\(\int \sinh (a+b x) \, dx\) [1]

Optimal result

Integrand size = 6, antiderivative size = 10 \[ \int \sinh (a+b x) \, dx=\frac {\cosh (a+b x)}{b} \]

[Out]

cosh(b*x+a)/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2718} \[ \int \sinh (a+b x) \, dx=\frac {\cosh (a+b x)}{b} \]

[In]

Int[Sinh[a + b*x],x]

[Out]

Cosh[a + b*x]/b

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\cosh (a+b x)}{b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(21\) vs. \(2(10)=20\).

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 2.10 \[ \int \sinh (a+b x) \, dx=\frac {\cosh (a) \cosh (b x)}{b}+\frac {\sinh (a) \sinh (b x)}{b} \]

[In]

Integrate[Sinh[a + b*x],x]

[Out]

(Cosh[a]*Cosh[b*x])/b + (Sinh[a]*Sinh[b*x])/b

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {\cosh \left (b x +a \right )}{b}\)	\(11\)
default	\(\frac {\cosh \left (b x +a \right )}{b}\)	\(11\)
parallelrisch	\(\frac {1+\cosh \left (b x +a \right )}{b}\)	\(13\)
risch	\(\frac {{\mathrm e}^{b x +a}}{2 b}+\frac {{\mathrm e}^{-b x -a}}{2 b}\)	\(27\)
meijerg	\(\frac {\sinh \left (a \right ) \sinh \left (b x \right )}{b}-\frac {\cosh \left (a \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (b x \right )}{\sqrt {\pi }}\right )}{b}\)	\(35\)

[In]

int(sinh(b*x+a),x,method=_RETURNVERBOSE)

[Out]

cosh(b*x+a)/b

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \sinh (a+b x) \, dx=\frac {\cosh \left (b x + a\right )}{b} \]

[In]

integrate(sinh(b*x+a),x, algorithm="fricas")

[Out]

cosh(b*x + a)/b

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.20 \[ \int \sinh (a+b x) \, dx=\begin {cases} \frac {\cosh {\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \sinh {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sinh(b*x+a),x)

[Out]

Piecewise((cosh(a + b*x)/b, Ne(b, 0)), (x*sinh(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \sinh (a+b x) \, dx=\frac {\cosh \left (b x + a\right )}{b} \]

[In]

integrate(sinh(b*x+a),x, algorithm="maxima")

[Out]

cosh(b*x + a)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.60 \[ \int \sinh (a+b x) \, dx=\frac {e^{\left (b x + a\right )}}{2 \, b} + \frac {e^{\left (-b x - a\right )}}{2 \, b} \]

[In]

integrate(sinh(b*x+a),x, algorithm="giac")

[Out]

1/2*e^(b*x + a)/b + 1/2*e^(-b*x - a)/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \sinh (a+b x) \, dx=\frac {\mathrm {cosh}\left (a+b\,x\right )}{b} \]

[In]

int(sinh(a + b*x),x)

[Out]

cosh(a + b*x)/b