3.228 \(\int \cosh (x) \cosh (m x) \, dx\)

Optimal. Leaf size=35 \[ \frac {\sinh ((1-m) x)}{2 (1-m)}+\frac {\sinh ((m+1) x)}{2 (m+1)} \]

[Out]

1/2*sinh((1-m)*x)/(1-m)+1/2*sinh((1+m)*x)/(1+m)

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5614, 2637} \[ \frac {\sinh ((1-m) x)}{2 (1-m)}+\frac {\sinh ((m+1) x)}{2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]*Cosh[m*x],x]

[Out]

Sinh[(1 - m)*x]/(2*(1 - m)) + Sinh[(1 + m)*x]/(2*(1 + m))

Rule 2637

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

Rule 5614

Int[Cosh[v_]^(p_.)*Cosh[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Cosh[v]^p*Cosh[w]^q, x], x] /; IGtQ[p, 0]
&& IGtQ[q, 0] && ((PolynomialQ[v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/
w], x]))

Rubi steps

\begin {align*} \int \cosh (x) \cosh (m x) \, dx &=\int \left (\frac {1}{2} \cosh ((1-m) x)+\frac {1}{2} \cosh ((1+m) x)\right ) \, dx\\ &=\frac {1}{2} \int \cosh ((1-m) x) \, dx+\frac {1}{2} \int \cosh ((1+m) x) \, dx\\ &=\frac {\sinh ((1-m) x)}{2 (1-m)}+\frac {\sinh ((1+m) x)}{2 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 25, normalized size = 0.71 \[ \frac {m \cosh (x) \sinh (m x)-\sinh (x) \cosh (m x)}{m^2-1} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]*Cosh[m*x],x]

[Out]

(-(Cosh[m*x]*Sinh[x]) + m*Cosh[x]*Sinh[m*x])/(-1 + m^2)

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fricas [A]  time = 0.42, size = 42, normalized size = 1.20 \[ \frac {m \cosh \relax (x) \sinh \left (m x\right ) - \cosh \left (m x\right ) \sinh \relax (x)}{{\left (m^{2} - 1\right )} \cosh \relax (x)^{2} - {\left (m^{2} - 1\right )} \sinh \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*cosh(m*x),x, algorithm="fricas")

[Out]

(m*cosh(x)*sinh(m*x) - cosh(m*x)*sinh(x))/((m^2 - 1)*cosh(x)^2 - (m^2 - 1)*sinh(x)^2)

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giac [B]  time = 0.11, size = 59, normalized size = 1.69 \[ \frac {e^{\left (m x + x\right )}}{4 \, {\left (m + 1\right )}} + \frac {e^{\left (m x - x\right )}}{4 \, {\left (m - 1\right )}} - \frac {e^{\left (-m x + x\right )}}{4 \, {\left (m - 1\right )}} - \frac {e^{\left (-m x - x\right )}}{4 \, {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*cosh(m*x),x, algorithm="giac")

[Out]

1/4*e^(m*x + x)/(m + 1) + 1/4*e^(m*x - x)/(m - 1) - 1/4*e^(-m*x + x)/(m - 1) - 1/4*e^(-m*x - x)/(m + 1)

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maple [A]  time = 0.18, size = 28, normalized size = 0.80 \[ \frac {\sinh \left (\left (-1+m \right ) x \right )}{-2+2 m}+\frac {\sinh \left (\left (1+m \right ) x \right )}{2+2 m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*cosh(m*x),x)

[Out]

1/2/(-1+m)*sinh((-1+m)*x)+1/2*sinh((1+m)*x)/(1+m)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*cosh(m*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(m-2>0)', see `assume?` for mor
e details)Is m-2 equal to -1?

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mupad [B]  time = 0.07, size = 26, normalized size = 0.74 \[ -\frac {\mathrm {cosh}\left (m\,x\right )\,\mathrm {sinh}\relax (x)-m\,\mathrm {sinh}\left (m\,x\right )\,\mathrm {cosh}\relax (x)}{m^2-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(m*x)*cosh(x),x)

[Out]

-(cosh(m*x)*sinh(x) - m*sinh(m*x)*cosh(x))/(m^2 - 1)

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sympy [A]  time = 0.92, size = 56, normalized size = 1.60 \[ \begin {cases} - \frac {x \sinh ^{2}{\relax (x )}}{2} + \frac {x \cosh ^{2}{\relax (x )}}{2} + \frac {\sinh {\relax (x )} \cosh {\relax (x )}}{2} & \text {for}\: m = -1 \vee m = 1 \\\frac {m \sinh {\left (m x \right )} \cosh {\relax (x )}}{m^{2} - 1} - \frac {\sinh {\relax (x )} \cosh {\left (m x \right )}}{m^{2} - 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*cosh(m*x),x)

[Out]

Piecewise((-x*sinh(x)**2/2 + x*cosh(x)**2/2 + sinh(x)*cosh(x)/2, Eq(m, -1) | Eq(m, 1)), (m*sinh(m*x)*cosh(x)/(
m**2 - 1) - sinh(x)*cosh(m*x)/(m**2 - 1), True))

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