Integrand size = 15, antiderivative size = 73 \[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {(1+m) x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-b^2 n^2}-\frac {b n x^{1+m} \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-b^2 n^2} \] Output:
(1+m)*x^(1+m)*cosh(a+b*ln(c*x^n))/((1+m)^2-b^2*n^2)-b*n*x^(1+m)*sinh(a+b*l n(c*x^n))/((1+m)^2-b^2*n^2)
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.74 \[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{1+m} \left ((1+m) \cosh \left (a+b \log \left (c x^n\right )\right )-b n \sinh \left (a+b \log \left (c x^n\right )\right )\right )}{(1+m-b n) (1+m+b n)} \] Input:
Integrate[x^m*Cosh[a + b*Log[c*x^n]],x]
Output:
(x^(1 + m)*((1 + m)*Cosh[a + b*Log[c*x^n]] - b*n*Sinh[a + b*Log[c*x^n]]))/ ((1 + m - b*n)*(1 + m + b*n))
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6054}
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.
\(\displaystyle \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 6054 |
\(\displaystyle \frac {(m+1) x^{m+1} \cosh \left (a+b \log \left (c x^n\right )\right )}{(-b n+m+1) (b n+m+1)}-\frac {b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2}\) |
Input:
Int[x^m*Cosh[a + b*Log[c*x^n]],x]
Output:
((1 + m)*x^(1 + m)*Cosh[a + b*Log[c*x^n]])/((1 + m - b*n)*(1 + m + b*n)) - (b*n*x^(1 + m)*Sinh[a + b*Log[c*x^n]])/((1 + m)^2 - b^2*n^2)
Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]*((e_.)*(x_))^(m_.), x _Symbol] :> Simp[(-(m + 1))*(e*x)^(m + 1)*(Cosh[d*(a + b*Log[c*x^n])]/(b^2* d^2*e*n^2 - e*(m + 1)^2)), x] + Simp[b*d*n*(e*x)^(m + 1)*(Sinh[d*(a + b*Log [c*x^n])]/(b^2*d^2*e*n^2 - e*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, e, m, n} , x] && NeQ[b^2*d^2*n^2 - (m + 1)^2, 0]
Time = 1.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \(\frac {x^{1+m} \left (b n \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )-\cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) m -\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{b^{2} n^{2}-m^{2}-2 m -1}\) | \(68\) |
Input:
int(x^m*cosh(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
Output:
x^(1+m)/(b^2*n^2-m^2-2*m-1)*(b*n*sinh(a+b*ln(c*x^n))-cosh(a+b*ln(c*x^n))*m -cosh(a+b*ln(c*x^n)))
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.36 \[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {{\left (m + 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \cosh \left (m \log \left (x\right )\right ) + {\left (m + 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (m \log \left (x\right )\right ) - {\left (b n x \cosh \left (m \log \left (x\right )\right ) + b n x \sinh \left (m \log \left (x\right )\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{b^{2} n^{2} - m^{2} - 2 \, m - 1} \] Input:
integrate(x^m*cosh(a+b*log(c*x^n)),x, algorithm="fricas")
Output:
-((m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + (m + 1)*x*cos h(b*n*log(x) + b*log(c) + a)*sinh(m*log(x)) - (b*n*x*cosh(m*log(x)) + b*n* x*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a))/(b^2*n^2 - m^2 - 2*m - 1)
\[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \log {\left (x \right )} \cosh {\left (a \right )} & \text {for}\: b = 0 \wedge m = -1 \\\int x^{m} \cosh {\left (- a + \frac {m \log {\left (c x^{n} \right )}}{n} + \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {m + 1}{n} \\\int x^{m} \cosh {\left (a + \frac {m \log {\left (c x^{n} \right )}}{n} + \frac {\log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {m + 1}{n} \\\frac {b n x x^{m} \sinh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} - m^{2} - 2 m - 1} - \frac {m x x^{m} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} - m^{2} - 2 m - 1} - \frac {x x^{m} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} - m^{2} - 2 m - 1} & \text {otherwise} \end {cases} \] Input:
integrate(x**m*cosh(a+b*ln(c*x**n)),x)
Output:
Piecewise((log(x)*cosh(a), Eq(b, 0) & Eq(m, -1)), (Integral(x**m*cosh(-a + m*log(c*x**n)/n + log(c*x**n)/n), x), Eq(b, -(m + 1)/n)), (Integral(x**m* cosh(a + m*log(c*x**n)/n + log(c*x**n)/n), x), Eq(b, (m + 1)/n)), (b*n*x*x **m*sinh(a + b*log(c*x**n))/(b**2*n**2 - m**2 - 2*m - 1) - m*x*x**m*cosh(a + b*log(c*x**n))/(b**2*n**2 - m**2 - 2*m - 1) - x*x**m*cosh(a + b*log(c*x **n))/(b**2*n**2 - m**2 - 2*m - 1), True))
Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{b} x e^{\left (b \log \left (x^{n}\right ) + m \log \left (x\right ) + a\right )}}{2 \, {\left (b n + m + 1\right )}} - \frac {x e^{\left (-b \log \left (x^{n}\right ) + m \log \left (x\right ) - a\right )}}{2 \, {\left (b c^{b} n - c^{b} {\left (m + 1\right )}\right )}} \] Input:
integrate(x^m*cosh(a+b*log(c*x^n)),x, algorithm="maxima")
Output:
1/2*c^b*x*e^(b*log(x^n) + m*log(x) + a)/(b*n + m + 1) - 1/2*x*e^(-b*log(x^ n) + m*log(x) - a)/(b*c^b*n - c^b*(m + 1))
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (75) = 150\).
Time = 0.12 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.22 \[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b c^{b} n x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {c^{b} m x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {c^{b} x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {b n x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} - \frac {m x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} - \frac {x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} \] Input:
integrate(x^m*cosh(a+b*log(c*x^n)),x, algorithm="giac")
Output:
1/2*b*c^b*n*x*x^(b*n)*x^m*e^a/(b^2*n^2 - m^2 - 2*m - 1) - 1/2*c^b*m*x*x^(b *n)*x^m*e^a/(b^2*n^2 - m^2 - 2*m - 1) - 1/2*c^b*x*x^(b*n)*x^m*e^a/(b^2*n^2 - m^2 - 2*m - 1) - 1/2*b*n*x*x^m*e^(-a)/((b^2*n^2 - m^2 - 2*m - 1)*c^b*x^ (b*n)) - 1/2*m*x*x^m*e^(-a)/((b^2*n^2 - m^2 - 2*m - 1)*c^b*x^(b*n)) - 1/2* x*x^m*e^(-a)/((b^2*n^2 - m^2 - 2*m - 1)*c^b*x^(b*n))
Time = 1.99 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (2\,m-2\,b\,n+2\right )}+\frac {x\,x^m\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{2\,m+2\,b\,n+2} \] Input:
int(x^m*cosh(a + b*log(c*x^n)),x)
Output:
(x*x^m*exp(-a))/((c*x^n)^b*(2*m - 2*b*n + 2)) + (x*x^m*exp(a)*(c*x^n)^b)/( 2*m + 2*b*n + 2)
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int x^m \cosh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{m} x \left (-\cosh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) m -\cosh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )+\sinh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) b n \right )}{b^{2} n^{2}-m^{2}-2 m -1} \] Input:
int(x^m*cosh(a+b*log(c*x^n)),x)
Output:
(x**m*x*( - cosh(log(x**n*c)*b + a)*m - cosh(log(x**n*c)*b + a) + sinh(log (x**n*c)*b + a)*b*n))/(b**2*n**2 - m**2 - 2*m - 1)