### 3.735 $$\int \frac{B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^3} \, dx$$

Optimal. Leaf size=71 $\frac{B c-b C}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{\sinh (x) (b B-c C)}{b \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}$

[Out]

(B*c - b*C)/(2*(b^2 - c^2)*(b*Cosh[x] + c*Sinh[x])^2) + ((b*B - c*C)*Sinh[x])/(b*(b^2 - c^2)*(b*Cosh[x] + c*Si
nh[x]))

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Rubi [A]  time = 0.0687751, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {3156, 12, 3075} $\frac{B c-b C}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{\sinh (x) (b B-c C)}{b \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}$

Antiderivative was successfully veriﬁed.

[In]

Int[(B*Cosh[x] + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^3,x]

[Out]

(B*c - b*C)/(2*(b^2 - c^2)*(b*Cosh[x] + c*Sinh[x])^2) + ((b*B - c*C)*Sinh[x])/(b*(b^2 - c^2)*(b*Cosh[x] + c*Si
nh[x]))

Rule 3156

Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_)*((A_.) + cos[(d_.) + (e_.)*(x
_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> -Simp[((c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B -
b*A)*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 + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B - c*C)
+ (n + 2)*(a*B - b*A)*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A,
B, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -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 3075

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

Rubi steps

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

Mathematica [A]  time = 0.158388, size = 70, normalized size = 0.99 $\frac{C \left (c^2-b^2\right )+b \sinh (2 x) (b B-c C)+c \cosh (2 x) (b B-c C)}{2 b (b-c) (b+c) (b \cosh (x)+c \sinh (x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(B*Cosh[x] + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^3,x]

[Out]

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

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Maple [A]  time = 0.083, size = 63, normalized size = 0.9 \begin{align*} -2\,{\frac{1}{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+2\,c\tanh \left ( x/2 \right ) +b \right ) ^{2}} \left ( -{\frac{B \left ( \tanh \left ( x/2 \right ) \right ) ^{3}}{b}}-{\frac{ \left ( Bc+bC \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}}{{b}^{2}}}-{\frac{B\tanh \left ( x/2 \right ) }{b}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x)

[Out]

-2*(-B/b*tanh(1/2*x)^3-(B*c+C*b)/b^2*tanh(1/2*x)^2-B/b*tanh(1/2*x))/(tanh(1/2*x)^2*b+2*c*tanh(1/2*x)+b)^2

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Maxima [B]  time = 1.11157, size = 455, normalized size = 6.41 \begin{align*} 2 \, B{\left (\frac{{\left (b - c\right )} e^{\left (-2 \, x\right )}}{b^{4} - 2 \, b^{2} c^{2} + c^{4} + 2 \,{\left (b^{4} - 2 \, b^{3} c + 2 \, b c^{3} - c^{4}\right )} e^{\left (-2 \, x\right )} +{\left (b^{4} - 4 \, b^{3} c + 6 \, b^{2} c^{2} - 4 \, b c^{3} + c^{4}\right )} e^{\left (-4 \, x\right )}} + \frac{b}{b^{4} - 2 \, b^{2} c^{2} + c^{4} + 2 \,{\left (b^{4} - 2 \, b^{3} c + 2 \, b c^{3} - c^{4}\right )} e^{\left (-2 \, x\right )} +{\left (b^{4} - 4 \, b^{3} c + 6 \, b^{2} c^{2} - 4 \, b c^{3} + c^{4}\right )} e^{\left (-4 \, x\right )}}\right )} - 2 \, C{\left (\frac{{\left (b - c\right )} e^{\left (-2 \, x\right )}}{b^{4} - 2 \, b^{2} c^{2} + c^{4} + 2 \,{\left (b^{4} - 2 \, b^{3} c + 2 \, b c^{3} - c^{4}\right )} e^{\left (-2 \, x\right )} +{\left (b^{4} - 4 \, b^{3} c + 6 \, b^{2} c^{2} - 4 \, b c^{3} + c^{4}\right )} e^{\left (-4 \, x\right )}} + \frac{c}{b^{4} - 2 \, b^{2} c^{2} + c^{4} + 2 \,{\left (b^{4} - 2 \, b^{3} c + 2 \, b c^{3} - c^{4}\right )} e^{\left (-2 \, x\right )} +{\left (b^{4} - 4 \, b^{3} c + 6 \, b^{2} c^{2} - 4 \, b c^{3} + c^{4}\right )} e^{\left (-4 \, x\right )}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="maxima")

[Out]

2*B*((b - c)*e^(-2*x)/(b^4 - 2*b^2*c^2 + c^4 + 2*(b^4 - 2*b^3*c + 2*b*c^3 - c^4)*e^(-2*x) + (b^4 - 4*b^3*c + 6
*b^2*c^2 - 4*b*c^3 + c^4)*e^(-4*x)) + b/(b^4 - 2*b^2*c^2 + c^4 + 2*(b^4 - 2*b^3*c + 2*b*c^3 - c^4)*e^(-2*x) +
(b^4 - 4*b^3*c + 6*b^2*c^2 - 4*b*c^3 + c^4)*e^(-4*x))) - 2*C*((b - c)*e^(-2*x)/(b^4 - 2*b^2*c^2 + c^4 + 2*(b^4
- 2*b^3*c + 2*b*c^3 - c^4)*e^(-2*x) + (b^4 - 4*b^3*c + 6*b^2*c^2 - 4*b*c^3 + c^4)*e^(-4*x)) + c/(b^4 - 2*b^2*
c^2 + c^4 + 2*(b^4 - 2*b^3*c + 2*b*c^3 - c^4)*e^(-2*x) + (b^4 - 4*b^3*c + 6*b^2*c^2 - 4*b*c^3 + c^4)*e^(-4*x))
)

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Fricas [B]  time = 1.98492, size = 556, normalized size = 7.83 \begin{align*} -\frac{2 \,{\left ({\left ({\left (2 \, B + C\right )} b + B c\right )} \cosh \left (x\right ) +{\left (C b +{\left (B + 2 \, C\right )} c\right )} \sinh \left (x\right )\right )}}{{\left (b^{4} + 4 \, b^{3} c + 6 \, b^{2} c^{2} + 4 \, b c^{3} + c^{4}\right )} \cosh \left (x\right )^{3} + 3 \,{\left (b^{4} + 4 \, b^{3} c + 6 \, b^{2} c^{2} + 4 \, b c^{3} + c^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{2} +{\left (b^{4} + 4 \, b^{3} c + 6 \, b^{2} c^{2} + 4 \, b c^{3} + c^{4}\right )} \sinh \left (x\right )^{3} +{\left (3 \, b^{4} + 4 \, b^{3} c - 2 \, b^{2} c^{2} - 4 \, b c^{3} - c^{4}\right )} \cosh \left (x\right ) +{\left (b^{4} + 4 \, b^{3} c + 2 \, b^{2} c^{2} - 4 \, b c^{3} - 3 \, c^{4} + 3 \,{\left (b^{4} + 4 \, b^{3} c + 6 \, b^{2} c^{2} + 4 \, b c^{3} + c^{4}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="fricas")

[Out]

-2*(((2*B + C)*b + B*c)*cosh(x) + (C*b + (B + 2*C)*c)*sinh(x))/((b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c^3 + c^4)*co
sh(x)^3 + 3*(b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c^3 + c^4)*cosh(x)*sinh(x)^2 + (b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c
^3 + c^4)*sinh(x)^3 + (3*b^4 + 4*b^3*c - 2*b^2*c^2 - 4*b*c^3 - c^4)*cosh(x) + (b^4 + 4*b^3*c + 2*b^2*c^2 - 4*b
*c^3 - 3*c^4 + 3*(b^4 + 4*b^3*c + 6*b^2*c^2 + 4*b*c^3 + c^4)*cosh(x)^2)*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((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20253, size = 95, normalized size = 1.34 \begin{align*} -\frac{2 \,{\left (B b e^{\left (2 \, x\right )} + C b e^{\left (2 \, x\right )} + B c e^{\left (2 \, x\right )} + C c e^{\left (2 \, x\right )} + B b - C c\right )}}{{\left (b^{2} + 2 \, b c + c^{2}\right )}{\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + b - c\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="giac")

[Out]

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