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

Optimal. Leaf size=325 \[ -\frac{2 f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{2 f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2}+\frac{2 f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^3}+\frac{f (e+f x) \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{a d^2}-\frac{f^2 \text{PolyLog}\left (3,e^{2 (c+d x)}\right )}{2 a d^3}-\frac{(e+f x)^2 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d}-\frac{(e+f x)^2 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d}+\frac{(e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d} \]

[Out]

-(((e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(a*d)) - ((e + f*x)^2*Log[1 + (b*E^(c + d*x))/(
a + Sqrt[a^2 + b^2])])/(a*d) + ((e + f*x)^2*Log[1 - E^(2*(c + d*x))])/(a*d) - (2*f*(e + f*x)*PolyLog[2, -((b*E
^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(a*d^2) - (2*f*(e + f*x)*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2
]))])/(a*d^2) + (f*(e + f*x)*PolyLog[2, E^(2*(c + d*x))])/(a*d^2) + (2*f^2*PolyLog[3, -((b*E^(c + d*x))/(a - S
qrt[a^2 + b^2]))])/(a*d^3) + (2*f^2*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(a*d^3) - (f^2*PolyL
og[3, E^(2*(c + d*x))])/(2*a*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.653868, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {5569, 3716, 2190, 2531, 2282, 6589, 5561} \[ -\frac{2 f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{2 f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2}+\frac{2 f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^3}+\frac{f (e+f x) \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{a d^2}-\frac{f^2 \text{PolyLog}\left (3,e^{2 (c+d x)}\right )}{2 a d^3}-\frac{(e+f x)^2 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d}-\frac{(e+f x)^2 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d}+\frac{(e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(a*d)) - ((e + f*x)^2*Log[1 + (b*E^(c + d*x))/(
a + Sqrt[a^2 + b^2])])/(a*d) + ((e + f*x)^2*Log[1 - E^(2*(c + d*x))])/(a*d) - (2*f*(e + f*x)*PolyLog[2, -((b*E
^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(a*d^2) - (2*f*(e + f*x)*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2
]))])/(a*d^2) + (f*(e + f*x)*PolyLog[2, E^(2*(c + d*x))])/(a*d^2) + (2*f^2*PolyLog[3, -((b*E^(c + d*x))/(a - S
qrt[a^2 + b^2]))])/(a*d^3) + (2*f^2*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(a*d^3) - (f^2*PolyL
og[3, E^(2*(c + d*x))])/(2*a*d^3)

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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)^2 \coth (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x)^2 \coth (c+d x) \, dx}{a}-\frac{b \int \frac{(e+f x)^2 \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=-\frac{2 \int \frac{e^{2 (c+d x)} (e+f x)^2}{1-e^{2 (c+d x)}} \, dx}{a}-\frac{b \int \frac{e^{c+d x} (e+f x)^2}{a-\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a}-\frac{b \int \frac{e^{c+d x} (e+f x)^2}{a+\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a}\\ &=-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d}-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d}+\frac{(e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d}+\frac{(2 f) \int (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{a d}+\frac{(2 f) \int (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{a d}-\frac{(2 f) \int (e+f x) \log \left (1-e^{2 (c+d x)}\right ) \, dx}{a d}\\ &=-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d}-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d}+\frac{(e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d}-\frac{2 f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{2 f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d^2}+\frac{f (e+f x) \text{Li}_2\left (e^{2 (c+d x)}\right )}{a d^2}-\frac{f^2 \int \text{Li}_2\left (e^{2 (c+d x)}\right ) \, dx}{a d^2}+\frac{\left (2 f^2\right ) \int \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{a d^2}+\frac{\left (2 f^2\right ) \int \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{a d^2}\\ &=-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d}-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d}+\frac{(e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d}-\frac{2 f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{2 f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d^2}+\frac{f (e+f x) \text{Li}_2\left (e^{2 (c+d x)}\right )}{a d^2}-\frac{f^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 a d^3}+\frac{\left (2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{b x}{-a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a d^3}+\frac{\left (2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a d^3}\\ &=-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d}-\frac{(e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d}+\frac{(e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d}-\frac{2 f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{2 f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d^2}+\frac{f (e+f x) \text{Li}_2\left (e^{2 (c+d x)}\right )}{a d^2}+\frac{2 f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^3}+\frac{2 f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{a d^3}-\frac{f^2 \text{Li}_3\left (e^{2 (c+d x)}\right )}{2 a d^3}\\ \end{align*}

Mathematica [B]  time = 14.1565, size = 1296, normalized size = 3.99 \[ \frac{2 e^{2 c} f^2 x^3+6 e e^{2 c} f x^2-\frac{3 e^{2 c} f^2 \log \left (\frac{e^{2 c+d x} b}{a e^c-\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x^2}{d}+\frac{3 f^2 \log \left (\frac{e^{2 c+d x} b}{a e^c-\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x^2}{d}-\frac{3 e^{2 c} f^2 \log \left (\frac{e^{2 c+d x} b}{e^c a+\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x^2}{d}+\frac{3 f^2 \log \left (\frac{e^{2 c+d x} b}{e^c a+\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x^2}{d}+6 e^2 e^{2 c} x-\frac{6 e e^{2 c} f \log \left (\frac{e^{2 c+d x} b}{a e^c-\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x}{d}+\frac{6 e f \log \left (\frac{e^{2 c+d x} b}{a e^c-\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x}{d}-\frac{6 e e^{2 c} f \log \left (\frac{e^{2 c+d x} b}{e^c a+\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x}{d}+\frac{6 e f \log \left (\frac{e^{2 c+d x} b}{e^c a+\sqrt{\left (a^2+b^2\right ) e^{2 c}}}+1\right ) x}{d}-\frac{2 (e+f x)^3}{f}+\frac{6 a \sqrt{-\left (a^2+b^2\right )^2} e^2 e^{2 c} \tan ^{-1}\left (\frac{a+b e^{c+d x}}{\sqrt{-a^2-b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}+\frac{6 a \sqrt{a^2+b^2} e^2 \tan ^{-1}\left (\frac{a+b e^{c+d x}}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-\left (a^2+b^2\right )^2} d}+\frac{6 a \sqrt{-\left (a^2+b^2\right )^2} e^2 e^{2 c} \tanh ^{-1}\left (\frac{a+b e^{c+d x}}{\sqrt{a^2+b^2}}\right )}{\left (-a^2-b^2\right )^{3/2} d}-\frac{6 a \sqrt{-\left (a^2+b^2\right )^2} e^2 \tanh ^{-1}\left (\frac{a+b e^{c+d x}}{\sqrt{a^2+b^2}}\right )}{\left (-a^2-b^2\right )^{3/2} d}+\frac{3 \left (-1+e^{2 c}\right ) (e+f x)^2 \log \left (1-e^{-c-d x}\right )}{d}+\frac{3 \left (-1+e^{2 c}\right ) (e+f x)^2 \log \left (1+e^{-c-d x}\right )}{d}-\frac{3 e^2 e^{2 c} \log \left (2 e^{c+d x} a+b \left (-1+e^{2 (c+d x)}\right )\right )}{d}+\frac{3 e^2 \log \left (2 e^{c+d x} a+b \left (-1+e^{2 (c+d x)}\right )\right )}{d}-\frac{6 \left (-1+e^{2 c}\right ) f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{2 c+d x}}{a e^c-\sqrt{\left (a^2+b^2\right ) e^{2 c}}}\right )}{d^2}-\frac{6 \left (-1+e^{2 c}\right ) f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{2 c+d x}}{e^c a+\sqrt{\left (a^2+b^2\right ) e^{2 c}}}\right )}{d^2}-\frac{6 \left (-1+e^{2 c}\right ) f \left (d (e+f x) \text{PolyLog}\left (2,-e^{-c-d x}\right )+f \text{PolyLog}\left (3,-e^{-c-d x}\right )\right )}{d^3}-\frac{6 \left (-1+e^{2 c}\right ) f \left (d (e+f x) \text{PolyLog}\left (2,e^{-c-d x}\right )+f \text{PolyLog}\left (3,e^{-c-d x}\right )\right )}{d^3}+\frac{6 e^{2 c} f^2 \text{PolyLog}\left (3,-\frac{b e^{2 c+d x}}{a e^c-\sqrt{\left (a^2+b^2\right ) e^{2 c}}}\right )}{d^3}-\frac{6 f^2 \text{PolyLog}\left (3,-\frac{b e^{2 c+d x}}{a e^c-\sqrt{\left (a^2+b^2\right ) e^{2 c}}}\right )}{d^3}+\frac{6 e^{2 c} f^2 \text{PolyLog}\left (3,-\frac{b e^{2 c+d x}}{e^c a+\sqrt{\left (a^2+b^2\right ) e^{2 c}}}\right )}{d^3}-\frac{6 f^2 \text{PolyLog}\left (3,-\frac{b e^{2 c+d x}}{e^c a+\sqrt{\left (a^2+b^2\right ) e^{2 c}}}\right )}{d^3}}{3 a \left (-1+e^{2 c}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*e^2*E^(2*c)*x + 6*e*E^(2*c)*f*x^2 + 2*E^(2*c)*f^2*x^3 - (2*(e + f*x)^3)/f + (6*a*Sqrt[a^2 + b^2]*e^2*ArcTan
[(a + b*E^(c + d*x))/Sqrt[-a^2 - b^2]])/(Sqrt[-(a^2 + b^2)^2]*d) + (6*a*Sqrt[-(a^2 + b^2)^2]*e^2*E^(2*c)*ArcTa
n[(a + b*E^(c + d*x))/Sqrt[-a^2 - b^2]])/((a^2 + b^2)^(3/2)*d) - (6*a*Sqrt[-(a^2 + b^2)^2]*e^2*ArcTanh[(a + b*
E^(c + d*x))/Sqrt[a^2 + b^2]])/((-a^2 - b^2)^(3/2)*d) + (6*a*Sqrt[-(a^2 + b^2)^2]*e^2*E^(2*c)*ArcTanh[(a + b*E
^(c + d*x))/Sqrt[a^2 + b^2]])/((-a^2 - b^2)^(3/2)*d) + (3*(-1 + E^(2*c))*(e + f*x)^2*Log[1 - E^(-c - d*x)])/d
+ (3*(-1 + E^(2*c))*(e + f*x)^2*Log[1 + E^(-c - d*x)])/d + (3*e^2*Log[2*a*E^(c + d*x) + b*(-1 + E^(2*(c + d*x)
))])/d - (3*e^2*E^(2*c)*Log[2*a*E^(c + d*x) + b*(-1 + E^(2*(c + d*x)))])/d + (6*e*f*x*Log[1 + (b*E^(2*c + d*x)
)/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e*E^(2*c)*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^
2)*E^(2*c)])])/d + (3*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (3*E^(2*c)*f
^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (6*e*f*x*Log[1 + (b*E^(2*c + d*x))/
(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e*E^(2*c)*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)
*E^(2*c)])])/d + (3*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (3*E^(2*c)*f^2
*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*(-1 + E^(2*c))*f*(e + f*x)*PolyLog
[2, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^2 - (6*(-1 + E^(2*c))*f*(e + f*x)*PolyLog[2,
-((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^2 - (6*(-1 + E^(2*c))*f*(d*(e + f*x)*PolyLog[2, -
E^(-c - d*x)] + f*PolyLog[3, -E^(-c - d*x)]))/d^3 - (6*(-1 + E^(2*c))*f*(d*(e + f*x)*PolyLog[2, E^(-c - d*x)]
+ f*PolyLog[3, E^(-c - d*x)]))/d^3 - (6*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])
)])/d^3 + (6*E^(2*c)*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 - (6*f^2*Po
lyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (6*E^(2*c)*f^2*PolyLog[3, -((b*E^(2*
c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3)/(3*a*(-1 + E^(2*c)))

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Maple [F]  time = 0.39, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2}{\rm coth} \left (dx+c\right )}{a+b\sinh \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -e^{2}{\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 )} + \frac{2 \,{\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac{2 \,{\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac{{\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac{{\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} - \frac{2 \,{\left (d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2}\right )}}{3 \, a d^{3}} + \int -\frac{2 \,{\left (b f^{2} x^{2} + 2 \, b e f x -{\left (a f^{2} x^{2} e^{c} + 2 \, a e f x e^{c}\right )} e^{\left (d x\right )}\right )}}{a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} e^{\left (d x + c\right )} - a b}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.61739, size = 2053, normalized size = 6.32 \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)^2*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(2*f^2*polylog(3, (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) + 2*f^2*polylog(3, (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) - 2*f^2*polylog(3, cosh(d*x + c) + sinh(d*x + c)) - 2*f^2*polylog(3, -cosh(d*x + c) - sinh(d*x
+ c)) - 2*(d*f^2*x + d*e*f)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqr
t((a^2 + b^2)/b^2) - b)/b + 1) - 2*(d*f^2*x + d*e*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*(d*f^2*x + d*e*f)*dilog(cosh(d*x + c) + sinh(d*x +
c)) + 2*(d*f^2*x + d*e*f)*dilog(-cosh(d*x + c) - sinh(d*x + c)) - (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*log(2*b*cos
h(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) - (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*log(2*b*co
sh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) - (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f -
c^2*f^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) - (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^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) + (d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)*log(cosh(d*x
+ c) + sinh(d*x + c) + 1) + (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + (d^2*f^2*
x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*log(-cosh(d*x + c) - sinh(d*x + c) + 1))/(a*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right )^{2} \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)**2*coth(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

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

[Out]

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