\(\int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) [255]

Optimal result

Integrand size = 27, antiderivative size = 73 \[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i (e+f x)^2}{2 a f}-\frac {2 i (e+f x) \log \left (1+i e^{c+d x}\right )}{a d}-\frac {2 i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2} \] Output:

1/2*I*(f*x+e)^2/a/f-2*I*(f*x+e)*ln(1+I*exp(d*x+c))/a/d-2*I*f*polylog(2,-I* 
exp(d*x+c))/a/d^2
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \left (d (e+f x) \left (d (e+f x)-4 f \log \left (1+i e^{c+d x}\right )\right )-4 f^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )\right )}{2 a d^2 f} \] Input:

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

Output:

((I/2)*(d*(e + f*x)*(d*(e + f*x) - 4*f*Log[1 + I*E^(c + d*x)]) - 4*f^2*Pol 
yLog[2, (-I)*E^(c + d*x)]))/(a*d^2*f)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6093, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx\)

\(\Big \downarrow \) 6093

\(\displaystyle 2 \int \frac {e^{c+d x} (e+f x)}{i e^{c+d x} a+a}dx+\frac {i (e+f x)^2}{2 a f}\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 \left (\frac {i f \int \log \left (1+i e^{c+d x}\right )dx}{a d}-\frac {i (e+f x) \log \left (1+i e^{c+d x}\right )}{a d}\right )+\frac {i (e+f x)^2}{2 a f}\)

\(\Big \downarrow \) 2715

\(\displaystyle 2 \left (\frac {i f \int e^{-c-d x} \log \left (1+i e^{c+d x}\right )de^{c+d x}}{a d^2}-\frac {i (e+f x) \log \left (1+i e^{c+d x}\right )}{a d}\right )+\frac {i (e+f x)^2}{2 a f}\)

\(\Big \downarrow \) 2838

\(\displaystyle 2 \left (-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}-\frac {i (e+f x) \log \left (1+i e^{c+d x}\right )}{a d}\right )+\frac {i (e+f x)^2}{2 a f}\)

Input:

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

Output:

((I/2)*(e + f*x)^2)/(a*f) + 2*(((-I)*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a* 
d) - (I*f*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^2))
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 6093

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin 
h[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[-(e + f*x)^(m + 1)/(b*f*(m + 1)), 
 x] + Simp[2   Int[(e + f*x)^m*(E^(c + d*x)/(a + b*E^(c + d*x))), x], x] /; 
 FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[a^2 + b^2, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (64 ) = 128\).

Time = 0.82 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.58

method	result	size
risch	\(\frac {i f \,x^{2}}{2 a}-\frac {i e x}{a}-\frac {2 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e}{d a}+\frac {2 i \ln \left ({\mathrm e}^{d x +c}\right ) e}{d a}+\frac {2 i f c x}{d a}+\frac {i f \,c^{2}}{d^{2} a}-\frac {2 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{d a}-\frac {2 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{d^{2} a}-\frac {2 i f \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {2 i c f \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{2} a}-\frac {2 i c f \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a}\)	\(188\)

Input:

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

Output:

1/2*I/a*f*x^2-I/a*e*x-2*I/d/a*ln(exp(d*x+c)-I)*e+2*I/d/a*ln(exp(d*x+c))*e+ 
2*I/d/a*f*c*x+I/d^2/a*f*c^2-2*I/d/a*f*ln(1+I*exp(d*x+c))*x-2*I/d^2/a*f*ln( 
1+I*exp(d*x+c))*c-2*I*f*polylog(2,-I*exp(d*x+c))/a/d^2+2*I/d^2/a*c*f*ln(ex 
p(d*x+c)-I)-2*I/d^2/a*c*f*ln(exp(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \, d^{2} f x^{2} + 2 i \, d^{2} e x + 4 i \, c d e - 2 i \, c^{2} f - 4 i \, f {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 4 \, {\left (i \, d e - i \, c f\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 4 \, {\left (i \, d f x + i \, c f\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{2 \, a d^{2}} \] Input:

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

Output:

1/2*(I*d^2*f*x^2 + 2*I*d^2*e*x + 4*I*c*d*e - 2*I*c^2*f - 4*I*f*dilog(-I*e^ 
(d*x + c)) - 4*(I*d*e - I*c*f)*log(e^(d*x + c) - I) - 4*(I*d*f*x + I*c*f)* 
log(I*e^(d*x + c) + 1))/(a*d^2)
 

Sympy [F]

\[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f x \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \] Input:

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

Output:

-I*(Integral(e*cosh(c + d*x)/(sinh(c + d*x) - I), x) + Integral(f*x*cosh(c 
 + d*x)/(sinh(c + d*x) - I), x))/a
 

Maxima [F]

\[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \cosh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

1/2*f*(-I*x^2/a + 4*integrate(x/(a*e^(d*x + c) - I*a), x)) - I*e*log(I*a*s 
inh(d*x + c) + a)/(a*d)
 

Giac [F]

\[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \cosh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (e+f\,x\right )}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\left (\int \frac {\cosh \left (d x +c \right ) x}{\sinh \left (d x +c \right ) i +1}d x \right ) d f -\mathrm {log}\left (\sinh \left (d x +c \right ) i +1\right ) e i}{a d} \] Input:

int((f*x+e)*cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x)
                                                                                    
                                                                                    
 

Output:

(int((cosh(c + d*x)*x)/(sinh(c + d*x)*i + 1),x)*d*f - log(sinh(c + d*x)*i 
+ 1)*e*i)/(a*d)