3.5.56 \(\int \frac {(e+f x) \coth ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) [456]

Optimal. Leaf size=294 \[ \frac {2 b (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}-\frac {(e+f x) \coth (c+d x)}{a d}+\frac {\sqrt {a^2+b^2} (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d}-\frac {\sqrt {a^2+b^2} (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d}+\frac {f \log (\sinh (c+d x))}{a d^2}+\frac {b f \text {PolyLog}\left (2,-e^{c+d x}\right )}{a^2 d^2}-\frac {b f \text {PolyLog}\left (2,e^{c+d x}\right )}{a^2 d^2}+\frac {\sqrt {a^2+b^2} f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {\sqrt {a^2+b^2} f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a^2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.49, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 14, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {5688, 3801, 3556, 5704, 5558, 3377, 2717, 4267, 2317, 2438, 5684, 3403, 2296, 2221} \begin {gather*} \frac {f \sqrt {a^2+b^2} \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {f \sqrt {a^2+b^2} \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) \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) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a^2 d}+\frac {b f \text {Li}_2\left (-e^{c+d x}\right )}{a^2 d^2}-\frac {b f \text {Li}_2\left (e^{c+d x}\right )}{a^2 d^2}+\frac {2 b (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a^2 d}+\frac {f \log (\sinh (c+d x))}{a d^2}-\frac {(e+f x) \coth (c+d x)}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2317

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 2438

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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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 3403

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 3556

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

Rule 3801

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 4267

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 5558

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 5684

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 5688

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 5704

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]

Rubi steps

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

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Mathematica [A]
time = 2.61, size = 364, normalized size = 1.24 \begin {gather*} \frac {-a d (e+f x) \coth \left (\frac {1}{2} (c+d x)\right )+2 a f \log (\sinh (c+d x))-2 b d e \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+2 b c f \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+2 b f \left (-\left ((c+d x) \left (\log \left (1-e^{-c-d x}\right )-\log \left (1+e^{-c-d x}\right )\right )\right )-\text {PolyLog}\left (2,-e^{-c-d x}\right )+\text {PolyLog}\left (2,e^{-c-d x}\right )\right )+2 \sqrt {a^2+b^2} \left (-2 d e \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )+2 c f \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )+f (c+d x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )-f (c+d x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+f \text {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )-f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )-a d (e+f x) \tanh \left (\frac {1}{2} (c+d x)\right )}{2 a^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1016\) vs. \(2(271)=542\).
time = 5.40, size = 1017, normalized size = 3.46

method	result	size
risch	\(-\frac {2 f \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {f \ln \left ({\mathrm e}^{d x +c}+1\right )}{a \,d^{2}}+\frac {f \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{2}}+\frac {b f \dilog \left ({\mathrm e}^{d x +c}+1\right )}{a^{2} d^{2}}-\frac {f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \sqrt {a^{2}+b^{2}}}+\frac {f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \sqrt {a^{2}+b^{2}}}-\frac {f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} \sqrt {a^{2}+b^{2}}}+\frac {f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} \sqrt {a^{2}+b^{2}}}+\frac {b f c \ln \left ({\mathrm e}^{d x +c}-1\right )}{a^{2} d^{2}}+\frac {b f \ln \left ({\mathrm e}^{d x +c}+1\right ) x}{a^{2} d}-\frac {2 b^{2} e \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} d \sqrt {a^{2}+b^{2}}}+\frac {b^{2} f \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{a^{2} d^{2} \sqrt {a^{2}+b^{2}}}-\frac {b^{2} f \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{a^{2} d^{2} \sqrt {a^{2}+b^{2}}}+\frac {b^{2} f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{a^{2} d \sqrt {a^{2}+b^{2}}}+\frac {b^{2} f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{a^{2} d^{2} \sqrt {a^{2}+b^{2}}}-\frac {b^{2} f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{a^{2} d \sqrt {a^{2}+b^{2}}}-\frac {b^{2} f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{a^{2} d^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 b^{2} f c \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} d^{2} \sqrt {a^{2}+b^{2}}}-\frac {2 e \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \sqrt {a^{2}+b^{2}}}-\frac {f \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} \sqrt {a^{2}+b^{2}}}+\frac {f \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} \sqrt {a^{2}+b^{2}}}+\frac {b e \ln \left ({\mathrm e}^{d x +c}+1\right )}{a^{2} d}-\frac {b e \ln \left ({\mathrm e}^{d x +c}-1\right )}{a^{2} d}-\frac {2 \left (f x +e \right )}{d a \left ({\mathrm e}^{2 d x +2 c}-1\right )}+\frac {b f \dilog \left ({\mathrm e}^{d x +c}\right )}{a^{2} d^{2}}+\frac {2 f c \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} \sqrt {a^{2}+b^{2}}}\)	\(1017\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1447 vs. \(2 (271) = 542\).
time = 0.41, size = 1447, normalized size = 4.92 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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