Integrand size = 13, antiderivative size = 88 \[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {2 b^2 n^2 x}{1-4 b^2 n^2}+\frac {x \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}-\frac {2 b n x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2} \] Output:
-2*b^2*n^2*x/(-4*b^2*n^2+1)+x*cosh(a+b*ln(c*x^n))^2/(-4*b^2*n^2+1)-2*b*n*x *cosh(a+b*ln(c*x^n))*sinh(a+b*ln(c*x^n))/(-4*b^2*n^2+1)
Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64 \[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (-1+4 b^2 n^2-\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+2 b n \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right )}{-2+8 b^2 n^2} \] Input:
Integrate[Cosh[a + b*Log[c*x^n]]^2,x]
Output:
(x*(-1 + 4*b^2*n^2 - Cosh[2*(a + b*Log[c*x^n])] + 2*b*n*Sinh[2*(a + b*Log[ c*x^n])]))/(-2 + 8*b^2*n^2)
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6046, 24}
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.
\(\displaystyle \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 6046 |
\(\displaystyle -\frac {2 b^2 n^2 \int 1dx}{1-4 b^2 n^2}+\frac {x \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}-\frac {2 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {x \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}-\frac {2 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}-\frac {2 b^2 n^2 x}{1-4 b^2 n^2}\) |
Input:
Int[Cosh[a + b*Log[c*x^n]]^2,x]
Output:
(-2*b^2*n^2*x)/(1 - 4*b^2*n^2) + (x*Cosh[a + b*Log[c*x^n]]^2)/(1 - 4*b^2*n ^2) - (2*b*n*x*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]])/(1 - 4*b^2*n ^2)
Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> Si mp[(-x)*(Cosh[d*(a + b*Log[c*x^n])]^p/(b^2*d^2*n^2*p^2 - 1)), x] + (Simp[b* d*n*p*x*Cosh[d*(a + b*Log[c*x^n])]^(p - 1)*(Sinh[d*(a + b*Log[c*x^n])]/(b^2 *d^2*n^2*p^2 - 1)), x] + Simp[b^2*d^2*n^2*p*((p - 1)/(b^2*d^2*n^2*p^2 - 1)) Int[Cosh[d*(a + b*Log[c*x^n])]^(p - 2), x], x]) /; FreeQ[{a, b, c, d, n} , x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 - 1, 0]
Time = 1.79 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.67
method | result | size |
parallelrisch | \(\frac {x \left (-\cosh \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-1+2 b n \sinh \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+4 b^{2} n^{2}\right )}{8 b^{2} n^{2}-2}\) | \(59\) |
Input:
int(cosh(a+b*ln(c*x^n))^2,x,method=_RETURNVERBOSE)
Output:
x*(-cosh(2*b*ln(c*x^n)+2*a)-1+2*b*n*sinh(2*b*ln(c*x^n)+2*a)+4*b^2*n^2)/(8* b^2*n^2-2)
Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 \, b n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - x \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + {\left (4 \, b^{2} n^{2} - 1\right )} x}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )}} \] Input:
integrate(cosh(a+b*log(c*x^n))^2,x, algorithm="fricas")
Output:
1/2*(4*b*n*x*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) - x*cosh(b*n*log(x) + b*log(c) + a)^2 - x*sinh(b*n*log(x) + b*log(c) + a)^2 + (4*b^2*n^2 - 1)*x)/(4*b^2*n^2 - 1)
\[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \int \cosh ^{2}{\left (a - \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {1}{2 n} \\\int \cosh ^{2}{\left (a + \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {1}{2 n} \\- \frac {2 b^{2} n^{2} x \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} + \frac {2 b^{2} n^{2} x \cosh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} + \frac {2 b n x \sinh {\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} - \frac {x \cosh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} & \text {otherwise} \end {cases} \] Input:
integrate(cosh(a+b*ln(c*x**n))**2,x)
Output:
Piecewise((Integral(cosh(a - log(c*x**n)/(2*n))**2, x), Eq(b, -1/(2*n))), (Integral(cosh(a + log(c*x**n)/(2*n))**2, x), Eq(b, 1/(2*n))), (-2*b**2*n* *2*x*sinh(a + b*log(c*x**n))**2/(4*b**2*n**2 - 1) + 2*b**2*n**2*x*cosh(a + b*log(c*x**n))**2/(4*b**2*n**2 - 1) + 2*b*n*x*sinh(a + b*log(c*x**n))*cos h(a + b*log(c*x**n))/(4*b**2*n**2 - 1) - x*cosh(a + b*log(c*x**n))**2/(4*b **2*n**2 - 1), True))
Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{2 \, b} x e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}}{4 \, {\left (2 \, b n + 1\right )}} + \frac {1}{2} \, x - \frac {x e^{\left (-2 \, a\right )}}{4 \, {\left (2 \, b c^{2 \, b} n - c^{2 \, b}\right )} {\left (x^{n}\right )}^{2 \, b}} \] Input:
integrate(cosh(a+b*log(c*x^n))^2,x, algorithm="maxima")
Output:
1/4*c^(2*b)*x*e^(2*b*log(x^n) + 2*a)/(2*b*n + 1) + 1/2*x - 1/4*x*e^(-2*a)/ ((2*b*c^(2*b)*n - c^(2*b))*(x^n)^(2*b))
Time = 0.13 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.92 \[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b c^{2 \, b} n x x^{2 \, b n} e^{\left (2 \, a\right )}}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )}} + \frac {2 \, b^{2} n^{2} x}{4 \, b^{2} n^{2} - 1} - \frac {c^{2 \, b} x x^{2 \, b n} e^{\left (2 \, a\right )}}{4 \, {\left (4 \, b^{2} n^{2} - 1\right )}} - \frac {b n x e^{\left (-2 \, a\right )}}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )} c^{2 \, b} x^{2 \, b n}} - \frac {x}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )}} - \frac {x e^{\left (-2 \, a\right )}}{4 \, {\left (4 \, b^{2} n^{2} - 1\right )} c^{2 \, b} x^{2 \, b n}} \] Input:
integrate(cosh(a+b*log(c*x^n))^2,x, algorithm="giac")
Output:
1/2*b*c^(2*b)*n*x*x^(2*b*n)*e^(2*a)/(4*b^2*n^2 - 1) + 2*b^2*n^2*x/(4*b^2*n ^2 - 1) - 1/4*c^(2*b)*x*x^(2*b*n)*e^(2*a)/(4*b^2*n^2 - 1) - 1/2*b*n*x*e^(- 2*a)/((4*b^2*n^2 - 1)*c^(2*b)*x^(2*b*n)) - 1/2*x/(4*b^2*n^2 - 1) - 1/4*x*e ^(-2*a)/((4*b^2*n^2 - 1)*c^(2*b)*x^(2*b*n))
Time = 1.95 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x}{2}-\frac {x\,{\mathrm {e}}^{-2\,a}}{{\left (c\,x^n\right )}^{2\,b}\,\left (8\,b\,n-4\right )}+\frac {x\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}}{8\,b\,n+4} \] Input:
int(cosh(a + b*log(c*x^n))^2,x)
Output:
x/2 - (x*exp(-2*a))/((c*x^n)^(2*b)*(8*b*n - 4)) + (x*exp(2*a)*(c*x^n)^(2*b ))/(8*b*n + 4)
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int \cosh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (2 x^{4 b n} e^{4 a} c^{4 b} b n -x^{4 b n} e^{4 a} c^{4 b}+8 x^{2 b n} e^{2 a} c^{2 b} b^{2} n^{2}-2 x^{2 b n} e^{2 a} c^{2 b}-2 b n -1\right )}{4 x^{2 b n} e^{2 a} c^{2 b} \left (4 b^{2} n^{2}-1\right )} \] Input:
int(cosh(a+b*log(c*x^n))^2,x)
Output:
(x*(2*x**(4*b*n)*e**(4*a)*c**(4*b)*b*n - x**(4*b*n)*e**(4*a)*c**(4*b) + 8* x**(2*b*n)*e**(2*a)*c**(2*b)*b**2*n**2 - 2*x**(2*b*n)*e**(2*a)*c**(2*b) - 2*b*n - 1))/(4*x**(2*b*n)*e**(2*a)*c**(2*b)*(4*b**2*n**2 - 1))