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

Optimal. Leaf size=439 \[ -\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a d^2 \left (a^2+b^2\right )}-\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a d^2 \left (a^2+b^2\right )}+\frac {b^2 f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 a d^2 \left (a^2+b^2\right )}+\frac {i b f \text {Li}_2\left (-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}-\frac {i b f \text {Li}_2\left (i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}-\frac {b^2 (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a d \left (a^2+b^2\right )}-\frac {b^2 (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a d \left (a^2+b^2\right )}+\frac {b^2 (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{a d \left (a^2+b^2\right )}-\frac {2 b (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )}-\frac {f \text {Li}_2\left (-e^{2 c+2 d x}\right )}{2 a d^2}+\frac {f \text {Li}_2\left (e^{2 c+2 d x}\right )}{2 a d^2}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.64, antiderivative size = 439, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 11, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {5589, 5461, 4182, 2279, 2391, 5573, 5561, 2190, 6742, 4180, 3718} \[ -\frac {b^2 f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a d^2 \left (a^2+b^2\right )}-\frac {b^2 f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a d^2 \left (a^2+b^2\right )}+\frac {b^2 f \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 a d^2 \left (a^2+b^2\right )}+\frac {i b f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}-\frac {i b f \text {PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}-\frac {f \text {PolyLog}\left (2,-e^{2 c+2 d x}\right )}{2 a d^2}+\frac {f \text {PolyLog}\left (2,e^{2 c+2 d x}\right )}{2 a d^2}-\frac {b^2 (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a d \left (a^2+b^2\right )}-\frac {b^2 (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a d \left (a^2+b^2\right )}+\frac {b^2 (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{a d \left (a^2+b^2\right )}-\frac {2 b (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(e + f*x)*ArcTan[E^(c + d*x)])/((a^2 + b^2)*d) - (2*(e + f*x)*ArcTanh[E^(2*c + 2*d*x)])/(a*d) - (b^2*(e
+ f*x)*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(a*(a^2 + b^2)*d) - (b^2*(e + f*x)*Log[1 + (b*E^(c + d*
x))/(a + Sqrt[a^2 + b^2])])/(a*(a^2 + b^2)*d) + (b^2*(e + f*x)*Log[1 + E^(2*(c + d*x))])/(a*(a^2 + b^2)*d) + (
I*b*f*PolyLog[2, (-I)*E^(c + d*x)])/((a^2 + b^2)*d^2) - (I*b*f*PolyLog[2, I*E^(c + d*x)])/((a^2 + b^2)*d^2) -
(b^2*f*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(a*(a^2 + b^2)*d^2) - (b^2*f*PolyLog[2, -((b*E^(c
+ d*x))/(a + Sqrt[a^2 + b^2]))])/(a*(a^2 + b^2)*d^2) + (b^2*f*PolyLog[2, -E^(2*(c + d*x))])/(2*a*(a^2 + b^2)*
d^2) - (f*PolyLog[2, -E^(2*c + 2*d*x)])/(2*a*d^2) + (f*PolyLog[2, E^(2*c + 2*d*x)])/(2*a*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 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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 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 5589

Int[(Csch[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(p_.))/((a_) + (b_.)*S
inh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[(e + f*x)^m*Sech[c + d*x]^p*Csch[c + d*x]^n, x], x] - Dis
t[b/a, Int[((e + f*x)^m*Sech[c + d*x]^p*Csch[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 {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x) \text {csch}(c+d x) \text {sech}(c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=\frac {2 \int (e+f x) \text {csch}(2 c+2 d x) \, dx}{a}-\frac {b \int (e+f x) \text {sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{a \left (a^2+b^2\right )}-\frac {b^3 \int \frac {(e+f x) \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=\frac {b^2 (e+f x)^2}{2 a \left (a^2+b^2\right ) f}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d}-\frac {b \int (a (e+f x) \text {sech}(c+d x)-b (e+f x) \tanh (c+d x)) \, dx}{a \left (a^2+b^2\right )}-\frac {b^3 \int \frac {e^{c+d x} (e+f x)}{a-\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a \left (a^2+b^2\right )}-\frac {b^3 \int \frac {e^{c+d x} (e+f x)}{a+\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a \left (a^2+b^2\right )}-\frac {f \int \log \left (1-e^{2 c+2 d x}\right ) \, dx}{a d}+\frac {f \int \log \left (1+e^{2 c+2 d x}\right ) \, dx}{a d}\\ &=\frac {b^2 (e+f x)^2}{2 a \left (a^2+b^2\right ) f}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}-\frac {b \int (e+f x) \text {sech}(c+d x) \, dx}{a^2+b^2}+\frac {b^2 \int (e+f x) \tanh (c+d x) \, dx}{a \left (a^2+b^2\right )}-\frac {f \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 c+2 d x}\right )}{2 a d^2}+\frac {f \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 c+2 d x}\right )}{2 a d^2}+\frac {\left (b^2 f\right ) \int \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) \, dx}{a \left (a^2+b^2\right ) d}+\frac {\left (b^2 f\right ) \int \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) \, dx}{a \left (a^2+b^2\right ) d}\\ &=-\frac {2 b (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}-\frac {f \text {Li}_2\left (-e^{2 c+2 d x}\right )}{2 a d^2}+\frac {f \text {Li}_2\left (e^{2 c+2 d x}\right )}{2 a d^2}+\frac {\left (2 b^2\right ) \int \frac {e^{2 (c+d x)} (e+f x)}{1+e^{2 (c+d x)}} \, dx}{a \left (a^2+b^2\right )}+\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a-\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a \left (a^2+b^2\right ) d^2}+\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a \left (a^2+b^2\right ) d^2}+\frac {(i b f) \int \log \left (1-i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d}-\frac {(i b f) \int \log \left (1+i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d}\\ &=-\frac {2 b (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}+\frac {b^2 (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{a \left (a^2+b^2\right ) d}-\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d^2}-\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d^2}-\frac {f \text {Li}_2\left (-e^{2 c+2 d x}\right )}{2 a d^2}+\frac {f \text {Li}_2\left (e^{2 c+2 d x}\right )}{2 a d^2}+\frac {(i b f) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}-\frac {(i b f) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}-\frac {\left (b^2 f\right ) \int \log \left (1+e^{2 (c+d x)}\right ) \, dx}{a \left (a^2+b^2\right ) d}\\ &=-\frac {2 b (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}+\frac {b^2 (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{a \left (a^2+b^2\right ) d}+\frac {i b f \text {Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}-\frac {i b f \text {Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}-\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d^2}-\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d^2}-\frac {f \text {Li}_2\left (-e^{2 c+2 d x}\right )}{2 a d^2}+\frac {f \text {Li}_2\left (e^{2 c+2 d x}\right )}{2 a d^2}-\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 a \left (a^2+b^2\right ) d^2}\\ &=-\frac {2 b (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{2 c+2 d x}\right )}{a d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}-\frac {b^2 (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d}+\frac {b^2 (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{a \left (a^2+b^2\right ) d}+\frac {i b f \text {Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}-\frac {i b f \text {Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}-\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d^2}-\frac {b^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right ) d^2}+\frac {b^2 f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 a \left (a^2+b^2\right ) d^2}-\frac {f \text {Li}_2\left (-e^{2 c+2 d x}\right )}{2 a d^2}+\frac {f \text {Li}_2\left (e^{2 c+2 d x}\right )}{2 a d^2}\\ \end {align*}

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Mathematica [B]  time = 2.73, size = 1541, normalized size = 3.51 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*(((-1/4*I)*(a^2 - b^2)*(d*e - c*f)*(c + d*x))/(a*(a^2 + b^2)*d^2) - ((I/8)*(a^2 - b^2)*f*(c + d*x)^2)/(a*(a^
2 + b^2)*d^2) - (e*ArcTanh[1 - (2*I)*Tanh[(c + d*x)/2]])/((a - I*b)*d) + (I*b*e*ArcTanh[1 - (2*I)*Tanh[(c + d*
x)/2]])/(a*(a - I*b)*d) + (c*f*ArcTanh[1 - (2*I)*Tanh[(c + d*x)/2]])/((a - I*b)*d^2) - (I*b*c*f*ArcTanh[1 - (2
*I)*Tanh[(c + d*x)/2]])/(a*(a - I*b)*d^2) + (e*Log[Cosh[(c + d*x)/2]])/(2*a*d) - (c*f*Log[Cosh[(c + d*x)/2]])/
(2*a*d^2) - (e*((-1/2*I)*(c + d*x) + Log[Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]]))/(2*(a + I*b)*d) + (c*f*((-
1/2*I)*(c + d*x) + Log[Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]]))/(2*(a + I*b)*d^2) - ((I/4)*b*e*((-I)*(c + d*
x) + 2*ArcTanh[1 - (2*I)*Tanh[(c + d*x)/2]] + Log[-1 + Cosh[c + d*x] + I*Sinh[c + d*x]]))/(a*(a - I*b)*d) + ((
I/4)*b*c*f*((-I)*(c + d*x) + 2*ArcTanh[1 - (2*I)*Tanh[(c + d*x)/2]] + Log[-1 + Cosh[c + d*x] + I*Sinh[c + d*x]
]))/(a*(a - I*b)*d^2) + (I*f*((-1/8*I)*(c + d*x)^2 - (I/2)*(c + d*x)*Log[1 + E^(-c - d*x)] + (I/2)*PolyLog[2,
-E^(-c - d*x)]))/(a*d^2) - ((I/2)*b*f*((-1/2*I)*(c + d*x)^2 + (I/4)*(3*Pi*(c + d*x) + (1 - I)*(c + d*x)^2 + 2*
(Pi - (2*I)*(c + d*x))*Log[1 + I*E^(-c - d*x)] - 4*Pi*Log[1 + E^(c + d*x)] - 2*Pi*Log[-Cos[(Pi + (2*I)*(c + d*
x))/4]] + 4*Pi*Log[Cosh[(c + d*x)/2]] + (4*I)*PolyLog[2, (-I)*E^(-c - d*x)])))/(a*(a - I*b)*d^2) + ((I/2)*f*((
c + d*x)^2/4 + (-3*Pi*(c + d*x) - (1 - I)*(c + d*x)^2 - 2*(Pi - (2*I)*(c + d*x))*Log[1 + I*E^(-c - d*x)] + 4*P
i*Log[1 + E^(c + d*x)] + 2*Pi*Log[-Cos[(Pi + (2*I)*(c + d*x))/4]] - 4*Pi*Log[Cosh[(c + d*x)/2]] - (4*I)*PolyLo
g[2, (-I)*E^(-c - d*x)])/4 - (I/2)*(-1/2*(c + d*x)^2 + 2*(c + d*x)*Log[1 - E^(c + d*x)] + 2*PolyLog[2, E^(c +
d*x)])))/((a - I*b)*d^2) + (b*f*((c + d*x)^2/4 + (-3*Pi*(c + d*x) - (1 - I)*(c + d*x)^2 - 2*(Pi - (2*I)*(c + d
*x))*Log[1 + I*E^(-c - d*x)] + 4*Pi*Log[1 + E^(c + d*x)] + 2*Pi*Log[-Cos[(Pi + (2*I)*(c + d*x))/4]] - 4*Pi*Log
[Cosh[(c + d*x)/2]] - (4*I)*PolyLog[2, (-I)*E^(-c - d*x)])/4 - (I/2)*(-1/2*(c + d*x)^2 + 2*(c + d*x)*Log[1 - E
^(c + d*x)] + 2*PolyLog[2, E^(c + d*x)])))/(2*a*(a - I*b)*d^2) - (b^2*(-1/2*(f*(c + d*x)^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])] + d*e*
Log[a + b*Sinh[c + d*x]] - c*f*Log[a + b*Sinh[c + d*x]] + 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*(a^2 + b^2)*d^2) + (I*f*((E^((I/4)*Pi)*(c + d
*x)^2)/4 - ((Pi*(c + d*x))/4 - Pi*Log[1 + E^(c + d*x)] - 2*(Pi/4 + (I/2)*(c + d*x))*Log[1 - E^((2*I)*(Pi/4 + (
I/2)*(c + d*x)))] + Pi*Log[Cosh[(c + d*x)/2]] + (Pi*Log[Sin[Pi/4 + (I/2)*(c + d*x)]])/2 + I*PolyLog[2, E^((2*I
)*(Pi/4 + (I/2)*(c + d*x)))])/Sqrt[2]))/(Sqrt[2]*(a + I*b)*d^2))

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fricas [B]  time = 0.52, size = 808, normalized size = 1.84 \[ -\frac {b^{2} f {\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 ) + b^{2} f {\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 ) - {\left (a^{2} + b^{2}\right )} f {\rm Li}_2\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - {\left (a^{2} + b^{2}\right )} f {\rm Li}_2\left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )\right ) + {\left (a^{2} f + i \, a b f\right )} {\rm Li}_2\left (i \, \cosh \left (d x + c\right ) + i \, \sinh \left (d x + c\right )\right ) + {\left (a^{2} f - i \, a b f\right )} {\rm Li}_2\left (-i \, \cosh \left (d x + c\right ) - i \, \sinh \left (d x + c\right )\right ) + {\left (b^{2} d e - b^{2} c f\right )} \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 ) + {\left (b^{2} d e - b^{2} c f\right )} \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 ) + {\left (b^{2} d f x + b^{2} c f\right )} \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 ) + {\left (b^{2} d f x + b^{2} c f\right )} \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 ) - {\left ({\left (a^{2} + b^{2}\right )} d f x + {\left (a^{2} + b^{2}\right )} d e\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) + {\left (a^{2} d e + i \, a b d e - a^{2} c f - i \, a b c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + i\right ) + {\left (a^{2} d e - i \, a b d e - a^{2} c f + i \, a b c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - i\right ) - {\left ({\left (a^{2} + b^{2}\right )} d e - {\left (a^{2} + b^{2}\right )} c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) + {\left (a^{2} d f x - i \, a b d f x + a^{2} c f - i \, a b c f\right )} \log \left (i \, \cosh \left (d x + c\right ) + i \, \sinh \left (d x + c\right ) + 1\right ) + {\left (a^{2} d f x + i \, a b d f x + a^{2} c f + i \, a b c f\right )} \log \left (-i \, \cosh \left (d x + c\right ) - i \, \sinh \left (d x + c\right ) + 1\right ) - {\left ({\left (a^{2} + b^{2}\right )} d f x + {\left (a^{2} + b^{2}\right )} c f\right )} \log \left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right ) + 1\right )}{{\left (a^{3} + a b^{2}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*f*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^2*f*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^2 + b^2)*f*dilog(cosh(d*x + c) + sinh(d*x + c)) - (a^2 + b^2)*f*dilog(-cosh(d*x + c
) - sinh(d*x + c)) + (a^2*f + I*a*b*f)*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) + (a^2*f - I*a*b*f)*dilog(-I*c
osh(d*x + c) - I*sinh(d*x + c)) + (b^2*d*e - b^2*c*f)*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^2*d*e - b^2*c*f)*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^2*d*f*x + b^2*c*f)*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^2*d*f*x + b^2*c*f)*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^2 + b^2)*d*f*x + (a^2 + b^2)*d*e)*log(cosh(d*x + c
) + sinh(d*x + c) + 1) + (a^2*d*e + I*a*b*d*e - a^2*c*f - I*a*b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) + I) +
(a^2*d*e - I*a*b*d*e - a^2*c*f + I*a*b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) - I) - ((a^2 + b^2)*d*e - (a^2 +
b^2)*c*f)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + (a^2*d*f*x - I*a*b*d*f*x + a^2*c*f - I*a*b*c*f)*log(I*cosh
(d*x + c) + I*sinh(d*x + c) + 1) + (a^2*d*f*x + I*a*b*d*f*x + a^2*c*f + I*a*b*c*f)*log(-I*cosh(d*x + c) - I*si
nh(d*x + c) + 1) - ((a^2 + b^2)*d*f*x + (a^2 + b^2)*c*f)*log(-cosh(d*x + c) - sinh(d*x + c) + 1))/((a^3 + a*b^
2)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.26, size = 1065, normalized size = 2.43 \[ \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 {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}-\frac {f \,b^{2} \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \left (a^{2}+b^{2}\right ) a}-\frac {f \,b^{2} \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} \left (a^{2}+b^{2}\right ) a}-\frac {f \,b^{2} \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \left (a^{2}+b^{2}\right ) a}-\frac {f \,b^{2} \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} \left (a^{2}+b^{2}\right ) a}+\frac {f c \,b^{2} \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}{d^{2} \left (a^{2}+b^{2}\right ) a}+\frac {4 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) b x}{d \left (4 a^{2}+4 b^{2}\right )}-\frac {4 f \ln \left (1+i {\mathrm e}^{d x +c}\right ) a x}{d \left (4 a^{2}+4 b^{2}\right )}-\frac {4 f \ln \left (1+i {\mathrm e}^{d x +c}\right ) a c}{d^{2} \left (4 a^{2}+4 b^{2}\right )}-\frac {4 f \ln \left (1-i {\mathrm e}^{d x +c}\right ) a x}{d \left (4 a^{2}+4 b^{2}\right )}-\frac {4 f \ln \left (1-i {\mathrm e}^{d x +c}\right ) a c}{d^{2} \left (4 a^{2}+4 b^{2}\right )}+\frac {4 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) b c}{d^{2} \left (4 a^{2}+4 b^{2}\right )}-\frac {4 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) b x}{d \left (4 a^{2}+4 b^{2}\right )}-\frac {4 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) b c}{d^{2} \left (4 a^{2}+4 b^{2}\right )}+\frac {8 f c b \arctan \left ({\mathrm e}^{d x +c}\right )}{d^{2} \left (4 a^{2}+4 b^{2}\right )}-\frac {e \,b^{2} \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}{d \left (a^{2}+b^{2}\right ) a}-\frac {f \,b^{2} \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} \left (a^{2}+b^{2}\right ) a}-\frac {f \,b^{2} \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} \left (a^{2}+b^{2}\right ) a}+\frac {4 i f \dilog \left (1+i {\mathrm e}^{d x +c}\right ) b}{d^{2} \left (4 a^{2}+4 b^{2}\right )}-\frac {4 i f \dilog \left (1-i {\mathrm e}^{d x +c}\right ) b}{d^{2} \left (4 a^{2}+4 b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{d x +c}+1\right ) f x}{a d}-\frac {f c \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{2}}-\frac {4 e a \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{d \left (4 a^{2}+4 b^{2}\right )}-\frac {8 e b \arctan \left ({\mathrm e}^{d x +c}\right )}{d \left (4 a^{2}+4 b^{2}\right )}+\frac {4 f c a \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{d^{2} \left (4 a^{2}+4 b^{2}\right )}-\frac {4 f \dilog \left (1+i {\mathrm e}^{d x +c}\right ) a}{d^{2} \left (4 a^{2}+4 b^{2}\right )}-\frac {4 f \dilog \left (1-i {\mathrm e}^{d x +c}\right ) a}{d^{2} \left (4 a^{2}+4 b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/d*e*ln(exp(d*x+c)+1)+1/a/d*e*ln(exp(d*x+c)-1)+1/d^2*f/a*dilog(exp(d*x+c)+1)-1/d^2*f*dilog(exp(d*x+c))/a-1/
d*f*b^2/(a^2+b^2)/a*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^2/(a^2+b^2)/a*ln((b*e
xp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*c-1/d*f*b^2/(a^2+b^2)/a*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^2/(a^2+b^2)/a*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*c*b^2/(a^2+b^2)/a*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)+4*I/d*f/(4*a^2+4*b^2)*ln(1+I*exp(d*x+c))*b*x+4
*I/d^2*f/(4*a^2+4*b^2)*ln(1+I*exp(d*x+c))*b*c-4*I/d*f/(4*a^2+4*b^2)*ln(1-I*exp(d*x+c))*b*x-4*I/d^2*f/(4*a^2+4*
b^2)*ln(1-I*exp(d*x+c))*b*c+8/d^2*f*c/(4*a^2+4*b^2)*b*arctan(exp(d*x+c))-1/d^2*f*b^2/(a^2+b^2)/a*dilog((-b*exp
(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))-1/d^2*f*b^2/(a^2+b^2)/a*dilog((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a
)/(a+(a^2+b^2)^(1/2)))-1/d*e*b^2/(a^2+b^2)/a*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)+4*I/d^2*f/(4*a^2+4*b^2)*dil
og(1+I*exp(d*x+c))*b-4*I/d^2*f/(4*a^2+4*b^2)*dilog(1-I*exp(d*x+c))*b+1/a/d*ln(exp(d*x+c)+1)*f*x-1/a/d^2*f*c*ln
(exp(d*x+c)-1)-4/d*e/(4*a^2+4*b^2)*a*ln(1+exp(2*d*x+2*c))-8/d*e/(4*a^2+4*b^2)*b*arctan(exp(d*x+c))-4/d^2*f/(4*
a^2+4*b^2)*dilog(1+I*exp(d*x+c))*a-4/d^2*f/(4*a^2+4*b^2)*dilog(1-I*exp(d*x+c))*a-4/d*f/(4*a^2+4*b^2)*ln(1+I*ex
p(d*x+c))*a*x-4/d^2*f/(4*a^2+4*b^2)*ln(1+I*exp(d*x+c))*a*c-4/d*f/(4*a^2+4*b^2)*ln(1-I*exp(d*x+c))*a*x-4/d^2*f/
(4*a^2+4*b^2)*ln(1-I*exp(d*x+c))*a*c+4/d^2*f*c/(4*a^2+4*b^2)*a*ln(1+exp(2*d*x+2*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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)*csch(d*x+c)*sech(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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