3.461 \(\int \frac{(e+f x) \cosh (c+d x) \coth ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=324 \[ \frac{f \left (a^2+b^2\right ) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a^2 b d^2}+\frac{f \left (a^2+b^2\right ) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a^2 b d^2}-\frac{b f \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{2 a^2 d^2}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a^2 b d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a^2 b d}-\frac{\left (a^2+b^2\right ) (e+f x)^2}{2 a^2 b f}-\frac{b (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a^2 d}+\frac{b (e+f x)^2}{2 a^2 f}-\frac{f \tanh ^{-1}(\cosh (c+d x))}{a d^2}-\frac{(e+f x) \text{csch}(c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.742632, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 28, number of rules used = 15, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.469, Rules used = {5585, 5450, 3296, 2638, 5452, 3770, 5446, 2635, 8, 3716, 2190, 2279, 2391, 5565, 5561} \[ \frac{f \left (a^2+b^2\right ) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a^2 b d^2}+\frac{f \left (a^2+b^2\right ) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a^2 b d^2}-\frac{b f \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{2 a^2 d^2}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a^2 b d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a^2 b d}-\frac{\left (a^2+b^2\right ) (e+f x)^2}{2 a^2 b f}-\frac{b (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a^2 d}+\frac{b (e+f x)^2}{2 a^2 f}-\frac{f \tanh ^{-1}(\cosh (c+d x))}{a d^2}-\frac{(e+f x) \text{csch}(c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5585

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

Rule 5450

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*Coth[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int
[(c + d*x)^m*Cosh[a + b*x]^n*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cosh[a + b*x]^(n - 2)*Coth[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && 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 5452

Int[Coth[(a_.) + (b_.)*(x_)]^(p_.)*Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Csch[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Csch[a + b*x]^n, x], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 3770

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

Rule 5446

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 5565

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

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) \cosh (c+d x) \coth ^2(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x) \cosh (c+d x) \coth ^2(c+d x) \, dx}{a}-\frac{b \int \frac{(e+f x) \cosh ^2(c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=\frac{\int (e+f x) \cosh (c+d x) \, dx}{a}+\frac{\int (e+f x) \coth (c+d x) \text{csch}(c+d x) \, dx}{a}-\frac{b \int (e+f x) \cosh ^2(c+d x) \coth (c+d x) \, dx}{a^2}+\frac{b^2 \int \frac{(e+f x) \cosh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2}\\ &=-\frac{(e+f x) \text{csch}(c+d x)}{a d}+\frac{(e+f x) \sinh (c+d x)}{a d}-\frac{\int (e+f x) \cosh (c+d x) \, dx}{a}-\frac{b \int (e+f x) \coth (c+d x) \, dx}{a^2}+\frac{\left (a^2+b^2\right ) \int \frac{(e+f x) \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2}+\frac{f \int \text{csch}(c+d x) \, dx}{a d}-\frac{f \int \sinh (c+d x) \, dx}{a d}\\ &=\frac{b (e+f x)^2}{2 a^2 f}-\frac{\left (a^2+b^2\right ) (e+f x)^2}{2 a^2 b f}-\frac{f \tanh ^{-1}(\cosh (c+d x))}{a d^2}-\frac{f \cosh (c+d x)}{a d^2}-\frac{(e+f x) \text{csch}(c+d x)}{a d}+\frac{(2 b) \int \frac{e^{2 (c+d x)} (e+f x)}{1-e^{2 (c+d x)}} \, dx}{a^2}+\frac{\left (a^2+b^2\right ) \int \frac{e^{c+d x} (e+f x)}{a-\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a^2}+\frac{\left (a^2+b^2\right ) \int \frac{e^{c+d x} (e+f x)}{a+\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a^2}+\frac{f \int \sinh (c+d x) \, dx}{a d}\\ &=\frac{b (e+f x)^2}{2 a^2 f}-\frac{\left (a^2+b^2\right ) (e+f x)^2}{2 a^2 b f}-\frac{f \tanh ^{-1}(\cosh (c+d x))}{a d^2}-\frac{(e+f x) \text{csch}(c+d x)}{a d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a^2 b d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a^2 b d}-\frac{b (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a^2 d}+\frac{(b f) \int \log \left (1-e^{2 (c+d x)}\right ) \, dx}{a^2 d}-\frac{\left (\left (a^2+b^2\right ) f\right ) \int \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{a^2 b d}-\frac{\left (\left (a^2+b^2\right ) f\right ) \int \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{a^2 b d}\\ &=\frac{b (e+f x)^2}{2 a^2 f}-\frac{\left (a^2+b^2\right ) (e+f x)^2}{2 a^2 b f}-\frac{f \tanh ^{-1}(\cosh (c+d x))}{a d^2}-\frac{(e+f x) \text{csch}(c+d x)}{a d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a^2 b d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a^2 b d}-\frac{b (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a^2 d}+\frac{(b f) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 a^2 d^2}-\frac{\left (\left (a^2+b^2\right ) f\right ) \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^2 b d^2}-\frac{\left (\left (a^2+b^2\right ) f\right ) \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^2 b d^2}\\ &=\frac{b (e+f x)^2}{2 a^2 f}-\frac{\left (a^2+b^2\right ) (e+f x)^2}{2 a^2 b f}-\frac{f \tanh ^{-1}(\cosh (c+d x))}{a d^2}-\frac{(e+f x) \text{csch}(c+d x)}{a d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a^2 b d}+\frac{\left (a^2+b^2\right ) (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a^2 b d}-\frac{b (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a^2 d}+\frac{\left (a^2+b^2\right ) f \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a^2 b d^2}+\frac{\left (a^2+b^2\right ) f \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a^2 b d^2}-\frac{b f \text{Li}_2\left (e^{2 (c+d x)}\right )}{2 a^2 d^2}\\ \end{align*}

Mathematica [A]  time = 2.5848, size = 313, normalized size = 0.97 \[ \frac{\frac{2 \left (a^2+b^2\right ) \left (f \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )+f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+f (c+d x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )+d e \log (a+b \sinh (c+d x))-c f \log (a+b \sinh (c+d x))-\frac{1}{2} f (c+d x)^2\right )}{b}+b f \left (\text{PolyLog}\left (2,e^{-2 (c+d x)}\right )-(c+d x) \left (2 \log \left (1-e^{-2 (c+d x)}\right )+c+d x\right )\right )+a d (e+f x) \tanh \left (\frac{1}{2} (c+d x)\right )-a d (e+f x) \coth \left (\frac{1}{2} (c+d x)\right )+2 a f \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )-2 b d e \log (\sinh (c+d x))+2 b c f \log (\sinh (c+d x))}{2 a^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.23, size = 938, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.47833, size = 4255, normalized size = 13.13 \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)*coth(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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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 )} \coth ^{2}{\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)*cosh(d*x+c)*coth(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Integral((e + f*x)*cosh(c + d*x)*coth(c + d*x)**2/(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 )} \cosh \left (d x + c\right ) \coth \left (d x + c\right )^{2}}{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)*cosh(d*x+c)*coth(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*cosh(d*x + c)*coth(d*x + c)^2/(b*sinh(d*x + c) + a), x)