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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2743, 2833, 12, 2738, 211} \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\frac {a \left (2 a^2+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {b \left (11 a^2+4 b^2\right ) \sinh (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \cosh (c+d x))}-\frac {5 a b \sinh (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \cosh (c+d x))^2}-\frac {b \sinh (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^3} \]

[In]

Int[(a + b*Cosh[c + d*x])^(-4),x]

[Out]

(a*(2*a^2 + 3*b^2)*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) - (b*
Sinh[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cosh[c + d*x])^3) - (5*a*b*Sinh[c + d*x])/(6*(a^2 - b^2)^2*d*(a + b*Cos
h[c + d*x])^2) - (b*(11*a^2 + 4*b^2)*Sinh[c + d*x])/(6*(a^2 - b^2)^3*d*(a + b*Cosh[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[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 2743

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] + Dist[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] && Integ
erQ[2*n]

Rule 2833

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {b \sinh (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cosh (c+d x))^3}-\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 \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 {\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}{6 \left (a^2-b^2\right )^2} \\ & = -\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))}-\frac {\int -\frac {3 a \left (2 a^2+3 b^2\right )}{a+b \cosh (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3} \\ & = -\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))}+\frac {\left (a \left (2 a^2+3 b^2\right )\right ) \int \frac {1}{a+b \cosh (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = -\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))}-\frac {\left (i a \left (2 a^2+3 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{\left (a^2-b^2\right )^3 d} \\ & = \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))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\frac {\frac {6 a \left (2 a^2+3 b^2\right ) \arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}-\frac {b \left (36 a^4+a^2 b^2+8 b^4+6 a b \left (9 a^2+b^2\right ) \cosh (c+d x)+\left (11 a^2 b^2+4 b^4\right ) \cosh (2 (c+d x))\right ) \sinh (c+d x)}{2 (a-b)^3 (a+b)^3 (a+b \cosh (c+d x))^3}}{6 d} \]

[In]

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

[Out]

((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)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.54

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

[In]

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

[Out]

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)*
tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b-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 = 0.32 (sec) , antiderivative size = 5705, normalized size of antiderivative = 31.01 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\frac {\frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \arctan \left (\frac {b e^{\left (d x + c\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, a^{3} b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 9 \, a b^{4} e^{\left (5 \, d x + 5 \, c\right )} + 30 \, a^{4} b e^{\left (4 \, d x + 4 \, c\right )} + 45 \, a^{2} b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 44 \, a^{5} e^{\left (3 \, d x + 3 \, c\right )} + 82 \, a^{3} b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 24 \, a b^{4} e^{\left (3 \, d x + 3 \, c\right )} + 102 \, a^{4} b e^{\left (2 \, d x + 2 \, c\right )} + 36 \, a^{2} b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{5} e^{\left (2 \, d x + 2 \, c\right )} + 60 \, a^{3} b^{2} e^{\left (d x + c\right )} + 15 \, a b^{4} e^{\left (d x + c\right )} + 11 \, a^{2} b^{3} + 4 \, b^{5}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} + b\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

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)*sqr
t(-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. \[ \int \frac {1}{(a+b \cosh (c+d x))^4} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\right )}^4} \,d x \]

[In]

int(1/(a + b*cosh(c + d*x))^4,x)

[Out]

int(1/(a + b*cosh(c + d*x))^4, x)