3.19 \(\int \frac {(e+f x) \cosh (c+d x)}{a+b \text {csch}(c+d x)} \, dx\)

Optimal. Leaf size=212 \[ -\frac {b f \text {Li}_2\left (-\frac {a e^{c+d x}}{b-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {b f \text {Li}_2\left (-\frac {a e^{c+d x}}{b+\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {b (e+f x) \log \left (\frac {a e^{c+d x}}{b-\sqrt {a^2+b^2}}+1\right )}{a^2 d}-\frac {b (e+f x) \log \left (\frac {a e^{c+d x}}{\sqrt {a^2+b^2}+b}+1\right )}{a^2 d}+\frac {b (e+f x)^2}{2 a^2 f}-\frac {f \cosh (c+d x)}{a d^2}+\frac {(e+f x) \sinh (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5594, 5579, 3296, 2638, 5561, 2190, 2279, 2391} \[ -\frac {b f \text {PolyLog}\left (2,-\frac {a e^{c+d x}}{b-\sqrt {a^2+b^2}}\right )}{a^2 d^2}-\frac {b f \text {PolyLog}\left (2,-\frac {a e^{c+d x}}{\sqrt {a^2+b^2}+b}\right )}{a^2 d^2}-\frac {b (e+f x) \log \left (\frac {a e^{c+d x}}{b-\sqrt {a^2+b^2}}+1\right )}{a^2 d}-\frac {b (e+f x) \log \left (\frac {a e^{c+d x}}{\sqrt {a^2+b^2}+b}+1\right )}{a^2 d}+\frac {b (e+f x)^2}{2 a^2 f}-\frac {f \cosh (c+d x)}{a d^2}+\frac {(e+f x) \sinh (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(e + f*x)^2)/(2*a^2*f) - (f*Cosh[c + d*x])/(a*d^2) - (b*(e + f*x)*Log[1 + (a*E^(c + d*x))/(b - Sqrt[a^2 + b
^2])])/(a^2*d) - (b*(e + f*x)*Log[1 + (a*E^(c + d*x))/(b + Sqrt[a^2 + b^2])])/(a^2*d) - (b*f*PolyLog[2, -((a*E
^(c + d*x))/(b - Sqrt[a^2 + b^2]))])/(a^2*d^2) - (b*f*PolyLog[2, -((a*E^(c + d*x))/(b + Sqrt[a^2 + b^2]))])/(a
^2*d^2) + ((e + f*x)*Sinh[c + d*x])/(a*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 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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 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 5579

Int[(Cosh[(c_.) + (d_.)*(x_)]^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*S
inh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[1/b, Int[(e + f*x)^m*Cosh[c + d*x]^p*Sinh[c + d*x]^(n - 1), x], x]
- Dist[a/b, Int[((e + f*x)^m*Cosh[c + d*x]^p*Sinh[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 5594

Int[(((e_.) + (f_.)*(x_))^(m_.)*(F_)[(c_.) + (d_.)*(x_)]^(n_.))/(Csch[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Sym
bol] :> Int[((e + f*x)^m*Sinh[c + d*x]*F[c + d*x]^n)/(b + a*Sinh[c + d*x]), x] /; FreeQ[{a, b, c, d, e, f}, x]
&& HyperbolicQ[F] && IntegersQ[m, n]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 1.39, size = 435, normalized size = 2.05 \[ \frac {\text {csch}(c+d x) (a \sinh (c+d x)+b) \left (i b f \left (i \left (\text {Li}_2\left (\frac {\left (b-\sqrt {a^2+b^2}\right ) e^{c+d x}}{a}\right )+\text {Li}_2\left (\frac {\left (b+\sqrt {a^2+b^2}\right ) e^{c+d x}}{a}\right )\right )-\frac {1}{2} \log \left (\frac {\left (\sqrt {a^2+b^2}-b\right ) e^{c+d x}}{a}+1\right ) \left (4 \sin ^{-1}\left (\frac {\sqrt {1+\frac {i b}{a}}}{\sqrt {2}}\right )-2 i c-2 i d x+\pi \right )-\frac {1}{2} \log \left (1-\frac {\left (\sqrt {a^2+b^2}+b\right ) e^{c+d x}}{a}\right ) \left (-4 \sin ^{-1}\left (\frac {\sqrt {1+\frac {i b}{a}}}{\sqrt {2}}\right )-2 i c-2 i d x+\pi \right )-4 i \sin ^{-1}\left (\frac {\sqrt {1+\frac {i b}{a}}}{\sqrt {2}}\right ) \tan ^{-1}\left (\frac {(b+i a) \cot \left (\frac {1}{4} (2 i c+2 i d x+\pi )\right )}{\sqrt {a^2+b^2}}\right )+\left (\frac {\pi }{2}-i (c+d x)\right ) \log (a \sinh (c+d x)+b)-\frac {1}{8} i (2 c+2 d x+i \pi )^2\right )+d e (a \sinh (c+d x)-b \log (a \sinh (c+d x)+b))-b f (c+d x) \log (a \sinh (c+d x)+b)+b c f \log \left (\frac {a \sinh (c+d x)}{b}+1\right )+a d f x \sinh (c+d x)-a f \cosh (c+d x)\right )}{a^2 d^2 (a+b \text {csch}(c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csch[c + d*x]*(b + a*Sinh[c + d*x])*(-(a*f*Cosh[c + d*x]) - b*f*(c + d*x)*Log[b + a*Sinh[c + d*x]] + b*c*f*Lo
g[1 + (a*Sinh[c + d*x])/b] + I*b*f*((-1/8*I)*(2*c + I*Pi + 2*d*x)^2 - (4*I)*ArcSin[Sqrt[1 + (I*b)/a]/Sqrt[2]]*
ArcTan[((I*a + b)*Cot[((2*I)*c + Pi + (2*I)*d*x)/4])/Sqrt[a^2 + b^2]] - (((-2*I)*c + Pi - (2*I)*d*x + 4*ArcSin
[Sqrt[1 + (I*b)/a]/Sqrt[2]])*Log[1 + ((-b + Sqrt[a^2 + b^2])*E^(c + d*x))/a])/2 - (((-2*I)*c + Pi - (2*I)*d*x
- 4*ArcSin[Sqrt[1 + (I*b)/a]/Sqrt[2]])*Log[1 - ((b + Sqrt[a^2 + b^2])*E^(c + d*x))/a])/2 + (Pi/2 - I*(c + d*x)
)*Log[b + a*Sinh[c + d*x]] + I*(PolyLog[2, ((b - Sqrt[a^2 + b^2])*E^(c + d*x))/a] + PolyLog[2, ((b + Sqrt[a^2
+ b^2])*E^(c + d*x))/a])) + a*d*f*x*Sinh[c + d*x] + d*e*(-(b*Log[b + a*Sinh[c + d*x]]) + a*Sinh[c + d*x])))/(a
^2*d^2*(a + b*Csch[c + d*x]))

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 692, normalized size = 3.26 \[ -\frac {a d f x + a d e - {\left (a d f x + a d e - a f\right )} \cosh \left (d x + c\right )^{2} - {\left (a d f x + a d e - a f\right )} \sinh \left (d x + c\right )^{2} + a f - {\left (b d^{2} f x^{2} + 2 \, b d^{2} e x + 4 \, b c d e - 2 \, b c^{2} f\right )} \cosh \left (d x + c\right ) + 2 \, {\left (b f \cosh \left (d x + c\right ) + b f \sinh \left (d x + c\right )\right )} {\rm Li}_2\left (\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + {\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a} + 1\right ) + 2 \, {\left (b f \cosh \left (d x + c\right ) + b f \sinh \left (d x + c\right )\right )} {\rm Li}_2\left (\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) - {\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a} + 1\right ) + 2 \, {\left ({\left (b d e - b c f\right )} \cosh \left (d x + c\right ) + {\left (b d e - b c f\right )} \sinh \left (d x + c\right )\right )} \log \left (2 \, a \cosh \left (d x + c\right ) + 2 \, a \sinh \left (d x + c\right ) + 2 \, a \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} + 2 \, b\right ) + 2 \, {\left ({\left (b d e - b c f\right )} \cosh \left (d x + c\right ) + {\left (b d e - b c f\right )} \sinh \left (d x + c\right )\right )} \log \left (2 \, a \cosh \left (d x + c\right ) + 2 \, a \sinh \left (d x + c\right ) - 2 \, a \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} + 2 \, b\right ) + 2 \, {\left ({\left (b d f x + b c f\right )} \cosh \left (d x + c\right ) + {\left (b d f x + b c f\right )} \sinh \left (d x + c\right )\right )} \log \left (-\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + {\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a}\right ) + 2 \, {\left ({\left (b d f x + b c f\right )} \cosh \left (d x + c\right ) + {\left (b d f x + b c f\right )} \sinh \left (d x + c\right )\right )} \log \left (-\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) - {\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a}\right ) - {\left (b d^{2} f x^{2} + 2 \, b d^{2} e x + 4 \, b c d e - 2 \, b c^{2} f + 2 \, {\left (a d f x + a d e - a f\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left (a^{2} d^{2} \cosh \left (d x + c\right ) + a^{2} d^{2} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.77, size = 483, normalized size = 2.28 \[ \frac {b f \,x^{2}}{2 a^{2}}-\frac {b e x}{a^{2}}+\frac {\left (d f x +d e -f \right ) {\mathrm e}^{d x +c}}{2 a \,d^{2}}-\frac {\left (d f x +d e +f \right ) {\mathrm e}^{-d x -c}}{2 a \,d^{2}}+\frac {2 b e \ln \left ({\mathrm e}^{d x +c}\right )}{d \,a^{2}}-\frac {b e \ln \left (a \,{\mathrm e}^{2 d x +2 c}+2 b \,{\mathrm e}^{d x +c}-a \right )}{d \,a^{2}}+\frac {2 b f c x}{d \,a^{2}}+\frac {b f \,c^{2}}{d^{2} a^{2}}-\frac {b f \ln \left (\frac {-a \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-b}{-b +\sqrt {a^{2}+b^{2}}}\right ) x}{d \,a^{2}}-\frac {b f \ln \left (\frac {-a \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-b}{-b +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} a^{2}}-\frac {b f \ln \left (\frac {a \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+b}{b +\sqrt {a^{2}+b^{2}}}\right ) x}{d \,a^{2}}-\frac {b f \ln \left (\frac {a \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+b}{b +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} a^{2}}-\frac {b f \dilog \left (\frac {a \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+b}{b +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} a^{2}}-\frac {b f \dilog \left (\frac {-a \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-b}{-b +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} a^{2}}-\frac {2 b f c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a^{2}}+\frac {b f c \ln \left (a \,{\mathrm e}^{2 d x +2 c}+2 b \,{\mathrm e}^{d x +c}-a \right )}{d^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________