\(\int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx\) [135]

Optimal result

Integrand size = 16, antiderivative size = 12 \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=-\frac {\cosh (x)}{b+a \sinh (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2833, 8} \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=-\frac {\cosh (x)}{a \sinh (x)+b} \]

[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 2833

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=-\frac {\cosh (x)}{b+a \sinh (x)} \]

[In]

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

[Out]

-(Cosh[x]/(b + a*Sinh[x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(25\) vs. \(2(12)=24\).

Time = 0.50 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.17

method	result	size
parallelrisch	\(\frac {-a \sinh \left (x \right )-b \left (\cosh \left (x \right )+1\right )}{b \left (b +a \sinh \left (x \right )\right )}\)	\(26\)
risch	\(-\frac {2 \left (-{\mathrm e}^{x} b +a \right )}{a \left ({\mathrm e}^{2 x} a +2 \,{\mathrm e}^{x} b -a \right )}\)	\(30\)
default	\(-\frac {2 \left (\frac {a \tanh \left (\frac {x}{2}\right )}{2 b}+\frac {1}{2}\right )}{-\frac {\tanh \left (\frac {x}{2}\right )^{2} b}{2}+a \tanh \left (\frac {x}{2}\right )+\frac {b}{2}}\)	\(36\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (12) = 24\).

Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 4.83 \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=\frac {2 \, {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) - a\right )}}{a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) - a^{2} + 2 \, {\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (12) = 24\).

Time = 0.28 (sec) , antiderivative size = 230, normalized size of antiderivative = 19.17 \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=-b {\left (\frac {a \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b^{2} e^{\left (-x\right )} + a b\right )}}{a^{4} + a^{2} b^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} e^{\left (-x\right )} - {\left (a^{4} + a^{2} b^{2}\right )} e^{\left (-2 \, x\right )}}\right )} + a {\left (\frac {b \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b e^{\left (-x\right )} + a\right )}}{a^{3} + a b^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} e^{\left (-x\right )} - {\left (a^{3} + a b^{2}\right )} e^{\left (-2 \, x\right )}}\right )} \]

[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)))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (12) = 24\).

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.50 \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=\frac {2 \, {\left (b e^{x} - a\right )}}{{\left (a e^{\left (2 \, x\right )} + 2 \, b e^{x} - a\right )} a} \]

[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)

Mupad [B] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 4.08 \[ \int \frac {a-b \sinh (x)}{(b+a \sinh (x))^2} \, dx=\frac {\frac {2\,{\mathrm {e}}^x\,\left (a^3\,b+a\,b^3\right )}{a\,\left (a^3+a\,b^2\right )}-2}{2\,b\,{\mathrm {e}}^x-a+a\,{\mathrm {e}}^{2\,x}} \]

[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))