3.736 \(\int \frac {A+B \cosh (x)+C \sinh (x)}{b \cosh (x)+c \sinh (x)} \, dx\)

Optimal. Leaf size=92 \[ \frac {A \tan ^{-1}\left (\frac {b \sinh (x)+c \cosh (x)}{\sqrt {b^2-c^2}}\right )}{\sqrt {b^2-c^2}}+\frac {x (b B-c C)}{b^2-c^2}-\frac {(B c-b C) \log (b \cosh (x)+c \sinh (x))}{b^2-c^2} \]

[Out]

(B*b-C*c)*x/(b^2-c^2)-(B*c-C*b)*ln(b*cosh(x)+c*sinh(x))/(b^2-c^2)+A*arctan((c*cosh(x)+b*sinh(x))/(b^2-c^2)^(1/
2))/(b^2-c^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3136, 3074, 206} \[ \frac {A \tan ^{-1}\left (\frac {b \sinh (x)+c \cosh (x)}{\sqrt {b^2-c^2}}\right )}{\sqrt {b^2-c^2}}+\frac {x (b B-c C)}{b^2-c^2}-\frac {(B c-b C) \log (b \cosh (x)+c \sinh (x))}{b^2-c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - c*C)*x)/(b^2 - c^2) + (A*ArcTan[(c*Cosh[x] + b*Sinh[x])/Sqrt[b^2 - c^2]])/Sqrt[b^2 - c^2] - ((B*c - b*
C)*Log[b*Cosh[x] + c*Sinh[x]])/(b^2 - c^2)

Rule 206

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

Rule 3074

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

Rule 3136

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + (Dist[(A*(b^2 + c^2
) - a*(b*B + c*C))/(b^2 + c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] + Simp[((c*B - b*C)*Log[a
+ b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x]) /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2 + c^
2, 0] && NeQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 90, normalized size = 0.98 \[ \frac {2 A \sqrt {b-c} \sqrt {b+c} \tan ^{-1}\left (\frac {b \tanh \left (\frac {x}{2}\right )+c}{\sqrt {b-c} \sqrt {b+c}}\right )+x (b B-c C)+(b C-B c) \log (b \cosh (x)+c \sinh (x))}{(b-c) (b+c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - c*C)*x + 2*A*Sqrt[b - c]*Sqrt[b + c]*ArcTan[(c + b*Tanh[x/2])/(Sqrt[b - c]*Sqrt[b + c])] + (-(B*c) + b
*C)*Log[b*Cosh[x] + c*Sinh[x]])/((b - c)*(b + c))

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fricas [A]  time = 0.53, size = 264, normalized size = 2.87 \[ \left [-\frac {\sqrt {-b^{2} + c^{2}} A \log \left (\frac {{\left (b + c\right )} \cosh \relax (x)^{2} + 2 \, {\left (b + c\right )} \cosh \relax (x) \sinh \relax (x) + {\left (b + c\right )} \sinh \relax (x)^{2} - 2 \, \sqrt {-b^{2} + c^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - b + c}{{\left (b + c\right )} \cosh \relax (x)^{2} + 2 \, {\left (b + c\right )} \cosh \relax (x) \sinh \relax (x) + {\left (b + c\right )} \sinh \relax (x)^{2} + b - c}\right ) - {\left ({\left (B - C\right )} b + {\left (B - C\right )} c\right )} x - {\left (C b - B c\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + c \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{b^{2} - c^{2}}, -\frac {2 \, \sqrt {b^{2} - c^{2}} A \arctan \left (\frac {\sqrt {b^{2} - c^{2}}}{{\left (b + c\right )} \cosh \relax (x) + {\left (b + c\right )} \sinh \relax (x)}\right ) - {\left ({\left (B - C\right )} b + {\left (B - C\right )} c\right )} x - {\left (C b - B c\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + c \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{b^{2} - c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 89, normalized size = 0.97 \[ \frac {2 \, A \arctan \left (\frac {b e^{x} + c e^{x}}{\sqrt {b^{2} - c^{2}}}\right )}{\sqrt {b^{2} - c^{2}}} + \frac {{\left (B - C\right )} x}{b - c} + \frac {{\left (C b - B c\right )} \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + b - c\right )}{b^{2} - c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.21, size = 253, normalized size = 2.75 \[ -\frac {2 B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 c}-\frac {2 C \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 c}+\frac {2 B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b -2 c}-\frac {2 C \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b -2 c}-\frac {B c \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 c \tanh \left (\frac {x}{2}\right )+b \right )}{\left (b -c \right ) \left (b +c \right )}+\frac {b C \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 c \tanh \left (\frac {x}{2}\right )+b \right )}{\left (b -c \right ) \left (b +c \right )}+\frac {2 \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 c}{2 \sqrt {b^{2}-c^{2}}}\right ) A \,b^{2}}{\left (b -c \right ) \left (b +c \right ) \sqrt {b^{2}-c^{2}}}-\frac {2 \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 c}{2 \sqrt {b^{2}-c^{2}}}\right ) A \,c^{2}}{\left (b -c \right ) \left (b +c \right ) \sqrt {b^{2}-c^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x)),x)

[Out]

-2*B/(2*b+2*c)*ln(tanh(1/2*x)-1)-2*C/(2*b+2*c)*ln(tanh(1/2*x)-1)+2*B/(2*b-2*c)*ln(tanh(1/2*x)+1)-2*C/(2*b-2*c)
*ln(tanh(1/2*x)+1)-1/(b-c)/(b+c)*B*c*ln(tanh(1/2*x)^2*b+2*c*tanh(1/2*x)+b)+1/(b-c)/(b+c)*b*C*ln(tanh(1/2*x)^2*
b+2*c*tanh(1/2*x)+b)+2/(b-c)/(b+c)/(b^2-c^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*c)/(b^2-c^2)^(1/2))*A*b^2-2/(
b-c)/(b+c)/(b^2-c^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*c)/(b^2-c^2)^(1/2))*A*c^2

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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(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x)),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*c^2-4*b^2>0)', see `assume?`
for more details)Is 4*c^2-4*b^2 positive or negative?

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mupad [B]  time = 3.81, size = 302, normalized size = 3.28 \[ \frac {2\,\mathrm {atan}\left (\frac {A\,{\mathrm {e}}^x\,\sqrt {b^2-c^2}}{b\,\sqrt {A^2}-c\,\sqrt {A^2}}\right )\,\sqrt {A^2}}{\sqrt {b^2-c^2}}+\frac {B\,x}{b-c}-\frac {C\,x}{b-c}+\frac {B\,c^3\,\ln \left (4\,A^2\,b-4\,A^2\,c+4\,A^2\,b\,{\mathrm {e}}^{2\,x}+4\,A^2\,c\,{\mathrm {e}}^{2\,x}\right )}{b^4-2\,b^2\,c^2+c^4}+\frac {C\,b^3\,\ln \left (4\,A^2\,b-4\,A^2\,c+4\,A^2\,b\,{\mathrm {e}}^{2\,x}+4\,A^2\,c\,{\mathrm {e}}^{2\,x}\right )}{b^4-2\,b^2\,c^2+c^4}-\frac {B\,b^2\,c\,\ln \left (4\,A^2\,b-4\,A^2\,c+4\,A^2\,b\,{\mathrm {e}}^{2\,x}+4\,A^2\,c\,{\mathrm {e}}^{2\,x}\right )}{b^4-2\,b^2\,c^2+c^4}-\frac {C\,b\,c^2\,\ln \left (4\,A^2\,b-4\,A^2\,c+4\,A^2\,b\,{\mathrm {e}}^{2\,x}+4\,A^2\,c\,{\mathrm {e}}^{2\,x}\right )}{b^4-2\,b^2\,c^2+c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(x) + C*sinh(x))/(b*cosh(x) + c*sinh(x)),x)

[Out]

(2*atan((A*exp(x)*(b^2 - c^2)^(1/2))/(b*(A^2)^(1/2) - c*(A^2)^(1/2)))*(A^2)^(1/2))/(b^2 - c^2)^(1/2) + (B*x)/(
b - c) - (C*x)/(b - c) + (B*c^3*log(4*A^2*b - 4*A^2*c + 4*A^2*b*exp(2*x) + 4*A^2*c*exp(2*x)))/(b^4 + c^4 - 2*b
^2*c^2) + (C*b^3*log(4*A^2*b - 4*A^2*c + 4*A^2*b*exp(2*x) + 4*A^2*c*exp(2*x)))/(b^4 + c^4 - 2*b^2*c^2) - (B*b^
2*c*log(4*A^2*b - 4*A^2*c + 4*A^2*b*exp(2*x) + 4*A^2*c*exp(2*x)))/(b^4 + c^4 - 2*b^2*c^2) - (C*b*c^2*log(4*A^2
*b - 4*A^2*c + 4*A^2*b*exp(2*x) + 4*A^2*c*exp(2*x)))/(b^4 + c^4 - 2*b^2*c^2)

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sympy [A]  time = 49.36, size = 643, normalized size = 6.99 \[ \begin {cases} \tilde {\infty } \left (A \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )} + B x - 2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} + B \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )} + C x\right ) & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2 A}{- 2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} + \frac {B x \sinh {\relax (x )}}{- 2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} - \frac {B x \cosh {\relax (x )}}{- 2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} - \frac {B \cosh {\relax (x )}}{- 2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} - \frac {C x \sinh {\relax (x )}}{- 2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} + \frac {C x \cosh {\relax (x )}}{- 2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} - \frac {C \cosh {\relax (x )}}{- 2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} & \text {for}\: b = - c \\- \frac {2 A}{2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} + \frac {B x \sinh {\relax (x )}}{2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} + \frac {B x \cosh {\relax (x )}}{2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} - \frac {B \cosh {\relax (x )}}{2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} + \frac {C x \sinh {\relax (x )}}{2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} + \frac {C x \cosh {\relax (x )}}{2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} + \frac {C \cosh {\relax (x )}}{2 c \sinh {\relax (x )} + 2 c \cosh {\relax (x )}} & \text {for}\: b = c \\\frac {A \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )} + B x - 2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} + B \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )} + C x}{c} & \text {for}\: b = 0 \\- \frac {A \sqrt {- b^{2} + c^{2}} \log {\left (\tanh {\left (\frac {x}{2} \right )} + \frac {c}{b} - \frac {\sqrt {- b^{2} + c^{2}}}{b} \right )}}{b^{2} - c^{2}} + \frac {A \sqrt {- b^{2} + c^{2}} \log {\left (\tanh {\left (\frac {x}{2} \right )} + \frac {c}{b} + \frac {\sqrt {- b^{2} + c^{2}}}{b} \right )}}{b^{2} - c^{2}} + \frac {B b x}{b^{2} - c^{2}} - \frac {B c x}{b^{2} - c^{2}} + \frac {2 B c \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )}}{b^{2} - c^{2}} - \frac {B c \log {\left (\tanh {\left (\frac {x}{2} \right )} + \frac {c}{b} - \frac {\sqrt {- b^{2} + c^{2}}}{b} \right )}}{b^{2} - c^{2}} - \frac {B c \log {\left (\tanh {\left (\frac {x}{2} \right )} + \frac {c}{b} + \frac {\sqrt {- b^{2} + c^{2}}}{b} \right )}}{b^{2} - c^{2}} + \frac {C b x}{b^{2} - c^{2}} - \frac {2 C b \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )}}{b^{2} - c^{2}} + \frac {C b \log {\left (\tanh {\left (\frac {x}{2} \right )} + \frac {c}{b} - \frac {\sqrt {- b^{2} + c^{2}}}{b} \right )}}{b^{2} - c^{2}} + \frac {C b \log {\left (\tanh {\left (\frac {x}{2} \right )} + \frac {c}{b} + \frac {\sqrt {- b^{2} + c^{2}}}{b} \right )}}{b^{2} - c^{2}} - \frac {C c x}{b^{2} - c^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x)),x)

[Out]

Piecewise((zoo*(A*log(tanh(x/2)) + B*x - 2*B*log(tanh(x/2) + 1) + B*log(tanh(x/2)) + C*x), Eq(b, 0) & Eq(c, 0)
), (-2*A/(-2*c*sinh(x) + 2*c*cosh(x)) + B*x*sinh(x)/(-2*c*sinh(x) + 2*c*cosh(x)) - B*x*cosh(x)/(-2*c*sinh(x) +
2*c*cosh(x)) - B*cosh(x)/(-2*c*sinh(x) + 2*c*cosh(x)) - C*x*sinh(x)/(-2*c*sinh(x) + 2*c*cosh(x)) + C*x*cosh(x
)/(-2*c*sinh(x) + 2*c*cosh(x)) - C*cosh(x)/(-2*c*sinh(x) + 2*c*cosh(x)), Eq(b, -c)), (-2*A/(2*c*sinh(x) + 2*c*
cosh(x)) + B*x*sinh(x)/(2*c*sinh(x) + 2*c*cosh(x)) + B*x*cosh(x)/(2*c*sinh(x) + 2*c*cosh(x)) - B*cosh(x)/(2*c*
sinh(x) + 2*c*cosh(x)) + C*x*sinh(x)/(2*c*sinh(x) + 2*c*cosh(x)) + C*x*cosh(x)/(2*c*sinh(x) + 2*c*cosh(x)) + C
*cosh(x)/(2*c*sinh(x) + 2*c*cosh(x)), Eq(b, c)), ((A*log(tanh(x/2)) + B*x - 2*B*log(tanh(x/2) + 1) + B*log(tan
h(x/2)) + C*x)/c, Eq(b, 0)), (-A*sqrt(-b**2 + c**2)*log(tanh(x/2) + c/b - sqrt(-b**2 + c**2)/b)/(b**2 - c**2)
+ A*sqrt(-b**2 + c**2)*log(tanh(x/2) + c/b + sqrt(-b**2 + c**2)/b)/(b**2 - c**2) + B*b*x/(b**2 - c**2) - B*c*x
/(b**2 - c**2) + 2*B*c*log(tanh(x/2) + 1)/(b**2 - c**2) - B*c*log(tanh(x/2) + c/b - sqrt(-b**2 + c**2)/b)/(b**
2 - c**2) - B*c*log(tanh(x/2) + c/b + sqrt(-b**2 + c**2)/b)/(b**2 - c**2) + C*b*x/(b**2 - c**2) - 2*C*b*log(ta
nh(x/2) + 1)/(b**2 - c**2) + C*b*log(tanh(x/2) + c/b - sqrt(-b**2 + c**2)/b)/(b**2 - c**2) + C*b*log(tanh(x/2)
+ c/b + sqrt(-b**2 + c**2)/b)/(b**2 - c**2) - C*c*x/(b**2 - c**2), True))

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