3.175 \(\int \frac{c+d x}{(a+b \cosh (e+f x))^2} \, dx\)

Optimal. Leaf size=274 \[ \frac{a d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac{a d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac{a (c+d x) \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2-b^2}}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac{a (c+d x) \log \left (\frac{b e^{e+f x}}{\sqrt{a^2-b^2}+a}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac{b (c+d x) \sinh (e+f x)}{f \left (a^2-b^2\right ) (a+b \cosh (e+f x))}+\frac{d \log (a+b \cosh (e+f x))}{f^2 \left (a^2-b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.456887, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3324, 3320, 2264, 2190, 2279, 2391, 2668, 31} \[ \frac{a d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac{a d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac{a (c+d x) \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2-b^2}}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac{a (c+d x) \log \left (\frac{b e^{e+f x}}{\sqrt{a^2-b^2}+a}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac{b (c+d x) \sinh (e+f x)}{f \left (a^2-b^2\right ) (a+b \cosh (e+f x))}+\frac{d \log (a+b \cosh (e+f x))}{f^2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*Cosh[e + f*x])^2,x]

[Out]

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

Rule 3324

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

Rule 3320

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol]
:> Dist[2, Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(E^(I*Pi*(k - 1/2))*(b + (2*a*E^(-(I*e) + f*fz*x))/E^(I*Pi*(k
- 1/2)) - (b*E^(2*(-(I*e) + f*fz*x)))/E^(2*I*k*Pi))), x], x] /; FreeQ[{a, b, c, d, e, f, fz}, x] && IntegerQ[
2*k] && NeQ[a^2 - b^2, 0] && 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 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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 4.9781, size = 509, normalized size = 1.86 \[ \frac{\frac{\left (a^2-b^2\right ) \left (-a d \sqrt{b^2-a^2} \text{PolyLog}\left (2,\frac{b e^{e+f x}}{\sqrt{a^2-b^2}-a}\right )+a d \sqrt{b^2-a^2} \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2-b^2}+a}\right )+2 a c f \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{a+b e^{e+f x}}{\sqrt{a^2-b^2}}\right )+d \sqrt{-\left (a^2-b^2\right )^2} (e+f x)-a d \sqrt{b^2-a^2} (e+f x) \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2-b^2}}+1\right )+a d \sqrt{b^2-a^2} (e+f x) \log \left (\frac{b e^{e+f x}}{\sqrt{a^2-b^2}+a}+1\right )-d \sqrt{-\left (a^2-b^2\right )^2} \log \left (2 a e^{e+f x}+b e^{2 (e+f x)}+b\right )-2 a d \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a+b e^{e+f x}}{\sqrt{b^2-a^2}}\right )-2 a d \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{a+b e^{e+f x}}{\sqrt{a^2-b^2}}\right )-2 a d e \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{a+b e^{e+f x}}{\sqrt{a^2-b^2}}\right )\right )}{\left (-\left (a^2-b^2\right )^2\right )^{3/2}}-\frac{b f (c+d x) \sinh (e+f x)}{(a-b) (a+b) (a+b \cosh (e+f x))}}{f^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)/(a + b*Cosh[e + f*x])^2,x]

[Out]

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

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Maple [B]  time = 0.096, size = 585, normalized size = 2.1 \begin{align*} 2\,{\frac{ \left ( dx+c \right ) \left ( a{{\rm e}^{fx+e}}+b \right ) }{f \left ({a}^{2}-{b}^{2} \right ) \left ( b{{\rm e}^{2\,fx+2\,e}}+2\,a{{\rm e}^{fx+e}}+b \right ) }}+2\,{\frac{ac}{f \left ({a}^{2}-{b}^{2} \right ) \sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b{{\rm e}^{fx+e}}+2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }+{\frac{adx}{f}\ln \left ({ \left ( -b{{\rm e}^{fx+e}}+\sqrt{{a}^{2}-{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{dae}{{f}^{2}}\ln \left ({ \left ( -b{{\rm e}^{fx+e}}+\sqrt{{a}^{2}-{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{adx}{f}\ln \left ({ \left ( b{{\rm e}^{fx+e}}+\sqrt{{a}^{2}-{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{dae}{{f}^{2}}\ln \left ({ \left ( b{{\rm e}^{fx+e}}+\sqrt{{a}^{2}-{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{da}{{f}^{2}}{\it dilog} \left ({ \left ( -b{{\rm e}^{fx+e}}+\sqrt{{a}^{2}-{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{da}{{f}^{2}}{\it dilog} \left ({ \left ( b{{\rm e}^{fx+e}}+\sqrt{{a}^{2}-{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{d\ln \left ({{\rm e}^{fx+e}} \right ) }{ \left ({a}^{2}-{b}^{2} \right ){f}^{2}}}+{\frac{d\ln \left ( b{{\rm e}^{2\,fx+2\,e}}+2\,a{{\rm e}^{fx+e}}+b \right ) }{ \left ({a}^{2}-{b}^{2} \right ){f}^{2}}}-2\,{\frac{dae}{ \left ({a}^{2}-{b}^{2} \right ){f}^{2}\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b{{\rm e}^{fx+e}}+2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(a+b*cosh(f*x+e))^2,x)

[Out]

2*(d*x+c)*(a*exp(f*x+e)+b)/f/(a^2-b^2)/(b*exp(2*f*x+2*e)+2*a*exp(f*x+e)+b)+2/(a^2-b^2)/f*a*c/(-a^2+b^2)^(1/2)*
arctan(1/2*(2*b*exp(f*x+e)+2*a)/(-a^2+b^2)^(1/2))+1/(a^2-b^2)^(3/2)/f*d*a*ln((-b*exp(f*x+e)+(a^2-b^2)^(1/2)-a)
/(-a+(a^2-b^2)^(1/2)))*x+1/(a^2-b^2)^(3/2)/f^2*d*a*ln((-b*exp(f*x+e)+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*
e-1/(a^2-b^2)^(3/2)/f*d*a*ln((b*exp(f*x+e)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*x-1/(a^2-b^2)^(3/2)/f^2*d*a
*ln((b*exp(f*x+e)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*e+1/(a^2-b^2)^(3/2)/f^2*d*a*dilog((-b*exp(f*x+e)+(a^
2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))-1/(a^2-b^2)^(3/2)/f^2*d*a*dilog((b*exp(f*x+e)+(a^2-b^2)^(1/2)+a)/(a+(a^2
-b^2)^(1/2)))-2/(a^2-b^2)/f^2*d*ln(exp(f*x+e))+1/(a^2-b^2)/f^2*d*ln(b*exp(2*f*x+2*e)+2*a*exp(f*x+e)+b)-2/(a^2-
b^2)/f^2*a*d*e/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*exp(f*x+e)+2*a)/(-a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cosh(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.32256, size = 4074, normalized size = 14.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cosh(f*x+e))^2,x, algorithm="fricas")

[Out]

-(2*(a^2*b - b^3)*d*e - 2*(a^2*b - b^3)*c*f + 2*((a^2*b - b^3)*d*f*x + (a^2*b - b^3)*d*e)*cosh(f*x + e)^2 + 2*
((a^2*b - b^3)*d*f*x + (a^2*b - b^3)*d*e)*sinh(f*x + e)^2 - (a*b^2*d*cosh(f*x + e)^2 + a*b^2*d*sinh(f*x + e)^2
+ 2*a^2*b*d*cosh(f*x + e) + a*b^2*d + 2*(a*b^2*d*cosh(f*x + e) + a^2*b*d)*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2
)*dilog(-(a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b
+ 1) + (a*b^2*d*cosh(f*x + e)^2 + a*b^2*d*sinh(f*x + e)^2 + 2*a^2*b*d*cosh(f*x + e) + a*b^2*d + 2*(a*b^2*d*co
sh(f*x + e) + a^2*b*d)*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cos
h(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - (a*b^2*d*f*x + a*b^2*d*e + (a*b^2*d*f*x + a*
b^2*d*e)*cosh(f*x + e)^2 + (a*b^2*d*f*x + a*b^2*d*e)*sinh(f*x + e)^2 + 2*(a^2*b*d*f*x + a^2*b*d*e)*cosh(f*x +
e) + 2*(a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*cosh(f*x + e))*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2
)*log((a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b) +
(a*b^2*d*f*x + a*b^2*d*e + (a*b^2*d*f*x + a*b^2*d*e)*cosh(f*x + e)^2 + (a*b^2*d*f*x + a*b^2*d*e)*sinh(f*x + e
)^2 + 2*(a^2*b*d*f*x + a^2*b*d*e)*cosh(f*x + e) + 2*(a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*cosh(
f*x + e))*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2)*log((a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f*x + e) + b*s
inh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b) + 2*((a^3 - a*b^2)*d*f*x + 2*(a^3 - a*b^2)*d*e - (a^3 - a*b^2)*c*f
)*cosh(f*x + e) - ((a^2*b - b^3)*d*cosh(f*x + e)^2 + (a^2*b - b^3)*d*sinh(f*x + e)^2 + 2*(a^3 - a*b^2)*d*cosh(
f*x + e) + (a^2*b - b^3)*d + 2*((a^2*b - b^3)*d*cosh(f*x + e) + (a^3 - a*b^2)*d)*sinh(f*x + e) + (a*b^2*d*e -
a*b^2*c*f + (a*b^2*d*e - a*b^2*c*f)*cosh(f*x + e)^2 + (a*b^2*d*e - a*b^2*c*f)*sinh(f*x + e)^2 + 2*(a^2*b*d*e -
a^2*b*c*f)*cosh(f*x + e) + 2*(a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e - a*b^2*c*f)*cosh(f*x + e))*sinh(f*x + e))*s
qrt((a^2 - b^2)/b^2))*log(2*b*cosh(f*x + e) + 2*b*sinh(f*x + e) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) - ((a^2*b -
b^3)*d*cosh(f*x + e)^2 + (a^2*b - b^3)*d*sinh(f*x + e)^2 + 2*(a^3 - a*b^2)*d*cosh(f*x + e) + (a^2*b - b^3)*d
+ 2*((a^2*b - b^3)*d*cosh(f*x + e) + (a^3 - a*b^2)*d)*sinh(f*x + e) - (a*b^2*d*e - a*b^2*c*f + (a*b^2*d*e - a*
b^2*c*f)*cosh(f*x + e)^2 + (a*b^2*d*e - a*b^2*c*f)*sinh(f*x + e)^2 + 2*(a^2*b*d*e - a^2*b*c*f)*cosh(f*x + e) +
2*(a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e - a*b^2*c*f)*cosh(f*x + e))*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2))*log(2
*b*cosh(f*x + e) + 2*b*sinh(f*x + e) - 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) + 2*((a^3 - a*b^2)*d*f*x + 2*(a^3 - a*
b^2)*d*e - (a^3 - a*b^2)*c*f + 2*((a^2*b - b^3)*d*f*x + (a^2*b - b^3)*d*e)*cosh(f*x + e))*sinh(f*x + e))/((a^4
*b - 2*a^2*b^3 + b^5)*f^2*cosh(f*x + e)^2 + (a^4*b - 2*a^2*b^3 + b^5)*f^2*sinh(f*x + e)^2 + 2*(a^5 - 2*a^3*b^2
+ a*b^4)*f^2*cosh(f*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*f^2 + 2*((a^4*b - 2*a^2*b^3 + b^5)*f^2*cosh(f*x + e) +
(a^5 - 2*a^3*b^2 + a*b^4)*f^2)*sinh(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cosh(f*x+e))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{{\left (b \cosh \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cosh(f*x+e))^2,x, algorithm="giac")

[Out]

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