### 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[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]))

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Rubi [A]  time = 0.0875916, antiderivative size = 90, normalized size of antiderivative = 1., 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 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]

Rule 12

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

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 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 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])

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*}

Mathematica [A]  time = 0.267378, 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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Maple [B]  time = 0.061, size = 191, normalized size = 2.1 \begin{align*} -2\,{\frac{1}{a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}- \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b-2\,c\tanh \left ( x/2 \right ) -a-b} \left ( -{\frac{ \left ( ab-{b}^{2}+{c}^{2} \right ) \tanh \left ( x/2 \right ) }{{a}^{3}-{a}^{2}b-a{b}^{2}+a{c}^{2}+{b}^{3}-b{c}^{2}}}-{\frac{ac}{{a}^{3}-{a}^{2}b-a{b}^{2}+a{c}^{2}+{b}^{3}-b{c}^{2}}} \right ) }-2\,{\frac{a}{ \left ({a}^{2}-{b}^{2}+{c}^{2} \right ) \sqrt{-{a}^{2}+{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tanh \left ( x/2 \right ) -2\,c}{\sqrt{-{a}^{2}+{b}^{2}-{c}^{2}}}} \right ) } \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.19876, size = 2912, normalized size = 32.36 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.13329, size = 150, normalized size = 1.67 \begin{align*} \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 )}} \end{align*}

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))