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

Optimal. Leaf size=517 \[ -\frac {2 f^2 \sqrt {a^2+b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^3}+\frac {2 f^2 \sqrt {a^2+b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d^3}+\frac {2 f \sqrt {a^2+b^2} (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {2 f \sqrt {a^2+b^2} (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d^2}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a^2 d}-\frac {2 b f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a^2 d^3}+\frac {2 b f^2 \text {Li}_3\left (e^{c+d x}\right )}{a^2 d^3}+\frac {2 b f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a^2 d^2}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}+\frac {f^2 \text {Li}_2\left (e^{2 (c+d x)}\right )}{a d^3}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}-\frac {(e+f x)^2 \coth (c+d x)}{a d}-\frac {(e+f x)^2}{a d} \]

[Out]

-(f*x+e)^2/a/d+2*b*(f*x+e)^2*arctanh(exp(d*x+c))/a^2/d-(f*x+e)^2*coth(d*x+c)/a/d+2*f*(f*x+e)*ln(1-exp(2*d*x+2*
c))/a/d^2+2*b*f*(f*x+e)*polylog(2,-exp(d*x+c))/a^2/d^2-2*b*f*(f*x+e)*polylog(2,exp(d*x+c))/a^2/d^2+f^2*polylog
(2,exp(2*d*x+2*c))/a/d^3-2*b*f^2*polylog(3,-exp(d*x+c))/a^2/d^3+2*b*f^2*polylog(3,exp(d*x+c))/a^2/d^3+(f*x+e)^
2*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/a^2/d-(f*x+e)^2*ln(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2))
)*(a^2+b^2)^(1/2)/a^2/d+2*f*(f*x+e)*polylog(2,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/a^2/d^2-2*f*(
f*x+e)*polylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/a^2/d^2-2*f^2*polylog(3,-b*exp(d*x+c)/(a-(
a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/a^2/d^3+2*f^2*polylog(3,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/a^
2/d^3

________________________________________________________________________________________

Rubi [A]  time = 1.30, antiderivative size = 517, normalized size of antiderivative = 1.00, number of steps used = 34, number of rules used = 18, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {5569, 3720, 3716, 2190, 2279, 2391, 32, 5585, 5450, 3296, 2638, 4182, 2531, 2282, 6589, 5565, 3322, 2264} \[ \frac {2 f \sqrt {a^2+b^2} (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {2 f \sqrt {a^2+b^2} (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a^2 d^2}-\frac {2 f^2 \sqrt {a^2+b^2} \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^3}+\frac {2 f^2 \sqrt {a^2+b^2} \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a^2 d^3}+\frac {2 b f (e+f x) \text {PolyLog}\left (2,-e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f (e+f x) \text {PolyLog}\left (2,e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f^2 \text {PolyLog}\left (3,-e^{c+d x}\right )}{a^2 d^3}+\frac {2 b f^2 \text {PolyLog}\left (3,e^{c+d x}\right )}{a^2 d^3}+\frac {f^2 \text {PolyLog}\left (2,e^{2 (c+d x)}\right )}{a d^3}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a^2 d}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}-\frac {(e+f x)^2 \coth (c+d x)}{a d}-\frac {(e+f x)^2}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 2264

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]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(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 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 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 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 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 2638

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

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 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[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]

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 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 4182

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

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \coth ^2(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x)^2 \coth ^2(c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^2 \cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=-\frac {(e+f x)^2 \coth (c+d x)}{a d}+\frac {\int (e+f x)^2 \, dx}{a}-\frac {b \int (e+f x)^2 \cosh (c+d x) \coth (c+d x) \, dx}{a^2}+\frac {b^2 \int \frac {(e+f x)^2 \cosh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2}+\frac {(2 f) \int (e+f x) \coth (c+d x) \, dx}{a d}\\ &=-\frac {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}-\frac {(e+f x)^2 \coth (c+d x)}{a d}-\frac {\int (e+f x)^2 \, dx}{a}-\frac {b \int (e+f x)^2 \text {csch}(c+d x) \, dx}{a^2}+\frac {\left (a^2+b^2\right ) \int \frac {(e+f x)^2}{a+b \sinh (c+d x)} \, dx}{a^2}-\frac {(4 f) \int \frac {e^{2 (c+d x)} (e+f x)}{1-e^{2 (c+d x)}} \, dx}{a d}\\ &=-\frac {(e+f x)^2}{a d}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}-\frac {(e+f x)^2 \coth (c+d x)}{a d}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {\left (2 \left (a^2+b^2\right )\right ) \int \frac {e^{c+d x} (e+f x)^2}{-b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{a^2}+\frac {(2 b f) \int (e+f x) \log \left (1-e^{c+d x}\right ) \, dx}{a^2 d}-\frac {(2 b f) \int (e+f x) \log \left (1+e^{c+d x}\right ) \, dx}{a^2 d}-\frac {\left (2 f^2\right ) \int \log \left (1-e^{2 (c+d x)}\right ) \, dx}{a d^2}\\ &=-\frac {(e+f x)^2}{a d}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}-\frac {(e+f x)^2 \coth (c+d x)}{a d}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a^2 d^2}+\frac {\left (2 b \sqrt {a^2+b^2}\right ) \int \frac {e^{c+d x} (e+f x)^2}{2 a-2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{a^2}-\frac {\left (2 b \sqrt {a^2+b^2}\right ) \int \frac {e^{c+d x} (e+f x)^2}{2 a+2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{a^2}-\frac {f^2 \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{a d^3}-\frac {\left (2 b f^2\right ) \int \text {Li}_2\left (-e^{c+d x}\right ) \, dx}{a^2 d^2}+\frac {\left (2 b f^2\right ) \int \text {Li}_2\left (e^{c+d x}\right ) \, dx}{a^2 d^2}\\ &=-\frac {(e+f x)^2}{a d}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}-\frac {(e+f x)^2 \coth (c+d x)}{a d}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a^2 d^2}+\frac {f^2 \text {Li}_2\left (e^{2 (c+d x)}\right )}{a d^3}-\frac {\left (2 \sqrt {a^2+b^2} f\right ) \int (e+f x) \log \left (1+\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{a^2 d}+\frac {\left (2 \sqrt {a^2+b^2} f\right ) \int (e+f x) \log \left (1+\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{a^2 d}-\frac {\left (2 b f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{c+d x}\right )}{a^2 d^3}+\frac {\left (2 b f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{c+d x}\right )}{a^2 d^3}\\ &=-\frac {(e+f x)^2}{a d}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}-\frac {(e+f x)^2 \coth (c+d x)}{a d}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a^2 d^2}+\frac {2 \sqrt {a^2+b^2} f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {2 \sqrt {a^2+b^2} f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d^2}+\frac {f^2 \text {Li}_2\left (e^{2 (c+d x)}\right )}{a d^3}-\frac {2 b f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a^2 d^3}+\frac {2 b f^2 \text {Li}_3\left (e^{c+d x}\right )}{a^2 d^3}-\frac {\left (2 \sqrt {a^2+b^2} f^2\right ) \int \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{a^2 d^2}+\frac {\left (2 \sqrt {a^2+b^2} f^2\right ) \int \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{a^2 d^2}\\ &=-\frac {(e+f x)^2}{a d}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}-\frac {(e+f x)^2 \coth (c+d x)}{a d}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a^2 d^2}+\frac {2 \sqrt {a^2+b^2} f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {2 \sqrt {a^2+b^2} f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d^2}+\frac {f^2 \text {Li}_2\left (e^{2 (c+d x)}\right )}{a d^3}-\frac {2 b f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a^2 d^3}+\frac {2 b f^2 \text {Li}_3\left (e^{c+d x}\right )}{a^2 d^3}-\frac {\left (2 \sqrt {a^2+b^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^2 d^3}+\frac {\left (2 \sqrt {a^2+b^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^2 d^3}\\ &=-\frac {(e+f x)^2}{a d}+\frac {2 b (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}-\frac {(e+f x)^2 \coth (c+d x)}{a d}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d}+\frac {2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a^2 d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a^2 d^2}+\frac {2 \sqrt {a^2+b^2} f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {2 \sqrt {a^2+b^2} f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d^2}+\frac {f^2 \text {Li}_2\left (e^{2 (c+d x)}\right )}{a d^3}-\frac {2 b f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a^2 d^3}+\frac {2 b f^2 \text {Li}_3\left (e^{c+d x}\right )}{a^2 d^3}-\frac {2 \sqrt {a^2+b^2} f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^3}+\frac {2 \sqrt {a^2+b^2} f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d^3}\\ \end {align*}

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Mathematica [A]  time = 7.88, size = 792, normalized size = 1.53 \[ \frac {\sqrt {a^2+b^2} \left (-2 d^2 e^2 \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )+2 d^2 e f x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )-2 d^2 e f x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )-d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+2 d f (e+f x) \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-2 d f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 f^2 \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{a^2 d^3}-\frac {2 f \text {Li}_2\left (-e^{-c-d x}\right ) (a f+b d e)+2 f \text {Li}_2\left (e^{-c-d x}\right ) (a f-b d e)+2 d f x \log \left (1-e^{-c-d x}\right ) (b d e-a f)-2 d f x \log \left (e^{-c-d x}+1\right ) (a f+b d e)-d e \left (d x-\log \left (1-e^{c+d x}\right )\right ) (b d e-2 a f)+d e \left (d x-\log \left (e^{c+d x}+1\right )\right ) (2 a f+b d e)+\frac {2 a d^2 (e+f x)^2}{e^{2 c}-1}+b d^2 f^2 x^2 \log \left (1-e^{-c-d x}\right )-b d^2 f^2 x^2 \log \left (e^{-c-d x}+1\right )+2 b f^2 \left (d x \text {Li}_2\left (-e^{-c-d x}\right )+\text {Li}_3\left (-e^{-c-d x}\right )\right )-2 b f^2 \left (d x \text {Li}_2\left (e^{-c-d x}\right )+\text {Li}_3\left (e^{-c-d x}\right )\right )}{a^2 d^3}+\frac {\text {csch}\left (\frac {c}{2}\right ) \text {csch}\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (e^2 \sinh \left (\frac {d x}{2}\right )+2 e f x \sinh \left (\frac {d x}{2}\right )+f^2 x^2 \sinh \left (\frac {d x}{2}\right )\right )}{2 a d}+\frac {\text {sech}\left (\frac {c}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (e^2 \left (-\sinh \left (\frac {d x}{2}\right )\right )-2 e f x \sinh \left (\frac {d x}{2}\right )-f^2 x^2 \sinh \left (\frac {d x}{2}\right )\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*a*d^2*(e + f*x)^2)/(-1 + E^(2*c)) + 2*d*f*(b*d*e - a*f)*x*Log[1 - E^(-c - d*x)] + b*d^2*f^2*x^2*Log[1 -
E^(-c - d*x)] - 2*d*f*(b*d*e + a*f)*x*Log[1 + E^(-c - d*x)] - b*d^2*f^2*x^2*Log[1 + E^(-c - d*x)] - d*e*(b*d*e
- 2*a*f)*(d*x - Log[1 - E^(c + d*x)]) + d*e*(b*d*e + 2*a*f)*(d*x - Log[1 + E^(c + d*x)]) + 2*f*(b*d*e + a*f)*
PolyLog[2, -E^(-c - d*x)] + 2*f*(-(b*d*e) + a*f)*PolyLog[2, E^(-c - d*x)] + 2*b*f^2*(d*x*PolyLog[2, -E^(-c - d
*x)] + PolyLog[3, -E^(-c - d*x)]) - 2*b*f^2*(d*x*PolyLog[2, E^(-c - d*x)] + PolyLog[3, E^(-c - d*x)]))/(a^2*d^
3)) + (Sqrt[a^2 + b^2]*(-2*d^2*e^2*ArcTanh[(a + b*E^(c + d*x))/Sqrt[a^2 + b^2]] + 2*d^2*e*f*x*Log[1 + (b*E^(c
+ d*x))/(a - Sqrt[a^2 + b^2])] + d^2*f^2*x^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*d^2*e*f*x*Log[
1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] - d^2*f^2*x^2*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] + 2*d*
f*(e + f*x)*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] - 2*d*f*(e + f*x)*PolyLog[2, -((b*E^(c + d*x))/
(a + Sqrt[a^2 + b^2]))] - 2*f^2*PolyLog[3, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] + 2*f^2*PolyLog[3, -((b*E^(
c + d*x))/(a + Sqrt[a^2 + b^2]))]))/(a^2*d^3) + (Sech[c/2]*Sech[c/2 + (d*x)/2]*(-(e^2*Sinh[(d*x)/2]) - 2*e*f*x
*Sinh[(d*x)/2] - f^2*x^2*Sinh[(d*x)/2]))/(2*a*d) + (Csch[c/2]*Csch[c/2 + (d*x)/2]*(e^2*Sinh[(d*x)/2] + 2*e*f*x
*Sinh[(d*x)/2] + f^2*x^2*Sinh[(d*x)/2]))/(2*a*d)

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fricas [C]  time = 0.58, size = 2729, normalized size = 5.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*d^2*e^2 - 4*a*c*d*e*f + 2*a*c^2*f^2 + 2*(a*d^2*f^2*x^2 + 2*a*d^2*e*f*x + 2*a*c*d*e*f - a*c^2*f^2)*cosh(d
*x + c)^2 + 4*(a*d^2*f^2*x^2 + 2*a*d^2*e*f*x + 2*a*c*d*e*f - a*c^2*f^2)*cosh(d*x + c)*sinh(d*x + c) + 2*(a*d^2
*f^2*x^2 + 2*a*d^2*e*f*x + 2*a*c*d*e*f - a*c^2*f^2)*sinh(d*x + c)^2 + 2*(b*d*f^2*x + b*d*e*f - (b*d*f^2*x + b*
d*e*f)*cosh(d*x + c)^2 - 2*(b*d*f^2*x + b*d*e*f)*cosh(d*x + c)*sinh(d*x + c) - (b*d*f^2*x + b*d*e*f)*sinh(d*x
+ c)^2)*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*s
qrt((a^2 + b^2)/b^2) - b)/b + 1) - 2*(b*d*f^2*x + b*d*e*f - (b*d*f^2*x + b*d*e*f)*cosh(d*x + c)^2 - 2*(b*d*f^2
*x + b*d*e*f)*cosh(d*x + c)*sinh(d*x + c) - (b*d*f^2*x + b*d*e*f)*sinh(d*x + c)^2)*sqrt((a^2 + b^2)/b^2)*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) -
(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2 - (b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*cosh(d*x + c)^2 - 2*(b*d^2*e^2 -
2*b*c*d*e*f + b*c^2*f^2)*cosh(d*x + c)*sinh(d*x + c) - (b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*sinh(d*x + c)^2)*
sqrt((a^2 + b^2)/b^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) + (b*d^2*e^
2 - 2*b*c*d*e*f + b*c^2*f^2 - (b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*cosh(d*x + c)^2 - 2*(b*d^2*e^2 - 2*b*c*d*e
*f + b*c^2*f^2)*cosh(d*x + c)*sinh(d*x + c) - (b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*sinh(d*x + c)^2)*sqrt((a^2
+ b^2)/b^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) + (b*d^2*f^2*x^2 + 2
*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2 - (b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*cosh(d*x +
c)^2 - 2*(b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*cosh(d*x + c)*sinh(d*x + c) - (b*d^2*f^2*x^
2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*sinh(d*x + c)^2)*sqrt((a^2 + b^2)/b^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) - (b*d^2*f^2*x^2 + 2*b*d^2*e
*f*x + 2*b*c*d*e*f - b*c^2*f^2 - (b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*cosh(d*x + c)^2 - 2
*(b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*cosh(d*x + c)*sinh(d*x + c) - (b*d^2*f^2*x^2 + 2*b*
d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*sinh(d*x + c)^2)*sqrt((a^2 + b^2)/b^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*f^2*cosh(d*x + c)^2 + 2*b*f^2
*cosh(d*x + c)*sinh(d*x + c) + b*f^2*sinh(d*x + c)^2 - b*f^2)*sqrt((a^2 + b^2)/b^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*(b*f^2*cosh(d*x + c)^2
+ 2*b*f^2*cosh(d*x + c)*sinh(d*x + c) + b*f^2*sinh(d*x + c)^2 - b*f^2)*sqrt((a^2 + b^2)/b^2)*polylog(3, (a*co
sh(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*(b*d*f^2*x +
b*d*e*f - a*f^2 - (b*d*f^2*x + b*d*e*f - a*f^2)*cosh(d*x + c)^2 - 2*(b*d*f^2*x + b*d*e*f - a*f^2)*cosh(d*x +
c)*sinh(d*x + c) - (b*d*f^2*x + b*d*e*f - a*f^2)*sinh(d*x + c)^2)*dilog(cosh(d*x + c) + sinh(d*x + c)) + 2*(b*
d*f^2*x + b*d*e*f + a*f^2 - (b*d*f^2*x + b*d*e*f + a*f^2)*cosh(d*x + c)^2 - 2*(b*d*f^2*x + b*d*e*f + a*f^2)*co
sh(d*x + c)*sinh(d*x + c) - (b*d*f^2*x + b*d*e*f + a*f^2)*sinh(d*x + c)^2)*dilog(-cosh(d*x + c) - sinh(d*x + c
)) + (b*d^2*f^2*x^2 + b*d^2*e^2 + 2*a*d*e*f - (b*d^2*f^2*x^2 + b*d^2*e^2 + 2*a*d*e*f + 2*(b*d^2*e*f + a*d*f^2)
*x)*cosh(d*x + c)^2 - 2*(b*d^2*f^2*x^2 + b*d^2*e^2 + 2*a*d*e*f + 2*(b*d^2*e*f + a*d*f^2)*x)*cosh(d*x + c)*sinh
(d*x + c) - (b*d^2*f^2*x^2 + b*d^2*e^2 + 2*a*d*e*f + 2*(b*d^2*e*f + a*d*f^2)*x)*sinh(d*x + c)^2 + 2*(b*d^2*e*f
+ a*d*f^2)*x)*log(cosh(d*x + c) + sinh(d*x + c) + 1) - (b*d^2*e^2 - 2*(b*c + a)*d*e*f + (b*c^2 + 2*a*c)*f^2 -
(b*d^2*e^2 - 2*(b*c + a)*d*e*f + (b*c^2 + 2*a*c)*f^2)*cosh(d*x + c)^2 - 2*(b*d^2*e^2 - 2*(b*c + a)*d*e*f + (b
*c^2 + 2*a*c)*f^2)*cosh(d*x + c)*sinh(d*x + c) - (b*d^2*e^2 - 2*(b*c + a)*d*e*f + (b*c^2 + 2*a*c)*f^2)*sinh(d*
x + c)^2)*log(cosh(d*x + c) + sinh(d*x + c) - 1) - (b*d^2*f^2*x^2 + 2*b*c*d*e*f - (b*c^2 + 2*a*c)*f^2 - (b*d^2
*f^2*x^2 + 2*b*c*d*e*f - (b*c^2 + 2*a*c)*f^2 + 2*(b*d^2*e*f - a*d*f^2)*x)*cosh(d*x + c)^2 - 2*(b*d^2*f^2*x^2 +
2*b*c*d*e*f - (b*c^2 + 2*a*c)*f^2 + 2*(b*d^2*e*f - a*d*f^2)*x)*cosh(d*x + c)*sinh(d*x + c) - (b*d^2*f^2*x^2 +
2*b*c*d*e*f - (b*c^2 + 2*a*c)*f^2 + 2*(b*d^2*e*f - a*d*f^2)*x)*sinh(d*x + c)^2 + 2*(b*d^2*e*f - a*d*f^2)*x)*l
og(-cosh(d*x + c) - sinh(d*x + c) + 1) - 2*(b*f^2*cosh(d*x + c)^2 + 2*b*f^2*cosh(d*x + c)*sinh(d*x + c) + b*f^
2*sinh(d*x + c)^2 - b*f^2)*polylog(3, cosh(d*x + c) + sinh(d*x + c)) + 2*(b*f^2*cosh(d*x + c)^2 + 2*b*f^2*cosh
(d*x + c)*sinh(d*x + c) + b*f^2*sinh(d*x + c)^2 - b*f^2)*polylog(3, -cosh(d*x + c) - sinh(d*x + c)))/(a^2*d^3*
cosh(d*x + c)^2 + 2*a^2*d^3*cosh(d*x + c)*sinh(d*x + c) + a^2*d^3*sinh(d*x + c)^2 - a^2*d^3)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 1.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \left (\coth ^{2}\left (d x +c \right )\right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{2} {\left (\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 {\sqrt {a^{2} + b^{2}} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{a^{2} d} + \frac {2}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d}\right )} - \frac {4 \, e f x}{a d} - \frac {2 \, {\left (f^{2} x^{2} + 2 \, e f x\right )}}{a d e^{\left (2 \, d x + 2 \, c\right )} - a d} + \frac {2 \, e f \log \left (e^{\left (d x + c\right )} + 1\right )}{a d^{2}} + \frac {2 \, e f \log \left (e^{\left (d x + c\right )} - 1\right )}{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 )} b f^{2}}{a^{2} 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 )} b f^{2}}{a^{2} d^{3}} + \frac {2 \, {\left (b d e f + a f^{2}\right )} {\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 )}}{a^{2} d^{3}} - \frac {2 \, {\left (b d e f - a f^{2}\right )} {\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 )}}{a^{2} d^{3}} - \frac {b d^{3} f^{2} x^{3} + 3 \, {\left (b d e f + a f^{2}\right )} d^{2} x^{2}}{3 \, a^{2} d^{3}} + \frac {b d^{3} f^{2} x^{3} + 3 \, {\left (b d e f - a f^{2}\right )} d^{2} x^{2}}{3 \, a^{2} d^{3}} + \int \frac {2 \, {\left ({\left (a^{2} f^{2} e^{c} + b^{2} f^{2} e^{c}\right )} x^{2} + 2 \, {\left (a^{2} e f e^{c} + b^{2} e f e^{c}\right )} x\right )} e^{\left (d x\right )}}{a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} e^{\left (d x + c\right )} - a^{2} b}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {coth}\left (c+d\,x\right )}^2\,{\left (e+f\,x\right )}^2}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{2} \coth ^{2}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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