\(\int \frac {c+d x}{a+b \sinh (e+f x)} \, dx\) [171]

Optimal result

Integrand size = 18, antiderivative size = 187 \[ \int \frac {c+d x}{a+b \sinh (e+f x)} \, dx=\frac {(c+d x) \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f}-\frac {(c+d x) \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f}+\frac {d \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^2}-\frac {d \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x}{a+b \sinh (e+f x)} \, dx=\frac {f (c+d x) \left (\log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )-\log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )\right )+d \operatorname {PolyLog}\left (2,\frac {b e^{e+f x}}{-a+\sqrt {a^2+b^2}}\right )-d \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3803

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(392\) vs. \(2(167)=334\).

Time = 0.19 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.10

method	result	size
risch	\(-\frac {2 c \,\operatorname {arctanh}\left (\frac {2 b \,{\mathrm e}^{f x +e}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{f \sqrt {a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{f \sqrt {a^{2}+b^{2}}}-\frac {d \ln \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{f \sqrt {a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {a^{2}+b^{2}}}-\frac {d \ln \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {a^{2}+b^{2}}}+\frac {d \operatorname {dilog}\left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{f^{2} \sqrt {a^{2}+b^{2}}}-\frac {d \operatorname {dilog}\left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{f^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 d e \,\operatorname {arctanh}\left (\frac {2 b \,{\mathrm e}^{f x +e}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{f^{2} \sqrt {a^{2}+b^{2}}}\)	\(393\)

Input:

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

Output:

-2/f*c/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(f*x+e)+2*a)/(a^2+b^2)^(1/2))+1 
/f*d/(a^2+b^2)^(1/2)*ln((-b*exp(f*x+e)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1 
/2)))*x-1/f*d/(a^2+b^2)^(1/2)*ln((b*exp(f*x+e)+(a^2+b^2)^(1/2)+a)/(a+(a^2+ 
b^2)^(1/2)))*x+1/f^2*d/(a^2+b^2)^(1/2)*ln((-b*exp(f*x+e)+(a^2+b^2)^(1/2)-a 
)/(-a+(a^2+b^2)^(1/2)))*e-1/f^2*d/(a^2+b^2)^(1/2)*ln((b*exp(f*x+e)+(a^2+b^ 
2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*e+1/f^2*d/(a^2+b^2)^(1/2)*dilog((-b*exp(f 
*x+e)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))-1/f^2*d/(a^2+b^2)^(1/2)*dil 
og((b*exp(f*x+e)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))+2/f^2*d*e/(a^2+b^ 
2)^(1/2)*arctanh(1/2*(2*b*exp(f*x+e)+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. 455 vs. \(2 (165) = 330\).

Time = 0.09 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.43 \[ \int \frac {c+d x}{a+b \sinh (e+f x)} \, dx=\frac {b d \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) + {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) - b d \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) - {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + {\left (b d e - b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (f x + e\right ) + 2 \, b \sinh \left (f x + e\right ) + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - {\left (b d e - b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (f x + e\right ) + 2 \, b \sinh \left (f x + e\right ) - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) + {\left (b d f x + b d e\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) + {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) - {\left (b d f x + b d e\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) - {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right )}{{\left (a^{2} + b^{2}\right )} f^{2}} \] Input:

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

Output:

(b*d*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(f*x + e) + a*sinh(f*x + e) + (b*c 
osh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - b*d*sq 
rt((a^2 + b^2)/b^2)*dilog((a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f*x 
 + e) + b*sinh(f*x + e))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + (b*d*e - b*c* 
f)*sqrt((a^2 + b^2)/b^2)*log(2*b*cosh(f*x + e) + 2*b*sinh(f*x + e) + 2*b*s 
qrt((a^2 + b^2)/b^2) + 2*a) - (b*d*e - b*c*f)*sqrt((a^2 + b^2)/b^2)*log(2* 
b*cosh(f*x + e) + 2*b*sinh(f*x + e) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + ( 
b*d*f*x + b*d*e)*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(f*x + e) + a*sinh(f*x 
+ e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 + b^2)/b^2) - b)/b) - 
 (b*d*f*x + b*d*e)*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(f*x + e) + a*sinh(f* 
x + e) - (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 + b^2)/b^2) - b)/b) 
)/((a^2 + b^2)*f^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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