3.208 \(\int \frac {A+B \cosh (d+e x)+C \sinh (d+e x)}{(a+b \cosh (d+e x))^3} \, dx\)

Optimal. Leaf size=187 \[ \frac {\left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{e (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sinh (d+e x)}{2 e \left (a^2-b^2\right )^2 (a+b \cosh (d+e x))}-\frac {(A b-a B) \sinh (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cosh (d+e x))^2}-\frac {C}{2 b e (a+b \cosh (d+e x))^2} \]

[Out]

(2*A*a^2+A*b^2-3*B*a*b)*arctanh((a-b)^(1/2)*tanh(1/2*e*x+1/2*d)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/e-1/2*C/b
/e/(a+b*cosh(e*x+d))^2-1/2*(A*b-B*a)*sinh(e*x+d)/(a^2-b^2)/e/(a+b*cosh(e*x+d))^2-1/2*(3*A*a*b-B*a^2-2*B*b^2)*s
inh(e*x+d)/(a^2-b^2)^2/e/(a+b*cosh(e*x+d))

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Rubi [A]  time = 0.26, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4377, 2754, 12, 2659, 205, 2668, 32} \[ \frac {\left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{e (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sinh (d+e x)}{2 e \left (a^2-b^2\right )^2 (a+b \cosh (d+e x))}-\frac {(A b-a B) \sinh (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cosh (d+e x))^2}-\frac {C}{2 b e (a+b \cosh (d+e x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[d + e*x] + C*Sinh[d + e*x])/(a + b*Cosh[d + e*x])^3,x]

[Out]

((2*a^2*A + A*b^2 - 3*a*b*B)*ArcTanh[(Sqrt[a - b]*Tanh[(d + e*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2
)*e) - C/(2*b*e*(a + b*Cosh[d + e*x])^2) - ((A*b - a*B)*Sinh[d + e*x])/(2*(a^2 - b^2)*e*(a + b*Cosh[d + e*x])^
2) - ((3*a*A*b - a^2*B - 2*b^2*B)*Sinh[d + e*x])/(2*(a^2 - b^2)^2*e*(a + b*Cosh[d + e*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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 4377

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

\begin {align*} \int \frac {A+B \cosh (d+e x)+C \sinh (d+e x)}{(a+b \cosh (d+e x))^3} \, dx &=C \int \frac {\sinh (d+e x)}{(a+b \cosh (d+e x))^3} \, dx+\int \frac {A+B \cosh (d+e x)}{(a+b \cosh (d+e x))^3} \, dx\\ &=-\frac {(A b-a B) \sinh (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cosh (d+e x))^2}-\frac {\int \frac {-2 (a A-b B)+(A b-a B) \cosh (d+e x)}{(a+b \cosh (d+e x))^2} \, dx}{2 \left (a^2-b^2\right )}+\frac {C \operatorname {Subst}\left (\int \frac {1}{(a+x)^3} \, dx,x,b \cosh (d+e x)\right )}{b e}\\ &=-\frac {C}{2 b e (a+b \cosh (d+e x))^2}-\frac {(A b-a B) \sinh (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cosh (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cosh (d+e x))}+\frac {\int \frac {2 a^2 A+A b^2-3 a b B}{a+b \cosh (d+e x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {C}{2 b e (a+b \cosh (d+e x))^2}-\frac {(A b-a B) \sinh (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cosh (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cosh (d+e x))}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \int \frac {1}{a+b \cosh (d+e x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {C}{2 b e (a+b \cosh (d+e x))^2}-\frac {(A b-a B) \sinh (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cosh (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cosh (d+e x))}-\frac {\left (i \left (2 a^2 A+A b^2-3 a b B\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (i d+i e x)\right )\right )}{\left (a^2-b^2\right )^2 e}\\ &=\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} e}-\frac {C}{2 b e (a+b \cosh (d+e x))^2}-\frac {(A b-a B) \sinh (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cosh (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sinh (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cosh (d+e x))}\\ \end {align*}

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Mathematica [A]  time = 0.85, size = 175, normalized size = 0.94 \[ \frac {\frac {C \left (b^2-a^2\right )-b (A b-a B) \sinh (d+e x)}{b (a-b) (a+b) (a+b \cosh (d+e x))^2}-\frac {2 \left (2 a^2 A-3 a b B+A b^2\right ) \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac {\left (a^2 B-3 a A b+2 b^2 B\right ) \sinh (d+e x)}{(a-b)^2 (a+b)^2 (a+b \cosh (d+e x))}}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[d + e*x] + C*Sinh[d + e*x])/(a + b*Cosh[d + e*x])^3,x]

[Out]

((-2*(2*a^2*A + A*b^2 - 3*a*b*B)*ArcTan[((a - b)*Tanh[(d + e*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + ((
-3*a*A*b + a^2*B + 2*b^2*B)*Sinh[d + e*x])/((a - b)^2*(a + b)^2*(a + b*Cosh[d + e*x])) + ((-a^2 + b^2)*C - b*(
A*b - a*B)*Sinh[d + e*x])/((a - b)*b*(a + b)*(a + b*Cosh[d + e*x])^2))/(2*e)

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fricas [B]  time = 0.66, size = 3636, normalized size = 19.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d))^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(e*x + d)^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(e*
x + d)^3 + 2*(2*(B + C)*a^6 - 6*A*a^5*b + 3*(B - 2*C)*a^4*b^2 + 3*A*a^3*b^3 - 3*(B - 2*C)*a^2*b^4 + 3*A*a*b^5
- 2*(B + C)*b^6)*cosh(e*x + d)^2 + 2*(2*(B + C)*a^6 - 6*A*a^5*b + 3*(B - 2*C)*a^4*b^2 + 3*A*a^3*b^3 - 3*(B - 2
*C)*a^2*b^4 + 3*A*a*b^5 - 2*(B + C)*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(e
*x + d))*sinh(e*x + d)^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(e*x + d)^
4 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*sinh(e*x + d)^4 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cosh(e*x + d)^
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(e*x + d))*sinh(e*x + d)^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(e*x + d)^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(e*x + d)^2 + 6*(2*A*a^3*b^2 - 3*B*
a^2*b^3 + A*a*b^4)*cosh(e*x + d))*sinh(e*x + d)^2 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cosh(e*x + d) + 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(e*x + d)^3 + 3*(2*A*a^3*b^2 - 3*
B*a^2*b^3 + A*a*b^4)*cosh(e*x + d)^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(e*x +
d))*sinh(e*x + d))*sqrt(a^2 - b^2)*log((b^2*cosh(e*x + d)^2 + b^2*sinh(e*x + d)^2 + 2*a*b*cosh(e*x + d) + 2*a^
2 - b^2 + 2*(b^2*cosh(e*x + d) + a*b)*sinh(e*x + d) - 2*sqrt(a^2 - b^2)*(b*cosh(e*x + d) + b*sinh(e*x + d) + a
))/(b*cosh(e*x + d)^2 + b*sinh(e*x + d)^2 + 2*a*cosh(e*x + d) + 2*(b*cosh(e*x + d) + a)*sinh(e*x + d) + 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(e*x + d) + 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(e*x + d)^2 + 2*(2*(B + C)*a^6 - 6*A*a^5*b + 3*(B - 2*C)*a^4*b^2 + 3*A*a^3*b^3 - 3*(B - 2*C)*a^
2*b^4 + 3*A*a*b^5 - 2*(B + C)*b^6)*cosh(e*x + d))*sinh(e*x + d))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*co
sh(e*x + d)^4 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*sinh(e*x + d)^4 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6
- a*b^8)*e*cosh(e*x + d)^3 + 2*(2*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9)*e*cosh(e*x + d)^2 + 4*((a^6*
b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*cosh(e*x + d) + (a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e)*sinh(e*x + d
)^3 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cosh(e*x + d) + 2*(3*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^
9)*e*cosh(e*x + d)^2 + 6*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cosh(e*x + d) + (2*a^8*b - 5*a^6*b^3 + 3*
a^4*b^5 + a^2*b^7 - b^9)*e)*sinh(e*x + d)^2 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e + 4*((a^6*b^3 - 3*a^4*
b^5 + 3*a^2*b^7 - b^9)*e*cosh(e*x + d)^3 + 3*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cosh(e*x + d)^2 + (2*
a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9)*e*cosh(e*x + d) + (a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e)*
sinh(e*x + d)), -(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(e*x + d)^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
(e*x + d)^3 + (2*(B + C)*a^6 - 6*A*a^5*b + 3*(B - 2*C)*a^4*b^2 + 3*A*a^3*b^3 - 3*(B - 2*C)*a^2*b^4 + 3*A*a*b^5
- 2*(B + C)*b^6)*cosh(e*x + d)^2 + (2*(B + C)*a^6 - 6*A*a^5*b + 3*(B - 2*C)*a^4*b^2 + 3*A*a^3*b^3 - 3*(B - 2*
C)*a^2*b^4 + 3*A*a*b^5 - 2*(B + C)*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(e*
x + d))*sinh(e*x + d)^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(e*x + d)^4
+ (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*sinh(e*x + d)^4 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cosh(e*x + d)^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(e*x + d))*sinh(e*x + d)^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(e*x + d)^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(e*x + d)^2 + 6*(2*A*a^3*b^2 - 3*B*a
^2*b^3 + A*a*b^4)*cosh(e*x + d))*sinh(e*x + d)^2 + 4*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cosh(e*x + d) + 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(e*x + d)^3 + 3*(2*A*a^3*b^2 - 3*B
*a^2*b^3 + A*a*b^4)*cosh(e*x + d)^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(e*x + d
))*sinh(e*x + d))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(e*x + d) + b*sinh(e*x + d) + 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(e*x + d) + (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(e*x + d)^2 + 2*(2*(B + C)*a^6 - 6*A*a^5*b + 3*(B - 2*C)*a^4*b^2 + 3*A*a^3*b^3 - 3*(B - 2*C)*a
^2*b^4 + 3*A*a*b^5 - 2*(B + C)*b^6)*cosh(e*x + d))*sinh(e*x + d))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*c
osh(e*x + d)^4 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*sinh(e*x + d)^4 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^
6 - a*b^8)*e*cosh(e*x + d)^3 + 2*(2*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9)*e*cosh(e*x + d)^2 + 4*((a^6
*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*cosh(e*x + d) + (a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e)*sinh(e*x +
d)^3 + 4*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cosh(e*x + d) + 2*(3*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b
^9)*e*cosh(e*x + d)^2 + 6*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cosh(e*x + d) + (2*a^8*b - 5*a^6*b^3 + 3
*a^4*b^5 + a^2*b^7 - b^9)*e)*sinh(e*x + d)^2 + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e + 4*((a^6*b^3 - 3*a^4
*b^5 + 3*a^2*b^7 - b^9)*e*cosh(e*x + d)^3 + 3*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cosh(e*x + d)^2 + (2
*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 + a^2*b^7 - b^9)*e*cosh(e*x + d) + (a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e)
*sinh(e*x + d))]

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giac [B]  time = 0.21, size = 387, normalized size = 2.07 \[ {\left (\frac {{\left (2 \, A a^{2} - 3 \, B a b + A b^{2}\right )} \arctan \left (\frac {b e^{\left (x e + d\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, A a^{2} b^{2} e^{\left (3 \, x e + 3 \, d\right )} - 3 \, B a b^{3} e^{\left (3 \, x e + 3 \, d\right )} + A b^{4} e^{\left (3 \, x e + 3 \, d\right )} - 2 \, B a^{4} e^{\left (2 \, x e + 2 \, d\right )} - 2 \, C a^{4} e^{\left (2 \, x e + 2 \, d\right )} + 6 \, A a^{3} b e^{\left (2 \, x e + 2 \, d\right )} - 5 \, B a^{2} b^{2} e^{\left (2 \, x e + 2 \, d\right )} + 4 \, C a^{2} b^{2} e^{\left (2 \, x e + 2 \, d\right )} + 3 \, A a b^{3} e^{\left (2 \, x e + 2 \, d\right )} - 2 \, B b^{4} e^{\left (2 \, x e + 2 \, d\right )} - 2 \, C b^{4} e^{\left (2 \, x e + 2 \, d\right )} - 4 \, B a^{3} b e^{\left (x e + d\right )} + 10 \, A a^{2} b^{2} e^{\left (x e + d\right )} - 5 \, B a b^{3} e^{\left (x e + d\right )} - A b^{4} e^{\left (x e + d\right )} - B a^{2} b^{2} + 3 \, A a b^{3} - 2 \, B b^{4}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b e^{\left (2 \, x e + 2 \, d\right )} + 2 \, a e^{\left (x e + d\right )} + b\right )}^{2}}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d))^3,x, algorithm="giac")

[Out]

((2*A*a^2 - 3*B*a*b + A*b^2)*arctan((b*e^(x*e + d) + 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*e + 3*d) - 3*B*a*b^3*e^(3*x*e + 3*d) + A*b^4*e^(3*x*e + 3*d) - 2*B*a^4*e^(2*x*e +
2*d) - 2*C*a^4*e^(2*x*e + 2*d) + 6*A*a^3*b*e^(2*x*e + 2*d) - 5*B*a^2*b^2*e^(2*x*e + 2*d) + 4*C*a^2*b^2*e^(2*x
*e + 2*d) + 3*A*a*b^3*e^(2*x*e + 2*d) - 2*B*b^4*e^(2*x*e + 2*d) - 2*C*b^4*e^(2*x*e + 2*d) - 4*B*a^3*b*e^(x*e +
d) + 10*A*a^2*b^2*e^(x*e + d) - 5*B*a*b^3*e^(x*e + d) - A*b^4*e^(x*e + d) - 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*e + 2*d) + 2*a*e^(x*e + d) + b)^2))*e^(-1)

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maple [A]  time = 0.17, size = 273, normalized size = 1.46 \[ \frac {-\frac {2 \left (-\frac {\left (4 A a b +A \,b^{2}-2 B \,a^{2}-b B a -2 B \,b^{2}\right ) \left (\tanh ^{3}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {C \left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{a -b}+\frac {\left (4 A a b -A \,b^{2}-2 B \,a^{2}+b B a -2 B \,b^{2}\right ) \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {a C}{a^{2}-2 a b +b^{2}}\right )}{\left (a \left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right ) b -a -b \right )^{2}}+\frac {\left (2 a^{2} A +A \,b^{2}-3 b B a \right ) \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d))^3,x)

[Out]

1/e*(-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*e*x+1/2*d)^3+C/(a-b)*tanh(1
/2*e*x+1/2*d)^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*e*x+1/2*d)-a*C/(a^2-2
*a*b+b^2))/(a*tanh(1/2*e*x+1/2*d)^2-tanh(1/2*e*x+1/2*d)^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*e*x+1/2*d)/((a+b)*(a-b))^(1/2)))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {cosh}\left (d+e\,x\right )+C\,\mathrm {sinh}\left (d+e\,x\right )}{{\left (a+b\,\mathrm {cosh}\left (d+e\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(d + e*x) + C*sinh(d + e*x))/(a + b*cosh(d + e*x))^3,x)

[Out]

int((A + B*cosh(d + e*x) + C*sinh(d + e*x))/(a + b*cosh(d + e*x))^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+b*cosh(e*x+d))**3,x)

[Out]

Timed out

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