Integrand size = 17, antiderivative size = 65 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\cosh \left (a+b \log \left (c x^n\right )\right )}{b n}-\frac {2 \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac {\cosh ^5\left (a+b \log \left (c x^n\right )\right )}{5 b n} \] Output:
cosh(a+b*ln(c*x^n))/b/n-2/3*cosh(a+b*ln(c*x^n))^3/b/n+1/5*cosh(a+b*ln(c*x^ n))^5/b/n
Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {5 \cosh \left (a+b \log \left (c x^n\right )\right )}{8 b n}-\frac {5 \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )}{48 b n}+\frac {\cosh \left (5 \left (a+b \log \left (c x^n\right )\right )\right )}{80 b n} \] Input:
Integrate[Sinh[a + b*Log[c*x^n]]^5/x,x]
Output:
(5*Cosh[a + b*Log[c*x^n]])/(8*b*n) - (5*Cosh[3*(a + b*Log[c*x^n])])/(48*b* n) + Cosh[5*(a + b*Log[c*x^n])]/(80*b*n)
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3039, 3042, 26, 3113, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle \frac {\int \sinh ^5\left (a+b \log \left (c x^n\right )\right )d\log \left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -i \sin \left (i a+i b \log \left (c x^n\right )\right )^5d\log \left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {i \int \sin \left (i a+i b \log \left (c x^n\right )\right )^5d\log \left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {\int \left (\cosh ^4\left (a+b \log \left (c x^n\right )\right )-2 \cosh ^2\left (a+b \log \left (c x^n\right )\right )+1\right )d\cosh \left (a+b \log \left (c x^n\right )\right )}{b n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} \cosh ^5\left (a+b \log \left (c x^n\right )\right )-\frac {2}{3} \cosh ^3\left (a+b \log \left (c x^n\right )\right )+\cosh \left (a+b \log \left (c x^n\right )\right )}{b n}\) |
Input:
Int[Sinh[a + b*Log[c*x^n]]^5/x,x]
Output:
(Cosh[a + b*Log[c*x^n]] - (2*Cosh[a + b*Log[c*x^n]]^3)/3 + Cosh[a + b*Log[ c*x^n]]^5/5)/(b*n)
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst [[3]] Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /; !FalseQ[lst]] /; NonsumQ[u]
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Time = 74.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\left (\frac {8}{15}+\frac {{\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{4}}{5}-\frac {4 {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{15}\right ) \cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}{n b}\) | \(51\) |
default | \(\frac {\left (\frac {8}{15}+\frac {{\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{4}}{5}-\frac {4 {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{15}\right ) \cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}{n b}\) | \(51\) |
parallelrisch | \(\frac {128-25 \cosh \left (3 b \ln \left (c \,x^{n}\right )+3 a \right )+150 \cosh \left (a +b \ln \left (c \,x^{n}\right )\right )+3 \cosh \left (5 b \ln \left (c \,x^{n}\right )+5 a \right )}{240 b n}\) | \(56\) |
Input:
int(sinh(a+b*ln(c*x^n))^5/x,x,method=_RETURNVERBOSE)
Output:
1/n/b*(8/15+1/5*sinh(a+b*ln(c*x^n))^4-4/15*sinh(a+b*ln(c*x^n))^2)*cosh(a+b *ln(c*x^n))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (61) = 122\).
Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.00 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {3 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{5} + 15 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{4} - 25 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 15 \, {\left (2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 5 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 150 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{240 \, b n} \] Input:
integrate(sinh(a+b*log(c*x^n))^5/x,x, algorithm="fricas")
Output:
1/240*(3*cosh(b*n*log(x) + b*log(c) + a)^5 + 15*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a)^4 - 25*cosh(b*n*log(x) + b*log(c) + a)^3 + 15*(2*cosh(b*n*log(x) + b*log(c) + a)^3 - 5*cosh(b*n*log(x) + b*log (c) + a))*sinh(b*n*log(x) + b*log(c) + a)^2 + 150*cosh(b*n*log(x) + b*log( c) + a))/(b*n)
Time = 9.59 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.62 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\begin {cases} \log {\left (x \right )} \sinh ^{5}{\left (a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log {\left (x \right )} \sinh ^{5}{\left (a + b \log {\left (c \right )} \right )} & \text {for}\: n = 0 \\\frac {\sinh ^{4}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b n} - \frac {4 \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{3 b n} + \frac {8 \cosh ^{5}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{15 b n} & \text {otherwise} \end {cases} \] Input:
integrate(sinh(a+b*ln(c*x**n))**5/x,x)
Output:
Piecewise((log(x)*sinh(a)**5, Eq(b, 0) & (Eq(b, 0) | Eq(n, 0))), (log(x)*s inh(a + b*log(c))**5, Eq(n, 0)), (sinh(a + b*log(c*x**n))**4*cosh(a + b*lo g(c*x**n))/(b*n) - 4*sinh(a + b*log(c*x**n))**2*cosh(a + b*log(c*x**n))**3 /(3*b*n) + 8*cosh(a + b*log(c*x**n))**5/(15*b*n), True))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (61) = 122\).
Time = 0.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.00 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {e^{\left (5 \, b \log \left (c x^{n}\right ) + 5 \, a\right )}}{160 \, b n} - \frac {5 \, e^{\left (3 \, b \log \left (c x^{n}\right ) + 3 \, a\right )}}{96 \, b n} + \frac {5 \, e^{\left (b \log \left (c x^{n}\right ) + a\right )}}{16 \, b n} + \frac {5 \, e^{\left (-b \log \left (c x^{n}\right ) - a\right )}}{16 \, b n} - \frac {5 \, e^{\left (-3 \, b \log \left (c x^{n}\right ) - 3 \, a\right )}}{96 \, b n} + \frac {e^{\left (-5 \, b \log \left (c x^{n}\right ) - 5 \, a\right )}}{160 \, b n} \] Input:
integrate(sinh(a+b*log(c*x^n))^5/x,x, algorithm="maxima")
Output:
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)
Time = 0.13 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.77 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {{\left (3 \, c^{10 \, b} x^{5 \, b n} e^{\left (10 \, a\right )} - 25 \, c^{8 \, b} x^{3 \, b n} e^{\left (8 \, a\right )} + 150 \, c^{6 \, b} x^{b n} e^{\left (6 \, a\right )} + \frac {150 \, c^{4 \, b} x^{4 \, b n} e^{\left (4 \, a\right )} - 25 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 3}{x^{5 \, b n}}\right )} e^{\left (-5 \, a\right )}}{480 \, b c^{5 \, b} n} \] Input:
integrate(sinh(a+b*log(c*x^n))^5/x,x, algorithm="giac")
Output:
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)
Time = 1.71 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\frac {{\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^5}{5}-\frac {2\,{\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{3}+\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}{b\,n} \] Input:
int(sinh(a + b*log(c*x^n))^5/x,x)
Output:
(cosh(a + b*log(c*x^n)) - (2*cosh(a + b*log(c*x^n))^3)/3 + cosh(a + b*log( c*x^n))^5/5)/(b*n)
Time = 0.16 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.88 \[ \int \frac {\sinh ^5\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {3 x^{10 b n} e^{10 a} c^{10 b}-25 x^{8 b n} e^{8 a} c^{8 b}+150 x^{6 b n} e^{6 a} c^{6 b}+150 x^{4 b n} e^{4 a} c^{4 b}-25 x^{2 b n} e^{2 a} c^{2 b}+3}{480 x^{5 b n} e^{5 a} c^{5 b} b n} \] Input:
int(sinh(a+b*log(c*x^n))^5/x,x)
Output:
(3*x**(10*b*n)*e**(10*a)*c**(10*b) - 25*x**(8*b*n)*e**(8*a)*c**(8*b) + 150 *x**(6*b*n)*e**(6*a)*c**(6*b) + 150*x**(4*b*n)*e**(4*a)*c**(4*b) - 25*x**( 2*b*n)*e**(2*a)*c**(2*b) + 3)/(480*x**(5*b*n)*e**(5*a)*c**(5*b)*b*n)