3.239 \(\int \cosh (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=54 \[ \frac {x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2}-\frac {b n x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2} \]

[Out]

x*cosh(a+b*ln(c*x^n))/(-b^2*n^2+1)-b*n*x*sinh(a+b*ln(c*x^n))/(-b^2*n^2+1)

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Rubi [A]  time = 0.01, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5518} \[ \frac {x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2}-\frac {b n x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*Log[c*x^n]],x]

[Out]

(x*Cosh[a + b*Log[c*x^n]])/(1 - b^2*n^2) - (b*n*x*Sinh[a + b*Log[c*x^n]])/(1 - b^2*n^2)

Rule 5518

Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> -Simp[(x*Cosh[d*(a + b*Log[c*x^n])])/(b^2*
d^2*n^2 - 1), x] + Simp[(b*d*n*x*Sinh[d*(a + b*Log[c*x^n])])/(b^2*d^2*n^2 - 1), x] /; FreeQ[{a, b, c, d, n}, x
] && NeQ[b^2*d^2*n^2 - 1, 0]

Rubi steps

\begin {align*} \int \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2}-\frac {b n x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-b^2 n^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 41, normalized size = 0.76 \[ \frac {x \left (b n \sinh \left (a+b \log \left (c x^n\right )\right )-\cosh \left (a+b \log \left (c x^n\right )\right )\right )}{b^2 n^2-1} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*Log[c*x^n]],x]

[Out]

(x*(-Cosh[a + b*Log[c*x^n]] + b*n*Sinh[a + b*Log[c*x^n]]))/(-1 + b^2*n^2)

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fricas [A]  time = 0.97, size = 44, normalized size = 0.81 \[ \frac {b n x \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) - x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{b^{2} n^{2} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

(b*n*x*sinh(b*n*log(x) + b*log(c) + a) - x*cosh(b*n*log(x) + b*log(c) + a))/(b^2*n^2 - 1)

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giac [A]  time = 0.12, size = 47, normalized size = 0.87 \[ \frac {c^{b} x x^{b n} e^{a}}{2 \, {\left (b n + 1\right )}} - \frac {x e^{\left (-a\right )}}{2 \, {\left (b n - 1\right )} c^{b} x^{b n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \cosh \left (a +b \ln \left (c \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a+b*ln(c*x^n)),x)

[Out]

int(cosh(a+b*ln(c*x^n)),x)

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maxima [A]  time = 0.34, size = 51, normalized size = 0.94 \[ \frac {c^{b} x e^{\left (b \log \left (x^{n}\right ) + a\right )}}{2 \, {\left (b n + 1\right )}} - \frac {x e^{\left (-a\right )}}{2 \, {\left (b c^{b} n - c^{b}\right )} {\left (x^{n}\right )}^{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.00, size = 44, normalized size = 0.81 \[ \frac {x\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{2\,b\,n+2}-\frac {x\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (2\,b\,n-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*log(c*x^n)),x)

[Out]

(x*exp(a)*(c*x^n)^b)/(2*b*n + 2) - (x*exp(-a))/((c*x^n)^b*(2*b*n - 2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \int \cosh {\left (a - \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {1}{n} \\\int \cosh {\left (a + \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {1}{n} \\\frac {b n x \sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{b^{2} n^{2} - 1} - \frac {x \cosh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{b^{2} n^{2} - 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*ln(c*x**n)),x)

[Out]

Piecewise((Integral(cosh(a - log(c*x**n)/n), x), Eq(b, -1/n)), (Integral(cosh(a + log(c*x**n)/n), x), Eq(b, 1/
n)), (b*n*x*sinh(a + b*n*log(x) + b*log(c))/(b**2*n**2 - 1) - x*cosh(a + b*n*log(x) + b*log(c))/(b**2*n**2 - 1
), True))

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