3.179 \(\int \frac {\tanh ^4(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=113 \[ \frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4}-\frac {b \tanh (x) \text {sech}(x)}{2 a^2}+\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}-\frac {\left (4 a^2-3 b^2\right ) \tanh (x)}{3 a^3}+\frac {\tanh (x) \text {sech}^2(x)}{3 a} \]

[Out]

1/2*b*(3*a^2-2*b^2)*arctan(sinh(x))/a^4+2*(a-b)^(3/2)*(a+b)^(3/2)*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))
/a^4-1/3*(4*a^2-3*b^2)*tanh(x)/a^3-1/2*b*sech(x)*tanh(x)/a^2+1/3*sech(x)^2*tanh(x)/a

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2725, 3055, 3001, 3770, 2659, 208} \[ -\frac {\left (4 a^2-3 b^2\right ) \tanh (x)}{3 a^3}+\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4}-\frac {b \tanh (x) \text {sech}(x)}{2 a^2}+\frac {\tanh (x) \text {sech}^2(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(a + b*Cosh[x]),x]

[Out]

(b*(3*a^2 - 2*b^2)*ArcTan[Sinh[x]])/(2*a^4) + (2*(a - b)^(3/2)*(a + b)^(3/2)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/S
qrt[a + b]])/a^4 - ((4*a^2 - 3*b^2)*Tanh[x])/(3*a^3) - (b*Sech[x]*Tanh[x])/(2*a^2) + (Sech[x]^2*Tanh[x])/(3*a)

Rule 208

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

Rule 2659

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 2725

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(3*a*f*Sin[e + f*x]^3), x] + (-Dist[1/(6*a^2), Int[((a + b*Sin[e + f*x])^m*Simp[8*
a^2 - b^2*(m - 1)*(m - 2) + a*b*m*Sin[e + f*x] - (6*a^2 - b^2*m*(m - 2))*Sin[e + f*x]^2, x])/Sin[e + f*x]^2, x
], x] - Simp[(b*(m - 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(6*a^2*f*Sin[e + f*x]^2), x]) /; FreeQ[{a,
b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1] && IntegerQ[2*m]

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\tanh ^4(x)}{a+b \cosh (x)} \, dx &=-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}-\frac {\int \frac {\left (2 \left (4 a^2-3 b^2\right )-a b \cosh (x)-3 \left (2 a^2-b^2\right ) \cosh ^2(x)\right ) \text {sech}^2(x)}{a+b \cosh (x)} \, dx}{6 a^2}\\ &=-\frac {\left (4 a^2-3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}-\frac {\int \frac {\left (-3 b \left (3 a^2-2 b^2\right )-3 a \left (2 a^2-b^2\right ) \cosh (x)\right ) \text {sech}(x)}{a+b \cosh (x)} \, dx}{6 a^3}\\ &=-\frac {\left (4 a^2-3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}+\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \int \text {sech}(x) \, dx}{2 a^4}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \cosh (x)} \, dx}{a^4}\\ &=\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}-\frac {\left (4 a^2-3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^4}\\ &=\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4}-\frac {\left (4 a^2-3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.44, size = 100, normalized size = 0.88 \[ \frac {-12 \left (b^2-a^2\right )^{3/2} \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )+6 b \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+a \tanh (x) \left (2 a^2 \text {sech}^2(x)-8 a^2-3 a b \text {sech}(x)+6 b^2\right )}{6 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(a + b*Cosh[x]),x]

[Out]

(6*b*(3*a^2 - 2*b^2)*ArcTan[Tanh[x/2]] - 12*(-a^2 + b^2)^(3/2)*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]] +
a*(-8*a^2 + 6*b^2 - 3*a*b*Sech[x] + 2*a^2*Sech[x]^2)*Tanh[x])/(6*a^4)

________________________________________________________________________________________

fricas [B]  time = 1.57, size = 2003, normalized size = 17.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*a^2*b*cosh(x)^5 + 3*a^2*b*sinh(x)^5 - 6*(2*a^3 - a*b^2)*cosh(x)^4 + 3*(5*a^2*b*cosh(x) - 4*a^3 + 2*a*
b^2)*sinh(x)^4 - 3*a^2*b*cosh(x) + 6*(5*a^2*b*cosh(x)^2 - 4*(2*a^3 - a*b^2)*cosh(x))*sinh(x)^3 - 8*a^3 + 6*a*b
^2 - 12*(a^3 - a*b^2)*cosh(x)^2 + 6*(5*a^2*b*cosh(x)^3 - 2*a^3 + 2*a*b^2 - 6*(2*a^3 - a*b^2)*cosh(x)^2)*sinh(x
)^2 + 3*((a^2 - b^2)*cosh(x)^6 + 6*(a^2 - b^2)*cosh(x)*sinh(x)^5 + (a^2 - b^2)*sinh(x)^6 + 3*(a^2 - b^2)*cosh(
x)^4 + 3*(5*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^4 + 4*(5*(a^2 - b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x))
*sinh(x)^3 + 3*(a^2 - b^2)*cosh(x)^2 + 3*(5*(a^2 - b^2)*cosh(x)^4 + 6*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(
x)^2 + a^2 - b^2 + 6*((a^2 - b^2)*cosh(x)^5 + 2*(a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*sqrt(a^2
- b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) + 2*s
qrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(
x) + b)) - 3*((3*a^2*b - 2*b^3)*cosh(x)^6 + 6*(3*a^2*b - 2*b^3)*cosh(x)*sinh(x)^5 + (3*a^2*b - 2*b^3)*sinh(x)^
6 + 3*(3*a^2*b - 2*b^3)*cosh(x)^4 + 3*(3*a^2*b - 2*b^3 + 5*(3*a^2*b - 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(3*a^
2*b - 2*b^3)*cosh(x)^3 + 3*(3*a^2*b - 2*b^3)*cosh(x))*sinh(x)^3 + 3*a^2*b - 2*b^3 + 3*(3*a^2*b - 2*b^3)*cosh(x
)^2 + 3*(5*(3*a^2*b - 2*b^3)*cosh(x)^4 + 3*a^2*b - 2*b^3 + 6*(3*a^2*b - 2*b^3)*cosh(x)^2)*sinh(x)^2 + 6*((3*a^
2*b - 2*b^3)*cosh(x)^5 + 2*(3*a^2*b - 2*b^3)*cosh(x)^3 + (3*a^2*b - 2*b^3)*cosh(x))*sinh(x))*arctan(cosh(x) +
sinh(x)) + 3*(5*a^2*b*cosh(x)^4 - 8*(2*a^3 - a*b^2)*cosh(x)^3 - a^2*b - 8*(a^3 - a*b^2)*cosh(x))*sinh(x))/(a^4
*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 + 3*a^4*cosh(x)^4 + 3*a^4*cosh(x)^2 + 3*(5*a^4*cosh(x)^2
+ a^4)*sinh(x)^4 + a^4 + 4*(5*a^4*cosh(x)^3 + 3*a^4*cosh(x))*sinh(x)^3 + 3*(5*a^4*cosh(x)^4 + 6*a^4*cosh(x)^2
+ a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 + 2*a^4*cosh(x)^3 + a^4*cosh(x))*sinh(x)), -1/3*(3*a^2*b*cosh(x)^5 + 3*a^2
*b*sinh(x)^5 - 6*(2*a^3 - a*b^2)*cosh(x)^4 + 3*(5*a^2*b*cosh(x) - 4*a^3 + 2*a*b^2)*sinh(x)^4 - 3*a^2*b*cosh(x)
+ 6*(5*a^2*b*cosh(x)^2 - 4*(2*a^3 - a*b^2)*cosh(x))*sinh(x)^3 - 8*a^3 + 6*a*b^2 - 12*(a^3 - a*b^2)*cosh(x)^2
+ 6*(5*a^2*b*cosh(x)^3 - 2*a^3 + 2*a*b^2 - 6*(2*a^3 - a*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((a^2 - b^2)*cosh(x)^6 +
6*(a^2 - b^2)*cosh(x)*sinh(x)^5 + (a^2 - b^2)*sinh(x)^6 + 3*(a^2 - b^2)*cosh(x)^4 + 3*(5*(a^2 - b^2)*cosh(x)^
2 + a^2 - b^2)*sinh(x)^4 + 4*(5*(a^2 - b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x))*sinh(x)^3 + 3*(a^2 - b^2)*cosh(
x)^2 + 3*(5*(a^2 - b^2)*cosh(x)^4 + 6*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 - b^2 + 6*((a^2 - b^2
)*cosh(x)^5 + 2*(a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2
)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) - 3*((3*a^2*b - 2*b^3)*cosh(x)^6 + 6*(3*a^2*b - 2*b^3)*cosh(x)*sinh
(x)^5 + (3*a^2*b - 2*b^3)*sinh(x)^6 + 3*(3*a^2*b - 2*b^3)*cosh(x)^4 + 3*(3*a^2*b - 2*b^3 + 5*(3*a^2*b - 2*b^3)
*cosh(x)^2)*sinh(x)^4 + 4*(5*(3*a^2*b - 2*b^3)*cosh(x)^3 + 3*(3*a^2*b - 2*b^3)*cosh(x))*sinh(x)^3 + 3*a^2*b -
2*b^3 + 3*(3*a^2*b - 2*b^3)*cosh(x)^2 + 3*(5*(3*a^2*b - 2*b^3)*cosh(x)^4 + 3*a^2*b - 2*b^3 + 6*(3*a^2*b - 2*b^
3)*cosh(x)^2)*sinh(x)^2 + 6*((3*a^2*b - 2*b^3)*cosh(x)^5 + 2*(3*a^2*b - 2*b^3)*cosh(x)^3 + (3*a^2*b - 2*b^3)*c
osh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) + 3*(5*a^2*b*cosh(x)^4 - 8*(2*a^3 - a*b^2)*cosh(x)^3 - a^2*b - 8*(a
^3 - a*b^2)*cosh(x))*sinh(x))/(a^4*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 + 3*a^4*cosh(x)^4 + 3*a
^4*cosh(x)^2 + 3*(5*a^4*cosh(x)^2 + a^4)*sinh(x)^4 + a^4 + 4*(5*a^4*cosh(x)^3 + 3*a^4*cosh(x))*sinh(x)^3 + 3*(
5*a^4*cosh(x)^4 + 6*a^4*cosh(x)^2 + a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 + 2*a^4*cosh(x)^3 + a^4*cosh(x))*sinh(x)
)]

________________________________________________________________________________________

giac [A]  time = 0.15, size = 144, normalized size = 1.27 \[ \frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \arctan \left (e^{x}\right )}{a^{4}} + \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{4}} - \frac {3 \, a b e^{\left (5 \, x\right )} - 12 \, a^{2} e^{\left (4 \, x\right )} + 6 \, b^{2} e^{\left (4 \, x\right )} - 12 \, a^{2} e^{\left (2 \, x\right )} + 12 \, b^{2} e^{\left (2 \, x\right )} - 3 \, a b e^{x} - 8 \, a^{2} + 6 \, b^{2}}{3 \, a^{3} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a^2*b - 2*b^3)*arctan(e^x)/a^4 + 2*(a^4 - 2*a^2*b^2 + b^4)*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2
+ b^2)*a^4) - 1/3*(3*a*b*e^(5*x) - 12*a^2*e^(4*x) + 6*b^2*e^(4*x) - 12*a^2*e^(2*x) + 12*b^2*e^(2*x) - 3*a*b*e^
x - 8*a^2 + 6*b^2)/(a^3*(e^(2*x) + 1)^3)

________________________________________________________________________________________

maple [B]  time = 0.10, size = 315, normalized size = 2.79 \[ \frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {4 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) b^{2}}{a^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 b^{4} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {\left (\tanh ^{5}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2 \left (\tanh ^{5}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {20 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{3 a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 \tanh \left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2 \tanh \left (\frac {x}{2}\right ) b^{2}}{a^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {\tanh \left (\frac {x}{2}\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) b}{a^{2}}-\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) b^{3}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(a+b*cosh(x)),x)

[Out]

2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-4/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
anh(1/2*x)/((a+b)*(a-b))^(1/2))*b^2+2/a^4/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))*b
^4-2/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5+1/a^2/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5*b+2/a^3/(tanh(1/2*x)^2+1)^3*t
anh(1/2*x)^5*b^2-20/3/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^3+4/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^3*b^2-2/a/(tan
h(1/2*x)^2+1)^3*tanh(1/2*x)+2/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)*b^2-1/a^2/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)*b+
3/a^2*arctan(tanh(1/2*x))*b-2/a^4*arctan(tanh(1/2*x))*b^3

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*cosh(x)),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 details)Is 4*a^2-4*b^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 5.94, size = 722, normalized size = 6.39 \[ \frac {8}{3\,a\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}-\frac {\frac {4}{a}-\frac {2\,b\,{\mathrm {e}}^x}{a^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {2\,\left (2\,a^2-b^2\right )}{a^3}-\frac {b\,{\mathrm {e}}^x}{a^2}}{{\mathrm {e}}^{2\,x}+1}+\frac {\ln \left (\frac {\left (\frac {32\,a^8-288\,{\mathrm {e}}^x\,a^7\,b-272\,a^6\,b^2+600\,{\mathrm {e}}^x\,a^5\,b^3+456\,a^4\,b^4-416\,{\mathrm {e}}^x\,a^3\,b^5-288\,a^2\,b^6+96\,{\mathrm {e}}^x\,a\,b^7+64\,b^8}{a^6\,b^4}-\frac {\left (\frac {32\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\,\left (4\,{\mathrm {e}}^x\,a^3+3\,a^2\,b-3\,{\mathrm {e}}^x\,a\,b^2-2\,b^3\right )}{a^2\,b^5}+\frac {16\,\left (a^2-b^2\right )\,\left (8\,{\mathrm {e}}^x\,a^3+4\,a^2\,b-7\,{\mathrm {e}}^x\,a\,b^2-4\,b^3\right )}{a\,b^5}\right )\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a^4}\right )\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a^4}-\frac {8\,{\left (a^2-b^2\right )}^2\,\left (3\,a^2-2\,b^2\right )\,\left (10\,{\mathrm {e}}^x\,a^3+6\,a^2\,b-7\,{\mathrm {e}}^x\,a\,b^2-4\,b^3\right )}{a^9\,b^3}\right )\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a^4}-\frac {\ln \left (-\frac {\left (\frac {32\,a^8-288\,{\mathrm {e}}^x\,a^7\,b-272\,a^6\,b^2+600\,{\mathrm {e}}^x\,a^5\,b^3+456\,a^4\,b^4-416\,{\mathrm {e}}^x\,a^3\,b^5-288\,a^2\,b^6+96\,{\mathrm {e}}^x\,a\,b^7+64\,b^8}{a^6\,b^4}-\frac {\left (\frac {32\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\,\left (4\,{\mathrm {e}}^x\,a^3+3\,a^2\,b-3\,{\mathrm {e}}^x\,a\,b^2-2\,b^3\right )}{a^2\,b^5}-\frac {16\,\left (a^2-b^2\right )\,\left (8\,{\mathrm {e}}^x\,a^3+4\,a^2\,b-7\,{\mathrm {e}}^x\,a\,b^2-4\,b^3\right )}{a\,b^5}\right )\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a^4}\right )\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a^4}-\frac {8\,{\left (a^2-b^2\right )}^2\,\left (3\,a^2-2\,b^2\right )\,\left (10\,{\mathrm {e}}^x\,a^3+6\,a^2\,b-7\,{\mathrm {e}}^x\,a\,b^2-4\,b^3\right )}{a^9\,b^3}\right )\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a^4}-\frac {b\,\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,\left (3\,a^2-2\,b^2\right )\,1{}\mathrm {i}}{2\,a^4}+\frac {b\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (3\,a^2-2\,b^2\right )\,1{}\mathrm {i}}{2\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(a + b*cosh(x)),x)

[Out]

8/(3*a*(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1)) - (4/a - (2*b*exp(x))/a^2)/(2*exp(2*x) + exp(4*x) + 1) + ((2*
(2*a^2 - b^2))/a^3 - (b*exp(x))/a^2)/(exp(2*x) + 1) + (log((((32*a^8 + 64*b^8 - 288*a^2*b^6 + 456*a^4*b^4 - 27
2*a^6*b^2 + 96*a*b^7*exp(x) - 288*a^7*b*exp(x) - 416*a^3*b^5*exp(x) + 600*a^5*b^3*exp(x))/(a^6*b^4) - (((32*((
a + b)^3*(a - b)^3)^(1/2)*(3*a^2*b - 2*b^3 + 4*a^3*exp(x) - 3*a*b^2*exp(x)))/(a^2*b^5) + (16*(a^2 - b^2)*(4*a^
2*b - 4*b^3 + 8*a^3*exp(x) - 7*a*b^2*exp(x)))/(a*b^5))*((a + b)^3*(a - b)^3)^(1/2))/a^4)*((a + b)^3*(a - b)^3)
^(1/2))/a^4 - (8*(a^2 - b^2)^2*(3*a^2 - 2*b^2)*(6*a^2*b - 4*b^3 + 10*a^3*exp(x) - 7*a*b^2*exp(x)))/(a^9*b^3))*
((a + b)^3*(a - b)^3)^(1/2))/a^4 - (log(- (((32*a^8 + 64*b^8 - 288*a^2*b^6 + 456*a^4*b^4 - 272*a^6*b^2 + 96*a*
b^7*exp(x) - 288*a^7*b*exp(x) - 416*a^3*b^5*exp(x) + 600*a^5*b^3*exp(x))/(a^6*b^4) - (((32*((a + b)^3*(a - b)^
3)^(1/2)*(3*a^2*b - 2*b^3 + 4*a^3*exp(x) - 3*a*b^2*exp(x)))/(a^2*b^5) - (16*(a^2 - b^2)*(4*a^2*b - 4*b^3 + 8*a
^3*exp(x) - 7*a*b^2*exp(x)))/(a*b^5))*((a + b)^3*(a - b)^3)^(1/2))/a^4)*((a + b)^3*(a - b)^3)^(1/2))/a^4 - (8*
(a^2 - b^2)^2*(3*a^2 - 2*b^2)*(6*a^2*b - 4*b^3 + 10*a^3*exp(x) - 7*a*b^2*exp(x)))/(a^9*b^3))*((a + b)^3*(a - b
)^3)^(1/2))/a^4 - (b*log(exp(x) - 1i)*(3*a^2 - 2*b^2)*1i)/(2*a^4) + (b*log(exp(x) + 1i)*(3*a^2 - 2*b^2)*1i)/(2
*a^4)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{4}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**4/(a+b*cosh(x)),x)

[Out]

Integral(tanh(x)**4/(a + b*cosh(x)), x)

________________________________________________________________________________________