3.135 \(\int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx\)

Optimal. Leaf size=12 \[ -\frac {\cosh (x)}{a \sinh (x)+b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2754, 8} \[ -\frac {\cosh (x)}{a \sinh (x)+b} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*Sinh[x])/(b + a*Sinh[x])^2,x]

[Out]

-(Cosh[x]/(b + a*Sinh[x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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]

Rubi steps

\begin {align*} \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx &=-\frac {\cosh (x)}{b+a \sinh (x)}-\frac {\int 0 \, dx}{a^2+b^2}\\ &=-\frac {\cosh (x)}{b+a \sinh (x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 12, normalized size = 1.00 \[ -\frac {\cosh (x)}{a \sinh (x)+b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cosh[x]/(b + a*Sinh[x]))

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fricas [B]  time = 0.44, size = 58, normalized size = 4.83 \[ \frac {2 \, {\left (b \cosh \relax (x) + b \sinh \relax (x) - a\right )}}{a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) - a^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.19, size = 30, normalized size = 2.50 \[ \frac {2 \, {\left (b e^{x} - a\right )}}{{\left (a e^{\left (2 \, x\right )} + 2 \, b e^{x} - a\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*e^x - a)/((a*e^(2*x) + 2*b*e^x - a)*a)

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maple [B]  time = 0.05, size = 36, normalized size = 3.00 \[ -\frac {2 \left (-\frac {a \tanh \left (\frac {x}{2}\right )}{b}-1\right )}{\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-a/b*tanh(1/2*x)-1)/(tanh(1/2*x)^2*b-2*a*tanh(1/2*x)-b)

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maxima [B]  time = 0.42, size = 230, normalized size = 19.17 \[ -b {\left (\frac {a \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b^{2} e^{\left (-x\right )} + a b\right )}}{a^{4} + a^{2} b^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} e^{\left (-x\right )} - {\left (a^{4} + a^{2} b^{2}\right )} e^{\left (-2 \, x\right )}}\right )} + a {\left (\frac {b \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b e^{\left (-x\right )} + a\right )}}{a^{3} + a b^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} e^{\left (-x\right )} - {\left (a^{3} + a b^{2}\right )} e^{\left (-2 \, x\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*(a*log((a*e^(-x) - b - sqrt(a^2 + b^2))/(a*e^(-x) - b + sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) + 2*(b^2*e^(-x)
+ a*b)/(a^4 + a^2*b^2 + 2*(a^3*b + a*b^3)*e^(-x) - (a^4 + a^2*b^2)*e^(-2*x))) + a*(b*log((a*e^(-x) - b - sqrt
(a^2 + b^2))/(a*e^(-x) - b + sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) - 2*(b*e^(-x) + a)/(a^3 + a*b^2 + 2*(a^2*b +
b^3)*e^(-x) - (a^3 + a*b^2)*e^(-2*x)))

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mupad [B]  time = 0.57, size = 49, normalized size = 4.08 \[ \frac {\frac {2\,{\mathrm {e}}^x\,\left (a^3\,b+a\,b^3\right )}{a\,\left (a^3+a\,b^2\right )}-2}{2\,b\,{\mathrm {e}}^x-a+a\,{\mathrm {e}}^{2\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*exp(x)*(a*b^3 + a^3*b))/(a*(a*b^2 + a^3)) - 2)/(2*b*exp(x) - a + a*exp(2*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*sinh(x))/(b+a*sinh(x))**2,x)

[Out]

Timed out

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