3.3.3 \(\int \frac {\cosh (x)}{(a+b \sinh (x))^2} \, dx\) [203]

Optimal. Leaf size=13 \[ -\frac {1}{b (a+b \sinh (x))} \]

[Out]

-1/b/(a+b*sinh(x))

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Rubi [A]
time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2747, 32} \begin {gather*} -\frac {1}{b (a+b \sinh (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(a + b*Sinh[x])^2,x]

[Out]

-(1/(b*(a + b*Sinh[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 2747

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]

Rubi steps

\begin {align*} \int \frac {\cosh (x)}{(a+b \sinh (x))^2} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{(a+x)^2} \, dx,x,b \sinh (x)\right )}{b}\\ &=-\frac {1}{b (a+b \sinh (x))}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} -\frac {1}{b (a+b \sinh (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/(a + b*Sinh[x])^2,x]

[Out]

-(1/(b*(a + b*Sinh[x])))

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Maple [A]
time = 0.30, size = 14, normalized size = 1.08

method	result	size
derivativedivides	\(-\frac {1}{b \left (a +b \sinh \left (x \right )\right )}\)	\(14\)
default	\(-\frac {1}{b \left (a +b \sinh \left (x \right )\right )}\)	\(14\)
risch	\(-\frac {2 \,{\mathrm e}^{x}}{b \left (b \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}-b \right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(a+b*sinh(x))^2,x,method=_RETURNVERBOSE)

[Out]

-1/b/(a+b*sinh(x))

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Maxima [A]
time = 0.26, size = 13, normalized size = 1.00 \begin {gather*} -\frac {1}{{\left (b \sinh \left (x\right ) + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

-1/((b*sinh(x) + a)*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (13) = 26\).
time = 0.42, size = 51, normalized size = 3.92 \begin {gather*} -\frac {2 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}}{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) - b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(cosh(x) + sinh(x))/(b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x))

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Sympy [A]
time = 0.26, size = 19, normalized size = 1.46 \begin {gather*} \begin {cases} - \frac {1}{a b + b^{2} \sinh {\left (x \right )}} & \text {for}\: b \neq 0 \\\frac {\sinh {\left (x \right )}}{a^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)/(a+b*sinh(x))**2,x)

[Out]

Piecewise((-1/(a*b + b**2*sinh(x)), Ne(b, 0)), (sinh(x)/a**2, True))

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Giac [A]
time = 0.41, size = 22, normalized size = 1.69 \begin {gather*} \frac {2}{{\left (b {\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/((b*(e^(-x) - e^x) - 2*a)*b)

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Mupad [B]
time = 0.53, size = 14, normalized size = 1.08 \begin {gather*} \frac {\mathrm {sinh}\left (x\right )}{a\,\left (a+b\,\mathrm {sinh}\left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(a + b*sinh(x))^2,x)

[Out]

sinh(x)/(a*(a + b*sinh(x)))

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