3.206 \(\int \frac {A+B \cosh (d+e x)+C \sinh (d+e x)}{a+b \cosh (d+e x)} \, dx\)

Optimal. Leaf size=86 \[ \frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b e \sqrt {a-b} \sqrt {a+b}}+\frac {C \log (a+b \cosh (d+e x))}{b e}+\frac {B x}{b} \]

[Out]

B*x/b+C*ln(a+b*cosh(e*x+d))/b/e+2*(A*b-B*a)*arctanh((a-b)^(1/2)*tanh(1/2*e*x+1/2*d)/(a+b)^(1/2))/b/e/(a-b)^(1/
2)/(a+b)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4377, 2735, 2659, 205, 2668, 31} \[ \frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b e \sqrt {a-b} \sqrt {a+b}}+\frac {C \log (a+b \cosh (d+e x))}{b e}+\frac {B x}{b} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[d + e*x] + C*Sinh[d + e*x])/(a + b*Cosh[d + e*x]),x]

[Out]

(B*x)/b + (2*(A*b - a*B)*ArcTanh[(Sqrt[a - b]*Tanh[(d + e*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*e) +
(C*Log[a + b*Cosh[d + e*x]])/(b*e)

Rule 31

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

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 2735

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

Rule 4377

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 81, normalized size = 0.94 \[ \frac {\frac {2 (a B-A b) \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+C \log (a+b \cosh (d+e x))+B (d+e x)}{b e} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[d + e*x] + C*Sinh[d + e*x])/(a + b*Cosh[d + e*x]),x]

[Out]

(B*(d + e*x) + (2*(-(A*b) + a*B)*ArcTan[((a - b)*Tanh[(d + e*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + C*Lo
g[a + b*Cosh[d + e*x]])/(b*e)

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 405, normalized size = 4.71 \[ \left [\frac {{\left ({\left (B - C\right )} a^{2} - {\left (B - C\right )} b^{2}\right )} e x - {\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \left (e x + d\right )^{2} + b^{2} \sinh \left (e x + d\right )^{2} + 2 \, a b \cosh \left (e x + d\right ) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \left (e x + d\right ) + a b\right )} \sinh \left (e x + d\right ) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \left (e x + d\right ) + b \sinh \left (e x + d\right ) + a\right )}}{b \cosh \left (e x + d\right )^{2} + b \sinh \left (e x + d\right )^{2} + 2 \, a \cosh \left (e x + d\right ) + 2 \, {\left (b \cosh \left (e x + d\right ) + a\right )} \sinh \left (e x + d\right ) + b}\right ) + {\left (C a^{2} - C b^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \left (e x + d\right ) + a\right )}}{\cosh \left (e x + d\right ) - \sinh \left (e x + d\right )}\right )}{{\left (a^{2} b - b^{3}\right )} e}, \frac {{\left ({\left (B - C\right )} a^{2} - {\left (B - C\right )} b^{2}\right )} e x + 2 \, {\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \left (e x + d\right ) + b \sinh \left (e x + d\right ) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (C a^{2} - C b^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \left (e x + d\right ) + a\right )}}{\cosh \left (e x + d\right ) - \sinh \left (e x + d\right )}\right )}{{\left (a^{2} b - b^{3}\right )} e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d)),x, algorithm="fricas")

[Out]

[(((B - C)*a^2 - (B - C)*b^2)*e*x - (B*a - A*b)*sqrt(a^2 - b^2)*log((b^2*cosh(e*x + d)^2 + b^2*sinh(e*x + d)^2
+ 2*a*b*cosh(e*x + d) + 2*a^2 - b^2 + 2*(b^2*cosh(e*x + d) + a*b)*sinh(e*x + d) - 2*sqrt(a^2 - b^2)*(b*cosh(e
*x + d) + b*sinh(e*x + d) + a))/(b*cosh(e*x + d)^2 + b*sinh(e*x + d)^2 + 2*a*cosh(e*x + d) + 2*(b*cosh(e*x + d
) + a)*sinh(e*x + d) + b)) + (C*a^2 - C*b^2)*log(2*(b*cosh(e*x + d) + a)/(cosh(e*x + d) - sinh(e*x + d))))/((a
^2*b - b^3)*e), (((B - C)*a^2 - (B - C)*b^2)*e*x + 2*(B*a - A*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*
cosh(e*x + d) + b*sinh(e*x + d) + a)/(a^2 - b^2)) + (C*a^2 - C*b^2)*log(2*(b*cosh(e*x + d) + a)/(cosh(e*x + d)
- sinh(e*x + d))))/((a^2*b - b^3)*e)]

________________________________________________________________________________________

giac [A]  time = 0.16, size = 97, normalized size = 1.13 \[ {\left (\frac {{\left (x e + d\right )} {\left (B - C\right )}}{b} + \frac {C \log \left (b e^{\left (2 \, x e + 2 \, d\right )} + 2 \, a e^{\left (x e + d\right )} + b\right )}{b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b e^{\left (x e + d\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d)),x, algorithm="giac")

[Out]

((x*e + d)*(B - C)/b + C*log(b*e^(2*x*e + 2*d) + 2*a*e^(x*e + d) + b)/b - 2*(B*a - A*b)*arctan((b*e^(x*e + d)
+ a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*b))*e^(-1)

________________________________________________________________________________________

maple [B]  time = 0.17, size = 276, normalized size = 3.21 \[ -\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) B}{e b}-\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) C}{e b}+\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) B}{e b}-\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) C}{e b}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right ) b -a -b \right ) a C}{e b \left (a -b \right )}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right ) b -a -b \right ) C}{e \left (a -b \right )}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) A}{e \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a B}{e b \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d)),x)

[Out]

-1/e/b*ln(tanh(1/2*e*x+1/2*d)-1)*B-1/e/b*ln(tanh(1/2*e*x+1/2*d)-1)*C+1/e/b*ln(tanh(1/2*e*x+1/2*d)+1)*B-1/e/b*l
n(tanh(1/2*e*x+1/2*d)+1)*C+1/e/b/(a-b)*ln(a*tanh(1/2*e*x+1/2*d)^2-tanh(1/2*e*x+1/2*d)^2*b-a-b)*a*C-1/e/(a-b)*l
n(a*tanh(1/2*e*x+1/2*d)^2-tanh(1/2*e*x+1/2*d)^2*b-a-b)*C+2/e/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*e*x+1/
2*d)/((a+b)*(a-b))^(1/2))*A-2/e/b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*e*x+1/2*d)/((a+b)*(a-b))^(1/2))*a
*B

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 2.19, size = 653, normalized size = 7.59 \[ \frac {2\,\mathrm {atan}\left (\frac {a\,\sqrt {b^4\,e^2-a^2\,b^2\,e^2}\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}{B\,e\,a^3\,b-A\,e\,a^2\,b^2-B\,e\,a\,b^3+A\,e\,b^4}+\frac {a^2\,b^2\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d\,\sqrt {b^4\,e^2-a^2\,b^2\,e^2}\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}{B\,e\,a^3\,b^4-A\,e\,a^2\,b^5-B\,e\,a\,b^6+A\,e\,b^7}+\frac {A\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d\,\sqrt {b^4\,e^2-a^2\,b^2\,e^2}}{b\,e\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}-\frac {B\,a\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d\,\sqrt {b^4\,e^2-a^2\,b^2\,e^2}}{b^2\,e\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}\right )\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}{\sqrt {b^4\,e^2-a^2\,b^2\,e^2}}+\frac {B\,x}{b}-\frac {C\,x}{b}+\frac {C\,b^3\,e\,\ln \left (4\,A^2\,b^3+4\,B^2\,a^2\,b-8\,A\,B\,a\,b^2+8\,B^2\,a^3\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,A^2\,b^3\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}+8\,A^2\,a\,b^2\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,B^2\,a^2\,b\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}-16\,A\,B\,a^2\,b\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d-8\,A\,B\,a\,b^2\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}\right )}{b^4\,e^2-a^2\,b^2\,e^2}-\frac {C\,a^2\,b\,e\,\ln \left (4\,A^2\,b^3+4\,B^2\,a^2\,b-8\,A\,B\,a\,b^2+8\,B^2\,a^3\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,A^2\,b^3\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}+8\,A^2\,a\,b^2\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,B^2\,a^2\,b\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}-16\,A\,B\,a^2\,b\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d-8\,A\,B\,a\,b^2\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}\right )}{b^4\,e^2-a^2\,b^2\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(d + e*x) + C*sinh(d + e*x))/(a + b*cosh(d + e*x)),x)

[Out]

(2*atan((a*(b^4*e^2 - a^2*b^2*e^2)^(1/2)*(A^2*b^2 + B^2*a^2 - 2*A*B*a*b)^(1/2))/(A*b^4*e - B*a*b^3*e + B*a^3*b
*e - A*a^2*b^2*e) + (a^2*b^2*exp(e*x)*exp(d)*(b^4*e^2 - a^2*b^2*e^2)^(1/2)*(A^2*b^2 + B^2*a^2 - 2*A*B*a*b)^(1/
2))/(A*b^7*e - B*a*b^6*e - A*a^2*b^5*e + B*a^3*b^4*e) + (A*exp(e*x)*exp(d)*(b^4*e^2 - a^2*b^2*e^2)^(1/2))/(b*e
*(A^2*b^2 + B^2*a^2 - 2*A*B*a*b)^(1/2)) - (B*a*exp(e*x)*exp(d)*(b^4*e^2 - a^2*b^2*e^2)^(1/2))/(b^2*e*(A^2*b^2
+ B^2*a^2 - 2*A*B*a*b)^(1/2)))*(A^2*b^2 + B^2*a^2 - 2*A*B*a*b)^(1/2))/(b^4*e^2 - a^2*b^2*e^2)^(1/2) + (B*x)/b
- (C*x)/b + (C*b^3*e*log(4*A^2*b^3 + 4*B^2*a^2*b - 8*A*B*a*b^2 + 8*B^2*a^3*exp(e*x)*exp(d) + 4*A^2*b^3*exp(2*d
)*exp(2*e*x) + 8*A^2*a*b^2*exp(e*x)*exp(d) + 4*B^2*a^2*b*exp(2*d)*exp(2*e*x) - 16*A*B*a^2*b*exp(e*x)*exp(d) -
8*A*B*a*b^2*exp(2*d)*exp(2*e*x)))/(b^4*e^2 - a^2*b^2*e^2) - (C*a^2*b*e*log(4*A^2*b^3 + 4*B^2*a^2*b - 8*A*B*a*b
^2 + 8*B^2*a^3*exp(e*x)*exp(d) + 4*A^2*b^3*exp(2*d)*exp(2*e*x) + 8*A^2*a*b^2*exp(e*x)*exp(d) + 4*B^2*a^2*b*exp
(2*d)*exp(2*e*x) - 16*A*B*a^2*b*exp(e*x)*exp(d) - 8*A*B*a*b^2*exp(2*d)*exp(2*e*x)))/(b^4*e^2 - a^2*b^2*e^2)

________________________________________________________________________________________

sympy [A]  time = 31.24, size = 695, normalized size = 8.08 \[ \begin {cases} \frac {\tilde {\infty } x \left (A + B \cosh {\relax (d )} + C \sinh {\relax (d )}\right )}{\cosh {\relax (d )}} & \text {for}\: a = 0 \wedge b = 0 \wedge e = 0 \\- \frac {A}{b e \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )}} + \frac {B x}{b} - \frac {B}{b e \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )}} + \frac {C x}{b} - \frac {2 C \log {\left (\tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{b e} + \frac {2 C \log {\left (\tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{b e} & \text {for}\: a = - b \\\frac {A x + \frac {B \sinh {\left (d + e x \right )}}{e} + \frac {C \cosh {\left (d + e x \right )}}{e}}{a} & \text {for}\: b = 0 \\\frac {x \left (A + B \cosh {\relax (d )} + C \sinh {\relax (d )}\right )}{a + b \cosh {\relax (d )}} & \text {for}\: e = 0 \\\frac {A \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{b e} + \frac {B x}{b} - \frac {B \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{b e} + \frac {C x}{b} - \frac {2 C \log {\left (\tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{b e} & \text {for}\: a = b \\- \frac {A b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {A b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {B a e x}{a b e + b^{2} e} + \frac {B a \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} - \frac {B a \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {B b e x}{a b e + b^{2} e} + \frac {C a e x}{a b e + b^{2} e} + \frac {C a \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {C a \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} - \frac {2 C a \log {\left (\tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{a b e + b^{2} e} + \frac {C b e x}{a b e + b^{2} e} + \frac {C b \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {C b \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} - \frac {2 C b \log {\left (\tanh {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{a b e + b^{2} e} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d)),x)

[Out]

Piecewise((zoo*x*(A + B*cosh(d) + C*sinh(d))/cosh(d), Eq(a, 0) & Eq(b, 0) & Eq(e, 0)), (-A/(b*e*tanh(d/2 + e*x
/2)) + B*x/b - B/(b*e*tanh(d/2 + e*x/2)) + C*x/b - 2*C*log(tanh(d/2 + e*x/2) + 1)/(b*e) + 2*C*log(tanh(d/2 + e
*x/2))/(b*e), Eq(a, -b)), ((A*x + B*sinh(d + e*x)/e + C*cosh(d + e*x)/e)/a, Eq(b, 0)), (x*(A + B*cosh(d) + C*s
inh(d))/(a + b*cosh(d)), Eq(e, 0)), (A*tanh(d/2 + e*x/2)/(b*e) + B*x/b - B*tanh(d/2 + e*x/2)/(b*e) + C*x/b - 2
*C*log(tanh(d/2 + e*x/2) + 1)/(b*e), Eq(a, b)), (-A*b*sqrt(a/(a - b) + b/(a - b))*log(-sqrt(a/(a - b) + b/(a -
b)) + tanh(d/2 + e*x/2))/(a*b*e + b**2*e) + A*b*sqrt(a/(a - b) + b/(a - b))*log(sqrt(a/(a - b) + b/(a - b)) +
tanh(d/2 + e*x/2))/(a*b*e + b**2*e) + B*a*e*x/(a*b*e + b**2*e) + B*a*sqrt(a/(a - b) + b/(a - b))*log(-sqrt(a/
(a - b) + b/(a - b)) + tanh(d/2 + e*x/2))/(a*b*e + b**2*e) - B*a*sqrt(a/(a - b) + b/(a - b))*log(sqrt(a/(a - b
) + b/(a - b)) + tanh(d/2 + e*x/2))/(a*b*e + b**2*e) + B*b*e*x/(a*b*e + b**2*e) + C*a*e*x/(a*b*e + b**2*e) + C
*a*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(d/2 + e*x/2))/(a*b*e + b**2*e) + C*a*log(sqrt(a/(a - b) + b/(a - b)
) + tanh(d/2 + e*x/2))/(a*b*e + b**2*e) - 2*C*a*log(tanh(d/2 + e*x/2) + 1)/(a*b*e + b**2*e) + C*b*e*x/(a*b*e +
b**2*e) + C*b*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(d/2 + e*x/2))/(a*b*e + b**2*e) + C*b*log(sqrt(a/(a - b)
+ b/(a - b)) + tanh(d/2 + e*x/2))/(a*b*e + b**2*e) - 2*C*b*log(tanh(d/2 + e*x/2) + 1)/(a*b*e + b**2*e), True)
)

________________________________________________________________________________________