3.68 \(\int \frac {1}{(a+b \cosh (c+d x))^2} \, dx\)

Optimal. Leaf size=86 \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac {b \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2664, 12, 2659, 205} \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac {b \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x])^(-2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 84, normalized size = 0.98 \[ \frac {\frac {2 a \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac {b \sinh (c+d x)}{(a-b) (a+b) (a+b \cosh (c+d x))}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x])^(-2),x]

[Out]

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

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fricas [B]  time = 0.73, size = 743, normalized size = 8.64 \[ \left [\frac {2 \, a^{2} b - 2 \, b^{3} - {\left (a b \cosh \left (d x + c\right )^{2} + a b \sinh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) + a b + 2 \, {\left (a b \cosh \left (d x + c\right ) + a^{2}\right )} \sinh \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) + b}\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \cosh \left (d x + c\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sinh \left (d x + c\right )^{2} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d + 2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )} \sinh \left (d x + c\right )}, \frac {2 \, {\left (a^{2} b - b^{3} - {\left (a b \cosh \left (d x + c\right )^{2} + a b \sinh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) + a b + 2 \, {\left (a b \cosh \left (d x + c\right ) + a^{2}\right )} \sinh \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (a^{3} - a b^{2}\right )} \cosh \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} \sinh \left (d x + c\right )\right )}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sinh \left (d x + c\right )^{2} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d + 2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )} \sinh \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c))^2,x, algorithm="fricas")

[Out]

[(2*a^2*b - 2*b^3 - (a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) + a*b + 2*(a*b*cosh(d*x +
c) + a^2)*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*cosh(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)*sinh(d*x + c) +
b)) + 2*(a^3 - a*b^2)*cosh(d*x + c) + 2*(a^3 - a*b^2)*sinh(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x +
c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*d*sinh(d*x + c)^2 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c) + (a^4*b - 2*
a^2*b^3 + b^5)*d + 2*((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d)*sinh(d*x + c)),
2*(a^2*b - b^3 - (a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) + a*b + 2*(a*b*cosh(d*x + c
) + a^2)*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)) + (a^3 - a*b^2)*cosh(d*x + c) + (a^3 - a*b^2)*sinh(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x +
c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*d*sinh(d*x + c)^2 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c) + (a^4*b - 2*
a^2*b^3 + b^5)*d + 2*((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d)*sinh(d*x + c))]

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giac [A]  time = 0.14, size = 99, normalized size = 1.15 \[ \frac {2 \, {\left (\frac {a \arctan \left (\frac {b e^{\left (d x + c\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {a e^{\left (d x + c\right )} + b}{{\left (a^{2} - b^{2}\right )} {\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} + b\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 118, normalized size = 1.37 \[ \frac {\frac {2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 a \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(d*x+c))^2,x)

[Out]

1/d*(2*b/(a^2-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*a/(a+b)/(a-b)/(
(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))^2,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 [B]  time = 1.30, size = 215, normalized size = 2.50 \[ \frac {\frac {2\,b^2}{d\,\left (a^2\,b-b^3\right )}+\frac {2\,a\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2\,b-b^3\right )}}{b+2\,a\,{\mathrm {e}}^{c+d\,x}+b\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {a\,\ln \left (-\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b\,\left (a^2-b^2\right )}-\frac {2\,a\,\left (b+a\,{\mathrm {e}}^{c+d\,x}\right )}{b\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {a\,\ln \left (\frac {2\,a\,\left (b+a\,{\mathrm {e}}^{c+d\,x}\right )}{b\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b\,\left (a^2-b^2\right )}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cosh(c + d*x))^2,x)

[Out]

((2*b^2)/(d*(a^2*b - b^3)) + (2*a*b*exp(c + d*x))/(d*(a^2*b - b^3)))/(b + 2*a*exp(c + d*x) + b*exp(2*c + 2*d*x
)) + (a*log(- (2*a*exp(c + d*x))/(b*(a^2 - b^2)) - (2*a*(b + a*exp(c + d*x)))/(b*(a + b)^(3/2)*(a - b)^(3/2)))
)/(d*(a + b)^(3/2)*(a - b)^(3/2)) - (a*log((2*a*(b + a*exp(c + d*x)))/(b*(a + b)^(3/2)*(a - b)^(3/2)) - (2*a*e
xp(c + d*x))/(b*(a^2 - b^2))))/(d*(a + b)^(3/2)*(a - b)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \cosh {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c))**2,x)

[Out]

Integral((a + b*cosh(c + d*x))**(-2), x)

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