3.93 \(\int \frac{\cosh ^4(x)}{a+b \text{csch}(x)} \, dx\)

Optimal. Leaf size=125 \[ \frac{x \left (12 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}+\frac{2 b \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^5}-\frac{\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2} \]

[Out]

((3*a^4 + 12*a^2*b^2 + 8*b^4)*x)/(8*a^5) + (2*b*(a^2 + b^2)^(3/2)*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/
a^5 - (Cosh[x]^3*(4*b - 3*a*Sinh[x]))/(12*a^2) - (Cosh[x]*(8*b*(a^2 + b^2) - a*(3*a^2 + 4*b^2)*Sinh[x]))/(8*a^
4)

________________________________________________________________________________________

Rubi [A]  time = 0.377696, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3872, 2865, 2735, 2660, 618, 204} \[ \frac{x \left (12 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}+\frac{2 b \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^5}-\frac{\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(a + b*Csch[x]),x]

[Out]

((3*a^4 + 12*a^2*b^2 + 8*b^4)*x)/(8*a^5) + (2*b*(a^2 + b^2)^(3/2)*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/
a^5 - (Cosh[x]^3*(4*b - 3*a*Sinh[x]))/(12*a^2) - (Cosh[x]*(8*b*(a^2 + b^2) - a*(3*a^2 + 4*b^2)*Sinh[x]))/(8*a^
4)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2865

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

Rule 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\cosh ^4(x)}{a+b \text{csch}(x)} \, dx &=i \int \frac{\cosh ^4(x) \sinh (x)}{i b+i a \sinh (x)} \, dx\\ &=-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}+\frac{\int \frac{\cosh ^2(x) \left (-i a b+i \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{i b+i a \sinh (x)} \, dx}{4 a^2}\\ &=-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac{\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}+\frac{\int \frac{-i a b \left (5 a^2+4 b^2\right )+i \left (3 a^4+12 a^2 b^2+8 b^4\right ) \sinh (x)}{i b+i a \sinh (x)} \, dx}{8 a^4}\\ &=\frac{\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac{\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}-\frac{\left (i b \left (a^2+b^2\right )^2\right ) \int \frac{1}{i b+i a \sinh (x)} \, dx}{a^5}\\ &=\frac{\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac{\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}-\frac{\left (2 i b \left (a^2+b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{i b+2 i a x-i b x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^5}\\ &=\frac{\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac{\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}+\frac{\left (4 i b \left (a^2+b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,2 i a-2 i b \tanh \left (\frac{x}{2}\right )\right )}{a^5}\\ &=\frac{\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac{2 b \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^5}-\frac{\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac{\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}\\ \end{align*}

Mathematica [A]  time = 1.14641, size = 180, normalized size = 1.44 \[ \frac{-24 a b \left (5 a^2+4 b^2\right ) \cosh (x)+3 \left (48 a^2 b^2 x+8 a^2 \left (a^2+b^2\right ) \sinh (2 x)+64 a^2 b \sqrt{-a^2-b^2} \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )+64 b^3 \sqrt{-a^2-b^2} \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )+12 a^4 x+a^4 \sinh (4 x)+32 b^4 x\right )-8 a^3 b \cosh (3 x)}{96 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a*b*(5*a^2 + 4*b^2)*Cosh[x] - 8*a^3*b*Cosh[3*x] + 3*(12*a^4*x + 48*a^2*b^2*x + 32*b^4*x + 64*a^2*b*Sqrt[-
a^2 - b^2]*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a^2 - b^2]] + 64*b^3*Sqrt[-a^2 - b^2]*ArcTan[(a - b*Tanh[x/2])/Sqrt[
-a^2 - b^2]] + 8*a^2*(a^2 + b^2)*Sinh[2*x] + a^4*Sinh[4*x]))/(96*a^5)

________________________________________________________________________________________

Maple [B]  time = 0.04, size = 486, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^5/a^5/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))-2/a*b/(a^2+b^2)^(1/2)*arctanh(1/
2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))-1/4/a/(tanh(1/2*x)+1)^4+1/2/a/(tanh(1/2*x)+1)^3-7/8/a/(tanh(1/2*x)+1)
^2+5/8/a/(tanh(1/2*x)+1)+3/8/a*ln(tanh(1/2*x)+1)+1/4/a/(tanh(1/2*x)-1)^4+1/2/a/(tanh(1/2*x)-1)^3+7/8/a/(tanh(1
/2*x)-1)^2+5/8/a/(tanh(1/2*x)-1)-3/8/a*ln(tanh(1/2*x)-1)-4*b^3/a^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1/2*x)*
b-2*a)/(a^2+b^2)^(1/2))+1/2/a^3/(tanh(1/2*x)-1)^2*b^2-3/2/a^3*ln(tanh(1/2*x)-1)*b^2-1/a^5*ln(tanh(1/2*x)-1)*b^
4+3/2/a^2/(tanh(1/2*x)-1)*b+1/2/a^3/(tanh(1/2*x)-1)*b^2+1/a^4/(tanh(1/2*x)-1)*b^3-3/2/a^2/(tanh(1/2*x)+1)*b+1/
2/a^3/(tanh(1/2*x)+1)*b^2-1/a^4/(tanh(1/2*x)+1)*b^3+1/3/a^2/(tanh(1/2*x)-1)^3*b+1/2/a^2/(tanh(1/2*x)-1)^2*b+3/
2/a^3*ln(tanh(1/2*x)+1)*b^2+1/a^5*ln(tanh(1/2*x)+1)*b^4-1/3/a^2/(tanh(1/2*x)+1)^3*b+1/2/a^2/(tanh(1/2*x)+1)^2*
b-1/2/a^3/(tanh(1/2*x)+1)^2*b^2

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.72759, size = 2414, normalized size = 19.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*a^4*cosh(x)^8 + 3*a^4*sinh(x)^8 - 8*a^3*b*cosh(x)^7 + 8*(3*a^4*cosh(x) - a^3*b)*sinh(x)^7 + 24*(a^4 +
a^2*b^2)*cosh(x)^6 + 4*(21*a^4*cosh(x)^2 - 14*a^3*b*cosh(x) + 6*a^4 + 6*a^2*b^2)*sinh(x)^6 + 24*(3*a^4 + 12*a
^2*b^2 + 8*b^4)*x*cosh(x)^4 - 24*(5*a^3*b + 4*a*b^3)*cosh(x)^5 + 24*(7*a^4*cosh(x)^3 - 7*a^3*b*cosh(x)^2 - 5*a
^3*b - 4*a*b^3 + 6*(a^4 + a^2*b^2)*cosh(x))*sinh(x)^5 - 8*a^3*b*cosh(x) + 2*(105*a^4*cosh(x)^4 - 140*a^3*b*cos
h(x)^3 + 180*(a^4 + a^2*b^2)*cosh(x)^2 + 12*(3*a^4 + 12*a^2*b^2 + 8*b^4)*x - 60*(5*a^3*b + 4*a*b^3)*cosh(x))*s
inh(x)^4 - 3*a^4 - 24*(5*a^3*b + 4*a*b^3)*cosh(x)^3 + 8*(21*a^4*cosh(x)^5 - 35*a^3*b*cosh(x)^4 - 15*a^3*b - 12
*a*b^3 + 60*(a^4 + a^2*b^2)*cosh(x)^3 + 12*(3*a^4 + 12*a^2*b^2 + 8*b^4)*x*cosh(x) - 30*(5*a^3*b + 4*a*b^3)*cos
h(x)^2)*sinh(x)^3 - 24*(a^4 + a^2*b^2)*cosh(x)^2 + 12*(7*a^4*cosh(x)^6 - 14*a^3*b*cosh(x)^5 + 30*(a^4 + a^2*b^
2)*cosh(x)^4 - 2*a^4 - 2*a^2*b^2 + 12*(3*a^4 + 12*a^2*b^2 + 8*b^4)*x*cosh(x)^2 - 20*(5*a^3*b + 4*a*b^3)*cosh(x
)^3 - 6*(5*a^3*b + 4*a*b^3)*cosh(x))*sinh(x)^2 + 192*((a^2*b + b^3)*cosh(x)^4 + 4*(a^2*b + b^3)*cosh(x)^3*sinh
(x) + 6*(a^2*b + b^3)*cosh(x)^2*sinh(x)^2 + 4*(a^2*b + b^3)*cosh(x)*sinh(x)^3 + (a^2*b + b^3)*sinh(x)^4)*sqrt(
a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) + a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) +
2*sqrt(a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*si
nh(x) - a)) + 8*(3*a^4*cosh(x)^7 - 7*a^3*b*cosh(x)^6 + 18*(a^4 + a^2*b^2)*cosh(x)^5 + 12*(3*a^4 + 12*a^2*b^2 +
8*b^4)*x*cosh(x)^3 - 15*(5*a^3*b + 4*a*b^3)*cosh(x)^4 - a^3*b - 9*(5*a^3*b + 4*a*b^3)*cosh(x)^2 - 6*(a^4 + a^
2*b^2)*cosh(x))*sinh(x))/(a^5*cosh(x)^4 + 4*a^5*cosh(x)^3*sinh(x) + 6*a^5*cosh(x)^2*sinh(x)^2 + 4*a^5*cosh(x)*
sinh(x)^3 + a^5*sinh(x)^4)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.17293, size = 298, normalized size = 2.38 \begin{align*} \frac{3 \, a^{3} e^{\left (4 \, x\right )} - 8 \, a^{2} b e^{\left (3 \, x\right )} + 24 \, a^{3} e^{\left (2 \, x\right )} + 24 \, a b^{2} e^{\left (2 \, x\right )} - 120 \, a^{2} b e^{x} - 96 \, b^{3} e^{x}}{192 \, a^{4}} + \frac{{\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x}{8 \, a^{5}} - \frac{{\left (8 \, a^{3} b e^{x} + 3 \, a^{4} + 24 \,{\left (5 \, a^{3} b + 4 \, a b^{3}\right )} e^{\left (3 \, x\right )} + 24 \,{\left (a^{4} + a^{2} b^{2}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-4 \, x\right )}}{192 \, a^{5}} - \frac{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (\frac{{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*a^3*e^(4*x) - 8*a^2*b*e^(3*x) + 24*a^3*e^(2*x) + 24*a*b^2*e^(2*x) - 120*a^2*b*e^x - 96*b^3*e^x)/a^4 +
1/8*(3*a^4 + 12*a^2*b^2 + 8*b^4)*x/a^5 - 1/192*(8*a^3*b*e^x + 3*a^4 + 24*(5*a^3*b + 4*a*b^3)*e^(3*x) + 24*(a^
4 + a^2*b^2)*e^(2*x))*e^(-4*x)/a^5 - (a^4*b + 2*a^2*b^3 + b^5)*log(abs(2*a*e^x + 2*b - 2*sqrt(a^2 + b^2))/abs(
2*a*e^x + 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^5)