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

Optimal. Leaf size=462 \[ \frac {2 f^2 \sqrt {a^2+b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a b d^3}-\frac {2 f^2 \sqrt {a^2+b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a b d^3}-\frac {2 f \sqrt {a^2+b^2} (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a b d^2}+\frac {2 f \sqrt {a^2+b^2} (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a b d^2}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a b d}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a b d}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {(e+f x)^3}{3 b f} \]

[Out]

1/3*(f*x+e)^3/b/f-2*(f*x+e)^2*arctanh(exp(d*x+c))/a/d-2*f*(f*x+e)*polylog(2,-exp(d*x+c))/a/d^2+2*f*(f*x+e)*pol
ylog(2,exp(d*x+c))/a/d^2+2*f^2*polylog(3,-exp(d*x+c))/a/d^3-2*f^2*polylog(3,exp(d*x+c))/a/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/b/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/b/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/b/d^2+2*f*(f*x+e)*p
olylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/a/b/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/b/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/b/d^3

________________________________________________________________________________________

Rubi [A]  time = 1.04, antiderivative size = 462, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 13, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.406, Rules used = {5585, 5450, 3296, 2638, 4182, 2531, 2282, 6589, 5565, 32, 3322, 2264, 2190} \[ -\frac {2 f \sqrt {a^2+b^2} (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a b d^2}+\frac {2 f \sqrt {a^2+b^2} (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a b d^2}+\frac {2 f^2 \sqrt {a^2+b^2} \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a b d^3}-\frac {2 f^2 \sqrt {a^2+b^2} \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a b d^3}-\frac {2 f (e+f x) \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac {2 f^2 \text {PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {PolyLog}\left (3,e^{c+d x}\right )}{a d^3}-\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a b d}+\frac {\sqrt {a^2+b^2} (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a b d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {(e+f x)^3}{3 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e + f*x)^3/(3*b*f) - (2*(e + f*x)^2*ArcTanh[E^(c + d*x)])/(a*d) - (Sqrt[a^2 + b^2]*(e + f*x)^2*Log[1 + (b*E^(
c + d*x))/(a - Sqrt[a^2 + b^2])])/(a*b*d) + (Sqrt[a^2 + b^2]*(e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2
+ b^2])])/(a*b*d) - (2*f*(e + f*x)*PolyLog[2, -E^(c + d*x)])/(a*d^2) + (2*f*(e + f*x)*PolyLog[2, E^(c + d*x)]
)/(a*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*b*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*b*d^2) + (2*f^2*PolyLog
[3, -E^(c + d*x)])/(a*d^3) - (2*f^2*PolyLog[3, E^(c + d*x)])/(a*d^3) + (2*Sqrt[a^2 + b^2]*f^2*PolyLog[3, -((b*
E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(a*b*d^3) - (2*Sqrt[a^2 + b^2]*f^2*PolyLog[3, -((b*E^(c + d*x))/(a + Sqr
t[a^2 + b^2]))])/(a*b*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 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 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 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 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 \cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x)^2 \cosh (c+d x) \coth (c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^2 \cosh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=\frac {\int (e+f x)^2 \text {csch}(c+d x) \, dx}{a}+\frac {\int (e+f x)^2 \, dx}{b}-\frac {\left (a^2+b^2\right ) \int \frac {(e+f x)^2}{a+b \sinh (c+d x)} \, dx}{a b}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\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 b}-\frac {(2 f) \int (e+f x) \log \left (1-e^{c+d x}\right ) \, dx}{a d}+\frac {(2 f) \int (e+f x) \log \left (1+e^{c+d x}\right ) \, dx}{a d}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {\left (2 \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}+\frac {\left (2 \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}+\frac {\left (2 f^2\right ) \int \text {Li}_2\left (-e^{c+d x}\right ) \, dx}{a d^2}-\frac {\left (2 f^2\right ) \int \text {Li}_2\left (e^{c+d x}\right ) \, dx}{a d^2}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{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 b 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 b d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\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 b 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 b d}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{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(x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{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 b 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 b d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a 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 b 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 b d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a 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 b 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 b d^2}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{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 b 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 b d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a 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 b 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 b d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a 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 b 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 b d^3}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{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 b 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 b d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a 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 b 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 b d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a 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 b 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 b d^3}\\ \end {align*}

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Mathematica [A]  time = 1.71, size = 489, normalized size = 1.06 \[ \frac {\sqrt {a^2+b^2} \left (2 d^2 e^2 \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-2 d^2 e f x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+2 d^2 e f x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-2 d f (e+f x) \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+2 d f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+2 f^2 \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{a b d^3}+\frac {-\frac {2 f \left (d (e+f x) \text {Li}_2\left (-e^{c+d x}\right )-f \text {Li}_3\left (-e^{c+d x}\right )\right )}{d^2}+\frac {2 f \left (d (e+f x) \text {Li}_2\left (e^{c+d x}\right )-f \text {Li}_3\left (e^{c+d x}\right )\right )}{d^2}+(e+f x)^2 \log \left (1-e^{c+d x}\right )-(e+f x)^2 \log \left (e^{c+d x}+1\right )}{a d}+\frac {x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.48, size = 992, normalized size = 2.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*d^3*f^2*x^3 + 3*a*d^3*e*f*x^2 + 3*a*d^3*e^2*x + 6*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) - 6*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) - 6*b*f^2*polylog(3, cosh(d*x + c) + sinh(d*x + c)) + 6*b*f^2*polylog(3, -cosh(d*x + c) - sinh(d*x + c)
) - 6*(b*d*f^2*x + b*d*e*f)*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) + 6*(b*d*f^2*x + b*d*e*f)*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) + 3*(b
*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^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) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^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) - 3*(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)*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))*sq
rt((a^2 + b^2)/b^2) - b)/b) + 3*(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)*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)
+ 6*(b*d*f^2*x + b*d*e*f)*dilog(cosh(d*x + c) + sinh(d*x + c)) - 6*(b*d*f^2*x + b*d*e*f)*dilog(-cosh(d*x + c)
- sinh(d*x + c)) - 3*(b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + b*d^2*e^2)*log(cosh(d*x + c) + sinh(d*x + c) + 1) + 3*(b
*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 3*(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)*log(-cosh(d*x + c) - sinh(d*x + c) + 1))/(a*b*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*cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 1.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \cosh \left (d x +c \right ) \coth \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{2} {\left (\frac {d x + c}{b d} - \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{a b d}\right )} + \frac {f^{2} x^{3} + 3 \, e f x^{2}}{3 \, b} - \frac {2 \, {\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac {2 \, {\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} - \frac {{\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac {{\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} - \int \frac {2 \, {\left ({\left (a^{2} f^{2} e^{c} + b^{2} f^{2} e^{c}\right )} x^{2} + 2 \, {\left (a^{2} e f e^{c} + b^{2} e f e^{c}\right )} x\right )} e^{\left (d x\right )}}{a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} b e^{\left (d x + c\right )} - a b^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cosh(c + d*x)*coth(c + d*x)*(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 \[ \int \frac {\left (e + f x\right )^{2} \cosh {\left (c + d x \right )} \coth {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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