3.793 \(\int \frac {A+B \cosh (x)}{(a+b \cosh (x)+c \sinh (x))^2} \, dx\)

Optimal. Leaf size=108 \[ -\frac {2 (a A-b B) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {\sinh (x) (A b-a B)+A c \cosh (x)+B c}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))} \]

[Out]

-2*(A*a-B*b)*arctanh((c-(a-b)*tanh(1/2*x))/(a^2-b^2+c^2)^(1/2))/(a^2-b^2+c^2)^(3/2)+(-B*c-A*c*cosh(x)-(A*b-B*a
)*sinh(x))/(a^2-b^2+c^2)/(a+b*cosh(x)+c*sinh(x))

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Rubi [A]  time = 0.13, antiderivative size = 108, 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 = {3155, 3124, 618, 206} \[ -\frac {2 (a A-b B) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {\sinh (x) (A b-a B)+A c \cosh (x)+B c}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x])/(a + b*Cosh[x] + c*Sinh[x])^2,x]

[Out]

(-2*(a*A - b*B)*ArcTanh[(c - (a - b)*Tanh[x/2])/Sqrt[a^2 - b^2 + c^2]])/(a^2 - b^2 + c^2)^(3/2) - (B*c + A*c*C
osh[x] + (A*b - a*B)*Sinh[x])/((a^2 - b^2 + c^2)*(a + b*Cosh[x] + c*Sinh[x]))

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 3155

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

Rubi steps

\begin {align*} \int \frac {A+B \cosh (x)}{(a+b \cosh (x)+c \sinh (x))^2} \, dx &=-\frac {B c+A c \cosh (x)+(A b-a B) \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}+\frac {(a A-b B) \int \frac {1}{a+b \cosh (x)+c \sinh (x)} \, dx}{a^2-b^2+c^2}\\ &=-\frac {B c+A c \cosh (x)+(A b-a B) \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}+\frac {(2 (a A-b B)) \operatorname {Subst}\left (\int \frac {1}{a+b+2 c x-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2+c^2}\\ &=-\frac {B c+A c \cosh (x)+(A b-a B) \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}-\frac {(4 (a A-b B)) \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 )}{a^2-b^2+c^2}\\ &=-\frac {2 (a A-b B) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {B c+A c \cosh (x)+(A b-a B) \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 125, normalized size = 1.16 \[ \frac {\sinh (x) \left (A \left (b^2-c^2\right )-a b B\right )-a A c+b B c}{b \left (-a^2+b^2-c^2\right ) (a+b \cosh (x)+c \sinh (x))}-\frac {2 (a A-b B) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2-c^2}}\right )}{\left (-a^2+b^2-c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[x])/(a + b*Cosh[x] + c*Sinh[x])^2,x]

[Out]

(-2*(a*A - b*B)*ArcTan[(c + (-a + b)*Tanh[x/2])/Sqrt[-a^2 + b^2 - c^2]])/(-a^2 + b^2 - c^2)^(3/2) + (-(a*A*c)
+ b*B*c + (-(a*b*B) + A*(b^2 - c^2))*Sinh[x])/(b*(-a^2 + b^2 - c^2)*(a + b*Cosh[x] + c*Sinh[x]))

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fricas [B]  time = 0.48, size = 2228, normalized size = 20.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x))^2,x, algorithm="fricas")

[Out]

[-(2*B*a^3*b - 2*A*a^2*b^2 - 2*B*a*b^3 + 2*A*b^4 + 2*A*c^4 + 2*(A*a^2 + B*a*b - 2*A*b^2)*c^2 + (A*a*b^2 - B*b^
3 - (A*a - B*b)*c^2 + (A*a*b^2 - B*b^3 + (A*a - B*b)*c^2 + 2*(A*a*b - B*b^2)*c)*cosh(x)^2 + (A*a*b^2 - B*b^3 +
(A*a - B*b)*c^2 + 2*(A*a*b - B*b^2)*c)*sinh(x)^2 + 2*(A*a^2*b - B*a*b^2 + (A*a^2 - B*a*b)*c)*cosh(x) + 2*(A*a
^2*b - B*a*b^2 + (A*a^2 - B*a*b)*c + (A*a*b^2 - B*b^3 + (A*a - B*b)*c^2 + 2*(A*a*b - B*b^2)*c)*cosh(x))*sinh(x
))*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)) + 2*(B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3 + B*c^4 - (A*a - B*b)*c^3 + (2*B*a^2 - A*a*
b - B*b^2)*c^2 - (A*a^3 - B*a^2*b - A*a*b^2 + B*b^3)*c)*cosh(x) + 2*(B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3 + B
*c^4 - (A*a - B*b)*c^3 + (2*B*a^2 - A*a*b - B*b^2)*c^2 - (A*a^3 - B*a^2*b - A*a*b^2 + B*b^3)*c)*sinh(x))/(a^4*
b^2 - 2*a^2*b^4 + b^6 - c^6 - (2*a^2 - 3*b^2)*c^4 - (a^4 - 4*a^2*b^2 + 3*b^4)*c^2 + (a^4*b^2 - 2*a^2*b^4 + b^6
+ 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*
cosh(x)^2 + (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)
*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*sinh(x)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + a*b*c^4 + a*c^5 + 2*(a^3 - a*
b^2)*c^3 + 2*(a^3*b - a*b^3)*c^2 + (a^5 - 2*a^3*b^2 + a*b^4)*c)*cosh(x) + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + a*b*c
^4 + a*c^5 + 2*(a^3 - a*b^2)*c^3 + 2*(a^3*b - a*b^3)*c^2 + (a^5 - 2*a^3*b^2 + a*b^4)*c + (a^4*b^2 - 2*a^2*b^4
+ b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5
)*c)*cosh(x))*sinh(x)), -2*(B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4 + A*c^4 + (A*a^2 + B*a*b - 2*A*b^2)*c^2 - (A
*a*b^2 - B*b^3 - (A*a - B*b)*c^2 + (A*a*b^2 - B*b^3 + (A*a - B*b)*c^2 + 2*(A*a*b - B*b^2)*c)*cosh(x)^2 + (A*a*
b^2 - B*b^3 + (A*a - B*b)*c^2 + 2*(A*a*b - B*b^2)*c)*sinh(x)^2 + 2*(A*a^2*b - B*a*b^2 + (A*a^2 - B*a*b)*c)*cos
h(x) + 2*(A*a^2*b - B*a*b^2 + (A*a^2 - B*a*b)*c + (A*a*b^2 - B*b^3 + (A*a - B*b)*c^2 + 2*(A*a*b - B*b^2)*c)*co
sh(x))*sinh(x))*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*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3 + B*c^4 - (A*a - B*b)*c^3 + (2*B*a^2 - A*a*b - B*b^
2)*c^2 - (A*a^3 - B*a^2*b - A*a*b^2 + B*b^3)*c)*cosh(x) + (B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3 + B*c^4 - (A*
a - B*b)*c^3 + (2*B*a^2 - A*a*b - B*b^2)*c^2 - (A*a^3 - B*a^2*b - A*a*b^2 + B*b^3)*c)*sinh(x))/(a^4*b^2 - 2*a^
2*b^4 + b^6 - c^6 - (2*a^2 - 3*b^2)*c^4 - (a^4 - 4*a^2*b^2 + 3*b^4)*c^2 + (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*b*c^5
+ c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*cosh(x)^2
+ (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(
a^4*b - 2*a^2*b^3 + b^5)*c)*sinh(x)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + a*b*c^4 + a*c^5 + 2*(a^3 - a*b^2)*c^3 +
2*(a^3*b - a*b^3)*c^2 + (a^5 - 2*a^3*b^2 + a*b^4)*c)*cosh(x) + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + a*b*c^4 + a*c^5
+ 2*(a^3 - a*b^2)*c^3 + 2*(a^3*b - a*b^3)*c^2 + (a^5 - 2*a^3*b^2 + a*b^4)*c + (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*
b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*cosh(
x))*sinh(x))]

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giac [A]  time = 0.12, size = 177, normalized size = 1.64 \[ \frac {2 \, {\left (A a - B b\right )} \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{{\left (a^{2} - b^{2} + c^{2}\right )} \sqrt {-a^{2} + b^{2} - c^{2}}} - \frac {2 \, {\left (B a^{2} e^{x} - A a b e^{x} - A a c e^{x} + B b c e^{x} + B c^{2} e^{x} + B a b - A b^{2} + A c^{2}\right )}}{{\left (a^{2} b - b^{3} + a^{2} c - b^{2} c + b c^{2} + c^{3}\right )} {\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x))^2,x, algorithm="giac")

[Out]

2*(A*a - B*b)*arctan((b*e^x + c*e^x + a)/sqrt(-a^2 + b^2 - c^2))/((a^2 - b^2 + c^2)*sqrt(-a^2 + b^2 - c^2)) -
2*(B*a^2*e^x - A*a*b*e^x - A*a*c*e^x + B*b*c*e^x + B*c^2*e^x + B*a*b - A*b^2 + A*c^2)/((a^2*b - b^3 + a^2*c -
b^2*c + b*c^2 + c^3)*(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c))

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maple [B]  time = 0.26, size = 287, normalized size = 2.66 \[ -\frac {2 \left (-\frac {\left (A a b -A \,b^{2}+A \,c^{2}-B \,a^{2}+b B a -B \,c^{2}\right ) \tanh \left (\frac {x}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}+a \,c^{2}+b^{3}-b \,c^{2}}-\frac {\left (A a -B b \right ) c}{a^{3}-a^{2} b -a \,b^{2}+a \,c^{2}+b^{3}-b \,c^{2}}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 c \tanh \left (\frac {x}{2}\right )-a -b}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) A a}{\left (a^{2}-b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}-c^{2}}}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) B b}{\left (a^{2}-b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}-c^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-(A*a*b-A*b^2+A*c^2-B*a^2+B*a*b-B*c^2)/(a^3-a^2*b-a*b^2+a*c^2+b^3-b*c^2)*tanh(1/2*x)-(A*a-B*b)*c/(a^3-a^2*
b-a*b^2+a*c^2+b^3-b*c^2))/(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-2*c*tanh(1/2*x)-a-b)-2/(a^2-b^2+c^2)/(-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*a+2/(a^2-b^2+c^2)/(-a^2+b^2-c^2)^(1/2)*arc
tan(1/2*(2*(a-b)*tanh(1/2*x)-2*c)/(-a^2+b^2-c^2)^(1/2))*B*b

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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))^2,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 [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {cosh}\relax (x)}{{\left (a+b\,\mathrm {cosh}\relax (x)+c\,\mathrm {sinh}\relax (x)\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))**2,x)

[Out]

Timed out

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