3.251 \(\int \frac {\cosh ^5(a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=65 \[ \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n}+\frac {2 \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac {\sinh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[Out]

sinh(a+b*ln(c*x^n))/b/n+2/3*sinh(a+b*ln(c*x^n))^3/b/n+1/5*sinh(a+b*ln(c*x^n))^5/b/n

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2633} \[ \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n}+\frac {2 \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac {\sinh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*Log[c*x^n]]^5/x,x]

[Out]

Sinh[a + b*Log[c*x^n]]/(b*n) + (2*Sinh[a + b*Log[c*x^n]]^3)/(3*b*n) + Sinh[a + b*Log[c*x^n]]^5/(5*b*n)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin {align*} \int \frac {\cosh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \cosh ^5(a+b x) \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=\frac {i \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-i \sinh \left (a+b \log \left (c x^n\right )\right )\right )}{b n}\\ &=\frac {\sinh \left (a+b \log \left (c x^n\right )\right )}{b n}+\frac {2 \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 65, normalized size = 1.00 \[ \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n}+\frac {2 \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac {\sinh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*Log[c*x^n]]^5/x,x]

[Out]

Sinh[a + b*Log[c*x^n]]/(b*n) + (2*Sinh[a + b*Log[c*x^n]]^3)/(3*b*n) + Sinh[a + b*Log[c*x^n]]^5/(5*b*n)

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fricas [A]  time = 0.55, size = 105, normalized size = 1.62 \[ \frac {3 \, \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{5} + 5 \, {\left (6 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 5\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{3} + 15 \, {\left (\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{4} + 5 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 10\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{240 \, b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*log(c*x^n))^5/x,x, algorithm="fricas")

[Out]

1/240*(3*sinh(b*n*log(x) + b*log(c) + a)^5 + 5*(6*cosh(b*n*log(x) + b*log(c) + a)^2 + 5)*sinh(b*n*log(x) + b*l
og(c) + a)^3 + 15*(cosh(b*n*log(x) + b*log(c) + a)^4 + 5*cosh(b*n*log(x) + b*log(c) + a)^2 + 10)*sinh(b*n*log(
x) + b*log(c) + a))/(b*n)

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giac [A]  time = 0.17, size = 116, normalized size = 1.78 \[ \frac {{\left (3 \, c^{10 \, b} x^{5 \, b n} e^{\left (10 \, a\right )} + 25 \, c^{8 \, b} x^{3 \, b n} e^{\left (8 \, a\right )} + 150 \, c^{6 \, b} x^{b n} e^{\left (6 \, a\right )} - \frac {150 \, c^{4 \, b} x^{4 \, b n} e^{\left (4 \, a\right )} + 25 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 3}{x^{5 \, b n}}\right )} e^{\left (-5 \, a\right )}}{480 \, b c^{5 \, b} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*log(c*x^n))^5/x,x, algorithm="giac")

[Out]

1/480*(3*c^(10*b)*x^(5*b*n)*e^(10*a) + 25*c^(8*b)*x^(3*b*n)*e^(8*a) + 150*c^(6*b)*x^(b*n)*e^(6*a) - (150*c^(4*
b)*x^(4*b*n)*e^(4*a) + 25*c^(2*b)*x^(2*b*n)*e^(2*a) + 3)/x^(5*b*n))*e^(-5*a)/(b*c^(5*b)*n)

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maple [A]  time = 0.22, size = 51, normalized size = 0.78 \[ \frac {\left (\frac {8}{15}+\frac {\left (\cosh ^{4}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{5}+\frac {4 \left (\cosh ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{15}\right ) \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}{n b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a+b*ln(c*x^n))^5/x,x)

[Out]

1/n/b*(8/15+1/5*cosh(a+b*ln(c*x^n))^4+4/15*cosh(a+b*ln(c*x^n))^2)*sinh(a+b*ln(c*x^n))

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maxima [B]  time = 0.32, size = 130, normalized size = 2.00 \[ \frac {e^{\left (5 \, b \log \left (c x^{n}\right ) + 5 \, a\right )}}{160 \, b n} + \frac {5 \, e^{\left (3 \, b \log \left (c x^{n}\right ) + 3 \, a\right )}}{96 \, b n} + \frac {5 \, e^{\left (b \log \left (c x^{n}\right ) + a\right )}}{16 \, b n} - \frac {5 \, e^{\left (-b \log \left (c x^{n}\right ) - a\right )}}{16 \, b n} - \frac {5 \, e^{\left (-3 \, b \log \left (c x^{n}\right ) - 3 \, a\right )}}{96 \, b n} - \frac {e^{\left (-5 \, b \log \left (c x^{n}\right ) - 5 \, a\right )}}{160 \, b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*log(c*x^n))^5/x,x, algorithm="maxima")

[Out]

1/160*e^(5*b*log(c*x^n) + 5*a)/(b*n) + 5/96*e^(3*b*log(c*x^n) + 3*a)/(b*n) + 5/16*e^(b*log(c*x^n) + a)/(b*n) -
5/16*e^(-b*log(c*x^n) - a)/(b*n) - 5/96*e^(-3*b*log(c*x^n) - 3*a)/(b*n) - 1/160*e^(-5*b*log(c*x^n) - 5*a)/(b*
n)

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mupad [B]  time = 1.17, size = 49, normalized size = 0.75 \[ \frac {\frac {{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^5}{5}+\frac {2\,{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{3}+\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}{b\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*log(c*x^n))^5/x,x)

[Out]

(sinh(a + b*log(c*x^n)) + (2*sinh(a + b*log(c*x^n))^3)/3 + sinh(a + b*log(c*x^n))^5/5)/(b*n)

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sympy [A]  time = 97.06, size = 128, normalized size = 1.97 \[ \begin {cases} \log {\relax (x )} \cosh ^{5}{\relax (a )} & \text {for}\: b = 0 \wedge n = 0 \\\log {\relax (x )} \cosh ^{5}{\left (a + b \log {\relax (c )} \right )} & \text {for}\: n = 0 \\\log {\relax (x )} \cosh ^{5}{\relax (a )} & \text {for}\: b = 0 \\\frac {8 \sinh ^{5}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{15 b n} - \frac {4 \sinh ^{3}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{3 b n} + \frac {\sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh ^{4}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*ln(c*x**n))**5/x,x)

[Out]

Piecewise((log(x)*cosh(a)**5, Eq(b, 0) & Eq(n, 0)), (log(x)*cosh(a + b*log(c))**5, Eq(n, 0)), (log(x)*cosh(a)*
*5, Eq(b, 0)), (8*sinh(a + b*n*log(x) + b*log(c))**5/(15*b*n) - 4*sinh(a + b*n*log(x) + b*log(c))**3*cosh(a +
b*n*log(x) + b*log(c))**2/(3*b*n) + sinh(a + b*n*log(x) + b*log(c))*cosh(a + b*n*log(x) + b*log(c))**4/(b*n),
True))

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