### 3.737 $$\int \frac{A+B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx$$

Optimal. Leaf size=88 $\frac{A b \sinh (x)+A c \cosh (x)-b C+B c}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}+\frac{(b B-c C) \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0686693, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {3153, 3074, 206} $\frac{A b \sinh (x)+A c \cosh (x)-b C+B c}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}+\frac{(b B-c C) \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3153

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - 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 - c*C)/(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, C}, x] && NeQ[
a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
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{A+B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx &=\frac{B c-b C+A c \cosh (x)+A b \sinh (x)}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}+\frac{(b B-c C) \int \frac{1}{b \cosh (x)+c \sinh (x)} \, dx}{b^2-c^2}\\ &=\frac{B c-b C+A c \cosh (x)+A b \sinh (x)}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}+\frac{(i (b B-c C)) \operatorname{Subst}\left (\int \frac{1}{b^2-c^2-x^2} \, dx,x,-i c \cosh (x)-i b \sinh (x)\right )}{b^2-c^2}\\ &=\frac{(b B-c C) \tan ^{-1}\left (\frac{c \cosh (x)+b \sinh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}+\frac{B c-b C+A c \cosh (x)+A b \sinh (x)}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.261468, size = 106, normalized size = 1.2 $\frac{A \left (b^2-c^2\right ) \sinh (x)+b (B c-b C)}{b (b-c) (b+c) (b \cosh (x)+c \sinh (x))}+\frac{2 (b B-c C) \tan ^{-1}\left (\frac{b \tanh \left (\frac{x}{2}\right )+c}{\sqrt{b-c} \sqrt{b+c}}\right )}{(b-c)^{3/2} (b+c)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.073, size = 167, normalized size = 1.9 \begin{align*} -2\,{\frac{1}{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+2\,c\tanh \left ( x/2 \right ) +b} \left ( -{\frac{ \left ( A{b}^{2}-A{c}^{2}+B{c}^{2}-Ccb \right ) \tanh \left ( x/2 \right ) }{b \left ({b}^{2}-{c}^{2} \right ) }}-{\frac{Bc-bC}{{b}^{2}-{c}^{2}}} \right ) }+2\,{\frac{Bb}{ \left ({b}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b+2\,c}{\sqrt{{b}^{2}-{c}^{2}}}} \right ) }-2\,{\frac{Cc}{ \left ({b}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b+2\,c}{\sqrt{{b}^{2}-{c}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-(A*b^2-A*c^2+B*c^2-C*b*c)/b/(b^2-c^2)*tanh(1/2*x)-(B*c-C*b)/(b^2-c^2))/(tanh(1/2*x)^2*b+2*c*tanh(1/2*x)+b
)+2*b*B/(b^2-c^2)^(3/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*c)/(b^2-c^2)^(1/2))-2*C*c/(b^2-c^2)^(3/2)*arctan(1/2*(2*
tanh(1/2*x)*b+2*c)/(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((A+B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.20108, size = 1897, normalized size = 21.56 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((A+B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20307, size = 128, normalized size = 1.45 \begin{align*} \frac{2 \,{\left (B b - C c\right )} \arctan \left (\frac{b e^{x} + c e^{x}}{\sqrt{b^{2} - c^{2}}}\right )}{{\left (b^{2} - c^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (C b e^{x} - B c e^{x} + A b - A c\right )}}{{\left (b^{2} - c^{2}\right )}{\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + b - c\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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