\(\int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx\) [247]

Optimal result

Integrand size = 15, antiderivative size = 25 \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=B \log (i+\sinh (x))-\frac {A \cosh (x)}{1-i \sinh (x)} \]

[Out]

B*ln(I+sinh(x))-A*cosh(x)/(1-I*sinh(x))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4486, 2727, 2746, 31} \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=B \log (\sinh (x)+i)-\frac {A \cosh (x)}{1-i \sinh (x)} \]

[In]

Int[(A + B*Cosh[x])/(I + Sinh[x]),x]

[Out]

B*Log[I + Sinh[x]] - (A*Cosh[x])/(1 - I*Sinh[x])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 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])

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=-2 i B \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+B \log (\cosh (x))-\frac {2 i A \sinh \left (\frac {x}{2}\right )}{\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )} \]

[In]

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

[Out]

(-2*I)*B*ArcTan[Tanh[x/2]] + B*Log[Cosh[x]] - ((2*I)*A*Sinh[x/2])/(Cosh[x/2] - I*Sinh[x/2])

Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
parts	\(-\frac {2 i A}{\tanh \left (\frac {x}{2}\right )+i}+B \ln \left (i+\sinh \left (x \right )\right )\)	\(23\)
risch	\(-B x -\frac {2 A}{{\mathrm e}^{x}+i}+2 B \ln \left ({\mathrm e}^{x}+i\right )\)	\(25\)
default	\(2 B \ln \left (\tanh \left (\frac {x}{2}\right )+i\right )-\frac {2 i A}{\tanh \left (\frac {x}{2}\right )+i}-B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=-\frac {B x e^{x} + i \, B x - 2 \, {\left (B e^{x} + i \, B\right )} \log \left (e^{x} + i\right ) + 2 \, A}{e^{x} + i} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=- \frac {2 A}{e^{x} + i} - B x + 2 B \log {\left (e^{x} + i \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=B \log \left (\sinh \left (x\right ) + i\right ) - \frac {2 \, A}{e^{\left (-x\right )} - i} \]

[In]

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

[Out]

B*log(sinh(x) + I) - 2*A/(e^(-x) - I)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=-B x + 2 \, B \log \left (e^{x} + i\right ) - \frac {2 \, A}{e^{x} + i} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx=-B\,x-\frac {2\,A}{{\mathrm {e}}^x+1{}\mathrm {i}}+2\,B\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right ) \]

[In]

int((A + B*cosh(x))/(sinh(x) + 1i),x)

[Out]

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