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

Optimal. Leaf size=81 \[ -\frac {2 (A c-a C) \tanh ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2+c^2}}\right )}{c e \sqrt {a^2+c^2}}+\frac {B \log (a+c \sinh (d+e x))}{c e}+\frac {C x}{c} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4376, 2735, 2660, 618, 204, 2668, 31} \[ -\frac {2 (A c-a C) \tanh ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2+c^2}}\right )}{c e \sqrt {a^2+c^2}}+\frac {B \log (a+c \sinh (d+e x))}{c e}+\frac {C x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 4376

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

Rubi steps

\begin {align*} \int \frac {A+B \cosh (d+e x)+C \sinh (d+e x)}{a+c \sinh (d+e x)} \, dx &=B \int \frac {\cosh (d+e x)}{a+c \sinh (d+e x)} \, dx+\int \frac {A+C \sinh (d+e x)}{a+c \sinh (d+e x)} \, dx\\ &=\frac {C x}{c}-\frac {(i (i A c-i a C)) \int \frac {1}{a+c \sinh (d+e x)} \, dx}{c}+\frac {B \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,c \sinh (d+e x)\right )}{c e}\\ &=\frac {C x}{c}+\frac {B \log (a+c \sinh (d+e x))}{c e}-\frac {(2 i (A c-a C)) \operatorname {Subst}\left (\int \frac {1}{a-2 i c x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (i d+i e x)\right )\right )}{c e}\\ &=\frac {C x}{c}+\frac {B \log (a+c \sinh (d+e x))}{c e}+\frac {(4 i (A c-a C)) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2+c^2\right )-x^2} \, dx,x,-2 i c+2 a \tan \left (\frac {1}{2} (i d+i e x)\right )\right )}{c e}\\ &=\frac {C x}{c}-\frac {2 (A c-a C) \tanh ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2+c^2}}\right )}{c \sqrt {a^2+c^2} e}+\frac {B \log (a+c \sinh (d+e x))}{c e}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 85, normalized size = 1.05 \[ \frac {\frac {2 (A c-a C) \tan ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2-c^2}}\right )}{\sqrt {-a^2-c^2}}+B \log (a+c \sinh (d+e x))+C (d+e x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(d + e*x) + (2*(A*c - a*C)*ArcTan[(c - a*Tanh[(d + e*x)/2])/Sqrt[-a^2 - c^2]])/Sqrt[-a^2 - c^2] + B*Log[a +
c*Sinh[d + e*x]])/(c*e)

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fricas [B]  time = 1.40, size = 249, normalized size = 3.07 \[ -\frac {{\left ({\left (B - C\right )} a^{2} + {\left (B - C\right )} c^{2}\right )} e x + {\left (C a - A c\right )} \sqrt {a^{2} + c^{2}} \log \left (\frac {c^{2} \cosh \left (e x + d\right )^{2} + c^{2} \sinh \left (e x + d\right )^{2} + 2 \, a c \cosh \left (e x + d\right ) + 2 \, a^{2} + c^{2} + 2 \, {\left (c^{2} \cosh \left (e x + d\right ) + a c\right )} \sinh \left (e x + d\right ) - 2 \, \sqrt {a^{2} + c^{2}} {\left (c \cosh \left (e x + d\right ) + c \sinh \left (e x + d\right ) + a\right )}}{c \cosh \left (e x + d\right )^{2} + c \sinh \left (e x + d\right )^{2} + 2 \, a \cosh \left (e x + d\right ) + 2 \, {\left (c \cosh \left (e x + d\right ) + a\right )} \sinh \left (e x + d\right ) - c}\right ) - {\left (B a^{2} + B c^{2}\right )} \log \left (\frac {2 \, {\left (c \sinh \left (e x + d\right ) + a\right )}}{\cosh \left (e x + d\right ) - \sinh \left (e x + d\right )}\right )}{{\left (a^{2} c + c^{3}\right )} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((B - C)*a^2 + (B - C)*c^2)*e*x + (C*a - A*c)*sqrt(a^2 + c^2)*log((c^2*cosh(e*x + d)^2 + c^2*sinh(e*x + d)^2
+ 2*a*c*cosh(e*x + d) + 2*a^2 + c^2 + 2*(c^2*cosh(e*x + d) + a*c)*sinh(e*x + d) - 2*sqrt(a^2 + c^2)*(c*cosh(e
*x + d) + c*sinh(e*x + d) + a))/(c*cosh(e*x + d)^2 + c*sinh(e*x + d)^2 + 2*a*cosh(e*x + d) + 2*(c*cosh(e*x + d
) + a)*sinh(e*x + d) - c)) - (B*a^2 + B*c^2)*log(2*(c*sinh(e*x + d) + a)/(cosh(e*x + d) - sinh(e*x + d))))/((a
^2*c + c^3)*e)

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giac [A]  time = 0.20, size = 131, normalized size = 1.62 \[ -{\left (\frac {{\left (x e + d\right )} {\left (B - C\right )}}{c} - \frac {B \log \left ({\left | c e^{\left (2 \, x e + 2 \, d\right )} + 2 \, a e^{\left (x e + d\right )} - c \right |}\right )}{c} + \frac {{\left (C a - A c\right )} \log \left (\frac {{\left | 2 \, c e^{\left (x e + d\right )} + 2 \, a - 2 \, \sqrt {a^{2} + c^{2}} \right |}}{{\left | 2 \, c e^{\left (x e + d\right )} + 2 \, a + 2 \, \sqrt {a^{2} + c^{2}} \right |}}\right )}{\sqrt {a^{2} + c^{2}} c}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 213, normalized size = 2.63 \[ -\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) B}{e c}-\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) C}{e c}-\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) B}{e c}+\frac {\ln \left (\tanh \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) C}{e c}+\frac {B \ln \left (a \left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-2 c \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-a \right )}{e c}+\frac {2 \arctanh \left (\frac {2 a \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-2 c}{2 \sqrt {a^{2}+c^{2}}}\right ) A}{e \sqrt {a^{2}+c^{2}}}-\frac {2 \arctanh \left (\frac {2 a \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-2 c}{2 \sqrt {a^{2}+c^{2}}}\right ) C a}{e c \sqrt {a^{2}+c^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 176, normalized size = 2.17 \[ -C {\left (\frac {a \log \left (\frac {c e^{\left (-e x - d\right )} - a - \sqrt {a^{2} + c^{2}}}{c e^{\left (-e x - d\right )} - a + \sqrt {a^{2} + c^{2}}}\right )}{\sqrt {a^{2} + c^{2}} c e} - \frac {e x + d}{c e}\right )} + \frac {A \log \left (\frac {c e^{\left (-e x - d\right )} - a - \sqrt {a^{2} + c^{2}}}{c e^{\left (-e x - d\right )} - a + \sqrt {a^{2} + c^{2}}}\right )}{\sqrt {a^{2} + c^{2}} e} + \frac {B \log \left (c \sinh \left (e x + d\right ) + a\right )}{c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-C*(a*log((c*e^(-e*x - d) - a - sqrt(a^2 + c^2))/(c*e^(-e*x - d) - a + sqrt(a^2 + c^2)))/(sqrt(a^2 + c^2)*c*e)
- (e*x + d)/(c*e)) + A*log((c*e^(-e*x - d) - a - sqrt(a^2 + c^2))/(c*e^(-e*x - d) - a + sqrt(a^2 + c^2)))/(sq
rt(a^2 + c^2)*e) + B*log(c*sinh(e*x + d) + a)/(c*e)

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mupad [B]  time = 1.85, size = 656, normalized size = 8.10 \[ \frac {C\,x}{c}-\frac {B\,x}{c}-\frac {2\,\mathrm {atan}\left (\frac {a\,\sqrt {-a^2\,c^2\,e^2-c^4\,e^2}\,\sqrt {A^2\,c^2-2\,A\,C\,a\,c+C^2\,a^2}}{-C\,e\,a^3\,c+A\,e\,a^2\,c^2-C\,e\,a\,c^3+A\,e\,c^4}-\frac {a^2\,c^2\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d\,\sqrt {-a^2\,c^2\,e^2-c^4\,e^2}\,\sqrt {A^2\,c^2-2\,A\,C\,a\,c+C^2\,a^2}}{-C\,e\,a^3\,c^4+A\,e\,a^2\,c^5-C\,e\,a\,c^6+A\,e\,c^7}+\frac {A\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d\,\sqrt {-a^2\,c^2\,e^2-c^4\,e^2}}{c\,e\,\sqrt {A^2\,c^2-2\,A\,C\,a\,c+C^2\,a^2}}-\frac {C\,a\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d\,\sqrt {-a^2\,c^2\,e^2-c^4\,e^2}}{c^2\,e\,\sqrt {A^2\,c^2-2\,A\,C\,a\,c+C^2\,a^2}}\right )\,\sqrt {A^2\,c^2-2\,A\,C\,a\,c+C^2\,a^2}}{\sqrt {-a^2\,c^2\,e^2-c^4\,e^2}}+\frac {B\,c^3\,e\,\ln \left (8\,A\,C\,a\,c^2-4\,C^2\,a^2\,c-4\,A^2\,c^3+8\,C^2\,a^3\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,A^2\,c^3\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}+8\,A^2\,a\,c^2\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,C^2\,a^2\,c\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}-16\,A\,C\,a^2\,c\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d-8\,A\,C\,a\,c^2\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}\right )}{a^2\,c^2\,e^2+c^4\,e^2}+\frac {B\,a^2\,c\,e\,\ln \left (8\,A\,C\,a\,c^2-4\,C^2\,a^2\,c-4\,A^2\,c^3+8\,C^2\,a^3\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,A^2\,c^3\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}+8\,A^2\,a\,c^2\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d+4\,C^2\,a^2\,c\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}-16\,A\,C\,a^2\,c\,{\mathrm {e}}^{e\,x}\,{\mathrm {e}}^d-8\,A\,C\,a\,c^2\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,e\,x}\right )}{a^2\,c^2\,e^2+c^4\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 40.01, size = 1318, normalized size = 16.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(A + B*cosh(d) + C*sinh(d))/sinh(d), Eq(a, 0) & Eq(c, 0) & Eq(e, 0)), (2*I*A/(c*e*tanh(d/2 +
e*x/2) - I*c*e) + B*e*x*tanh(d/2 + e*x/2)/(c*e*tanh(d/2 + e*x/2) - I*c*e) - I*B*e*x/(c*e*tanh(d/2 + e*x/2) - I
*c*e) - 2*B*log(tanh(d/2 + e*x/2) + 1)*tanh(d/2 + e*x/2)/(c*e*tanh(d/2 + e*x/2) - I*c*e) + 2*I*B*log(tanh(d/2
+ e*x/2) + 1)/(c*e*tanh(d/2 + e*x/2) - I*c*e) + 2*B*log(tanh(d/2 + e*x/2) - I)*tanh(d/2 + e*x/2)/(c*e*tanh(d/2
+ e*x/2) - I*c*e) - 2*I*B*log(tanh(d/2 + e*x/2) - I)/(c*e*tanh(d/2 + e*x/2) - I*c*e) + C*e*x*tanh(d/2 + e*x/2
)/(c*e*tanh(d/2 + e*x/2) - I*c*e) - I*C*e*x/(c*e*tanh(d/2 + e*x/2) - I*c*e) - 2*C/(c*e*tanh(d/2 + e*x/2) - I*c
*e), Eq(a, -I*c)), (-2*I*A/(c*e*tanh(d/2 + e*x/2) + I*c*e) + B*e*x*tanh(d/2 + e*x/2)/(c*e*tanh(d/2 + e*x/2) +
I*c*e) + I*B*e*x/(c*e*tanh(d/2 + e*x/2) + I*c*e) - 2*B*log(tanh(d/2 + e*x/2) + 1)*tanh(d/2 + e*x/2)/(c*e*tanh(
d/2 + e*x/2) + I*c*e) - 2*I*B*log(tanh(d/2 + e*x/2) + 1)/(c*e*tanh(d/2 + e*x/2) + I*c*e) + 2*B*log(tanh(d/2 +
e*x/2) + I)*tanh(d/2 + e*x/2)/(c*e*tanh(d/2 + e*x/2) + I*c*e) + 2*I*B*log(tanh(d/2 + e*x/2) + I)/(c*e*tanh(d/2
+ e*x/2) + I*c*e) + C*e*x*tanh(d/2 + e*x/2)/(c*e*tanh(d/2 + e*x/2) + I*c*e) + I*C*e*x/(c*e*tanh(d/2 + e*x/2)
+ I*c*e) - 2*C/(c*e*tanh(d/2 + e*x/2) + I*c*e), Eq(a, I*c)), ((A*x + B*sinh(d + e*x)/e + C*cosh(d + e*x)/e)/a,
Eq(c, 0)), (x*(A + B*cosh(d) + C*sinh(d))/(a + c*sinh(d)), Eq(e, 0)), ((A*log(tanh(d/2 + e*x/2))/e + B*x - 2*
B*log(tanh(d/2 + e*x/2) + 1)/e + B*log(tanh(d/2 + e*x/2))/e + C*x)/c, Eq(a, 0)), (-A*c*sqrt(a**2 + c**2)*log(t
anh(d/2 + e*x/2) - c/a - sqrt(a**2 + c**2)/a)/(a**2*c*e + c**3*e) + A*c*sqrt(a**2 + c**2)*log(tanh(d/2 + e*x/2
) - c/a + sqrt(a**2 + c**2)/a)/(a**2*c*e + c**3*e) + B*a**2*e*x/(a**2*c*e + c**3*e) - 2*B*a**2*log(tanh(d/2 +
e*x/2) + 1)/(a**2*c*e + c**3*e) + B*a**2*log(tanh(d/2 + e*x/2) - c/a - sqrt(a**2 + c**2)/a)/(a**2*c*e + c**3*e
) + B*a**2*log(tanh(d/2 + e*x/2) - c/a + sqrt(a**2 + c**2)/a)/(a**2*c*e + c**3*e) + B*c**2*e*x/(a**2*c*e + c**
3*e) - 2*B*c**2*log(tanh(d/2 + e*x/2) + 1)/(a**2*c*e + c**3*e) + B*c**2*log(tanh(d/2 + e*x/2) - c/a - sqrt(a**
2 + c**2)/a)/(a**2*c*e + c**3*e) + B*c**2*log(tanh(d/2 + e*x/2) - c/a + sqrt(a**2 + c**2)/a)/(a**2*c*e + c**3*
e) + C*a**2*e*x/(a**2*c*e + c**3*e) + C*a*sqrt(a**2 + c**2)*log(tanh(d/2 + e*x/2) - c/a - sqrt(a**2 + c**2)/a)
/(a**2*c*e + c**3*e) - C*a*sqrt(a**2 + c**2)*log(tanh(d/2 + e*x/2) - c/a + sqrt(a**2 + c**2)/a)/(a**2*c*e + c*
*3*e) + C*c**2*e*x/(a**2*c*e + c**3*e), True))

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