\(\int x \cosh (a+b x^2) \, dx\) [3]

Optimal result

Integrand size = 10, antiderivative size = 15 \[ \int x \cosh \left (a+b x^2\right ) \, dx=\frac {\sinh \left (a+b x^2\right )}{2 b} \]

[Out]

1/2*sinh(b*x^2+a)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5429, 2717} \[ \int x \cosh \left (a+b x^2\right ) \, dx=\frac {\sinh \left (a+b x^2\right )}{2 b} \]

[In]

Int[x*Cosh[a + b*x^2],x]

[Out]

Sinh[a + b*x^2]/(2*b)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 5429

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Cosh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(31\) vs. \(2(15)=30\).

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07 \[ \int x \cosh \left (a+b x^2\right ) \, dx=\frac {\cosh \left (b x^2\right ) \sinh (a)}{2 b}+\frac {\cosh (a) \sinh \left (b x^2\right )}{2 b} \]

[In]

Integrate[x*Cosh[a + b*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\sinh \left (b \,x^{2}+a \right )}{2 b}\)	\(14\)
default	\(\frac {\sinh \left (b \,x^{2}+a \right )}{2 b}\)	\(14\)
parallelrisch	\(\frac {\sinh \left (b \,x^{2}+a \right )}{2 b}\)	\(14\)
risch	\(\frac {{\mathrm e}^{b \,x^{2}+a}}{4 b}-\frac {{\mathrm e}^{-b \,x^{2}-a}}{4 b}\)	\(31\)
meijerg	\(\frac {\cosh \left (a \right ) \sinh \left (b \,x^{2}\right )}{2 b}-\frac {\sinh \left (a \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (b \,x^{2}\right )}{\sqrt {\pi }}\right )}{2 b}\)	\(40\)

[In]

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

[Out]

1/2*sinh(b*x^2+a)/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int x \cosh \left (a+b x^2\right ) \, dx=\frac {\sinh \left (b x^{2} + a\right )}{2 \, b} \]

[In]

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

[Out]

1/2*sinh(b*x^2 + a)/b

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int x \cosh \left (a+b x^2\right ) \, dx=\begin {cases} \frac {\sinh {\left (a + b x^{2} \right )}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \cosh {\left (a \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int x \cosh \left (a+b x^2\right ) \, dx=\frac {\sinh \left (b x^{2} + a\right )}{2 \, b} \]

[In]

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

[Out]

1/2*sinh(b*x^2 + a)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.80 \[ \int x \cosh \left (a+b x^2\right ) \, dx=\frac {e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}}{4 \, b} \]

[In]

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

[Out]

1/4*(e^(b*x^2 + a) - e^(-b*x^2 - a))/b

Mupad [B] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int x \cosh \left (a+b x^2\right ) \, dx=\frac {\mathrm {sinh}\left (b\,x^2+a\right )}{2\,b} \]

[In]

int(x*cosh(a + b*x^2),x)

[Out]

sinh(a + b*x^2)/(2*b)