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

Optimal. Leaf size=286 \[ -\frac {f \sqrt {a^2+b^2} \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a b d^2}+\frac {f \sqrt {a^2+b^2} \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a b d^2}-\frac {\sqrt {a^2+b^2} (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a b d}+\frac {\sqrt {a^2+b^2} (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a b d}-\frac {f \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {f \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {e x}{b}+\frac {f x^2}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.58, antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {5585, 5450, 3296, 2637, 4182, 2279, 2391, 5565, 3322, 2264, 2190} \[ -\frac {f \sqrt {a^2+b^2} \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a b d^2}+\frac {f \sqrt {a^2+b^2} \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a b d^2}-\frac {f \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac {f \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}-\frac {\sqrt {a^2+b^2} (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a b d}+\frac {\sqrt {a^2+b^2} (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a b d}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {e x}{b}+\frac {f x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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]

Rubi steps

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

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Mathematica [A]  time = 1.78, size = 339, normalized size = 1.19 \[ \frac {2 \sqrt {a^2+b^2} \left (2 d e \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-f \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-f (c+d x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-2 c f \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )\right )-a (c+d x) (c f-d (2 e+f x))+2 b d e \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+2 b f \left (\text {Li}_2\left (-e^{-c-d x}\right )-\text {Li}_2\left (e^{-c-d x}\right )+(c+d x) \left (\log \left (1-e^{-c-d x}\right )-\log \left (e^{-c-d x}+1\right )\right )\right )-2 b c f \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{2 a b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*(c + d*x)*(c*f - d*(2*e + f*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]))]))/(2*a*b*d^2)

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fricas [B]  time = 0.52, size = 598, normalized size = 2.09 \[ \frac {a d^{2} f x^{2} + 2 \, a d^{2} e x - 2 \, b f \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + 2 \, b f \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + 2 \, b f {\rm Li}_2\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - 2 \, b f {\rm Li}_2\left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )\right ) + 2 \, {\left (b d e - b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - 2 \, {\left (b d e - b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - 2 \, {\left (b d f x + b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + 2 \, {\left (b d f x + b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) - 2 \, {\left (b d f x + b d e\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) + 2 \, {\left (b d e - b c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) + 2 \, {\left (b d f x + b c f\right )} \log \left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right ) + 1\right )}{2 \, a b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*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 [B]  time = 0.28, size = 970, normalized size = 3.39 \[ \frac {f \,x^{2}}{2 b}+\frac {e x}{b}-\frac {2 a f c \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {a^{2}+b^{2}}}-\frac {a 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 b \sqrt {a^{2}+b^{2}}}-\frac {a 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} b \sqrt {a^{2}+b^{2}}}+\frac {a 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 b \sqrt {a^{2}+b^{2}}}+\frac {a 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} b \sqrt {a^{2}+b^{2}}}+\frac {2 a e \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d b \sqrt {a^{2}+b^{2}}}-\frac {2 f c b \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {a^{2}+b^{2}}}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right ) f x}{a d}-\frac {f b \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {a^{2}+b^{2}}}+\frac {f b \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {a^{2}+b^{2}}}-\frac {f c \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{2}}-\frac {f b \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d a \sqrt {a^{2}+b^{2}}}-\frac {f b \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} a \sqrt {a^{2}+b^{2}}}+\frac {2 e b \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d a \sqrt {a^{2}+b^{2}}}-\frac {e \ln \left ({\mathrm e}^{d x +c}+1\right )}{a d}+\frac {e \ln \left ({\mathrm e}^{d x +c}-1\right )}{a d}-\frac {a 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} b \sqrt {a^{2}+b^{2}}}+\frac {a 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} b \sqrt {a^{2}+b^{2}}}+\frac {f b \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d a \sqrt {a^{2}+b^{2}}}+\frac {f b \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} a \sqrt {a^{2}+b^{2}}}-\frac {f \dilog \left ({\mathrm e}^{d x +c}+1\right )}{d^{2} a}-\frac {f \dilog \left ({\mathrm e}^{d x +c}\right )}{d^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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