### 3.758 $$\int \frac{1}{(\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x))^2} \, dx$$

Optimal. Leaf size=100 $\frac{b \sinh (x)+c \cosh (x)}{3 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2}-\frac{\sqrt{b^2-c^2} \sinh (x)+c}{3 c \sqrt{b^2-c^2} (b \sinh (x)+c \cosh (x))}$

[Out]

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

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Rubi [A]  time = 0.0806866, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {3116, 3114} $\frac{b \sinh (x)+c \cosh (x)}{3 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2}-\frac{\sqrt{b^2-c^2} \sinh (x)+c}{3 c \sqrt{b^2-c^2} (b \sinh (x)+c \cosh (x))}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3116

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)/(a*e*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n +
1)), Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 -
c^2, 0] && LtQ[n, -1]

Rule 3114

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

Rubi steps

\begin{align*} \int \frac{1}{\left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2} \, dx &=\frac{c \cosh (x)+b \sinh (x)}{3 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2}+\frac{\int \frac{1}{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)} \, dx}{3 \sqrt{b^2-c^2}}\\ &=\frac{c \cosh (x)+b \sinh (x)}{3 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2}-\frac{c+\sqrt{b^2-c^2} \sinh (x)}{3 c \sqrt{b^2-c^2} (c \cosh (x)+b \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.143531, size = 68, normalized size = 0.68 $-\frac{-2 c \sqrt{b^2-c^2}+b^2 \sinh ^3(x)+2 b c \cosh ^3(x)+2 c^2 \sinh (x)+c^2 \sinh (x) \cosh ^2(x)}{3 c (b \sinh (x)+c \cosh (x))^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.086, size = 217, normalized size = 2.2 \begin{align*} 2\,{\frac{\sqrt{{b}^{2}-{c}^{2}}+b}{{c}^{2}} \left ({\frac{ \left ( \sqrt{{b}^{2}-{c}^{2}}+b \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}}{{c}^{2}}}+{\frac{ \left ( 2\,{b}^{2}-{c}^{2}+2\,\sqrt{{b}^{2}-{c}^{2}}b \right ) \tanh \left ( x/2 \right ) }{{c}^{3}}}+2/3\,{\frac{2\,\sqrt{{b}^{2}-{c}^{2}}{b}^{2}-\sqrt{{b}^{2}-{c}^{2}}{c}^{2}+2\,{b}^{3}-2\,b{c}^{2}}{{c}^{4}}} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+2\,{\frac{\sqrt{ \left ( b-c \right ) \left ( b+c \right ) }\tanh \left ( x/2 \right ) }{c}}+2\,{\frac{\tanh \left ( x/2 \right ) b}{c}}+2\,{\frac{\sqrt{ \left ( b-c \right ) \left ( b+c \right ) }b}{{c}^{2}}}+2\,{\frac{{b}^{2}}{{c}^{2}}}-1 \right ) ^{-1} \left ( \tanh \left ( x/2 \right ) +{\frac{\sqrt{ \left ( b-c \right ) \left ( b+c \right ) }}{c}}+{\frac{b}{c}} \right ) ^{-1}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((b^2-c^2)^(1/2)+b)/c^2*(((b^2-c^2)^(1/2)+b)/c^2*tanh(1/2*x)^2+(2*b^2-c^2+2*(b^2-c^2)^(1/2)*b)/c^3*tanh(1/2*
x)+2/3*(2*(b^2-c^2)^(1/2)*b^2-(b^2-c^2)^(1/2)*c^2+2*b^3-2*b*c^2)/c^4)/(tanh(1/2*x)^2+2/c*((b-c)*(b+c))^(1/2)*t
anh(1/2*x)+2*b/c*tanh(1/2*x)+2/c^2*((b-c)*(b+c))^(1/2)*b+2/c^2*b^2-1)/(tanh(1/2*x)+1/c*((b-c)*(b+c))^(1/2)+b/c
)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.05014, size = 1639, normalized size = 16.39 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*(b^2 + 2*b*c + c^2)*cosh(x)^4 + 12*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x)^3 + 3*(b^2 + 2*b*c + c^2)*sinh(
x)^4 + 6*(b^2 - c^2)*cosh(x)^2 + 6*(3*(b^2 + 2*b*c + c^2)*cosh(x)^2 + b^2 - c^2)*sinh(x)^2 - b^2 + 2*b*c - c^2
+ 12*((b^2 + 2*b*c + c^2)*cosh(x)^3 + (b^2 - c^2)*cosh(x))*sinh(x) - 8*((b + c)*cosh(x)^3 + 3*(b + c)*cosh(x)
^2*sinh(x) + 3*(b + c)*cosh(x)*sinh(x)^2 + (b + c)*sinh(x)^3)*sqrt(b^2 - c^2))/((b^4 + 4*b^3*c + 6*b^2*c^2 + 4
*b*c^3 + c^4)*cosh(x)^6 + 6*(b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c^3 + c^4)*cosh(x)*sinh(x)^5 + (b^4 + 4*b^3*c + 6
*b^2*c^2 + 4*b*c^3 + c^4)*sinh(x)^6 - 3*(b^4 + 2*b^3*c - 2*b*c^3 - c^4)*cosh(x)^4 - 3*(b^4 + 2*b^3*c - 2*b*c^3
- c^4 - 5*(b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c^3 + c^4)*cosh(x)^2)*sinh(x)^4 - b^4 + 2*b^3*c - 2*b*c^3 + c^4 +
4*(5*(b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c^3 + c^4)*cosh(x)^3 - 3*(b^4 + 2*b^3*c - 2*b*c^3 - c^4)*cosh(x))*sinh(x
)^3 + 3*(b^4 - 2*b^2*c^2 + c^4)*cosh(x)^2 + 3*(5*(b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c^3 + c^4)*cosh(x)^4 + b^4 -
2*b^2*c^2 + c^4 - 6*(b^4 + 2*b^3*c - 2*b*c^3 - c^4)*cosh(x)^2)*sinh(x)^2 + 6*((b^4 + 4*b^3*c + 6*b^2*c^2 + 4*
b*c^3 + c^4)*cosh(x)^5 - 2*(b^4 + 2*b^3*c - 2*b*c^3 - c^4)*cosh(x)^3 + (b^4 - 2*b^2*c^2 + c^4)*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/(b*cosh(x)+c*sinh(x)+(b**2-c**2)**(1/2))**2,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError