\(\int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx\) [184]

Optimal result

Integrand size = 15, antiderivative size = 26 \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=-i \log (i+\sinh (x))-\frac {2 i}{1-i \sinh (x)} \]

[Out]

-I*ln(I+sinh(x))-2*I/(1-I*sinh(x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2746, 45} \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=-\frac {2 i}{1-i \sinh (x)}-i \log (\sinh (x)+i) \]

[In]

Int[Cosh[x]^3/(1 - I*Sinh[x])^3,x]

[Out]

(-I)*Log[I + Sinh[x]] - (2*I)/(1 - I*Sinh[x])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = i \text {Subst}\left (\int \frac {1-x}{(1+x)^2} \, dx,x,-i \sinh (x)\right ) \\ & = i \text {Subst}\left (\int \left (\frac {1}{-1-x}+\frac {2}{(1+x)^2}\right ) \, dx,x,-i \sinh (x)\right ) \\ & = -i \log (i+\sinh (x))-\frac {2 i}{1-i \sinh (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=\frac {\cosh ^4(x) (2+\log (i+\sinh (x))-i \log (i+\sinh (x)) \sinh (x))}{(-i+\sinh (x))^2 (i+\sinh (x))^3} \]

[In]

Integrate[Cosh[x]^3/(1 - I*Sinh[x])^3,x]

[Out]

(Cosh[x]^4*(2 + Log[I + Sinh[x]] - I*Log[I + Sinh[x]]*Sinh[x]))/((-I + Sinh[x])^2*(I + Sinh[x])^3)

Maple [A] (verified)

Time = 59.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {2}{i+\sinh \left (x \right )}-\frac {i \ln \left (\sinh \left (x \right )^{2}+1\right )}{2}-\arctan \left (\sinh \left (x \right )\right )\)	\(26\)
default	\(\frac {2}{i+\sinh \left (x \right )}-\frac {i \ln \left (\sinh \left (x \right )^{2}+1\right )}{2}-\arctan \left (\sinh \left (x \right )\right )\)	\(26\)
risch	\(i x +\frac {4 \,{\mathrm e}^{x}}{\left ({\mathrm e}^{x}+i\right )^{2}}-2 i \ln \left ({\mathrm e}^{x}+i\right )\)	\(26\)

[In]

int(cosh(x)^3/(1-I*sinh(x))^3,x,method=_RETURNVERBOSE)

[Out]

2/(I+sinh(x))-1/2*I*ln(sinh(x)^2+1)-arctan(sinh(x))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (18) = 36\).

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=\frac {i \, x e^{\left (2 \, x\right )} - 2 \, {\left (x - 2\right )} e^{x} - 2 \, {\left (i \, e^{\left (2 \, x\right )} - 2 \, e^{x} - i\right )} \log \left (e^{x} + i\right ) - i \, x}{e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1} \]

[In]

integrate(cosh(x)^3/(1-I*sinh(x))^3,x, algorithm="fricas")

[Out]

(I*x*e^(2*x) - 2*(x - 2)*e^x - 2*(I*e^(2*x) - 2*e^x - I)*log(e^x + I) - I*x)/(e^(2*x) + 2*I*e^x - 1)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=i x - 2 i \log {\left (e^{x} + i \right )} + \frac {4 e^{x}}{e^{2 x} + 2 i e^{x} - 1} \]

[In]

integrate(cosh(x)**3/(1-I*sinh(x))**3,x)

[Out]

I*x - 2*I*log(exp(x) + I) + 4*exp(x)/(exp(2*x) + 2*I*exp(x) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=-i \, x - \frac {4 \, e^{\left (-x\right )}}{-2 i \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 1} - 2 i \, \log \left (e^{\left (-x\right )} - i\right ) \]

[In]

integrate(cosh(x)^3/(1-I*sinh(x))^3,x, algorithm="maxima")

[Out]

-I*x - 4*e^(-x)/(-2*I*e^(-x) + e^(-2*x) - 1) - 2*I*log(e^(-x) - I)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=i \, x + \frac {4 \, e^{x}}{{\left (e^{x} + i\right )}^{2}} - 2 i \, \log \left (e^{x} + i\right ) \]

[In]

integrate(cosh(x)^3/(1-I*sinh(x))^3,x, algorithm="giac")

[Out]

I*x + 4*e^x/(e^x + I)^2 - 2*I*log(e^x + I)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {\cosh ^3(x)}{(1-i \sinh (x))^3} \, dx=x\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,2{}\mathrm {i}-\frac {4{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}}+\frac {4}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

[In]

int(-cosh(x)^3/(sinh(x)*1i - 1)^3,x)

[Out]

x*1i - log(exp(x) + 1i)*2i - 4i/(exp(2*x) + exp(x)*2i - 1) + 4/(exp(x) + 1i)