3.355 \(\int \frac {(e+f x) \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=335 \[ -\frac {a b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )^{3/2}}+\frac {a b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )^{3/2}}+\frac {a f \tan ^{-1}(\sinh (c+d x))}{d^2 \left (a^2+b^2\right )}+\frac {a^2 f \log (\cosh (c+d x))}{b d^2 \left (a^2+b^2\right )}-\frac {a b (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )^{3/2}}+\frac {a b (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {a^2 (e+f x) \tanh (c+d x)}{b d \left (a^2+b^2\right )}-\frac {a (e+f x) \text {sech}(c+d x)}{d \left (a^2+b^2\right )}-\frac {f \log (\cosh (c+d x))}{b d^2}+\frac {(e+f x) \tanh (c+d x)}{b d} \]

[Out]

a*f*arctan(sinh(d*x+c))/(a^2+b^2)/d^2-f*ln(cosh(d*x+c))/b/d^2+a^2*f*ln(cosh(d*x+c))/b/(a^2+b^2)/d^2-a*b*(f*x+e
)*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)^(3/2)/d+a*b*(f*x+e)*ln(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/
(a^2+b^2)^(3/2)/d-a*b*f*polylog(2,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)^(3/2)/d^2+a*b*f*polylog(2,-b*ex
p(d*x+c)/(a+(a^2+b^2)^(1/2)))/(a^2+b^2)^(3/2)/d^2-a*(f*x+e)*sech(d*x+c)/(a^2+b^2)/d+(f*x+e)*tanh(d*x+c)/b/d-a^
2*(f*x+e)*tanh(d*x+c)/b/(a^2+b^2)/d

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Rubi [A]  time = 0.66, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5583, 4184, 3475, 5573, 3322, 2264, 2190, 2279, 2391, 6742, 5451, 3770} \[ -\frac {a b f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )^{3/2}}+\frac {a b f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )^{3/2}}+\frac {a f \tan ^{-1}(\sinh (c+d x))}{d^2 \left (a^2+b^2\right )}+\frac {a^2 f \log (\cosh (c+d x))}{b d^2 \left (a^2+b^2\right )}-\frac {a b (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )^{3/2}}+\frac {a b (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {a^2 (e+f x) \tanh (c+d x)}{b d \left (a^2+b^2\right )}-\frac {a (e+f x) \text {sech}(c+d x)}{d \left (a^2+b^2\right )}-\frac {f \log (\cosh (c+d x))}{b d^2}+\frac {(e+f x) \tanh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3475

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

Rule 3770

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

Rule 4184

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

Rule 5451

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Sech[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 5573

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[b^2/(a^2 + b^2), Int[((e + f*x)^m*Sech[c + d*x]^(n - 2))/(a + b*Sinh[c + d*x]), x], x] + Dist[1/(
a^2 + b^2), Int[(e + f*x)^m*Sech[c + d*x]^n*(a - b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && I
GtQ[m, 0] && NeQ[a^2 + b^2, 0] && IGtQ[n, 0]

Rule 5583

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(p_.)*Tanh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*S
inh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[1/b, Int[(e + f*x)^m*Sech[c + d*x]^(p + 1)*Tanh[c + d*x]^(n - 1),
x], x] - Dist[a/b, Int[((e + f*x)^m*Sech[c + d*x]^(p + 1)*Tanh[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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 2.90, size = 285, normalized size = 0.85 \[ \frac {\frac {a b \left (2 d e \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-f \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-f (c+d x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-2 c f \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {d (e+f x) \text {sech}(c+d x) (b \sinh (c+d x)-a)}{a^2+b^2}+\frac {2 a f \tan ^{-1}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{a^2+b^2}-\frac {b f \log (\cosh (c+d x))}{a^2+b^2}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*f*ArcTan[Tanh[(c + d*x)/2]])/(a^2 + b^2) - (b*f*Log[Cosh[c + d*x]])/(a^2 + b^2) + (a*b*(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*PolyLo
g[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^2 +
b^2)^(3/2) + (d*(e + f*x)*Sech[c + d*x]*(-a + b*Sinh[c + d*x]))/(a^2 + b^2))/d^2

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fricas [B]  time = 0.54, size = 1338, normalized size = 3.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a^2*b + b^3)*d*f*x*cosh(d*x + c)^2 + 2*(a^2*b + b^3)*d*f*x*sinh(d*x + c)^2 - 2*(a^2*b + b^3)*d*e - (a*b^2*
f*cosh(d*x + c)^2 + 2*a*b^2*f*cosh(d*x + c)*sinh(d*x + c) + a*b^2*f*sinh(d*x + c)^2 + a*b^2*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) + (a*b^2*f*cosh(d*x + c)^2 + 2*a*b^2*f*cosh(d*x + c)*sinh(d*x + c) + a*b^2*f*sinh(d*x + c)^2 + a*b^2
*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) + (a*b^2*d*e - a*b^2*c*f + (a*b^2*d*e - a*b^2*c*f)*cosh(d*x + c)^2 + 2*(a*b^2*d*e
- a*b^2*c*f)*cosh(d*x + c)*sinh(d*x + c) + (a*b^2*d*e - a*b^2*c*f)*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) - (a*b^2*d*e - a*b^2*c*f + (a*b^2*d*
e - a*b^2*c*f)*cosh(d*x + c)^2 + 2*(a*b^2*d*e - a*b^2*c*f)*cosh(d*x + c)*sinh(d*x + c) + (a*b^2*d*e - a*b^2*c*
f)*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) - (a*b^2*d*f*x + a*b^2*c*f + (a*b^2*d*f*x + a*b^2*c*f)*cosh(d*x + c)^2 + 2*(a*b^2*d*f*x + a*b^2*c*f)*
cosh(d*x + c)*sinh(d*x + c) + (a*b^2*d*f*x + a*b^2*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) + (a*b^2*d*f*x +
a*b^2*c*f + (a*b^2*d*f*x + a*b^2*c*f)*cosh(d*x + c)^2 + 2*(a*b^2*d*f*x + a*b^2*c*f)*cosh(d*x + c)*sinh(d*x + c
) + (a*b^2*d*f*x + a*b^2*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) + 2*((a^3 + a*b^2)*f*cosh(d*x + c)^2 + 2*(a
^3 + a*b^2)*f*cosh(d*x + c)*sinh(d*x + c) + (a^3 + a*b^2)*f*sinh(d*x + c)^2 + (a^3 + a*b^2)*f)*arctan(cosh(d*x
+ c) + sinh(d*x + c)) - 2*((a^3 + a*b^2)*d*f*x + (a^3 + a*b^2)*d*e)*cosh(d*x + c) - ((a^2*b + b^3)*f*cosh(d*x
+ c)^2 + 2*(a^2*b + b^3)*f*cosh(d*x + c)*sinh(d*x + c) + (a^2*b + b^3)*f*sinh(d*x + c)^2 + (a^2*b + b^3)*f)*l
og(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 2*(2*(a^2*b + b^3)*d*f*x*cosh(d*x + c) - (a^3 + a*b^2)*d
*f*x - (a^3 + a*b^2)*d*e)*sinh(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d^2*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + b
^4)*d^2*cosh(d*x + c)*sinh(d*x + c) + (a^4 + 2*a^2*b^2 + b^4)*d^2*sinh(d*x + c)^2 + (a^4 + 2*a^2*b^2 + b^4)*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)*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.25, size = 1858, normalized size = 5.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

2/(a^2+b^2)^(3/2)/d*a*b^3*f/(2*a^2+2*b^2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*x+2/(a^2+b^
2)^(3/2)/d^2*a*b^3*f/(2*a^2+2*b^2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*c-2/(a^2+b^2)^(3/2
)/d^2*a^3*b*f/(2*a^2+2*b^2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*c-2/(a^2+b^2)^(3/2)/d^2
*a*b^3*f/(2*a^2+2*b^2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*c+2/(a^2+b^2)^(3/2)/d^2*a^3*
b*f/(2*a^2+2*b^2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*c-2/(a^2+b^2)^(3/2)/d*a^3*b*f/(2*a^
2+2*b^2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*x-2/(a^2+b^2)^(3/2)/d*a*b^3*f/(2*a^2+2*b^2
)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*x+2/(a^2+b^2)^(3/2)/d*a^3*b*f/(2*a^2+2*b^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+b^2)/d^2*f*b^3/(2*a^2+2*b^2)*ln(b*exp(2*d*x+2*c)+
2*a*exp(d*x+c)-b)-2/(a^2+b^2)/d^2*f*b^3/(2*a^2+2*b^2)*ln(1+exp(2*d*x+2*c))+4/(a^2+b^2)/d^2*a^3*f/(2*a^2+2*b^2)
*arctan(exp(d*x+c))-1/(a^2+b^2)^2/d^2*f*b*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)*a^2-2/(a^2+b^2)^(3/2)/d^2*a*b^
3*f/(2*a^2+2*b^2)*dilog((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))-2/(a^2+b^2)^(1/2)/d^2*a*b*f/(2
*a^2+2*b^2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))-2/(a^2+b^2)^(3/2)/d^2*a^3*b*f/(2*a^2+2*b^2)*arct
anh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+2/(a^2+b^2)^(3/2)/d^2*a^3*b*f/(2*a^2+2*b^2)*dilog((b*exp(d*x+c)+
(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))-2/(a^2+b^2)^(3/2)/d^2*a^3*b*f/(2*a^2+2*b^2)*dilog((-b*exp(d*x+c)+(a^2+
b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))-2/(a^2+b^2)^(3/2)/d^2*a*b^3*f/(2*a^2+2*b^2)*arctanh(1/2*(2*b*exp(d*x+c)+2*
a)/(a^2+b^2)^(1/2))+2/(a^2+b^2)/d^2*a^2*b*f/(2*a^2+2*b^2)*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)-2/(a^2+b^2)/d^
2*a^2*b*f/(2*a^2+2*b^2)*ln(1+exp(2*d*x+2*c))+2/(a^2+b^2)^(5/2)/d^2*f*b*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b
^2)^(1/2))*a^3+2/(a^2+b^2)^(5/2)/d^2*f*b^3*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))*a+4/(a^2+b^2)/d^2
*f*b^2/(2*a^2+2*b^2)*a*arctan(exp(d*x+c))-2*(f*x+e)*(a*exp(d*x+c)+b)/d/(a^2+b^2)/(1+exp(2*d*x+2*c))+2/(a^2+b^2
)/d^2*b*f*ln(exp(d*x+c))-1/2/(a^2+b^2)^2/d^2*f*b^3*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)+2/(a^2+b^2)^(3/2)/d^2
*a*b^3*f/(2*a^2+2*b^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)^(1/2)*a*b*e/(
2*a^2+2*b^2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+2/d/(a^2+b^2)^(3/2)*a*b^3*e/(2*a^2+2*b^2)*arcta
nh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+2/d/(a^2+b^2)^(3/2)*a^3*b*e/(2*a^2+2*b^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)^(1/2)*a*b*f*c/(2*a^2+2*b^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)^(3/2)*a*b^3*f*c/(2*a^2+2*b^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)^(3/2)*a^3*b*f*c/(2*a^2+2*b^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 \[ -{\left (2 \, a b \int -\frac {x e^{\left (d x + c\right )}}{a^{2} b + b^{3} - {\left (a^{2} b e^{\left (2 \, c\right )} + b^{3} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - 2 \, {\left (a^{3} e^{c} + a b^{2} e^{c}\right )} e^{\left (d x\right )}}\,{d x} + \frac {2 \, {\left (a x e^{\left (d x + c\right )} + b x\right )}}{a^{2} d + b^{2} d + {\left (a^{2} d e^{\left (2 \, c\right )} + b^{2} d e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}} - \frac {2 \, b x}{{\left (a^{2} + b^{2}\right )} d} - \frac {2 \, a \arctan \left (e^{\left (d x + c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d^{2}} + \frac {b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d^{2}}\right )} f - e {\left (\frac {a b \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, {\left (a e^{\left (-d x - c\right )} - b\right )}}{{\left (a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral((e + f*x)*tanh(c + d*x)*sech(c + d*x)/(a + b*sinh(c + d*x)), x)

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