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

Optimal. Leaf size=137 \[ -\frac {2 \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right ) \left (-a b B+a c C+A b^2-A c^2\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {x (b B-c C)}{b^2-c^2} \]

[Out]

(B*b-C*c)*x/(b^2-c^2)-(B*c-C*b)*ln(a+b*cosh(x)+c*sinh(x))/(b^2-c^2)-2*(A*b^2-A*c^2-B*a*b+C*a*c)*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 = 0.22, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3136, 3124, 618, 206} \[ -\frac {2 \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right ) \left (-a b B+a c C+A b^2-A c^2\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {x (b B-c C)}{b^2-c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.34, size = 119, normalized size = 0.87 \[ \frac {\frac {2 \left (-a b B+a c C+A b^2-A c^2\right ) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2-c^2}}\right )}{\sqrt {-a^2+b^2-c^2}}+(b C-B c) \log (a+b \cosh (x)+c \sinh (x))+x (b B-c C)}{(b-c) (b+c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 605, normalized size = 4.42 \[ \left [\frac {{\left (B a b - A b^{2} - C a c + 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 ({\left (B - C\right )} a^{2} b - {\left (B - C\right )} b^{3} + {\left (B - C\right )} b c^{2} + {\left (B - C\right )} c^{3} + {\left ({\left (B - C\right )} a^{2} - {\left (B - C\right )} b^{2}\right )} c\right )} x + {\left (C a^{2} b - C b^{3} + C b c^{2} - 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} - C a c + 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 ({\left (B - C\right )} a^{2} b - {\left (B - C\right )} b^{3} + {\left (B - C\right )} b c^{2} + {\left (B - C\right )} c^{3} + {\left ({\left (B - C\right )} a^{2} - {\left (B - C\right )} b^{2}\right )} c\right )} x - {\left (C a^{2} b - C b^{3} + C b c^{2} - 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)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="fricas")

[Out]

[((B*a*b - A*b^2 - C*a*c + 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*sqrt(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 - C)*a^2*b - (B - C)*b^3 + (B - C)*b*c^2 + (B - C
)*c^3 + ((B - C)*a^2 - (B - C)*b^2)*c)*x + (C*a^2*b - C*b^3 + C*b*c^2 - B*c^3 - (B*a^2 - B*b^2)*c)*log(2*(b*co
sh(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 - C
*a*c + A*c^2)*sqrt(-a^2 + b^2 - c^2)*arctan(sqrt(-a^2 + b^2 - c^2)*((b + c)*cosh(x) + (b + c)*sinh(x) + a)/(a^
2 - b^2 + c^2)) - ((B - C)*a^2*b - (B - C)*b^3 + (B - C)*b*c^2 + (B - C)*c^3 + ((B - C)*a^2 - (B - C)*b^2)*c)*
x - (C*a^2*b - C*b^3 + C*b*c^2 - 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 = 136, normalized size = 0.99 \[ \frac {{\left (B - C\right )} x}{b - c} + \frac {{\left (C b - B c\right )} \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}{b^{2} - c^{2}} - \frac {2 \, {\left (B a b - A b^{2} - C a c + A c^{2}\right )} \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt {-a^{2} + b^{2} - c^{2}} {\left (b^{2} - c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B - C)*x/(b - c) + (C*b - B*c)*log(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c)/(b^2 - c^2) - 2*(B*a*b - A*b^2 -
C*a*c + 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 = 1009, normalized size = 7.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x)+C*sinh(x))/(a+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)-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+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)*C*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*b^2+2/(b-c)/(b+c)/(-a^2+b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*ta
nh(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))*a*c*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))*C*c
*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)*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-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/(a-b)*a*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))*c/(a-b)*C*b^2+2*B/(
2*b-2*c)*ln(tanh(1/2*x)+1)-2*C/(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)+C*sinh(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 = 2.45, size = 454, normalized size = 3.31 \[ \frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}+a^2\,{\mathrm {e}}^x-b^2\,{\mathrm {e}}^x+c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3+C\,b^3-A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c-C\,a^2\,b+A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c-C\,b\,c^2+B\,a\,b\,\sqrt {a^2-b^2+c^2}-C\,a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}-a^2\,{\mathrm {e}}^x+b^2\,{\mathrm {e}}^x-c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3+C\,b^3+A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c-C\,a^2\,b-A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c-C\,b\,c^2-B\,a\,b\,\sqrt {a^2-b^2+c^2}+C\,a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {x\,\left (B-C\right )}{b-c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(x) + C*sinh(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 + C*b^3 - A*b^2*(a^2 - b^2 + c^2)^(1/2) + B*a^2*c - C*a^2*b + A*c^2*(a^2 - b^2
+ c^2)^(1/2) - B*b^2*c - C*b*c^2 + B*a*b*(a^2 - b^2 + c^2)^(1/2) - C*a*c*(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 + C*b^3 + A*b^2*(a^2 - b^2 + c^2)^(1/2) + B*
a^2*c - C*a^2*b - A*c^2*(a^2 - b^2 + c^2)^(1/2) - B*b^2*c - C*b*c^2 - B*a*b*(a^2 - b^2 + c^2)^(1/2) + C*a*c*(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) + (x*(B - C))/(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)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x)

[Out]

Timed out

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