3.112 \(\int \frac{A+B \cosh (x)}{(a+b \cosh (x))^3} \, dx\)
Optimal. Leaf size=135 \[ \frac{\left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac{\sinh (x) \left (a^2 (-B)+3 a A b-2 b^2 B\right )}{2 \left (a^2-b^2\right )^2 (a+b \cosh (x))}-\frac{\sinh (x) (A b-a B)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))^2} \]
[Out]
((2*a^2*A + A*b^2 - 3*a*b*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) - ((A
*b - a*B)*Sinh[x])/(2*(a^2 - b^2)*(a + b*Cosh[x])^2) - ((3*a*A*b - a^2*B - 2*b^2*B)*Sinh[x])/(2*(a^2 - b^2)^2*
(a + b*Cosh[x]))
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Rubi [A] time = 0.16936, antiderivative size = 135, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used =
{2754, 12, 2659, 208} \[ \frac{\left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac{\sinh (x) \left (a^2 (-B)+3 a A b-2 b^2 B\right )}{2 \left (a^2-b^2\right )^2 (a+b \cosh (x))}-\frac{\sinh (x) (A b-a B)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Cosh[x])/(a + b*Cosh[x])^3,x]
[Out]
((2*a^2*A + A*b^2 - 3*a*b*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) - ((A
*b - a*B)*Sinh[x])/(2*(a^2 - b^2)*(a + b*Cosh[x])^2) - ((3*a*A*b - a^2*B - 2*b^2*B)*Sinh[x])/(2*(a^2 - b^2)^2*
(a + b*Cosh[x]))
Rule 2754
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 2659
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{A+B \cosh (x)}{(a+b \cosh (x))^3} \, dx &=-\frac{(A b-a B) \sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))^2}-\frac{\int \frac{-2 (a A-b B)+(A b-a B) \cosh (x)}{(a+b \cosh (x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{(A b-a B) \sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (x)}{2 \left (a^2-b^2\right )^2 (a+b \cosh (x))}+\frac{\int \frac{2 a^2 A+A b^2-3 a b B}{a+b \cosh (x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{(A b-a B) \sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (x)}{2 \left (a^2-b^2\right )^2 (a+b \cosh (x))}+\frac{\left (2 a^2 A+A b^2-3 a b B\right ) \int \frac{1}{a+b \cosh (x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{(A b-a B) \sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (x)}{2 \left (a^2-b^2\right )^2 (a+b \cosh (x))}+\frac{\left (2 a^2 A+A b^2-3 a b B\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac{\left (2 a^2 A+A b^2-3 a b B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac{(A b-a B) \sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (x)}{2 \left (a^2-b^2\right )^2 (a+b \cosh (x))}\\ \end{align*}
Mathematica [A] time = 0.395397, size = 134, normalized size = 0.99 \[ \frac{1}{2} \left (-\frac{2 \left (2 a^2 A-3 a b B+A b^2\right ) \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{\sinh (x) \left (a^2 B-3 a A b+2 b^2 B\right )}{(a-b)^2 (a+b)^2 (a+b \cosh (x))}+\frac{\sinh (x) (a B-A b)}{(a-b) (a+b) (a+b \cosh (x))^2}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Cosh[x])/(a + b*Cosh[x])^3,x]
[Out]
((-2*(2*a^2*A + A*b^2 - 3*a*b*B)*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + ((-(A*b) +
a*B)*Sinh[x])/((a - b)*(a + b)*(a + b*Cosh[x])^2) + ((-3*a*A*b + a^2*B + 2*b^2*B)*Sinh[x])/((a - b)^2*(a + b)
^2*(a + b*Cosh[x])))/2
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Maple [A] time = 0.023, size = 207, normalized size = 1.5 \begin{align*} -2\,{\frac{1}{ \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}- \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b-a-b \right ) ^{2}} \left ( -1/2\,{\frac{ \left ( 4\,Aab+A{b}^{2}-2\,{a}^{2}B-Bab-2\,B{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}}{ \left ( a-b \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+1/2\,{\frac{ \left ( 4\,Aab-A{b}^{2}-2\,{a}^{2}B+Bab-2\,B{b}^{2} \right ) \tanh \left ( x/2 \right ) }{ \left ( a+b \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }} \right ) }+{\frac{2\,A{a}^{2}+A{b}^{2}-3\,Bab}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}{\it Artanh} \left ({(a-b)\tanh \left ({\frac{x}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*cosh(x))/(a+b*cosh(x))^3,x)
[Out]
-2*(-1/2*(4*A*a*b+A*b^2-2*B*a^2-B*a*b-2*B*b^2)/(a-b)/(a^2+2*a*b+b^2)*tanh(1/2*x)^3+1/2*(4*A*a*b-A*b^2-2*B*a^2+
B*a*b-2*B*b^2)/(a+b)/(a^2-2*a*b+b^2)*tanh(1/2*x))/(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)^2+(2*A*a^2+A*b^2-3*B*a
*b)/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(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))/(a+b*cosh(x))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.67215, size = 7004, normalized size = 51.88 \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))/(a+b*cosh(x))^3,x, algorithm="fricas")
[Out]
[-1/2*(2*B*a^4*b^2 - 6*A*a^3*b^3 + 2*B*a^2*b^4 + 6*A*a*b^5 - 4*B*b^6 - 2*(2*A*a^4*b^2 - 3*B*a^3*b^3 - A*a^2*b^
4 + 3*B*a*b^5 - A*b^6)*cosh(x)^3 - 2*(2*A*a^4*b^2 - 3*B*a^3*b^3 - A*a^2*b^4 + 3*B*a*b^5 - A*b^6)*sinh(x)^3 + 2
*(2*B*a^6 - 6*A*a^5*b + 3*B*a^4*b^2 + 3*A*a^3*b^3 - 3*B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6)*cosh(x)^2 + 2*(2*B*a^6
- 6*A*a^5*b + 3*B*a^4*b^2 + 3*A*a^3*b^3 - 3*B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6 - 3*(2*A*a^4*b^2 - 3*B*a^3*b^3 - A
*a^2*b^4 + 3*B*a*b^5 - A*b^6)*cosh(x))*sinh(x)^2 - (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5 + (2*A*a^2*b^3 - 3*B*a*b^4
+ A*b^5)*cosh(x)^4 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*sinh(x)^4 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*co
sh(x)^3 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x))*sinh(x)^3 + 2*(4
*A*a^4*b - 6*B*a^3*b^2 + 4*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x)^2 + 2*(4*A*a^4*b - 6*B*a^3*b^2 + 4*A*a^2*b^3
- 3*B*a*b^4 + A*b^5 + 3*(2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x)^2 + 6*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)
*cosh(x))*sinh(x)^2 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cosh(x) + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4
+ (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x)^3 + 3*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cosh(x)^2 + (4*A*a^4*
b - 6*B*a^3*b^2 + 4*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x))*sinh(x))*sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*
sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sin
h(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) + 2*(4*B*a^5*b - 10*A*a^
4*b^2 + B*a^3*b^3 + 11*A*a^2*b^4 - 5*B*a*b^5 - A*b^6)*cosh(x) + 2*(4*B*a^5*b - 10*A*a^4*b^2 + B*a^3*b^3 + 11*A
*a^2*b^4 - 5*B*a*b^5 - A*b^6 - 3*(2*A*a^4*b^2 - 3*B*a^3*b^3 - A*a^2*b^4 + 3*B*a*b^5 - A*b^6)*cosh(x)^2 + 2*(2*
B*a^6 - 6*A*a^5*b + 3*B*a^4*b^2 + 3*A*a^3*b^3 - 3*B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6)*cosh(x))*sinh(x))/(a^6*b^3
- 3*a^4*b^5 + 3*a^2*b^7 - b^9 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cosh(x)^4 + (a^6*b^3 - 3*a^4*b^5 + 3*a
^2*b^7 - b^9)*sinh(x)^4 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cosh(x)^3 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a
^3*b^6 - a*b^8 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cosh(x))*sinh(x)^3 + 2*(2*a^8*b - 5*a^6*b^3 + 3*a^4*b
^5 + a^2*b^7 - b^9)*cosh(x)^2 + 2*(2*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9 + 3*(a^6*b^3 - 3*a^4*b^5 +
3*a^2*b^7 - b^9)*cosh(x)^2 + 6*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cosh(x))*sinh(x)^2 + 4*(a^7*b^2 - 3*a
^5*b^4 + 3*a^3*b^6 - a*b^8)*cosh(x) + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8 + (a^6*b^3 - 3*a^4*b^5 + 3*a^
2*b^7 - b^9)*cosh(x)^3 + 3*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cosh(x)^2 + (2*a^8*b - 5*a^6*b^3 + 3*a^4*
b^5 + a^2*b^7 - b^9)*cosh(x))*sinh(x)), -(B*a^4*b^2 - 3*A*a^3*b^3 + B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6 - (2*A*a^4
*b^2 - 3*B*a^3*b^3 - A*a^2*b^4 + 3*B*a*b^5 - A*b^6)*cosh(x)^3 - (2*A*a^4*b^2 - 3*B*a^3*b^3 - A*a^2*b^4 + 3*B*a
*b^5 - A*b^6)*sinh(x)^3 + (2*B*a^6 - 6*A*a^5*b + 3*B*a^4*b^2 + 3*A*a^3*b^3 - 3*B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6
)*cosh(x)^2 + (2*B*a^6 - 6*A*a^5*b + 3*B*a^4*b^2 + 3*A*a^3*b^3 - 3*B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6 - 3*(2*A*a^
4*b^2 - 3*B*a^3*b^3 - A*a^2*b^4 + 3*B*a*b^5 - A*b^6)*cosh(x))*sinh(x)^2 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5 + (
2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x)^4 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*sinh(x)^4 + 4*(2*A*a^3*b^2 - 3*
B*a^2*b^3 + A*a*b^4)*cosh(x)^3 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*co
sh(x))*sinh(x)^3 + 2*(4*A*a^4*b - 6*B*a^3*b^2 + 4*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x)^2 + 2*(4*A*a^4*b - 6*
B*a^3*b^2 + 4*A*a^2*b^3 - 3*B*a*b^4 + A*b^5 + 3*(2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x)^2 + 6*(2*A*a^3*b^2 -
3*B*a^2*b^3 + A*a*b^4)*cosh(x))*sinh(x)^2 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cosh(x) + 4*(2*A*a^3*b^2
- 3*B*a^2*b^3 + A*a*b^4 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x)^3 + 3*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4
)*cosh(x)^2 + (4*A*a^4*b - 6*B*a^3*b^2 + 4*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*a
rctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (4*B*a^5*b - 10*A*a^4*b^2 + B*a^3*b^3 + 11*
A*a^2*b^4 - 5*B*a*b^5 - A*b^6)*cosh(x) + (4*B*a^5*b - 10*A*a^4*b^2 + B*a^3*b^3 + 11*A*a^2*b^4 - 5*B*a*b^5 - A*
b^6 - 3*(2*A*a^4*b^2 - 3*B*a^3*b^3 - A*a^2*b^4 + 3*B*a*b^5 - A*b^6)*cosh(x)^2 + 2*(2*B*a^6 - 6*A*a^5*b + 3*B*a
^4*b^2 + 3*A*a^3*b^3 - 3*B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6)*cosh(x))*sinh(x))/(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 -
b^9 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cosh(x)^4 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*sinh(x)^4 +
4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cosh(x)^3 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8 + (a^6*b^3
- 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cosh(x))*sinh(x)^3 + 2*(2*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9)*cosh(
x)^2 + 2*(2*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9 + 3*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cosh(x)^
2 + 6*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cosh(x))*sinh(x)^2 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^
8)*cosh(x) + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cosh(x)^3 +
3*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cosh(x)^2 + (2*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9)*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))/(a+b*cosh(x))**3,x)
[Out]
Timed out
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Giac [B] time = 1.20303, size = 336, normalized size = 2.49 \begin{align*} \frac{{\left (2 \, A a^{2} - 3 \, B a b + A b^{2}\right )} \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \, A a^{2} b^{2} e^{\left (3 \, x\right )} - 3 \, B a b^{3} e^{\left (3 \, x\right )} + A b^{4} e^{\left (3 \, x\right )} - 2 \, B a^{4} e^{\left (2 \, x\right )} + 6 \, A a^{3} b e^{\left (2 \, x\right )} - 5 \, B a^{2} b^{2} e^{\left (2 \, x\right )} + 3 \, A a b^{3} e^{\left (2 \, x\right )} - 2 \, B b^{4} e^{\left (2 \, x\right )} - 4 \, B a^{3} b e^{x} + 10 \, A a^{2} b^{2} e^{x} - 5 \, B a b^{3} e^{x} - A b^{4} e^{x} - B a^{2} b^{2} + 3 \, A a b^{3} - 2 \, B b^{4}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )}{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cosh(x))/(a+b*cosh(x))^3,x, algorithm="giac")
[Out]
(2*A*a^2 - 3*B*a*b + A*b^2)*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) +
(2*A*a^2*b^2*e^(3*x) - 3*B*a*b^3*e^(3*x) + A*b^4*e^(3*x) - 2*B*a^4*e^(2*x) + 6*A*a^3*b*e^(2*x) - 5*B*a^2*b^2*e
^(2*x) + 3*A*a*b^3*e^(2*x) - 2*B*b^4*e^(2*x) - 4*B*a^3*b*e^x + 10*A*a^2*b^2*e^x - 5*B*a*b^3*e^x - A*b^4*e^x -
B*a^2*b^2 + 3*A*a*b^3 - 2*B*b^4)/((a^4*b - 2*a^2*b^3 + b^5)*(b*e^(2*x) + 2*a*e^x + b)^2)