\(\int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx\) [115]

Optimal result

Integrand size = 21, antiderivative size = 158 \[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=-\frac {2 d \log \left (\cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )\right )}{3 a^2 f^2}+\frac {d \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f^2}+\frac {(c+d x) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f} \] Output:

-2/3*d*ln(cosh(1/2*e+1/4*I*Pi+1/2*f*x))/a^2/f^2+1/6*d*sech(1/2*e+1/4*I*Pi+ 
1/2*f*x)^2/a^2/f^2+1/3*(d*x+c)*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/a^2/f+1/6*(d*x 
+c)*sech(1/2*e+1/4*I*Pi+1/2*f*x)^2*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/a^2/f
 

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.53 \[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=\frac {\left (-i \cosh \left (\frac {1}{2} (e+f x)\right )+\sinh \left (\frac {1}{2} (e+f x)\right )\right ) \left (d \cosh \left (\frac {1}{2} (e+f x)\right ) \left (-2 i+3 e+3 f x-6 \arctan \left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )+3 i \log (\cosh (e+f x))\right )+\cosh \left (\frac {3}{2} (e+f x)\right ) \left (-d e+2 c f+d f x+2 d \arctan \left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )-i d \log (\cosh (e+f x))\right )+2 i \left (-i d+2 d e-3 c f-d f x-4 d \arctan \left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )+d \cosh (e+f x) \left (e+f x-2 \arctan \left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )+i \log (\cosh (e+f x))\right )+2 i d \log (\cosh (e+f x))\right ) \sinh \left (\frac {1}{2} (e+f x)\right )\right )}{6 a^2 f^2 (-i+\sinh (e+f x))^2} \] Input:

Integrate[(c + d*x)/(a + I*a*Sinh[e + f*x])^2,x]
 

Output:

(((-I)*Cosh[(e + f*x)/2] + Sinh[(e + f*x)/2])*(d*Cosh[(e + f*x)/2]*(-2*I + 
 3*e + 3*f*x - 6*ArcTan[Tanh[(e + f*x)/2]] + (3*I)*Log[Cosh[e + f*x]]) + C 
osh[(3*(e + f*x))/2]*(-(d*e) + 2*c*f + d*f*x + 2*d*ArcTan[Tanh[(e + f*x)/2 
]] - I*d*Log[Cosh[e + f*x]]) + (2*I)*((-I)*d + 2*d*e - 3*c*f - d*f*x - 4*d 
*ArcTan[Tanh[(e + f*x)/2]] + d*Cosh[e + f*x]*(e + f*x - 2*ArcTan[Tanh[(e + 
 f*x)/2]] + I*Log[Cosh[e + f*x]]) + (2*I)*d*Log[Cosh[e + f*x]])*Sinh[(e + 
f*x)/2]))/(6*a^2*f^2*(-I + Sinh[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3799, 3042, 4673, 3042, 4672, 26, 3042, 26, 3956}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3799

\(\displaystyle \frac {\int (c+d x) \text {csch}^4\left (\frac {e}{2}+\frac {f x}{2}-\frac {i \pi }{4}\right )dx}{4 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (c+d x) \csc \left (\frac {i e}{2}+\frac {i f x}{2}+\frac {\pi }{4}\right )^4dx}{4 a^2}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\frac {2}{3} \int (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )dx+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f}+\frac {2 d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f^2}}{4 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2}{3} \int (c+d x) \csc \left (\frac {i e}{2}+\frac {i f x}{2}+\frac {\pi }{4}\right )^2dx+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f}+\frac {2 d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f^2}}{4 a^2}\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {\frac {2}{3} \left (\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f}-\frac {2 i d \int -i \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )dx}{f}\right )+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f}+\frac {2 d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f^2}}{4 a^2}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {2}{3} \left (\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f}-\frac {2 d \int \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )dx}{f}\right )+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f}+\frac {2 d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f^2}}{4 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2}{3} \left (\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f}-\frac {2 d \int -i \tan \left (\frac {i e}{2}+\frac {i f x}{2}-\frac {\pi }{4}\right )dx}{f}\right )+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f}+\frac {2 d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f^2}}{4 a^2}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {2}{3} \left (\frac {2 i d \int \tan \left (\frac {i e}{2}+\frac {i f x}{2}-\frac {\pi }{4}\right )dx}{f}+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f}\right )+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f}+\frac {2 d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f^2}}{4 a^2}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {2}{3} \left (\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f}-\frac {4 d \log \left (\cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )\right )}{f^2}\right )+\frac {2 (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f}+\frac {2 d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 f^2}}{4 a^2}\)

Input:

Int[(c + d*x)/(a + I*a*Sinh[e + f*x])^2,x]
 

Output:

((2*d*Sech[e/2 + (I/4)*Pi + (f*x)/2]^2)/(3*f^2) + (2*(c + d*x)*Sech[e/2 + 
(I/4)*Pi + (f*x)/2]^2*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(3*f) + (2*((-4*d*Lo 
g[Cosh[e/2 + (I/4)*Pi + (f*x)/2]])/f^2 + (2*(c + d*x)*Tanh[e/2 + (I/4)*Pi 
+ (f*x)/2])/f))/3)/(4*a^2)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3799

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

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 4673

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> 
 Simp[(-b^2)*(c + d*x)*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), 
 x] + (-Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x] + S 
imp[b^2*((n - 2)/(n - 1))   Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2), x], x]) 
 /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2]
 

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {2 d x}{3 f \,a^{2}}+\frac {2 d e}{3 f^{2} a^{2}}-\frac {2 i \left (3 i f d x \,{\mathrm e}^{f x +e}+3 i f c \,{\mathrm e}^{f x +e}-i d \,{\mathrm e}^{f x +e}+d x f +{\mathrm e}^{2 f x +2 e} d +c f \right )}{3 \left ({\mathrm e}^{f x +e}-i\right )^{3} f^{2} a^{2}}-\frac {2 d \ln \left ({\mathrm e}^{f x +e}-i\right )}{3 f^{2} a^{2}}\)	\(113\)
parallelrisch	\(\frac {18 \left (i \cosh \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {i \cosh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )}{3}-\frac {\sinh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )}{3}-\sinh \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d \ln \left (1-\tanh \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-18 \left (i \cosh \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {i \cosh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )}{3}-\frac {\sinh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )}{3}-\sinh \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d \ln \left (-i+\tanh \left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\left (3 i+\left (-9+9 i\right ) x f \right ) d -9 c f \right ) \cosh \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (\left (i+\left (-3-3 i\right ) x f \right ) d -3 c f \right ) \cosh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (\left (-3+\left (-9+9 i\right ) x f \right ) d +9 i c f \right ) \sinh \left (\frac {f x}{2}+\frac {e}{2}\right )-3 \left (\left (-\frac {1}{3}+\left (1+i\right ) x f \right ) d +i c f \right ) \sinh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )}{9 f^{2} a^{2} \left (-\sinh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )-3 \sinh \left (\frac {f x}{2}+\frac {e}{2}\right )-i \cosh \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+3 i \cosh \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}\)	\(291\)

Input:

int((d*x+c)/(a+I*a*sinh(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

2/3*d/f/a^2*x+2/3*d/f^2/a^2*e-2/3*I*(3*I*f*d*x*exp(f*x+e)+3*I*f*c*exp(f*x+ 
e)-I*d*exp(f*x+e)+d*x*f+exp(2*f*x+2*e)*d+c*f)/(exp(f*x+e)-I)^3/f^2/a^2-2/3 
*d/f^2/a^2*ln(exp(f*x+e)-I)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=\frac {2 \, {\left (d f x e^{\left (3 \, f x + 3 \, e\right )} - i \, c f - {\left (3 i \, d f x + i \, d\right )} e^{\left (2 \, f x + 2 \, e\right )} + {\left (3 \, c f - d\right )} e^{\left (f x + e\right )} - {\left (d e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, d e^{\left (2 \, f x + 2 \, e\right )} - 3 \, d e^{\left (f x + e\right )} + i \, d\right )} \log \left (e^{\left (f x + e\right )} - i\right )\right )}}{3 \, {\left (a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, a^{2} f^{2} e^{\left (f x + e\right )} + i \, a^{2} f^{2}\right )}} \] Input:

integrate((d*x+c)/(a+I*a*sinh(f*x+e))^2,x, algorithm="fricas")
 

Output:

2/3*(d*f*x*e^(3*f*x + 3*e) - I*c*f - (3*I*d*f*x + I*d)*e^(2*f*x + 2*e) + ( 
3*c*f - d)*e^(f*x + e) - (d*e^(3*f*x + 3*e) - 3*I*d*e^(2*f*x + 2*e) - 3*d* 
e^(f*x + e) + I*d)*log(e^(f*x + e) - I))/(a^2*f^2*e^(3*f*x + 3*e) - 3*I*a^ 
2*f^2*e^(2*f*x + 2*e) - 3*a^2*f^2*e^(f*x + e) + I*a^2*f^2)
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=\frac {- 2 i c f - 2 i d f x - 2 i d e^{2 e} e^{2 f x} + \left (6 c f e^{e} + 6 d f x e^{e} - 2 d e^{e}\right ) e^{f x}}{3 a^{2} f^{2} e^{3 e} e^{3 f x} - 9 i a^{2} f^{2} e^{2 e} e^{2 f x} - 9 a^{2} f^{2} e^{e} e^{f x} + 3 i a^{2} f^{2}} + \frac {2 d x}{3 a^{2} f} - \frac {2 d \log {\left (e^{f x} - i e^{- e} \right )}}{3 a^{2} f^{2}} \] Input:

integrate((d*x+c)/(a+I*a*sinh(f*x+e))**2,x)
 

Output:

(-2*I*c*f - 2*I*d*f*x - 2*I*d*exp(2*e)*exp(2*f*x) + (6*c*f*exp(e) + 6*d*f* 
x*exp(e) - 2*d*exp(e))*exp(f*x))/(3*a**2*f**2*exp(3*e)*exp(3*f*x) - 9*I*a* 
*2*f**2*exp(2*e)*exp(2*f*x) - 9*a**2*f**2*exp(e)*exp(f*x) + 3*I*a**2*f**2) 
 + 2*d*x/(3*a**2*f) - 2*d*log(exp(f*x) - I*exp(-e))/(3*a**2*f**2)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (110) = 220\).

Time = 0.07 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.61 \[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=\frac {2}{3} \, d {\left (\frac {f x e^{\left (3 \, f x + 3 \, e\right )} - {\left (3 i \, f x e^{\left (2 \, e\right )} + i \, e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )} - e^{\left (f x + e\right )}}{a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, a^{2} f^{2} e^{\left (f x + e\right )} + i \, a^{2} f^{2}} - \frac {\log \left (-i \, {\left (i \, e^{\left (f x + e\right )} + 1\right )} e^{\left (-e\right )}\right )}{a^{2} f^{2}}\right )} + \frac {2}{3} \, c {\left (\frac {3 \, e^{\left (-f x - e\right )}}{{\left (3 \, a^{2} e^{\left (-f x - e\right )} - 3 i \, a^{2} e^{\left (-2 \, f x - 2 \, e\right )} - a^{2} e^{\left (-3 \, f x - 3 \, e\right )} + i \, a^{2}\right )} f} + \frac {i}{{\left (3 \, a^{2} e^{\left (-f x - e\right )} - 3 i \, a^{2} e^{\left (-2 \, f x - 2 \, e\right )} - a^{2} e^{\left (-3 \, f x - 3 \, e\right )} + i \, a^{2}\right )} f}\right )} \] Input:

integrate((d*x+c)/(a+I*a*sinh(f*x+e))^2,x, algorithm="maxima")
 

Output:

2/3*d*((f*x*e^(3*f*x + 3*e) - (3*I*f*x*e^(2*e) + I*e^(2*e))*e^(2*f*x) - e^ 
(f*x + e))/(a^2*f^2*e^(3*f*x + 3*e) - 3*I*a^2*f^2*e^(2*f*x + 2*e) - 3*a^2* 
f^2*e^(f*x + e) + I*a^2*f^2) - log(-I*(I*e^(f*x + e) + 1)*e^(-e))/(a^2*f^2 
)) + 2/3*c*(3*e^(-f*x - e)/((3*a^2*e^(-f*x - e) - 3*I*a^2*e^(-2*f*x - 2*e) 
 - a^2*e^(-3*f*x - 3*e) + I*a^2)*f) + I/((3*a^2*e^(-f*x - e) - 3*I*a^2*e^( 
-2*f*x - 2*e) - a^2*e^(-3*f*x - 3*e) + I*a^2)*f))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.23 \[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=\frac {2 \, {\left (d f x e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, d f x e^{\left (2 \, f x + 2 \, e\right )} + 3 \, c f e^{\left (f x + e\right )} - d e^{\left (3 \, f x + 3 \, e\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + 3 i \, d e^{\left (2 \, f x + 2 \, e\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + 3 \, d e^{\left (f x + e\right )} \log \left (e^{\left (f x + e\right )} - i\right ) - i \, c f - i \, d e^{\left (2 \, f x + 2 \, e\right )} - d e^{\left (f x + e\right )} - i \, d \log \left (e^{\left (f x + e\right )} - i\right )\right )}}{3 \, {\left (a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, a^{2} f^{2} e^{\left (f x + e\right )} + i \, a^{2} f^{2}\right )}} \] Input:

integrate((d*x+c)/(a+I*a*sinh(f*x+e))^2,x, algorithm="giac")
 

Output:

2/3*(d*f*x*e^(3*f*x + 3*e) - 3*I*d*f*x*e^(2*f*x + 2*e) + 3*c*f*e^(f*x + e) 
 - d*e^(3*f*x + 3*e)*log(e^(f*x + e) - I) + 3*I*d*e^(2*f*x + 2*e)*log(e^(f 
*x + e) - I) + 3*d*e^(f*x + e)*log(e^(f*x + e) - I) - I*c*f - I*d*e^(2*f*x 
 + 2*e) - d*e^(f*x + e) - I*d*log(e^(f*x + e) - I))/(a^2*f^2*e^(3*f*x + 3* 
e) - 3*I*a^2*f^2*e^(2*f*x + 2*e) - 3*a^2*f^2*e^(f*x + e) + I*a^2*f^2)
 

Mupad [B] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=-\frac {\frac {2\,d\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )}{3}+{\mathrm {e}}^{e+f\,x}\,\left (-\frac {d\,2{}\mathrm {i}}{3}+c\,f\,2{}\mathrm {i}+d\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )\,2{}\mathrm {i}\right )+\frac {2\,d\,{\mathrm {e}}^{2\,e+2\,f\,x}}{3}+f\,\left (\frac {2\,c}{3}+2\,d\,x\,{\mathrm {e}}^{2\,e+2\,f\,x}+\frac {d\,x\,{\mathrm {e}}^{3\,e+3\,f\,x}\,2{}\mathrm {i}}{3}\right )-2\,d\,{\mathrm {e}}^{2\,e+2\,f\,x}\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )-\frac {d\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )\,2{}\mathrm {i}}{3}}{a^2\,f^2\,{\left (1+{\mathrm {e}}^{e+f\,x}\,1{}\mathrm {i}\right )}^3} \] Input:

int((c + d*x)/(a + a*sinh(e + f*x)*1i)^2,x)
 

Output:

-((2*d*log(exp(f*x)*exp(e) - 1i))/3 + exp(e + f*x)*(c*f*2i - (d*2i)/3 + d* 
log(exp(f*x)*exp(e) - 1i)*2i) + (2*d*exp(2*e + 2*f*x))/3 + f*((2*c)/3 + 2* 
d*x*exp(2*e + 2*f*x) + (d*x*exp(3*e + 3*f*x)*2i)/3) - 2*d*exp(2*e + 2*f*x) 
*log(exp(f*x)*exp(e) - 1i) - (d*exp(3*e + 3*f*x)*log(exp(f*x)*exp(e) - 1i) 
*2i)/3)/(a^2*f^2*(exp(e + f*x)*1i + 1)^3)
 

Reduce [F]

\[ \int \frac {c+d x}{(a+i a \sinh (e+f x))^2} \, dx=\frac {-\left (\int \frac {x}{\sinh \left (f x +e \right )^{2}-2 \sinh \left (f x +e \right ) i -1}d x \right ) d -\left (\int \frac {1}{\sinh \left (f x +e \right )^{2}-2 \sinh \left (f x +e \right ) i -1}d x \right ) c}{a^{2}} \] Input:

int((d*x+c)/(a+I*a*sinh(f*x+e))^2,x)
 

Output:

( - (int(x/(sinh(e + f*x)**2 - 2*sinh(e + f*x)*i - 1),x)*d + int(1/(sinh(e 
 + f*x)**2 - 2*sinh(e + f*x)*i - 1),x)*c))/a**2