\(\int \frac {1}{(a+b \cosh (c+d x))^4} \, dx\) [70]

Optimal result

Integrand size = 12, antiderivative size = 184 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\frac {a \left (2 a^2+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {b \sinh (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cosh (c+d x))^3}-\frac {5 a b \sinh (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cosh (c+d x))^2}-\frac {b \left (11 a^2+4 b^2\right ) \sinh (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cosh (c+d x))} \] Output:

a*(2*a^2+3*b^2)*arctanh((a-b)^(1/2)*tanh(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b) 
^(7/2)/(a+b)^(7/2)/d-1/3*b*sinh(d*x+c)/(a^2-b^2)/d/(a+b*cosh(d*x+c))^3-5/6 
*a*b*sinh(d*x+c)/(a^2-b^2)^2/d/(a+b*cosh(d*x+c))^2-1/6*b*(11*a^2+4*b^2)*si 
nh(d*x+c)/(a^2-b^2)^3/d/(a+b*cosh(d*x+c))
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\frac {\frac {6 a \left (2 a^2+3 b^2\right ) \arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}-\frac {b \left (36 a^4+a^2 b^2+8 b^4+6 a b \left (9 a^2+b^2\right ) \cosh (c+d x)+\left (11 a^2 b^2+4 b^4\right ) \cosh (2 (c+d x))\right ) \sinh (c+d x)}{2 (a-b)^3 (a+b)^3 (a+b \cosh (c+d x))^3}}{6 d} \] Input:

Integrate[(a + b*Cosh[c + d*x])^(-4),x]
 

Output:

((6*a*(2*a^2 + 3*b^2)*ArcTan[((a - b)*Tanh[(c + d*x)/2])/Sqrt[-a^2 + b^2]] 
)/(-a^2 + b^2)^(7/2) - (b*(36*a^4 + a^2*b^2 + 8*b^4 + 6*a*b*(9*a^2 + b^2)* 
Cosh[c + d*x] + (11*a^2*b^2 + 4*b^4)*Cosh[2*(c + d*x)])*Sinh[c + d*x])/(2* 
(a - b)^3*(a + b)^3*(a + b*Cosh[c + d*x])^3))/(6*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.23, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 3143, 25, 3042, 3233, 25, 3042, 3233, 27, 3042, 3138, 218}

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.

\(\displaystyle \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3143

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

\(\displaystyle \frac {-\frac {\int -\frac {2 \left (3 a^2+2 b^2\right )-5 a b \cosh (c+d x)}{(a+b \cosh (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {2 \left (3 a^2+2 b^2\right )-5 a b \cosh (c+d x)}{(a+b \cosh (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}+\frac {-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}+\frac {\int \frac {2 \left (3 a^2+2 b^2\right )-5 a b \sin \left (i c+i d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {\frac {-\frac {\int -\frac {3 a \left (2 a^2+3 b^2\right )}{a+b \cosh (c+d x)}dx}{a^2-b^2}-\frac {b \left (11 a^2+4 b^2\right ) \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 a \left (2 a^2+3 b^2\right ) \int \frac {1}{a+b \cosh (c+d x)}dx}{a^2-b^2}-\frac {b \left (11 a^2+4 b^2\right ) \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}+\frac {-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}+\frac {-\frac {b \left (11 a^2+4 b^2\right ) \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))}+\frac {3 a \left (2 a^2+3 b^2\right ) \int \frac {1}{a+b \sin \left (i c+i d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}}{2 \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}+\frac {-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}+\frac {-\frac {b \left (11 a^2+4 b^2\right ) \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))}-\frac {6 i a \left (2 a^2+3 b^2\right ) \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {1}{2} (c+d x)\right )\right )+a+b}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (a^2-b^2\right )}}{2 \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {6 a \left (2 a^2+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {b \left (11 a^2+4 b^2\right ) \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3}\)

Input:

Int[(a + b*Cosh[c + d*x])^(-4),x]
 

Output:

-1/3*(b*Sinh[c + d*x])/((a^2 - b^2)*d*(a + b*Cosh[c + d*x])^3) + ((-5*a*b* 
Sinh[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cosh[c + d*x])^2) + ((6*a*(2*a^2 + 
3*b^2)*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]* 
Sqrt[a + b]*(a^2 - b^2)*d) - (b*(11*a^2 + 4*b^2)*Sinh[c + d*x])/((a^2 - b^ 
2)*d*(a + b*Cosh[c + d*x])))/(2*(a^2 - b^2)))/(3*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3143

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp 
[1/((n + 1)*(a^2 - b^2))   Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) 
- b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3233

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.54

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (6 a^{2}+3 a b +2 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {2 \left (9 a^{2}+b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-3 a b +2 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b \right )^{3}}+\frac {a \left (2 a^{2}+3 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(284\)
default	\(\frac {-\frac {2 \left (-\frac {\left (6 a^{2}+3 a b +2 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {2 \left (9 a^{2}+b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-3 a b +2 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b \right )^{3}}+\frac {a \left (2 a^{2}+3 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(284\)
risch	\(\frac {6 \,{\mathrm e}^{5 d x +5 c} a^{3} b^{2}+9 a \,b^{4} {\mathrm e}^{5 d x +5 c}+30 a^{4} b \,{\mathrm e}^{4 d x +4 c}+45 a^{2} b^{3} {\mathrm e}^{4 d x +4 c}+44 a^{5} {\mathrm e}^{3 d x +3 c}+82 a^{3} b^{2} {\mathrm e}^{3 d x +3 c}+24 a \,b^{4} {\mathrm e}^{3 d x +3 c}+102 a^{4} b \,{\mathrm e}^{2 d x +2 c}+36 a^{2} b^{3} {\mathrm e}^{2 d x +2 c}+12 b^{5} {\mathrm e}^{2 d x +2 c}+60 \,{\mathrm e}^{d x +c} a^{3} b^{2}+15 \,{\mathrm e}^{d x +c} a \,b^{4}+11 a^{2} b^{3}+4 b^{5}}{3 d \left (a^{2}-b^{2}\right )^{3} \left ({\mathrm e}^{2 d x +2 c} b +2 \,{\mathrm e}^{d x +c} a +b \right )^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 a \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {a^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {3 a \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(543\)

Input:

int(1/(a+b*cosh(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2*(-1/2*(6*a^2+3*a*b+2*b^2)*b/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tanh(1 
/2*d*x+1/2*c)^5+2/3*(9*a^2+b^2)*b/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tanh(1/2 
*d*x+1/2*c)^3-1/2*(6*a^2-3*a*b+2*b^2)*b/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*ta 
nh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^2*a-b*tanh(1/2*d*x+1/2*c)^2-a-b)^3 
+a*(2*a^2+3*b^2)/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh 
((a-b)*tanh(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2798 vs. \(2 (169) = 338\).

Time = 0.15 (sec) , antiderivative size = 5705, normalized size of antiderivative = 31.01 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*cosh(d*x+c))^4,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*cosh(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*cosh(d*x+c))^4,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\frac {\frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \arctan \left (\frac {b e^{\left (d x + c\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, a^{3} b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 9 \, a b^{4} e^{\left (5 \, d x + 5 \, c\right )} + 30 \, a^{4} b e^{\left (4 \, d x + 4 \, c\right )} + 45 \, a^{2} b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 44 \, a^{5} e^{\left (3 \, d x + 3 \, c\right )} + 82 \, a^{3} b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 24 \, a b^{4} e^{\left (3 \, d x + 3 \, c\right )} + 102 \, a^{4} b e^{\left (2 \, d x + 2 \, c\right )} + 36 \, a^{2} b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{5} e^{\left (2 \, d x + 2 \, c\right )} + 60 \, a^{3} b^{2} e^{\left (d x + c\right )} + 15 \, a b^{4} e^{\left (d x + c\right )} + 11 \, a^{2} b^{3} + 4 \, b^{5}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} + b\right )}^{3}}}{3 \, d} \] Input:

integrate(1/(a+b*cosh(d*x+c))^4,x, algorithm="giac")
 

Output:

1/3*(3*(2*a^3 + 3*a*b^2)*arctan((b*e^(d*x + c) + a)/sqrt(-a^2 + b^2))/((a^ 
6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) + (6*a^3*b^2*e^(5*d*x + 
 5*c) + 9*a*b^4*e^(5*d*x + 5*c) + 30*a^4*b*e^(4*d*x + 4*c) + 45*a^2*b^3*e^ 
(4*d*x + 4*c) + 44*a^5*e^(3*d*x + 3*c) + 82*a^3*b^2*e^(3*d*x + 3*c) + 24*a 
*b^4*e^(3*d*x + 3*c) + 102*a^4*b*e^(2*d*x + 2*c) + 36*a^2*b^3*e^(2*d*x + 2 
*c) + 12*b^5*e^(2*d*x + 2*c) + 60*a^3*b^2*e^(d*x + c) + 15*a*b^4*e^(d*x + 
c) + 11*a^2*b^3 + 4*b^5)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(b*e^(2*d*x 
+ 2*c) + 2*a*e^(d*x + c) + b)^3))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\right )}^4} \,d x \] Input:

int(1/(a + b*cosh(c + d*x))^4,x)
 

Output:

int(1/(a + b*cosh(c + d*x))^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1720, normalized size of antiderivative = 9.35 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*cosh(d*x+c))^4,x)
 

Output:

( - 12*e**(6*c + 6*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqr 
t( - a**2 + b**2))*a**3*b**3 - 18*e**(6*c + 6*d*x)*sqrt( - a**2 + b**2)*at 
an((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a*b**5 - 72*e**(5*c + 5*d*x) 
*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**4 
*b**2 - 108*e**(5*c + 5*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a 
)/sqrt( - a**2 + b**2))*a**2*b**4 - 144*e**(4*c + 4*d*x)*sqrt( - a**2 + b* 
*2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**5*b - 252*e**(4*c + 
 4*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2 
))*a**3*b**3 - 54*e**(4*c + 4*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x) 
*b + a)/sqrt( - a**2 + b**2))*a*b**5 - 96*e**(3*c + 3*d*x)*sqrt( - a**2 + 
b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**6 - 288*e**(3*c + 
 3*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2 
))*a**4*b**2 - 216*e**(3*c + 3*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x 
)*b + a)/sqrt( - a**2 + b**2))*a**2*b**4 - 144*e**(2*c + 2*d*x)*sqrt( - a* 
*2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**5*b - 252*e* 
*(2*c + 2*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 
 + b**2))*a**3*b**3 - 54*e**(2*c + 2*d*x)*sqrt( - a**2 + b**2)*atan((e**(c 
 + d*x)*b + a)/sqrt( - a**2 + b**2))*a*b**5 - 72*e**(c + d*x)*sqrt( - a**2 
 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**4*b**2 - 108*e 
**(c + d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2...