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

Optimal. Leaf size=121 \[ \frac {2 \left (a b B-A \left (b^2-c^2\right )\right ) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {B c \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {b B x}{b^2-c^2} \]

[Out]

b*B*x/(b^2-c^2)-B*c*ln(a+b*cosh(x)+c*sinh(x))/(b^2-c^2)+2*(a*b*B-A*(b^2-c^2))*arctanh((c-(a-b)*tanh(1/2*x))/(a
^2-b^2+c^2)^(1/2))/(b^2-c^2)/(a^2-b^2+c^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3138, 3124, 618, 206} \[ \frac {2 \left (a b B-A \left (b^2-c^2\right )\right ) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {B c \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {b B x}{b^2-c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*B*x)/(b^2 - c^2) + (2*(a*b*B - A*(b^2 - c^2))*ArcTanh[(c - (a - b)*Tanh[x/2])/Sqrt[a^2 - b^2 + c^2]])/((b^2
- c^2)*Sqrt[a^2 - b^2 + c^2]) - (B*c*Log[a + 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 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 3124

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

Rule 3138

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

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 104, normalized size = 0.86 \[ \frac {B (b x-c \log (a+b \cosh (x)+c \sinh (x)))-\frac {2 \left (a b B+A \left (c^2-b^2\right )\right ) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2-c^2}}\right )}{\sqrt {-a^2+b^2-c^2}}}{(b-c) (b+c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 508, normalized size = 4.20 \[ \left [-\frac {{\left (B a b - A b^{2} + A c^{2}\right )} \sqrt {a^{2} - b^{2} + c^{2}} \log \left (\frac {{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \relax (x)^{2} + {\left (b^{2} + 2 \, b c + c^{2}\right )} \sinh \relax (x)^{2} + 2 \, a^{2} - b^{2} + c^{2} + 2 \, {\left (a b + a c\right )} \cosh \relax (x) + 2 \, {\left (a b + a c + {\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2} + c^{2}} {\left ({\left (b + c\right )} \cosh \relax (x) + {\left (b + c\right )} \sinh \relax (x) + a\right )}}{{\left (b + c\right )} \cosh \relax (x)^{2} + {\left (b + c\right )} \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left ({\left (b + c\right )} \cosh \relax (x) + a\right )} \sinh \relax (x) + b - c}\right ) - {\left (B a^{2} b - B b^{3} + B b c^{2} + B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} x + {\left (B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + c \sinh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}, -\frac {2 \, {\left (B a b - A b^{2} + A c^{2}\right )} \sqrt {-a^{2} + b^{2} - c^{2}} \arctan \left (\frac {\sqrt {-a^{2} + b^{2} - c^{2}} {\left ({\left (b + c\right )} \cosh \relax (x) + {\left (b + c\right )} \sinh \relax (x) + a\right )}}{a^{2} - b^{2} + c^{2}}\right ) - {\left (B a^{2} b - B b^{3} + B b c^{2} + B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} x + {\left (B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + c \sinh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((B*a*b - A*b^2 + A*c^2)*sqrt(a^2 - b^2 + c^2)*log(((b^2 + 2*b*c + c^2)*cosh(x)^2 + (b^2 + 2*b*c + c^2)*sinh
(x)^2 + 2*a^2 - b^2 + c^2 + 2*(a*b + a*c)*cosh(x) + 2*(a*b + a*c + (b^2 + 2*b*c + c^2)*cosh(x))*sinh(x) - 2*sq
rt(a^2 - b^2 + c^2)*((b + c)*cosh(x) + (b + c)*sinh(x) + a))/((b + c)*cosh(x)^2 + (b + c)*sinh(x)^2 + 2*a*cosh
(x) + 2*((b + c)*cosh(x) + a)*sinh(x) + b - c)) - (B*a^2*b - B*b^3 + B*b*c^2 + B*c^3 + (B*a^2 - B*b^2)*c)*x +
(B*c^3 + (B*a^2 - B*b^2)*c)*log(2*(b*cosh(x) + c*sinh(x) + a)/(cosh(x) - sinh(x))))/(a^2*b^2 - b^4 - c^4 - (a^
2 - 2*b^2)*c^2), -(2*(B*a*b - A*b^2 + A*c^2)*sqrt(-a^2 + b^2 - c^2)*arctan(sqrt(-a^2 + b^2 - c^2)*((b + c)*cos
h(x) + (b + c)*sinh(x) + a)/(a^2 - b^2 + c^2)) - (B*a^2*b - B*b^3 + B*b*c^2 + B*c^3 + (B*a^2 - B*b^2)*c)*x + (
B*c^3 + (B*a^2 - B*b^2)*c)*log(2*(b*cosh(x) + c*sinh(x) + a)/(cosh(x) - sinh(x))))/(a^2*b^2 - b^4 - c^4 - (a^2
- 2*b^2)*c^2)]

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giac [A]  time = 0.14, size = 122, normalized size = 1.01 \[ -\frac {B c \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}{b^{2} - c^{2}} + \frac {B x}{b - c} - \frac {2 \, {\left (B a b - A b^{2} + A c^{2}\right )} \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt {-a^{2} + b^{2} - c^{2}} {\left (b^{2} - c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.20, size = 574, normalized size = 4.74 \[ -\frac {2 B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 c}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 c \tanh \left (\frac {x}{2}\right )-a -b \right ) a B c}{\left (b -c \right ) \left (b +c \right ) \left (a -b \right )}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 c \tanh \left (\frac {x}{2}\right )-a -b \right ) b B c}{\left (b -c \right ) \left (b +c \right ) \left (a -b \right )}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) A \,b^{2}}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) A \,c^{2}}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) b B a}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) B \,c^{2}}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) c^{2} a B}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}\, \left (a -b \right )}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) c^{2} b B}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}\, \left (a -b \right )}+\frac {2 B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b -2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/(a+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(c^2-b^2+a^2>0)', see `assume?`
for more details)Is c^2-b^2+a^2 positive or negative?

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mupad [B]  time = 0.67, size = 375, normalized size = 3.10 \[ \frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}+a^2\,{\mathrm {e}}^x-b^2\,{\mathrm {e}}^x+c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3-A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c+A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c+B\,a\,b\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}-a^2\,{\mathrm {e}}^x+b^2\,{\mathrm {e}}^x-c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3+A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c-A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c-B\,a\,b\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {B\,x}{b-c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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