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

Optimal. Leaf size=205 \[ -\frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2}+\frac{f \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{2 a d^2}-\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d}-\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d}+\frac{(e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.380062, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5569, 3716, 2190, 2279, 2391, 5561} \[ -\frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2}+\frac{f \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{2 a d^2}-\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d}-\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d}+\frac{(e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5569

Int[(Coth[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Coth[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Cosh[c + d*x]*Coth[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]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391

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 5561

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

Rubi steps

\begin{align*} \int \frac{(e+f x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x) \coth (c+d x) \, dx}{a}-\frac{b \int \frac{(e+f x) \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=-\frac{2 \int \frac{e^{2 (c+d x)} (e+f x)}{1-e^{2 (c+d x)}} \, dx}{a}-\frac{b \int \frac{e^{c+d x} (e+f x)}{a-\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a}-\frac{b \int \frac{e^{c+d x} (e+f x)}{a+\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a}\\ &=-\frac{(e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d}-\frac{(e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d}+\frac{(e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d}+\frac{f \int \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{a d}+\frac{f \int \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{a d}-\frac{f \int \log \left (1-e^{2 (c+d x)}\right ) \, dx}{a d}\\ &=-\frac{(e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d}-\frac{(e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d}+\frac{(e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d}-\frac{f \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 a d^2}+\frac{f \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{b x}{a-\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a d^2}+\frac{f \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{b x}{a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a d^2}\\ &=-\frac{(e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d}-\frac{(e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d}+\frac{(e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d}-\frac{f \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{f \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d^2}+\frac{f \text{Li}_2\left (e^{2 (c+d x)}\right )}{2 a d^2}\\ \end{align*}

Mathematica [A]  time = 0.945634, size = 236, normalized size = 1.15 \[ -\frac{f \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )+f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+\frac{1}{2} f \text{PolyLog}\left (2,e^{-2 (c+d x)}\right )+f (c+d x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )+d e \log (a+b \sinh (c+d x))-c f \log (a+b \sinh (c+d x))-d e \log (\sinh (c+d x))-f (c+d x)^2-f (c+d x) \log \left (1-e^{-2 (c+d x)}\right )+c f \log (\sinh (c+d x))}{a d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.134, size = 451, normalized size = 2.2 \begin{align*}{\frac{e\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{da}}-{\frac{e\ln \left ( b{{\rm e}^{2\,dx+2\,c}}+2\,a{{\rm e}^{dx+c}}-b \right ) }{da}}+{\frac{e\ln \left ({{\rm e}^{dx+c}}+1 \right ) }{da}}-{\frac{f{\it dilog} \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-{\frac{f}{a{d}^{2}}{\it dilog} \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{f}{a{d}^{2}}{\it dilog} \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{f{\it dilog} \left ({{\rm e}^{dx+c}}+1 \right ) }{a{d}^{2}}}-{\frac{fx}{da}\ln \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{cf}{a{d}^{2}}\ln \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{fx}{da}\ln \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{cf}{a{d}^{2}}\ln \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{f\ln \left ({{\rm e}^{dx+c}}+1 \right ) x}{da}}-{\frac{fc\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{a{d}^{2}}}+{\frac{cf\ln \left ( b{{\rm e}^{2\,dx+2\,c}}+2\,a{{\rm e}^{dx+c}}-b \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -e{\left (\frac{\log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a d} - \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d}\right )} + f \int \frac{2 \, x{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{{\left (b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.48589, size = 1243, normalized size = 6.06 \begin{align*} -\frac{f{\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 ) + f{\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 ) - f{\rm Li}_2\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - f{\rm Li}_2\left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )\right ) +{\left (d e - c f\right )} \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 ) +{\left (d e - c f\right )} \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 ) +{\left (d f x + c f\right )} \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 ) +{\left (d f x + c f\right )} \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 ) -{\left (d f x + d e\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) -{\left (d e - c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) -{\left (d f x + c f\right )} \log \left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right ) + 1\right )}{a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right ) \coth{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \coth \left (d x + c\right )}{b \sinh \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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