+1/2*c)-1)^3-1/2/a^3/(tanh(1/2*d*x+1/2*c)-1)^2-1/2/a^4*( 
-a-6*b)/(tanh(1/2*d*x+1/2*c)-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4793 vs. \(2 (138) = 276\).

Time = 0.43 (sec) , antiderivative size = 8667, normalized size of antiderivative = 56.28 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6\,{\mathrm {sinh}\left (c+d\,x\right )}^3}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^3} \,d x \] Input:

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

Output:

int((cosh(c + d*x)^6*sinh(c + d*x)^3)/(b + a*cosh(c + d*x)^2)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 5982, normalized size of antiderivative = 38.84 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(90*e**(11*c + 11*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + 
 b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + 
 a + 2*b)))*a**3 + 210*e**(11*c + 11*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt 
(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sq 
rt(b)*sqrt(a + b) + a + 2*b)))*a**2*b + 360*e**(9*c + 9*d*x)*sqrt(b)*sqrt( 
a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a) 
/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**3 + 1560*e**(9*c + 9* 
d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*ata 
n((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**2*b 
 + 1680*e**(9*c + 9*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a 
 + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) 
 + a + 2*b)))*a*b**2 + 540*e**(7*c + 7*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sq 
rt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2* 
sqrt(b)*sqrt(a + b) + a + 2*b)))*a**3 + 2700*e**(7*c + 7*d*x)*sqrt(b)*sqrt 
(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a 
)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**2*b + 4800*e**(7*c + 
 7*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)* 
atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a*b 
**2 + 3360*e**(7*c + 7*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqr 
t(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(...