3.69 \(\int \frac {1}{(a+b \cosh (c+d x))^3} \, dx\)

Optimal. Leaf size=133 \[ \frac {\left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {3 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cosh (c+d x))}-\frac {b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2} \]

[Out]

(2*a^2+b^2)*arctanh((a-b)^(1/2)*tanh(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d-1/2*b*sinh(d*x+c)/(
a^2-b^2)/d/(a+b*cosh(d*x+c))^2-3/2*a*b*sinh(d*x+c)/(a^2-b^2)^2/d/(a+b*cosh(d*x+c))

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Rubi [A]  time = 0.15, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2664, 2754, 12, 2659, 205} \[ \frac {\left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {3 a b \sinh (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cosh (c+d x))}-\frac {b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x])^(-3),x]

[Out]

((2*a^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*d) - (b*Sinh
[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cosh[c + d*x])^2) - (3*a*b*Sinh[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Cosh[c
+ d*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 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 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

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]

Rubi steps

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

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Mathematica [A]  time = 0.41, size = 113, normalized size = 0.85 \[ \frac {\frac {b \sinh (c+d x) \left (-4 a^2-3 a b \cosh (c+d x)+b^2\right )}{(a-b)^2 (a+b)^2 (a+b \cosh (c+d x))^2}-\frac {2 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x])^(-3),x]

[Out]

((-2*(2*a^2 + b^2)*ArcTan[((a - b)*Tanh[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (b*(-4*a^2 + b^2
- 3*a*b*Cosh[c + d*x])*Sinh[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cosh[c + d*x])^2))/(2*d)

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fricas [B]  time = 0.55, size = 2591, normalized size = 19.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/2*(6*a^3*b^2 - 6*a*b^4 + 2*(2*a^4*b - a^2*b^3 - b^5)*cosh(d*x + c)^3 + 2*(2*a^4*b - a^2*b^3 - b^5)*sinh(d*x
+ c)^3 + 6*(2*a^5 - a^3*b^2 - a*b^4)*cosh(d*x + c)^2 + 6*(2*a^5 - a^3*b^2 - a*b^4 + (2*a^4*b - a^2*b^3 - b^5)
*cosh(d*x + c))*sinh(d*x + c)^2 + ((2*a^2*b^2 + b^4)*cosh(d*x + c)^4 + (2*a^2*b^2 + b^4)*sinh(d*x + c)^4 + 2*a
^2*b^2 + b^4 + 4*(2*a^3*b + a*b^3)*cosh(d*x + c)^3 + 4*(2*a^3*b + a*b^3 + (2*a^2*b^2 + b^4)*cosh(d*x + c))*sin
h(d*x + c)^3 + 2*(4*a^4 + 4*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(4*a^4 + 4*a^2*b^2 + b^4 + 3*(2*a^2*b^2 + b^4)*
cosh(d*x + c)^2 + 6*(2*a^3*b + a*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 4*(2*a^3*b + a*b^3)*cosh(d*x + c) + 4*(
2*a^3*b + a*b^3 + (2*a^2*b^2 + b^4)*cosh(d*x + c)^3 + 3*(2*a^3*b + a*b^3)*cosh(d*x + c)^2 + (4*a^4 + 4*a^2*b^2
+ b^4)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 - b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*c
osh(d*x + c) + 2*a^2 - b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) - 2*sqrt(a^2 - b^2)*(b*cosh(d*x + c) +
b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sin
h(d*x + c) + b)) + 2*(10*a^4*b - 11*a^2*b^3 + b^5)*cosh(d*x + c) + 2*(10*a^4*b - 11*a^2*b^3 + b^5 + 3*(2*a^4*b
- a^2*b^3 - b^5)*cosh(d*x + c)^2 + 6*(2*a^5 - a^3*b^2 - a*b^4)*cosh(d*x + c))*sinh(d*x + c))/((a^6*b^2 - 3*a^
4*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c)^4 + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*sinh(d*x + c)^4 + 4*(a^
7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cosh(d*x + c)^3 + 2*(2*a^8 - 5*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 - b^8)*d*c
osh(d*x + c)^2 + 4*((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c) + (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 -
a*b^7)*d)*sinh(d*x + c)^3 + 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cosh(d*x + c) + 2*(3*(a^6*b^2 - 3*a^4
*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c)^2 + 6*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cosh(d*x + c) + (2*a^8
- 5*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 - b^8)*d)*sinh(d*x + c)^2 + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d + 4*(
(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c)^3 + 3*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cosh(d
*x + c)^2 + (2*a^8 - 5*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 - b^8)*d*cosh(d*x + c) + (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 -
a*b^7)*d)*sinh(d*x + c)), (3*a^3*b^2 - 3*a*b^4 + (2*a^4*b - a^2*b^3 - b^5)*cosh(d*x + c)^3 + (2*a^4*b - a^2*b
^3 - b^5)*sinh(d*x + c)^3 + 3*(2*a^5 - a^3*b^2 - a*b^4)*cosh(d*x + c)^2 + 3*(2*a^5 - a^3*b^2 - a*b^4 + (2*a^4*
b - a^2*b^3 - b^5)*cosh(d*x + c))*sinh(d*x + c)^2 - ((2*a^2*b^2 + b^4)*cosh(d*x + c)^4 + (2*a^2*b^2 + b^4)*sin
h(d*x + c)^4 + 2*a^2*b^2 + b^4 + 4*(2*a^3*b + a*b^3)*cosh(d*x + c)^3 + 4*(2*a^3*b + a*b^3 + (2*a^2*b^2 + b^4)*
cosh(d*x + c))*sinh(d*x + c)^3 + 2*(4*a^4 + 4*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(4*a^4 + 4*a^2*b^2 + b^4 + 3*
(2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 6*(2*a^3*b + a*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 4*(2*a^3*b + a*b^3)*c
osh(d*x + c) + 4*(2*a^3*b + a*b^3 + (2*a^2*b^2 + b^4)*cosh(d*x + c)^3 + 3*(2*a^3*b + a*b^3)*cosh(d*x + c)^2 +
(4*a^4 + 4*a^2*b^2 + b^4)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(d*x
+ c) + b*sinh(d*x + c) + a)/(a^2 - b^2)) + (10*a^4*b - 11*a^2*b^3 + b^5)*cosh(d*x + c) + (10*a^4*b - 11*a^2*b^
3 + b^5 + 3*(2*a^4*b - a^2*b^3 - b^5)*cosh(d*x + c)^2 + 6*(2*a^5 - a^3*b^2 - a*b^4)*cosh(d*x + c))*sinh(d*x +
c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c)^4 + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*sin
h(d*x + c)^4 + 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cosh(d*x + c)^3 + 2*(2*a^8 - 5*a^6*b^2 + 3*a^4*b^4
+ a^2*b^6 - b^8)*d*cosh(d*x + c)^2 + 4*((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c) + (a^7*b - 3*a
^5*b^3 + 3*a^3*b^5 - a*b^7)*d)*sinh(d*x + c)^3 + 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cosh(d*x + c) + 2
*(3*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c)^2 + 6*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*co
sh(d*x + c) + (2*a^8 - 5*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 - b^8)*d)*sinh(d*x + c)^2 + (a^6*b^2 - 3*a^4*b^4 + 3*a^
2*b^6 - b^8)*d + 4*((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cosh(d*x + c)^3 + 3*(a^7*b - 3*a^5*b^3 + 3*a^3*b
^5 - a*b^7)*d*cosh(d*x + c)^2 + (2*a^8 - 5*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 - b^8)*d*cosh(d*x + c) + (a^7*b - 3*a
^5*b^3 + 3*a^3*b^5 - a*b^7)*d)*sinh(d*x + c))]

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giac [A]  time = 0.15, size = 195, normalized size = 1.47 \[ \frac {\frac {{\left (2 \, a^{2} + b^{2}\right )} \arctan \left (\frac {b e^{\left (d x + c\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} + b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 10 \, a^{2} b e^{\left (d x + c\right )} - b^{3} e^{\left (d x + c\right )} + 3 \, a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} + b\right )}^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c))^3,x, algorithm="giac")

[Out]

((2*a^2 + b^2)*arctan((b*e^(d*x + c) + a)/sqrt(-a^2 + b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + (2*a^
2*b*e^(3*d*x + 3*c) + b^3*e^(3*d*x + 3*c) + 6*a^3*e^(2*d*x + 2*c) + 3*a*b^2*e^(2*d*x + 2*c) + 10*a^2*b*e^(d*x
+ c) - b^3*e^(d*x + c) + 3*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(b*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) + b)^2))/d

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maple [A]  time = 0.09, size = 186, normalized size = 1.40 \[ \frac {-\frac {2 \left (-\frac {\left (4 a +b \right ) b \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a -b \right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(d*x+c))^3,x)

[Out]

1/d*(-2*(-1/2*(4*a+b)*b/(a-b)/(a^2+2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+1/2*(4*a-b)*b/(a+b)/(a^2-2*a*b+b^2)*tanh(1
/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b-a-b)^2+(2*a^2+b^2)/(a^4-2*a^2*b^2+b^4)/((a+b)*
(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*d*x+1/2*c)/((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(1/(a+b*cosh(d*x+c))^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 \[ \int \frac {1}{{\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cosh(c + d*x))^3,x)

[Out]

int(1/(a + b*cosh(c + d*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(1/(a+b*cosh(d*x+c))**3,x)

[Out]

Timed out

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