\(\int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx\) [115]

Optimal result

Integrand size = 15, antiderivative size = 23 \[ \int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx=B x-\frac {(i A+B) \cosh (x)}{i+\sinh (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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 [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(23)=46\).

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx=\cosh (x) \left (\frac {2 i B \arcsin \left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right )}{\sqrt {\cosh ^2(x)}}-\frac {i A+B}{i+\sinh (x)}\right ) \]

[In]

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

[Out]

Cosh[x]*(((2*I)*B*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]])/Sqrt[Cosh[x]^2] - (I*A + B)/(I + Sinh[x]))

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13

method	result	size
risch	\(B x -\frac {2 A}{{\mathrm e}^{x}+i}+\frac {2 i B}{{\mathrm e}^{x}+i}\)	\(26\)
parallelrisch	\(\frac {i B x +x \tanh \left (\frac {x}{2}\right ) B -2 i A -2 B}{\tanh \left (\frac {x}{2}\right )+i}\)	\(31\)
default	\(B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {2 i \left (-i B +A \right )}{\tanh \left (\frac {x}{2}\right )+i}-B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )\)	\(39\)

[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 = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx=\frac {B x e^{x} + i \, B x - 2 \, A + 2 i \, B}{e^{x} + i} \]

[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 = 0.65 \[ \int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx=B x + \frac {- 2 A + 2 i B}{e^{x} + i} \]

[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 = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx=B {\left (x + \frac {2 i}{e^{\left (-x\right )} - i}\right )} - \frac {2 \, A}{e^{\left (-x\right )} - i} \]

[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 = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx=B x - \frac {2 \, {\left (A - i \, B\right )}}{e^{x} + i} \]

[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 = 0.91 \[ \int \frac {A+B \sinh (x)}{i+\sinh (x)} \, dx=B\,x-\frac {2\,A-B\,2{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

[In]

int((A + B*sinh(x))/(sinh(x) + 1i),x)

[Out]

B*x - (2*A - B*2i)/(exp(x) + 1i)