3.35 \(\int y \cosh (a y) \, dy\)

Optimal. Leaf size=19 \[ \frac {y \sinh (a y)}{a}-\frac {\cosh (a y)}{a^2} \]

[Out]

-cosh(a*y)/a^2+y*sinh(a*y)/a

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3296, 2638} \[ \frac {y \sinh (a y)}{a}-\frac {\cosh (a y)}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[y*Cosh[a*y],y]

[Out]

-(Cosh[a*y]/a^2) + (y*Sinh[a*y])/a

Rule 2638

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rubi steps

\begin {align*} \int y \cosh (a y) \, dy &=\frac {y \sinh (a y)}{a}-\frac {\int \sinh (a y) \, dy}{a}\\ &=-\frac {\cosh (a y)}{a^2}+\frac {y \sinh (a y)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ \frac {y \sinh (a y)}{a}-\frac {\cosh (a y)}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[y*Cosh[a*y],y]

[Out]

-(Cosh[a*y]/a^2) + (y*Sinh[a*y])/a

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fricas [A]  time = 0.40, size = 18, normalized size = 0.95 \[ \frac {a y \sinh \left (a y\right ) - \cosh \left (a y\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(y*cosh(a*y),y, algorithm="fricas")

[Out]

(a*y*sinh(a*y) - cosh(a*y))/a^2

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giac [A]  time = 0.77, size = 30, normalized size = 1.58 \[ \frac {{\left (a y - 1\right )} e^{\left (a y\right )}}{2 \, a^{2}} - \frac {{\left (a y + 1\right )} e^{\left (-a y\right )}}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(y*cosh(a*y),y, algorithm="giac")

[Out]

1/2*(a*y - 1)*e^(a*y)/a^2 - 1/2*(a*y + 1)*e^(-a*y)/a^2

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maple [A]  time = 0.02, size = 19, normalized size = 1.00 \[ \frac {a y \sinh \left (a y \right )-\cosh \left (a y \right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(y*cosh(a*y),y)

[Out]

1/a^2*(y*a*sinh(a*y)-cosh(a*y))

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maxima [B]  time = 0.44, size = 57, normalized size = 3.00 \[ \frac {1}{2} \, y^{2} \cosh \left (a y\right ) - \frac {1}{4} \, a {\left (\frac {{\left (a^{2} y^{2} - 2 \, a y + 2\right )} e^{\left (a y\right )}}{a^{3}} + \frac {{\left (a^{2} y^{2} + 2 \, a y + 2\right )} e^{\left (-a y\right )}}{a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(y*cosh(a*y),y, algorithm="maxima")

[Out]

1/2*y^2*cosh(a*y) - 1/4*a*((a^2*y^2 - 2*a*y + 2)*e^(a*y)/a^3 + (a^2*y^2 + 2*a*y + 2)*e^(-a*y)/a^3)

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mupad [B]  time = 0.06, size = 18, normalized size = 0.95 \[ -\frac {\mathrm {cosh}\left (a\,y\right )-a\,y\,\mathrm {sinh}\left (a\,y\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(y*cosh(a*y),y)

[Out]

-(cosh(a*y) - a*y*sinh(a*y))/a^2

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sympy [A]  time = 0.23, size = 20, normalized size = 1.05 \[ \begin {cases} \frac {y \sinh {\left (a y \right )}}{a} - \frac {\cosh {\left (a y \right )}}{a^{2}} & \text {for}\: a \neq 0 \\\frac {y^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(y*cosh(a*y),y)

[Out]

Piecewise((y*sinh(a*y)/a - cosh(a*y)/a**2, Ne(a, 0)), (y**2/2, True))

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