\(\int \frac {(e+f x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [225]

Optimal result

Integrand size = 24, antiderivative size = 220 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {(e+f x)^2}{2 b f}-\frac {a (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {a (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}-\frac {a f \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {a f \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.86 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {x (2 e+f x)}{2 b}+\frac {a \left (d \left (2 e \text {arctanh}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-f x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )+f x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )-f \operatorname {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )+f \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{b \sqrt {a^2+b^2} d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6091, 17, 3042, 3803, 25, 2694, 27, 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.

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

\(\Big \downarrow \) 6091

\(\displaystyle \frac {\int (e+f x)dx}{b}-\frac {a \int \frac {e+f x}{a+b \sinh (c+d x)}dx}{b}\)

\(\Big \downarrow \) 17

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3803

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && 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 3042

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

rule 3803

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])* 
(f_.)*(x_)]), x_Symbol] :> Simp[2   Int[(c + d*x)^m*(E^((-I)*e + f*fz*x)/(( 
-I)*b + 2*a*E^((-I)*e + f*fz*x) + I*b*E^(2*((-I)*e + f*fz*x)))), x], x] /; 
FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 6091

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(439\) vs. \(2(198)=396\).

Time = 0.22 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.00

method	result	size
risch	\(\frac {f \,x^{2}}{2 b}+\frac {e x}{b}+\frac {2 a e \,\operatorname {arctanh}\left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d b \sqrt {a^{2}+b^{2}}}-\frac {a f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d b \sqrt {a^{2}+b^{2}}}+\frac {a f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d b \sqrt {a^{2}+b^{2}}}-\frac {a f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} b \sqrt {a^{2}+b^{2}}}+\frac {a f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} b \sqrt {a^{2}+b^{2}}}-\frac {a f \operatorname {dilog}\left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {a^{2}+b^{2}}}+\frac {a f \operatorname {dilog}\left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {a^{2}+b^{2}}}-\frac {2 a f c \,\operatorname {arctanh}\left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {a^{2}+b^{2}}}\)	\(440\)

Input:

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

Output:

1/2/b*f*x^2+1/b*e*x+2/d/b*a*e/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+ 
2*a)/(a^2+b^2)^(1/2))-1/d/b*a*f/(a^2+b^2)^(1/2)*ln((-b*exp(d*x+c)+(a^2+b^2 
)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*x+1/d/b*a*f/(a^2+b^2)^(1/2)*ln((b*exp(d*x 
+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*x-1/d^2/b*a*f/(a^2+b^2)^(1/2)* 
ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*c+1/d^2/b*a*f/( 
a^2+b^2)^(1/2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*c- 
1/d^2/b*a*f/(a^2+b^2)^(1/2)*dilog((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a 
^2+b^2)^(1/2)))+1/d^2/b*a*f/(a^2+b^2)^(1/2)*dilog((b*exp(d*x+c)+(a^2+b^2)^ 
(1/2)+a)/(a+(a^2+b^2)^(1/2)))-2/d^2/b*a*f*c/(a^2+b^2)^(1/2)*arctanh(1/2*(2 
*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 500 vs. \(2 (196) = 392\).

Time = 0.10 (sec) , antiderivative size = 500, normalized size of antiderivative = 2.27 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\left (a^{2} + b^{2}\right )} d^{2} f x^{2} + 2 \, {\left (a^{2} + b^{2}\right )} d^{2} e x - 2 \, a b f \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + 2 \, a b f \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + 2 \, {\left (a b d e - a b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - 2 \, {\left (a b d e - a b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - 2 \, {\left (a b d f x + a b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + 2 \, {\left (a b d f x + a b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right )}{2 \, {\left (a^{2} b + b^{3}\right )} d^{2}} \] Input:

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

Output:

1/2*((a^2 + b^2)*d^2*f*x^2 + 2*(a^2 + b^2)*d^2*e*x - 2*a*b*f*sqrt((a^2 + b 
^2)/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*s 
inh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + 2*a*b*f*sqrt((a^2 + b^2) 
/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh 
(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + 2*(a*b*d*e - a*b*c*f)*sqrt( 
(a^2 + b^2)/b^2)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 
 + b^2)/b^2) + 2*a) - 2*(a*b*d*e - a*b*c*f)*sqrt((a^2 + b^2)/b^2)*log(2*b* 
cosh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) - 2*( 
a*b*d*f*x + a*b*c*f)*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(d*x + c) + a*sinh( 
d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/ 
b) + 2*(a*b*d*f*x + a*b*c*f)*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(d*x + c) + 
 a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^ 
2) - b)/b))/((a^2*b + b^3)*d^2)
 

Sympy [F]

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

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

Output:

Integral((e + f*x)*sinh(c + d*x)/(a + b*sinh(c + d*x)), x)
 

Maxima [F]

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

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

Output:

-1/2*(4*a*integrate(x*e^(d*x + c)/(b^2*e^(2*d*x + 2*c) + 2*a*b*e^(d*x + c) 
 - b^2), x) - x^2/b)*f - e*(a*log((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/( 
b*e^(-d*x - c) - a + sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b*d) - (d*x + c)/( 
b*d))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e+f x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-4 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a e i -4 e^{c} \left (\int \frac {e^{d x} x}{e^{2 d x +2 c} b +2 e^{d x +c} a -b}d x \right ) a^{3} d f -4 e^{c} \left (\int \frac {e^{d x} x}{e^{2 d x +2 c} b +2 e^{d x +c} a -b}d x \right ) a \,b^{2} d f +2 a^{2} d e x +a^{2} d f \,x^{2}+2 b^{2} d e x +b^{2} d f \,x^{2}}{2 b d \left (a^{2}+b^{2}\right )} \] Input:

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

Output:

( - 4*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*a 
*e*i - 4*e**c*int((e**(d*x)*x)/(e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b) 
,x)*a**3*d*f - 4*e**c*int((e**(d*x)*x)/(e**(2*c + 2*d*x)*b + 2*e**(c + d*x 
)*a - b),x)*a*b**2*d*f + 2*a**2*d*e*x + a**2*d*f*x**2 + 2*b**2*d*e*x + b** 
2*d*f*x**2)/(2*b*d*(a**2 + b**2))