Integrand size = 13, antiderivative size = 36 \[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=-\frac {3 x}{2}-2 i \cosh (x)+\frac {3}{2} \cosh (x) \sinh (x)-\frac {\cosh (x) \sinh ^2(x)}{i+\sinh (x)} \] Output:
-3/2*x-2*I*cosh(x)+3/2*cosh(x)*sinh(x)-cosh(x)*sinh(x)^2/(I+sinh(x))
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=\frac {1}{2} \cosh (x) \left (-\frac {3 \text {arcsinh}(\sinh (x))}{\sqrt {\cosh ^2(x)}}+\frac {4-i \sinh (x)+\sinh ^2(x)}{i+\sinh (x)}\right ) \] Input:
Integrate[Sinh[x]^3/(I + Sinh[x]),x]
Output:
(Cosh[x]*((-3*ArcSinh[Sinh[x]])/Sqrt[Cosh[x]^2] + (4 - I*Sinh[x] + Sinh[x] ^2)/(I + Sinh[x])))/2
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 26, 26, 3246, 26, 3042, 26, 3213}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sinh ^3(x)}{\sinh (x)+i} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \sin (i x)^3}{i-i \sin (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int -\frac {i \sin (i x)^3}{1-\sin (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\sin (i x)^3}{1-\sin (i x)}dx\) |
\(\Big \downarrow \) 3246 |
\(\displaystyle \int -i (3 i \sinh (x)+2) \sinh (x)dx+\frac {i \sinh ^2(x) \cosh (x)}{1-i \sinh (x)}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \sinh ^2(x) \cosh (x)}{1-i \sinh (x)}-i \int (3 i \sinh (x)+2) \sinh (x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \sinh ^2(x) \cosh (x)}{1-i \sinh (x)}-i \int -i \sin (i x) (3 \sin (i x)+2)dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \sinh ^2(x) \cosh (x)}{1-i \sinh (x)}-\int \sin (i x) (3 \sin (i x)+2)dx\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle -\frac {3 x}{2}-2 i \cosh (x)+\frac {i \sinh ^2(x) \cosh (x)}{1-i \sinh (x)}+\frac {3}{2} \sinh (x) \cosh (x)\) |
Input:
Int[Sinh[x]^3/(I + Sinh[x]),x]
Output:
(-3*x)/2 - (2*I)*Cosh[x] + (3*Cosh[x]*Sinh[x])/2 + (I*Cosh[x]*Sinh[x]^2)/( 1 - I*Sinh[x])
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(a + b*Sin[e + f*x]))), x] - Simp[d/(a*b) Int[(c + d* Sin[e + f*x])^(n - 2)*Simp[b*d*(n - 1) - a*c*n + (b*c*(n - 1) - a*d*n)*Sin[ e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] & & EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && (IntegerQ[2*n] || EqQ[c, 0])
Time = 0.46 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08
method | result | size |
risch | \(-\frac {3 x}{2}+\frac {{\mathrm e}^{2 x}}{8}-\frac {i {\mathrm e}^{x}}{2}-\frac {i {\mathrm e}^{-x}}{2}-\frac {{\mathrm e}^{-2 x}}{8}-\frac {2 i}{{\mathrm e}^{x}+i}\) | \(39\) |
default | \(\frac {2}{\tanh \left (\frac {x}{2}\right )+i}+\frac {\frac {1}{2}-i}{\tanh \left (\frac {x}{2}\right )+1}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}+\frac {\frac {1}{2}+i}{\tanh \left (\frac {x}{2}\right )-1}+\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}\) | \(75\) |
parallelrisch | \(\frac {\left (12 i \cosh \left (\frac {x}{2}\right )+12 \sinh \left (\frac {x}{2}\right )\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\left (-12 i \cosh \left (\frac {x}{2}\right )-12 \sinh \left (\frac {x}{2}\right )\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )+16 i \sinh \left (\frac {x}{2}\right )-3 i \sinh \left (\frac {3 x}{2}\right )+i \sinh \left (\frac {5 x}{2}\right )+8 \cosh \left (\frac {x}{2}\right )+3 \cosh \left (\frac {3 x}{2}\right )+\cosh \left (\frac {5 x}{2}\right )}{8 i \cosh \left (\frac {x}{2}\right )+8 \sinh \left (\frac {x}{2}\right )}\) | \(100\) |
Input:
int(sinh(x)^3/(I+sinh(x)),x,method=_RETURNVERBOSE)
Output:
-3/2*x+1/8*exp(x)^2-1/2*I*exp(x)-1/2*I/exp(x)-1/8/exp(x)^2-2*I/(exp(x)+I)
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.53 \[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=-\frac {4 \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} + 4 \, {\left (3 i \, x + 5 i\right )} e^{\left (2 \, x\right )} - e^{\left (5 \, x\right )} + 3 i \, e^{\left (4 \, x\right )} - 3 \, e^{x} + i}{8 \, {\left (e^{\left (3 \, x\right )} + i \, e^{\left (2 \, x\right )}\right )}} \] Input:
integrate(sinh(x)^3/(I+sinh(x)),x, algorithm="fricas")
Output:
-1/8*(4*(3*x - 1)*e^(3*x) + 4*(3*I*x + 5*I)*e^(2*x) - e^(5*x) + 3*I*e^(4*x ) - 3*e^x + I)/(e^(3*x) + I*e^(2*x))
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=- \frac {3 x}{2} + \frac {e^{2 x}}{8} - \frac {i e^{x}}{2} - \frac {i e^{- x}}{2} - \frac {e^{- 2 x}}{8} - \frac {2 i}{e^{x} + i} \] Input:
integrate(sinh(x)**3/(I+sinh(x)),x)
Output:
-3*x/2 + exp(2*x)/8 - I*exp(x)/2 - I*exp(-x)/2 - exp(-2*x)/8 - 2*I/(exp(x) + I)
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=-\frac {3}{2} \, x - \frac {3 \, e^{\left (-x\right )} + 20 i \, e^{\left (-2 \, x\right )} + i}{8 \, {\left (-i \, e^{\left (-2 \, x\right )} + e^{\left (-3 \, x\right )}\right )}} - \frac {1}{2} i \, e^{\left (-x\right )} - \frac {1}{8} \, e^{\left (-2 \, x\right )} \] Input:
integrate(sinh(x)^3/(I+sinh(x)),x, algorithm="maxima")
Output:
-3/2*x - 1/8*(3*e^(-x) + 20*I*e^(-2*x) + I)/(-I*e^(-2*x) + e^(-3*x)) - 1/2 *I*e^(-x) - 1/8*e^(-2*x)
Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=-\frac {3}{2} \, x - \frac {{\left (20 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} + i\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (e^{x} + i\right )}} + \frac {1}{8} \, e^{\left (2 \, x\right )} - \frac {1}{2} i \, e^{x} \] Input:
integrate(sinh(x)^3/(I+sinh(x)),x, algorithm="giac")
Output:
-3/2*x - 1/8*(20*I*e^(2*x) - 3*e^x + I)*e^(-2*x)/(e^x + I) + 1/8*e^(2*x) - 1/2*I*e^x
Time = 1.73 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8}-\frac {{\mathrm {e}}^{-x}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{-2\,x}}{8}-\frac {3\,x}{2}-\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{2}-\frac {2{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}} \] Input:
int(sinh(x)^3/(sinh(x) + 1i),x)
Output:
exp(2*x)/8 - (exp(-x)*1i)/2 - exp(-2*x)/8 - (3*x)/2 - (exp(x)*1i)/2 - 2i/( exp(x) + 1i)
\[ \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx=\int \frac {\sinh \left (x \right )^{3}}{\sinh \left (x \right )+i}d x \] Input:
int(sinh(x)^3/(I+sinh(x)),x)
Output:
int(sinh(x)**3/(sinh(x) + i),x)