\(\int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) [256]

Optimal result

Integrand size = 22, antiderivative size = 23 \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \log (i-\sinh (c+d x))}{a d} \]

[Out]

-I*ln(I-sinh(d*x+c))/a/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2746, 31} \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \log (-\sinh (c+d x)+i)}{a d} \]

[In]

Int[Cosh[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

((-I)*Log[I - Sinh[c + d*x]])/(a*d)

Rule 31

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

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}& = -\frac {i \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,i a \sinh (c+d x)\right )}{a d} \\ & = -\frac {i \log (i-\sinh (c+d x))}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \log (i-\sinh (c+d x))}{a d} \]

[In]

Integrate[Cosh[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

((-I)*Log[I - Sinh[c + d*x]])/(a*d)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65

method	result	size
risch	\(\frac {i x}{a}+\frac {2 i c}{a d}-\frac {2 i \ln \left ({\mathrm e}^{d x +c}-i\right )}{a d}\)	\(38\)
derivativedivides	\(-\frac {i \ln \left (a^{2} \sinh \left (d x +c \right )^{2}+a^{2}\right )}{2 d a}+\frac {\arctan \left (\sinh \left (d x +c \right )\right )}{a d}\)	\(42\)
default	\(-\frac {i \ln \left (a^{2} \sinh \left (d x +c \right )^{2}+a^{2}\right )}{2 d a}+\frac {\arctan \left (\sinh \left (d x +c \right )\right )}{a d}\)	\(42\)

[In]

int(cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

I*x/a+2*I/a/d*c-2*I/a/d*ln(exp(d*x+c)-I)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \, d x - 2 i \, \log \left (e^{\left (d x + c\right )} - i\right )}{a d} \]

[In]

integrate(cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(I*d*x - 2*I*log(e^(d*x + c) - I))/(a*d)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i x}{a} - \frac {2 i \log {\left (e^{d x} - i e^{- c} \right )}}{a d} \]

[In]

integrate(cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

I*x/a - 2*I*log(exp(d*x) - I*exp(-c))/(a*d)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \, \log \left (i \, a \sinh \left (d x + c\right ) + a\right )}{a d} \]

[In]

integrate(cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-I*log(I*a*sinh(d*x + c) + a)/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {-\frac {i \, {\left (d x + c\right )}}{a} + \frac {2 i \, \log \left (e^{\left (d x + c\right )} - i\right )}{a}}{d} \]

[In]

integrate(cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

-(-I*(d*x + c)/a + 2*I*log(e^(d*x + c) - I)/a)/d

Mupad [B] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\ln \left (\mathrm {sinh}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d} \]

[In]

int(cosh(c + d*x)/(a + a*sinh(c + d*x)*1i),x)

[Out]

-(log(sinh(c + d*x) - 1i)*1i)/(a*d)