\(\int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) [25]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 118 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\left (3 a^2-6 a b-b^2\right ) x}{8 (a+b)^3}+\frac {a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{(a+b)^3 d}-\frac {(5 a+b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac {\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d} \] Output:

1/8*(3*a^2-6*a*b-b^2)*x/(a+b)^3+a^(3/2)*b^(1/2)*arctan(b^(1/2)*tanh(d*x+c) 
/a^(1/2))/(a+b)^3/d-1/8*(5*a+b)*cosh(d*x+c)*sinh(d*x+c)/(a+b)^2/d+1/4*cosh 
(d*x+c)^3*sinh(d*x+c)/(a+b)/d
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {4 \left (3 a^2-6 a b-b^2\right ) (c+d x)+32 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )-8 a (a+b) \sinh (2 (c+d x))+(a+b)^2 \sinh (4 (c+d x))}{32 (a+b)^3 d} \] Input:

Integrate[Sinh[c + d*x]^4/(a + b*Tanh[c + d*x]^2),x]
 

Output:

(4*(3*a^2 - 6*a*b - b^2)*(c + d*x) + 32*a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Ta 
nh[c + d*x])/Sqrt[a]] - 8*a*(a + b)*Sinh[2*(c + d*x)] + (a + b)^2*Sinh[4*( 
c + d*x)])/(32*(a + b)^3*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4146, 372, 402, 25, 397, 218, 219}

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.

\(\displaystyle \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (i c+i d x)^4}{a-b \tan (i c+i d x)^2}dx\)

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {a (3 a-b)-b (5 a+b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 (a+b)}+\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}}{4 (a+b)}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\int \frac {a (3 a-b)-b (5 a+b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 (a+b)}}{4 (a+b)}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {\left (3 a^2-6 a b-b^2\right ) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {8 a^2 b \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}}{2 (a+b)}}{4 (a+b)}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {\left (3 a^2-6 a b-b^2\right ) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {8 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a+b}}{2 (a+b)}}{4 (a+b)}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {8 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a+b}+\frac {\left (3 a^2-6 a b-b^2\right ) \text {arctanh}(\tanh (c+d x))}{a+b}}{2 (a+b)}}{4 (a+b)}}{d}\)

Input:

Int[Sinh[c + d*x]^4/(a + b*Tanh[c + d*x]^2),x]
 

Output:

(Tanh[c + d*x]/(4*(a + b)*(1 - Tanh[c + d*x]^2)^2) - (-1/2*((8*a^(3/2)*Sqr 
t[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a + b) + ((3*a^2 - 6*a*b - 
b^2)*ArcTanh[Tanh[c + d*x]])/(a + b))/(a + b) + ((5*a + b)*Tanh[c + d*x])/ 
(2*(a + b)*(1 - Tanh[c + d*x]^2)))/(4*(a + b)))/d
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 372
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 397
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 402
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4146
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ 
)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[c*(ff^(m + 1)/f)   Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 
2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[m/2]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(214\) vs. \(2(104)=208\).

Time = 20.98 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.82

method result size
risch \(\frac {3 x \,a^{2}}{8 \left (a +b \right )^{3}}-\frac {3 x a b}{4 \left (a +b \right )^{3}}-\frac {x \,b^{2}}{8 \left (a +b \right )^{3}}+\frac {{\mathrm e}^{4 d x +4 c}}{64 \left (a +b \right ) d}-\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 \left (a +b \right )^{2} d}+\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {{\mathrm e}^{-4 d x -4 c}}{64 \left (a +b \right ) d}+\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{2 \left (a +b \right )^{3} d}-\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{2 \left (a +b \right )^{3} d}\) \(215\)
derivativedivides \(\frac {\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a -3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-3 a^{2}+6 a b +b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}-\frac {2 a^{3} b \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{\left (a +b \right )^{3}}-\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a +3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (3 a^{2}-6 a b -b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a +b \right )^{3}}}{d}\) \(434\)
default \(\frac {\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a -3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-3 a^{2}+6 a b +b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}-\frac {2 a^{3} b \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{\left (a +b \right )^{3}}-\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a +3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (3 a^{2}-6 a b -b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a +b \right )^{3}}}{d}\) \(434\)

Input:

int(sinh(d*x+c)^4/(a+tanh(d*x+c)^2*b),x,method=_RETURNVERBOSE)
 

Output:

3/8*x/(a+b)^3*a^2-3/4*x/(a+b)^3*a*b-1/8*x/(a+b)^3*b^2+1/64/(a+b)/d*exp(4*d 
*x+4*c)-1/8*a/(a+b)^2/d*exp(2*d*x+2*c)+1/8*a/(a^2+2*a*b+b^2)/d*exp(-2*d*x- 
2*c)-1/64/(a+b)/d*exp(-4*d*x-4*c)+1/2*(-a*b)^(1/2)*a/(a+b)^3/d*ln(exp(2*d* 
x+2*c)+(2*(-a*b)^(1/2)+a-b)/(a+b))-1/2*(-a*b)^(1/2)*a/(a+b)^3/d*ln(exp(2*d 
*x+2*c)-(2*(-a*b)^(1/2)-a+b)/(a+b))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 860 vs. \(2 (104) = 208\).

Time = 0.16 (sec) , antiderivative size = 2024, normalized size of antiderivative = 17.15 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/64*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d* 
x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^8 + 8*(3*a^2 - 
6*a*b - b^2)*d*x*cosh(d*x + c)^4 - 8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7*(a 
^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x + c)^6 + 8*(7* 
(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d* 
x + c)^5 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(3*a^2 - 6*a*b - 
b^2)*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2 
*a*b + b^2)*cosh(d*x + c)^5 + 4*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c) - 
20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*b)*cosh(d*x + 
 c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 12*(3*a^2 - 6*a*b - b^2 
)*d*x*cosh(d*x + c)^2 - 30*(a^2 + a*b)*cosh(d*x + c)^4 + 2*a^2 + 2*a*b)*si 
nh(d*x + c)^2 + 32*(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)^3*sinh(d*x + c) 
+ 6*a*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 
+ a*sinh(d*x + c)^4)*sqrt(-a*b)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 
 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2) 
*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2 
)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a 
^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + 
c) + 4*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + 
(a + b)*sinh(d*x + c)^2 + a - b)*sqrt(-a*b))/((a + b)*cosh(d*x + c)^4 +...
 

Sympy [F]

\[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int \frac {\sinh ^{4}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \] Input:

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

Output:

Integral(sinh(c + d*x)**4/(a + b*tanh(c + d*x)**2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 514 vs. \(2 (104) = 208\).

Time = 0.19 (sec) , antiderivative size = 514, normalized size of antiderivative = 4.36 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {{\left (a b - b^{2}\right )} {\left (d x + c\right )}}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac {{\left (8 \, b e^{\left (-2 \, d x - 2 \, c\right )} + a + b\right )} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} - \frac {b \log \left ({\left (a + b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a - b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {b \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} - \frac {{\left (a^{2} b - 6 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b} d} - \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} - \frac {3 \, b \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, \sqrt {a b} {\left (a + b\right )} d} - \frac {8 \, b e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {3 \, {\left (d x + c\right )}}{8 \, {\left (a + b\right )} d} - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, {\left (a + b\right )} d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, {\left (a + b\right )} d} \] Input:

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

Output:

-1/2*(a*b - b^2)*(d*x + c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) + 1/64*(8*b 
*e^(-2*d*x - 2*c) + a + b)*e^(4*d*x + 4*c)/((a^2 + 2*a*b + b^2)*d) - 1/4*b 
*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b)/((a^2 + 
2*a*b + b^2)*d) + 1/4*b*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x 
 - 4*c) + a + b)/((a^2 + 2*a*b + b^2)*d) + 1/4*(a*b - b^2)*arctan(1/2*((a 
+ b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)*d) 
 - 1/8*(a^2*b - 6*a*b^2 + b^3)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - 
b)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)*d) - 1/4*(a*b - b 
^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/((a^2 + 2*a*b 
 + b^2)*sqrt(a*b)*d) - 3/8*b*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b) 
/sqrt(a*b))/(sqrt(a*b)*(a + b)*d) - 1/64*(8*b*e^(-2*d*x - 2*c) + (a + b)*e 
^(-4*d*x - 4*c))/((a^2 + 2*a*b + b^2)*d) + 3/8*(d*x + c)/((a + b)*d) - 1/8 
*e^(2*d*x + 2*c)/((a + b)*d) + 1/8*e^(-2*d*x - 2*c)/((a + b)*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (104) = 208\).

Time = 1.05 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.34 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\frac {64 \, a^{2} b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b}} + \frac {8 \, {\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} {\left (d x + c\right )}}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 36 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )}}{a^{2} + 2 \, a b + b^{2}}}{64 \, d} \] Input:

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

Output:

1/64*(64*a^2*b*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/ 
sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)) + 8*(3*a^2 - 6*a*b 
- b^2)*(d*x + c)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (18*a^2*e^(4*d*x + 4*c) 
 - 36*a*b*e^(4*d*x + 4*c) - 6*b^2*e^(4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) 
- 8*a*b*e^(2*d*x + 2*c) + a^2 + 2*a*b + b^2)*e^(-4*d*x - 4*c)/(a^3 + 3*a^2 
*b + 3*a*b^2 + b^3) + (a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) - 8*a*e^(2*d* 
x + 2*c))/(a^2 + 2*a*b + b^2))/d
 

Mupad [B] (verification not implemented)

Time = 1.72 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.12 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d\,\left (a+b\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d\,\left (a+b\right )}-\frac {x\,\left (-3\,a^2+6\,a\,b+b^2\right )}{8\,{\left (a+b\right )}^3}+\frac {a\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d\,{\left (a+b\right )}^2}-\frac {a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d\,{\left (a+b\right )}^2}+\frac {{\left (-a\right )}^{3/2}\,\sqrt {b}\,\ln \left ({\left (-a\right )}^{3/2}\,b^{3/2}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )-2\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\left (-a\right )}^{5/2}\,\sqrt {b}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\right )}{2\,d\,{\left (a+b\right )}^3}-\frac {{\left (-a\right )}^{3/2}\,\sqrt {b}\,\ln \left (2\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\left (-a\right )}^{3/2}\,b^{3/2}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )+{\left (-a\right )}^{5/2}\,\sqrt {b}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\right )}{2\,d\,{\left (a+b\right )}^3} \] Input:

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

Output:

exp(4*c + 4*d*x)/(64*d*(a + b)) - exp(- 4*c - 4*d*x)/(64*d*(a + b)) - (x*( 
6*a*b - 3*a^2 + b^2))/(8*(a + b)^3) + (a*exp(- 2*c - 2*d*x))/(8*d*(a + b)^ 
2) - (a*exp(2*c + 2*d*x))/(8*d*(a + b)^2) + ((-a)^(3/2)*b^(1/2)*log((-a)^( 
3/2)*b^(3/2)*(exp(2*c + 2*d*x) - 1) - 2*a^2*b*exp(2*c + 2*d*x) + (-a)^(5/2 
)*b^(1/2)*(exp(2*c + 2*d*x) + 1)))/(2*d*(a + b)^3) - ((-a)^(3/2)*b^(1/2)*l 
og(2*a^2*b*exp(2*c + 2*d*x) + (-a)^(3/2)*b^(3/2)*(exp(2*c + 2*d*x) - 1) + 
(-a)^(5/2)*b^(1/2)*(exp(2*c + 2*d*x) + 1)))/(2*d*(a + b)^3)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.38 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {64 e^{4 d x +4 c} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {e^{d x +c} \sqrt {a +b}-\sqrt {b}}{\sqrt {a}}\right ) a -64 e^{4 d x +4 c} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {e^{d x +c} \sqrt {a +b}+\sqrt {b}}{\sqrt {a}}\right ) a +e^{8 d x +8 c} a^{2}+2 e^{8 d x +8 c} a b +e^{8 d x +8 c} b^{2}-8 e^{6 d x +6 c} a^{2}-8 e^{6 d x +6 c} a b +24 e^{4 d x +4 c} a^{2} d x -48 e^{4 d x +4 c} a b d x -8 e^{4 d x +4 c} b^{2} d x +8 e^{2 d x +2 c} a^{2}+8 e^{2 d x +2 c} a b -a^{2}-2 a b -b^{2}}{64 e^{4 d x +4 c} d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )} \] Input:

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

Output:

(64*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt 
(b))/sqrt(a))*a - 64*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*s 
qrt(a + b) + sqrt(b))/sqrt(a))*a + e**(8*c + 8*d*x)*a**2 + 2*e**(8*c + 8*d 
*x)*a*b + e**(8*c + 8*d*x)*b**2 - 8*e**(6*c + 6*d*x)*a**2 - 8*e**(6*c + 6* 
d*x)*a*b + 24*e**(4*c + 4*d*x)*a**2*d*x - 48*e**(4*c + 4*d*x)*a*b*d*x - 8* 
e**(4*c + 4*d*x)*b**2*d*x + 8*e**(2*c + 2*d*x)*a**2 + 8*e**(2*c + 2*d*x)*a 
*b - a**2 - 2*a*b - b**2)/(64*e**(4*c + 4*d*x)*d*(a**3 + 3*a**2*b + 3*a*b* 
*2 + b**3))