3.743 \(\int \frac {1}{(a+b \cosh (x)+c \sinh (x))^2} \, dx\)

Optimal. Leaf size=90 \[ -\frac {2 a \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {b \sinh (x)+c \cosh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3129, 12, 3124, 618, 206} \[ -\frac {2 a \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {b \sinh (x)+c \cosh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3129

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[((-(c*Cos[d
+ e*x]) + b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] +
Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c*(n + 2)*Sin[d + e*x])*(a + b*C
os[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n
, -1] && NeQ[n, -3/2]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cosh (x)+c \sinh (x))^2} \, dx &=-\frac {c \cosh (x)+b \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}+\frac {\int \frac {a}{a+b \cosh (x)+c \sinh (x)} \, dx}{a^2-b^2+c^2}\\ &=-\frac {c \cosh (x)+b \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}+\frac {a \int \frac {1}{a+b \cosh (x)+c \sinh (x)} \, dx}{a^2-b^2+c^2}\\ &=-\frac {c \cosh (x)+b \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}+\frac {(2 a) \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 {c \cosh (x)+b \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}-\frac {(4 a) \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 \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 {c \cosh (x)+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 = 105, normalized size = 1.17 \[ \frac {\left (b^2-c^2\right ) \sinh (x)-a c}{b \left (-a^2+b^2-c^2\right ) (a+b \cosh (x)+c \sinh (x))}-\frac {2 a \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2-c^2}}\right )}{\left (-a^2+b^2-c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*ArcTan[(c + (-a + b)*Tanh[x/2])/Sqrt[-a^2 + b^2 - c^2]])/(-a^2 + b^2 - c^2)^(3/2) + (-(a*c) + (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 = 0.47, size = 1268, normalized size = 14.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)+c*sinh(x))^2,x, algorithm="fricas")

[Out]

[(2*a^2*b - 2*b^3 + 2*b*c^2 - 2*c^3 + (2*a^2*cosh(x) + (a*b + a*c)*cosh(x)^2 + (a*b + a*c)*sinh(x)^2 + a*b - a
*c + 2*(a^2 + (a*b + a*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)*sin
h(x)^2 + 2*a*cosh(x) + 2*((b + c)*cosh(x) + a)*sinh(x) + b - c)) - 2*(a^2 - b^2)*c + 2*(a^3 - a*b^2 + a*c^2)*c
osh(x) + 2*(a^3 - a*b^2 + a*c^2)*sinh(x))/(a^4*b - 2*a^2*b^3 + b^5 + b*c^4 - c^5 - 2*(a^2 - b^2)*c^3 + 2*(a^2*
b - b^3)*c^2 + (a^4*b - 2*a^2*b^3 + b^5 + b*c^4 + c^5 + 2*(a^2 - b^2)*c^3 + 2*(a^2*b - b^3)*c^2 + (a^4 - 2*a^2
*b^2 + b^4)*c)*cosh(x)^2 + (a^4*b - 2*a^2*b^3 + b^5 + b*c^4 + c^5 + 2*(a^2 - b^2)*c^3 + 2*(a^2*b - b^3)*c^2 +
(a^4 - 2*a^2*b^2 + b^4)*c)*sinh(x)^2 - (a^4 - 2*a^2*b^2 + b^4)*c + 2*(a^5 - 2*a^3*b^2 + a*b^4 + a*c^4 + 2*(a^3
- a*b^2)*c^2)*cosh(x) + 2*(a^5 - 2*a^3*b^2 + a*b^4 + a*c^4 + 2*(a^3 - a*b^2)*c^2 + (a^4*b - 2*a^2*b^3 + b^5 +
b*c^4 + c^5 + 2*(a^2 - b^2)*c^3 + 2*(a^2*b - b^3)*c^2 + (a^4 - 2*a^2*b^2 + b^4)*c)*cosh(x))*sinh(x)), 2*(a^2*
b - b^3 + b*c^2 - c^3 + (2*a^2*cosh(x) + (a*b + a*c)*cosh(x)^2 + (a*b + a*c)*sinh(x)^2 + a*b - a*c + 2*(a^2 +
(a*b + a*c)*cosh(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)) - (a^2 - b^2)*c + (a^3 - a*b^2 + a*c^2)*cosh(x) + (a^3 - a*b^2 + a*c^2)*sinh(x
))/(a^4*b - 2*a^2*b^3 + b^5 + b*c^4 - c^5 - 2*(a^2 - b^2)*c^3 + 2*(a^2*b - b^3)*c^2 + (a^4*b - 2*a^2*b^3 + b^5
+ b*c^4 + c^5 + 2*(a^2 - b^2)*c^3 + 2*(a^2*b - b^3)*c^2 + (a^4 - 2*a^2*b^2 + b^4)*c)*cosh(x)^2 + (a^4*b - 2*a
^2*b^3 + b^5 + b*c^4 + c^5 + 2*(a^2 - b^2)*c^3 + 2*(a^2*b - b^3)*c^2 + (a^4 - 2*a^2*b^2 + b^4)*c)*sinh(x)^2 -
(a^4 - 2*a^2*b^2 + b^4)*c + 2*(a^5 - 2*a^3*b^2 + a*b^4 + a*c^4 + 2*(a^3 - a*b^2)*c^2)*cosh(x) + 2*(a^5 - 2*a^3
*b^2 + a*b^4 + a*c^4 + 2*(a^3 - a*b^2)*c^2 + (a^4*b - 2*a^2*b^3 + b^5 + b*c^4 + c^5 + 2*(a^2 - b^2)*c^3 + 2*(a
^2*b - b^3)*c^2 + (a^4 - 2*a^2*b^2 + b^4)*c)*cosh(x))*sinh(x))]

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giac [A]  time = 0.12, size = 111, normalized size = 1.23 \[ \frac {2 \, a \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{{\left (a^{2} - b^{2} + c^{2}\right )} \sqrt {-a^{2} + b^{2} - c^{2}}} + \frac {2 \, {\left (a e^{x} + b - c\right )}}{{\left (a^{2} - b^{2} + c^{2}\right )} {\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)+c*sinh(x))^2,x, algorithm="giac")

[Out]

2*a*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*(a*e^x +
b - c)/((a^2 - b^2 + c^2)*(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c))

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maple [B]  time = 0.26, size = 191, normalized size = 2.12 \[ -\frac {2 \left (-\frac {\left (a b -b^{2}+c^{2}\right ) \tanh \left (\frac {x}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}+a \,c^{2}+b^{3}-b \,c^{2}}-\frac {a c}{a^{3}-a^{2} b -a \,b^{2}+a \,c^{2}+b^{3}-b \,c^{2}}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 c \tanh \left (\frac {x}{2}\right )-a -b}-\frac {2 a \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right )}{\left (a^{2}-b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}-c^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(x)+c*sinh(x))^2,x)

[Out]

-2*(-(a*b-b^2+c^2)/(a^3-a^2*b-a*b^2+a*c^2+b^3-b*c^2)*tanh(1/2*x)-a*c/(a^3-a^2*b-a*b^2+a*c^2+b^3-b*c^2))/(a*tan
h(1/2*x)^2-tanh(1/2*x)^2*b-2*c*tanh(1/2*x)-a-b)-2*a/(a^2-b^2+c^2)/(-a^2+b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tan
h(1/2*x)-2*c)/(-a^2+b^2-c^2)^(1/2))

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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(1/(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 \[ \int \frac {1}{{\left (a+b\,\mathrm {cosh}\relax (x)+c\,\mathrm {sinh}\relax (x)\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cosh(x) + c*sinh(x))^2,x)

[Out]

int(1/(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(1/(a+b*cosh(x)+c*sinh(x))**2,x)

[Out]

Timed out

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