3.274 \(\int \frac {\sinh (a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=18 \[ \frac {\cosh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2638} \[ \frac {\cosh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*Log[c*x^n]]/x,x]

[Out]

Cosh[a + b*Log[c*x^n]]/(b*n)

Rule 2638

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

Rubi steps

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

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Mathematica [B]  time = 0.01, size = 37, normalized size = 2.06 \[ \frac {\sinh (a) \sinh \left (b \log \left (c x^n\right )\right )}{b n}+\frac {\cosh (a) \cosh \left (b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*Log[c*x^n]]/x,x]

[Out]

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

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fricas [A]  time = 0.61, size = 19, normalized size = 1.06 \[ \frac {\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cosh(b*n*log(x) + b*log(c) + a)/(b*n)

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giac [B]  time = 0.19, size = 40, normalized size = 2.22 \[ \frac {{\left (c^{2 \, b} x^{b n} e^{\left (2 \, a\right )} + \frac {1}{x^{b n}}\right )} e^{\left (-a\right )}}{2 \, b c^{b} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 19, normalized size = 1.06 \[ \frac {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 18, normalized size = 1.00 \[ \frac {\cosh \left (b \log \left (c x^{n}\right ) + a\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cosh(b*log(c*x^n) + a)/(b*n)

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mupad [B]  time = 0.66, size = 18, normalized size = 1.00 \[ \frac {\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}{b\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*log(c*x^n))/x,x)

[Out]

cosh(a + b*log(c*x^n))/(b*n)

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sympy [A]  time = 0.99, size = 37, normalized size = 2.06 \[ \begin {cases} \log {\relax (x )} \sinh {\relax (a )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log {\relax (x )} \sinh {\left (a + b \log {\relax (c )} \right )} & \text {for}\: n = 0 \\\frac {\cosh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((log(x)*sinh(a), Eq(b, 0) & (Eq(b, 0) | Eq(n, 0))), (log(x)*sinh(a + b*log(c)), Eq(n, 0)), (cosh(a +
b*n*log(x) + b*log(c))/(b*n), True))

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