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

Optimal. Leaf size=334 \[ -\frac{i a f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{i a f \text{PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}-\frac{b f \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac{b (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac{2 a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.59629, antiderivative size = 334, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5573, 5561, 2190, 2279, 2391, 6742, 4180, 3718} \[ -\frac{i a f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{i a f \text{PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}-\frac{b f \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac{b (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac{2 a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 6742

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

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 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]

Rubi steps

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

Mathematica [A]  time = 2.73318, size = 439, normalized size = 1.31 \[ \frac{2 b f \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )+2 b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )-2 i a f \text{PolyLog}(2,-i (\sinh (c+d x)+\cosh (c+d x)))+2 i a f \text{PolyLog}(2,i (\sinh (c+d x)+\cosh (c+d x)))-b f \text{PolyLog}(2,-\sinh (2 (c+d x))-\cosh (2 (c+d x)))+2 b c f \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )+2 b c f \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )+2 b d f x \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )+2 b d f x \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )+2 b d e \log (a+b \sinh (c+d x))-2 b c f \log (a+b \sinh (c+d x))+4 a d e \tan ^{-1}(\sinh (c+d x)+\cosh (c+d x))+4 a d f x \tan ^{-1}(\sinh (c+d x)+\cosh (c+d x))-2 b c^2 f-2 b d e \log (\sinh (2 (c+d x))+\cosh (2 (c+d x))+1)+2 b c d e-2 b c d f x-2 b d f x \log (\sinh (2 (c+d x))+\cosh (2 (c+d x))+1)+2 b d^2 e x}{2 d^2 \left (a^2+b^2\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*b*c*d*e - 2*b*c^2*f + 2*b*d^2*e*x - 2*b*c*d*f*x + 4*a*d*e*ArcTan[Cosh[c + d*x] + Sinh[c + d*x]] + 4*a*d*f*x
*ArcTan[Cosh[c + d*x] + Sinh[c + d*x]] + 2*b*c*f*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] + 2*b*d*f*x*Lo
g[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] + 2*b*c*f*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] + 2*b*d*
f*x*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] + 2*b*d*e*Log[a + b*Sinh[c + d*x]] - 2*b*c*f*Log[a + b*Sinh
[c + d*x]] - 2*b*d*e*Log[1 + Cosh[2*(c + d*x)] + Sinh[2*(c + d*x)]] - 2*b*d*f*x*Log[1 + Cosh[2*(c + d*x)] + Si
nh[2*(c + d*x)]] + 2*b*f*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] + 2*b*f*PolyLog[2, -((b*E^(c + d*x
))/(a + Sqrt[a^2 + b^2]))] - (2*I)*a*f*PolyLog[2, (-I)*(Cosh[c + d*x] + Sinh[c + d*x])] + (2*I)*a*f*PolyLog[2,
I*(Cosh[c + d*x] + Sinh[c + d*x])] - b*f*PolyLog[2, -Cosh[2*(c + d*x)] - Sinh[2*(c + d*x)]])/(2*(a^2 + b^2)*d
^2)

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Maple [B]  time = 0.128, size = 954, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -e{\left (\frac{2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac{b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d}\right )} + 2 \, 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 )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.41935, size = 1547, normalized size = 4.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*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*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) + (I*a*f - b*f)*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) + (-I*a*f - b*f)*dilog(-I*cosh(d*x +
c) - I*sinh(d*x + c)) + (b*d*e - b*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*d*e - b*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*d*
f*x + b*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*d*f*x + b*c*f)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*s
qrt((a^2 + b^2)/b^2) - b)/b) + (I*a*d*e - b*d*e - I*a*c*f + b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) + I) + (-
I*a*d*e - b*d*e + I*a*c*f + b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) - I) + (-I*a*d*f*x - b*d*f*x - I*a*c*f -
b*c*f)*log(I*cosh(d*x + c) + I*sinh(d*x + c) + 1) + (I*a*d*f*x - b*d*f*x + I*a*c*f - b*c*f)*log(-I*cosh(d*x +
c) - I*sinh(d*x + c) + 1))/((a^2 + b^2)*d^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right ) \operatorname{sech}{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \operatorname{sech}\left (d x + c\right )}{b \sinh \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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