3.84 \(\int \frac {\cosh ^4(x)}{i+\text {csch}(x)} \, dx\)

Optimal. Leaf size=38 \[ \frac {i x}{8}+\frac {\cosh ^3(x)}{3}-\frac {1}{4} i \sinh (x) \cosh ^3(x)+\frac {1}{8} i \sinh (x) \cosh (x) \]

[Out]

1/8*I*x+1/3*cosh(x)^3+1/8*I*cosh(x)*sinh(x)-1/4*I*cosh(x)^3*sinh(x)

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Rubi [A]  time = 0.13, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3872, 2839, 2565, 30, 2568, 2635, 8} \[ \frac {i x}{8}+\frac {\cosh ^3(x)}{3}-\frac {1}{4} i \sinh (x) \cosh ^3(x)+\frac {1}{8} i \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(I + Csch[x]),x]

[Out]

(I/8)*x + Cosh[x]^3/3 + (I/8)*Cosh[x]*Sinh[x] - (I/4)*Cosh[x]^3*Sinh[x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2839

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 32, normalized size = 0.84 \[ \frac {i x}{8}-\frac {1}{32} i \sinh (4 x)+\frac {\cosh (x)}{4}+\frac {1}{12} \cosh (3 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^4/(I + Csch[x]),x]

[Out]

(I/8)*x + Cosh[x]/4 + Cosh[3*x]/12 - (I/32)*Sinh[4*x]

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fricas [A]  time = 0.41, size = 43, normalized size = 1.13 \[ \frac {1}{192} \, {\left (24 i \, x e^{\left (4 \, x\right )} - 3 i \, e^{\left (8 \, x\right )} + 8 \, e^{\left (7 \, x\right )} + 24 \, e^{\left (5 \, x\right )} + 24 \, e^{\left (3 \, x\right )} + 8 \, e^{x} + 3 i\right )} e^{\left (-4 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(I+csch(x)),x, algorithm="fricas")

[Out]

1/192*(24*I*x*e^(4*x) - 3*I*e^(8*x) + 8*e^(7*x) + 24*e^(5*x) + 24*e^(3*x) + 8*e^x + 3*I)*e^(-4*x)

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giac [A]  time = 0.14, size = 38, normalized size = 1.00 \[ \frac {1}{192} \, {\left (24 \, e^{\left (3 \, x\right )} + 8 \, e^{x} + 3 i\right )} e^{\left (-4 \, x\right )} + \frac {1}{8} i \, x - \frac {1}{64} i \, e^{\left (4 \, x\right )} + \frac {1}{24} \, e^{\left (3 \, x\right )} + \frac {1}{8} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(I+csch(x)),x, algorithm="giac")

[Out]

1/192*(24*e^(3*x) + 8*e^x + 3*I)*e^(-4*x) + 1/8*I*x - 1/64*I*e^(4*x) + 1/24*e^(3*x) + 1/8*e^x

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maple [B]  time = 0.16, size = 170, normalized size = 4.47 \[ -\frac {i}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {i}{4 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 i}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {i}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {3 i}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8}+\frac {1}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {1}{2 \tanh \left (\frac {x}{2}\right )+2}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {i}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(I+csch(x)),x)

[Out]

-1/8*I/(tanh(1/2*x)-1)+1/4*I/(tanh(1/2*x)+1)^4-1/3/(tanh(1/2*x)-1)^3-1/8*I*ln(tanh(1/2*x)-1)-1/2/(tanh(1/2*x)-
1)^2-3/8*I/(tanh(1/2*x)-1)^2-1/2/(tanh(1/2*x)-1)-1/8*I/(tanh(1/2*x)+1)+3/8*I/(tanh(1/2*x)+1)^2+1/8*I*ln(tanh(1
/2*x)+1)+1/3/(tanh(1/2*x)+1)^3-1/2*I/(tanh(1/2*x)+1)^3+1/2/(tanh(1/2*x)+1)-1/2*I/(tanh(1/2*x)-1)^3-1/2/(tanh(1
/2*x)+1)^2-1/4*I/(tanh(1/2*x)-1)^4

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maxima [A]  time = 0.30, size = 42, normalized size = 1.11 \[ \frac {1}{192} \, {\left (8 \, e^{\left (-x\right )} + 24 \, e^{\left (-3 \, x\right )} - 3 i\right )} e^{\left (4 \, x\right )} + \frac {1}{8} i \, x + \frac {1}{8} \, e^{\left (-x\right )} + \frac {1}{24} \, e^{\left (-3 \, x\right )} + \frac {1}{64} i \, e^{\left (-4 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(I+csch(x)),x, algorithm="maxima")

[Out]

1/192*(8*e^(-x) + 24*e^(-3*x) - 3*I)*e^(4*x) + 1/8*I*x + 1/8*e^(-x) + 1/24*e^(-3*x) + 1/64*I*e^(-4*x)

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mupad [B]  time = 1.55, size = 41, normalized size = 1.08 \[ \frac {{\mathrm {e}}^{-x}}{8}+\frac {{\mathrm {e}}^{-3\,x}}{24}+\frac {{\mathrm {e}}^{3\,x}}{24}+\frac {{\mathrm {e}}^x}{8}+\frac {x\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-4\,x}\,1{}\mathrm {i}}{64}-\frac {{\mathrm {e}}^{4\,x}\,1{}\mathrm {i}}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(1/sinh(x) + 1i),x)

[Out]

(x*1i)/8 + exp(-x)/8 + exp(-3*x)/24 + exp(3*x)/24 + (exp(-4*x)*1i)/64 - (exp(4*x)*1i)/64 + exp(x)/8

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**4/(I+csch(x)),x)

[Out]

Timed out

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