\(\int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx\) [119]

Optimal result

Integrand size = 17, antiderivative size = 27 \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=-B x+\frac {(i A-B) \cosh (x)}{i-\sinh (x)} \]

[Out]

-B*x+(I*A-B)*cosh(x)/(I-sinh(x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2814, 2727} \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=-B x+\frac {(-B+i A) \cosh (x)}{-\sinh (x)+i} \]

[In]

Int[(A + B*Sinh[x])/(I - Sinh[x]),x]

[Out]

-(B*x) + ((I*A - B)*Cosh[x])/(I - Sinh[x])

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2814

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]

Rubi steps \begin{align*} \text {integral}& = -B x+(A+i B) \int \frac {1}{i-\sinh (x)} \, dx \\ & = -B x+\frac {(i A-B) \cosh (x)}{i-\sinh (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=\cosh (x) \left (-\frac {B \text {arcsinh}(\sinh (x))}{\sqrt {\cosh ^2(x)}}+\frac {-i A+B}{-i+\sinh (x)}\right ) \]

[In]

Integrate[(A + B*Sinh[x])/(I - Sinh[x]),x]

[Out]

Cosh[x]*(-((B*ArcSinh[Sinh[x]])/Sqrt[Cosh[x]^2]) + ((-I)*A + B)/(-I + Sinh[x]))

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
risch	\(-B x +\frac {2 A}{{\mathrm e}^{x}-i}+\frac {2 i B}{{\mathrm e}^{x}-i}\)	\(27\)
parallelrisch	\(\frac {i B x -x \tanh \left (\frac {x}{2}\right ) B -2 i A +2 B}{-i+\tanh \left (\frac {x}{2}\right )}\)	\(32\)
default	\(-B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {2 i \left (i B +A \right )}{-i+\tanh \left (\frac {x}{2}\right )}+B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )\)	\(39\)

[In]

int((A+B*sinh(x))/(I-sinh(x)),x,method=_RETURNVERBOSE)

[Out]

-B*x+2/(exp(x)-I)*A+2*I/(exp(x)-I)*B

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=-\frac {B x e^{x} - i \, B x - 2 \, A - 2 i \, B}{e^{x} - i} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x)),x, algorithm="fricas")

[Out]

-(B*x*e^x - I*B*x - 2*A - 2*I*B)/(e^x - I)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=- B x + \frac {2 A + 2 i B}{e^{x} - i} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x)),x)

[Out]

-B*x + (2*A + 2*I*B)/(exp(x) - I)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=-B {\left (x - \frac {2 i}{e^{\left (-x\right )} + i}\right )} + \frac {2 \, A}{e^{\left (-x\right )} + i} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x)),x, algorithm="maxima")

[Out]

-B*(x - 2*I/(e^(-x) + I)) + 2*A/(e^(-x) + I)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=-B x + \frac {2 \, {\left (A + i \, B\right )}}{e^{x} - i} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x)),x, algorithm="giac")

[Out]

-B*x + 2*(A + I*B)/(e^x - I)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {A+B \sinh (x)}{i-\sinh (x)} \, dx=-B\,x+\frac {2\,A+B\,2{}\mathrm {i}}{{\mathrm {e}}^x-\mathrm {i}} \]

[In]

int(-(A + B*sinh(x))/(sinh(x) - 1i),x)

[Out]

(2*A + B*2i)/(exp(x) - 1i) - B*x