3.19 \(\int \frac{(e+f x) \cosh (c+d x)}{a+b \text{csch}(c+d x)} \, dx\)

Optimal. Leaf size=212 \[ -\frac{b f \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}\right )}{a^2 d^2}-\frac{b f \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}\right )}{a^2 d^2}-\frac{b (e+f x) \log \left (\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}+1\right )}{a^2 d}-\frac{b (e+f x) \log \left (\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}+1\right )}{a^2 d}+\frac{b (e+f x)^2}{2 a^2 f}-\frac{f \cosh (c+d x)}{a d^2}+\frac{(e+f x) \sinh (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.351254, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5594, 5579, 3296, 2638, 5561, 2190, 2279, 2391} \[ -\frac{b f \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}\right )}{a^2 d^2}-\frac{b f \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}\right )}{a^2 d^2}-\frac{b (e+f x) \log \left (\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}+1\right )}{a^2 d}-\frac{b (e+f x) \log \left (\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}+1\right )}{a^2 d}+\frac{b (e+f x)^2}{2 a^2 f}-\frac{f \cosh (c+d x)}{a d^2}+\frac{(e+f x) \sinh (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5594

Int[(((e_.) + (f_.)*(x_))^(m_.)*(F_)[(c_.) + (d_.)*(x_)]^(n_.))/(Csch[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Sym
bol] :> Int[((e + f*x)^m*Sinh[c + d*x]*F[c + d*x]^n)/(b + a*Sinh[c + d*x]), x] /; FreeQ[{a, b, c, d, e, f}, x]
&& HyperbolicQ[F] && IntegersQ[m, n]

Rule 5579

Int[(Cosh[(c_.) + (d_.)*(x_)]^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*S
inh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[1/b, Int[(e + f*x)^m*Cosh[c + d*x]^p*Sinh[c + d*x]^(n - 1), x], x]
- Dist[a/b, Int[((e + f*x)^m*Cosh[c + d*x]^p*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] && IGtQ[p, 0]

Rule 3296

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

Rule 2638

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

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]

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]

Rubi steps

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

Mathematica [C]  time = 1.35671, size = 435, normalized size = 2.05 \[ \frac{\text{csch}(c+d x) (a \sinh (c+d x)+b) \left (i b f \left (i \left (\text{PolyLog}\left (2,\frac{\left (b-\sqrt{a^2+b^2}\right ) e^{c+d x}}{a}\right )+\text{PolyLog}\left (2,\frac{\left (\sqrt{a^2+b^2}+b\right ) e^{c+d x}}{a}\right )\right )-\frac{1}{2} \log \left (\frac{\left (\sqrt{a^2+b^2}-b\right ) e^{c+d x}}{a}+1\right ) \left (4 \sin ^{-1}\left (\frac{\sqrt{1+\frac{i b}{a}}}{\sqrt{2}}\right )-2 i c-2 i d x+\pi \right )-\frac{1}{2} \log \left (1-\frac{\left (\sqrt{a^2+b^2}+b\right ) e^{c+d x}}{a}\right ) \left (-4 \sin ^{-1}\left (\frac{\sqrt{1+\frac{i b}{a}}}{\sqrt{2}}\right )-2 i c-2 i d x+\pi \right )-4 i \sin ^{-1}\left (\frac{\sqrt{1+\frac{i b}{a}}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(b+i a) \cot \left (\frac{1}{4} (2 i c+2 i d x+\pi )\right )}{\sqrt{a^2+b^2}}\right )+\left (\frac{\pi }{2}-i (c+d x)\right ) \log (a \sinh (c+d x)+b)-\frac{1}{8} i (2 c+2 d x+i \pi )^2\right )+d e (a \sinh (c+d x)-b \log (a \sinh (c+d x)+b))-b f (c+d x) \log (a \sinh (c+d x)+b)+b c f \log \left (\frac{a \sinh (c+d x)}{b}+1\right )+a d f x \sinh (c+d x)-a f \cosh (c+d x)\right )}{a^2 d^2 (a+b \text{csch}(c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csch[c + d*x]*(b + a*Sinh[c + d*x])*(-(a*f*Cosh[c + d*x]) - b*f*(c + d*x)*Log[b + a*Sinh[c + d*x]] + b*c*f*Lo
g[1 + (a*Sinh[c + d*x])/b] + I*b*f*((-I/8)*(2*c + I*Pi + 2*d*x)^2 - (4*I)*ArcSin[Sqrt[1 + (I*b)/a]/Sqrt[2]]*Ar
cTan[((I*a + b)*Cot[((2*I)*c + Pi + (2*I)*d*x)/4])/Sqrt[a^2 + b^2]] - (((-2*I)*c + Pi - (2*I)*d*x + 4*ArcSin[S
qrt[1 + (I*b)/a]/Sqrt[2]])*Log[1 + ((-b + Sqrt[a^2 + b^2])*E^(c + d*x))/a])/2 - (((-2*I)*c + Pi - (2*I)*d*x -
4*ArcSin[Sqrt[1 + (I*b)/a]/Sqrt[2]])*Log[1 - ((b + Sqrt[a^2 + b^2])*E^(c + d*x))/a])/2 + (Pi/2 - I*(c + d*x))*
Log[b + a*Sinh[c + d*x]] + I*(PolyLog[2, ((b - Sqrt[a^2 + b^2])*E^(c + d*x))/a] + PolyLog[2, ((b + Sqrt[a^2 +
b^2])*E^(c + d*x))/a])) + a*d*f*x*Sinh[c + d*x] + d*e*(-(b*Log[b + a*Sinh[c + d*x]]) + a*Sinh[c + d*x])))/(a^2
*d^2*(a + b*Csch[c + d*x]))

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Maple [B]  time = 0.159, size = 483, normalized size = 2.3 \begin{align*}{\frac{bf{x}^{2}}{2\,{a}^{2}}}-{\frac{bex}{{a}^{2}}}+{\frac{ \left ( dfx+de-f \right ){{\rm e}^{dx+c}}}{2\,a{d}^{2}}}-{\frac{ \left ( dfx+de+f \right ){{\rm e}^{-dx-c}}}{2\,a{d}^{2}}}-{\frac{be\ln \left ( a{{\rm e}^{2\,dx+2\,c}}+2\,b{{\rm e}^{dx+c}}-a \right ) }{{a}^{2}d}}+2\,{\frac{be\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{2}d}}-{\frac{bfx}{{a}^{2}d}\ln \left ({ \left ( -a{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-b \right ) \left ( -b+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{bfc}{{a}^{2}{d}^{2}}\ln \left ({ \left ( -a{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-b \right ) \left ( -b+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{bfx}{{a}^{2}d}\ln \left ({ \left ( a{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+b \right ) \left ( b+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{bfc}{{a}^{2}{d}^{2}}\ln \left ({ \left ( a{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+b \right ) \left ( b+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{bf}{{a}^{2}{d}^{2}}{\it dilog} \left ({ \left ( -a{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-b \right ) \left ( -b+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{bf}{{a}^{2}{d}^{2}}{\it dilog} \left ({ \left ( a{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+b \right ) \left ( b+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+2\,{\frac{bfcx}{{a}^{2}d}}+{\frac{bf{c}^{2}}{{a}^{2}{d}^{2}}}+{\frac{bfc\ln \left ( a{{\rm e}^{2\,dx+2\,c}}+2\,b{{\rm e}^{dx+c}}-a \right ) }{{a}^{2}{d}^{2}}}-2\,{\frac{bfc\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{2}{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, e{\left (\frac{2 \,{\left (d x + c\right )} b}{a^{2} d} - \frac{e^{\left (d x + c\right )}}{a d} + \frac{e^{\left (-d x - c\right )}}{a d} + \frac{2 \, b \log \left (-2 \, b e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )}{a^{2} d}\right )} - \frac{1}{2} \, f{\left (\frac{{\left (b d^{2} x^{2} e^{c} -{\left (a d x e^{\left (2 \, c\right )} - a e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} +{\left (a d x + a\right )} e^{\left (-d x\right )}\right )} e^{\left (-c\right )}}{a^{2} d^{2}} - \int \frac{4 \,{\left (b^{2} x e^{\left (d x + c\right )} - a b x\right )}}{a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} b e^{\left (d x + c\right )} - a^{3}}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e*(2*(d*x + c)*b/(a^2*d) - e^(d*x + c)/(a*d) + e^(-d*x - c)/(a*d) + 2*b*log(-2*b*e^(-d*x - c) + a*e^(-2*d
*x - 2*c) - a)/(a^2*d)) - 1/2*f*((b*d^2*x^2*e^c - (a*d*x*e^(2*c) - a*e^(2*c))*e^(d*x) + (a*d*x + a)*e^(-d*x))*
e^(-c)/(a^2*d^2) - integrate(4*(b^2*x*e^(d*x + c) - a*b*x)/(a^3*e^(2*d*x + 2*c) + 2*a^2*b*e^(d*x + c) - a^3),
x))

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Fricas [B]  time = 1.73429, size = 1760, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*d*f*x + a*d*e - (a*d*f*x + a*d*e - a*f)*cosh(d*x + c)^2 - (a*d*f*x + a*d*e - a*f)*sinh(d*x + c)^2 + a*
f - (b*d^2*f*x^2 + 2*b*d^2*e*x + 4*b*c*d*e - 2*b*c^2*f)*cosh(d*x + c) + 2*(b*f*cosh(d*x + c) + b*f*sinh(d*x +
c))*dilog((b*cosh(d*x + c) + b*sinh(d*x + c) + (a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt((a^2 + b^2)/a^2) - a)/
a + 1) + 2*(b*f*cosh(d*x + c) + b*f*sinh(d*x + c))*dilog((b*cosh(d*x + c) + b*sinh(d*x + c) - (a*cosh(d*x + c)
+ a*sinh(d*x + c))*sqrt((a^2 + b^2)/a^2) - a)/a + 1) + 2*((b*d*e - b*c*f)*cosh(d*x + c) + (b*d*e - b*c*f)*sin
h(d*x + c))*log(2*a*cosh(d*x + c) + 2*a*sinh(d*x + c) + 2*a*sqrt((a^2 + b^2)/a^2) + 2*b) + 2*((b*d*e - b*c*f)*
cosh(d*x + c) + (b*d*e - b*c*f)*sinh(d*x + c))*log(2*a*cosh(d*x + c) + 2*a*sinh(d*x + c) - 2*a*sqrt((a^2 + b^2
)/a^2) + 2*b) + 2*((b*d*f*x + b*c*f)*cosh(d*x + c) + (b*d*f*x + b*c*f)*sinh(d*x + c))*log(-(b*cosh(d*x + c) +
b*sinh(d*x + c) + (a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt((a^2 + b^2)/a^2) - a)/a) + 2*((b*d*f*x + b*c*f)*cos
h(d*x + c) + (b*d*f*x + b*c*f)*sinh(d*x + c))*log(-(b*cosh(d*x + c) + b*sinh(d*x + c) - (a*cosh(d*x + c) + a*s
inh(d*x + c))*sqrt((a^2 + b^2)/a^2) - a)/a) - (b*d^2*f*x^2 + 2*b*d^2*e*x + 4*b*c*d*e - 2*b*c^2*f + 2*(a*d*f*x
+ a*d*e - a*f)*cosh(d*x + c))*sinh(d*x + c))/(a^2*d^2*cosh(d*x + c) + a^2*d^2*sinh(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e + f*x)*cosh(c + d*x)/(a + b*csch(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 )} \cosh \left (d x + c\right )}{b \operatorname{csch}\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)*cosh(d*x+c)/(a+b*csch(d*x+c)),x, algorithm="giac")

[Out]

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