### 3.729 $$\int \frac{A+B \cosh (x)}{(b \cosh (x)+c \sinh (x))^3} \, dx$$

Optimal. Leaf size=120 $\frac{A b \sinh (x)+A c \cosh (x)+B c}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{A \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{2 \left (b^2-c^2\right )^{3/2}}+\frac{b^2 B \sinh (x)+b B c \cosh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}$

[Out]

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

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Rubi [A]  time = 0.118954, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {3158, 3153, 3074, 206} $\frac{A b \sinh (x)+A c \cosh (x)+B c}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{A \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{2 \left (b^2-c^2\right )^{3/2}}+\frac{b^2 B \sinh (x)+b B c \cosh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3158

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

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)}{(b \cosh (x)+c \sinh (x))^3} \, dx &=\frac{B c+A c \cosh (x)+A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{\int \frac{2 b B+A b \cosh (x)+A c \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx}{2 \left (b^2-c^2\right )}\\ &=\frac{B c+A c \cosh (x)+A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{b B c \cosh (x)+b^2 B \sinh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}+\frac{A \int \frac{1}{b \cosh (x)+c \sinh (x)} \, dx}{2 \left (b^2-c^2\right )}\\ &=\frac{B c+A c \cosh (x)+A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{b B c \cosh (x)+b^2 B \sinh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}+\frac{(i A) \operatorname{Subst}\left (\int \frac{1}{b^2-c^2-x^2} \, dx,x,-i c \cosh (x)-i b \sinh (x)\right )}{2 \left (b^2-c^2\right )}\\ &=\frac{A \tan ^{-1}\left (\frac{c \cosh (x)+b \sinh (x)}{\sqrt{b^2-c^2}}\right )}{2 \left (b^2-c^2\right )^{3/2}}+\frac{B c+A c \cosh (x)+A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac{b B c \cosh (x)+b^2 B \sinh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 1.03738, size = 134, normalized size = 1.12 $\frac{1}{2} \left (\frac{A \left (b^2-c^2\right ) \sinh (x)+b B c}{b (b-c) (b+c) (b \cosh (x)+c \sinh (x))^2}+\frac{A c+2 b B \sinh (x)}{b (b-c) (b+c) (b \cosh (x)+c \sinh (x))}+\frac{2 A \tan ^{-1}\left (\frac{b \tanh \left (\frac{x}{2}\right )+c}{\sqrt{b-c} \sqrt{b+c}}\right )}{(b-c)^{3/2} (b+c)^{3/2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.094, size = 214, normalized size = 1.8 \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 ( -1/2\,{\frac{ \left ( A{b}^{2}-2\,A{c}^{2}-2\,B{b}^{2}+2\,B{c}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}}{ \left ({b}^{2}-{c}^{2} \right ) b}}+1/2\,{\frac{c \left ( A{b}^{2}+2\,A{c}^{2}+2\,B{b}^{2}-2\,B{c}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}}{ \left ({b}^{2}-{c}^{2} \right ){b}^{2}}}+1/2\,{\frac{ \left ( A{b}^{2}+2\,A{c}^{2}+2\,B{b}^{2}-2\,B{c}^{2} \right ) \tanh \left ( x/2 \right ) }{ \left ({b}^{2}-{c}^{2} \right ) b}}+1/2\,{\frac{Ac}{{b}^{2}-{c}^{2}}} \right ) }+{A\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( x/2 \right ) b+2\,c \right ){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}} \right ) \left ({b}^{2}-{c}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.7187, size = 4246, normalized size = 35.38 \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))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="fricas")

[Out]

[-1/2*(4*B*b^3 - 8*B*b^2*c + 4*B*b*c^2 - 2*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^3 - 2*(A*b^3 + A*b^2*c
- A*b*c^2 - A*c^3)*sinh(x)^3 + 4*(B*b^3 - B*b^2*c - B*b*c^2 + B*c^3)*cosh(x)^2 + 2*(2*B*b^3 - 2*B*b^2*c - 2*B*
b*c^2 + 2*B*c^3 - 3*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x))*sinh(x)^2 - ((A*b^2 + 2*A*b*c + A*c^2)*cosh(x
)^4 + 4*(A*b^2 + 2*A*b*c + A*c^2)*cosh(x)*sinh(x)^3 + (A*b^2 + 2*A*b*c + A*c^2)*sinh(x)^4 + A*b^2 - 2*A*b*c +
A*c^2 + 2*(A*b^2 - A*c^2)*cosh(x)^2 + 2*(A*b^2 - A*c^2 + 3*(A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^2)*sinh(x)^2 + 4*
((A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^3 + (A*b^2 - A*c^2)*cosh(x))*sinh(x))*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*(A*b^3 - A*b^2*c - A*b*c^2 + A*c^3)*co
sh(x) + 2*(A*b^3 - A*b^2*c - A*b*c^2 + A*c^3 - 3*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^2 + 4*(B*b^3 - B*
b^2*c - B*b*c^2 + B*c^3)*cosh(x))*sinh(x))/(b^6 - 2*b^5*c - b^4*c^2 + 4*b^3*c^3 - b^2*c^4 - 2*b*c^5 + c^6 + (b
^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)^4 + 4*(b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c
^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)*sinh(x)^3 + (b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c
^6)*sinh(x)^4 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6)*cosh(x)^2 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6 + 3*(b^
6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)^2)*sinh(x)^2 + 4*((b^6 + 2*b^5*c - b^4*c^
2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)^3 + (b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6)*cosh(x))*sinh(x)), -(
2*B*b^3 - 4*B*b^2*c + 2*B*b*c^2 - (A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^3 - (A*b^3 + A*b^2*c - A*b*c^2 -
A*c^3)*sinh(x)^3 + 2*(B*b^3 - B*b^2*c - B*b*c^2 + B*c^3)*cosh(x)^2 + (2*B*b^3 - 2*B*b^2*c - 2*B*b*c^2 + 2*B*c
^3 - 3*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x))*sinh(x)^2 + ((A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^4 + 4*(A*b^
2 + 2*A*b*c + A*c^2)*cosh(x)*sinh(x)^3 + (A*b^2 + 2*A*b*c + A*c^2)*sinh(x)^4 + A*b^2 - 2*A*b*c + A*c^2 + 2*(A*
b^2 - A*c^2)*cosh(x)^2 + 2*(A*b^2 - A*c^2 + 3*(A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^2)*sinh(x)^2 + 4*((A*b^2 + 2*A
*b*c + A*c^2)*cosh(x)^3 + (A*b^2 - A*c^2)*cosh(x))*sinh(x))*sqrt(b^2 - c^2)*arctan(sqrt(b^2 - c^2)/((b + c)*co
sh(x) + (b + c)*sinh(x))) + (A*b^3 - A*b^2*c - A*b*c^2 + A*c^3)*cosh(x) + (A*b^3 - A*b^2*c - A*b*c^2 + A*c^3 -
3*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^2 + 4*(B*b^3 - B*b^2*c - B*b*c^2 + B*c^3)*cosh(x))*sinh(x))/(b^
6 - 2*b^5*c - b^4*c^2 + 4*b^3*c^3 - b^2*c^4 - 2*b*c^5 + c^6 + (b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 +
2*b*c^5 + c^6)*cosh(x)^4 + 4*(b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)*sinh(x)^
3 + (b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*sinh(x)^4 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4
- c^6)*cosh(x)^2 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6 + 3*(b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 +
2*b*c^5 + c^6)*cosh(x)^2)*sinh(x)^2 + 4*((b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(
x)^3 + (b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6)*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((A+B*cosh(x))/(b*cosh(x)+c*sinh(x))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.16551, size = 205, normalized size = 1.71 \begin{align*} \frac{A \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{A b^{2} e^{\left (3 \, x\right )} + 2 \, A b c e^{\left (3 \, x\right )} + A c^{2} e^{\left (3 \, x\right )} - 2 \, B b^{2} e^{\left (2 \, x\right )} + 2 \, B c^{2} e^{\left (2 \, x\right )} - A b^{2} e^{x} + A c^{2} e^{x} - 2 \, B b^{2} + 2 \, B b c}{{\left (b^{3} + b^{2} c - b c^{2} - c^{3}\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((A+B*cosh(x))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="giac")

[Out]

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