3.244 \(\int x^m \cosh ^2(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=120 \[ \frac {(m+1) x^{m+1} \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}-\frac {2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}-\frac {2 b^2 n^2 x^{m+1}}{(m+1) \left ((m+1)^2-4 b^2 n^2\right )} \]

[Out]

-2*b^2*n^2*x^(1+m)/(1+m)/((1+m)^2-4*b^2*n^2)+(1+m)*x^(1+m)*cosh(a+b*ln(c*x^n))^2/((1+m)^2-4*b^2*n^2)-2*b*n*x^(
1+m)*cosh(a+b*ln(c*x^n))*sinh(a+b*ln(c*x^n))/((1+m)^2-4*b^2*n^2)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5530, 30} \[ \frac {(m+1) x^{m+1} \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}-\frac {2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}-\frac {2 b^2 n^2 x^{m+1}}{(m+1) \left ((m+1)^2-4 b^2 n^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Cosh[a + b*Log[c*x^n]]^2,x]

[Out]

(-2*b^2*n^2*x^(1 + m))/((1 + m)*((1 + m)^2 - 4*b^2*n^2)) + ((1 + m)*x^(1 + m)*Cosh[a + b*Log[c*x^n]]^2)/((1 +
m)^2 - 4*b^2*n^2) - (2*b*n*x^(1 + m)*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]])/((1 + m)^2 - 4*b^2*n^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5530

Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> -Simp[((m + 1)*(e*
x)^(m + 1)*Cosh[d*(a + b*Log[c*x^n])]^p)/(b^2*d^2*e*n^2*p^2 - e*(m + 1)^2), x] + (Dist[(b^2*d^2*n^2*p*(p - 1))
/(b^2*d^2*n^2*p^2 - (m + 1)^2), Int[(e*x)^m*Cosh[d*(a + b*Log[c*x^n])]^(p - 2), x], x] + Simp[(b*d*n*p*(e*x)^(
m + 1)*Sinh[d*(a + b*Log[c*x^n])]*Cosh[d*(a + b*Log[c*x^n])]^(p - 1))/(b^2*d^2*e*n^2*p^2 - e*(m + 1)^2), x]) /
; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 - (m + 1)^2, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 87, normalized size = 0.72 \[ \frac {x^{m+1} \left (-2 b (m+1) n \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+(m+1)^2 \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-4 b^2 n^2+m^2+2 m+1\right )}{2 (m+1) (-2 b n+m+1) (2 b n+m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*Cosh[a + b*Log[c*x^n]]^2,x]

[Out]

(x^(1 + m)*(1 + 2*m + m^2 - 4*b^2*n^2 + (1 + m)^2*Cosh[2*(a + b*Log[c*x^n])] - 2*b*(1 + m)*n*Sinh[2*(a + b*Log
[c*x^n])]))/(2*(1 + m)*(1 + m - 2*b*n)*(1 + m + 2*b*n))

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 250, normalized size = 2.08 \[ \frac {{\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} \cosh \left (m \log \relax (x)\right ) - {\left (4 \, b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} x \cosh \left (m \log \relax (x)\right ) + {\left ({\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (m \log \relax (x)\right ) + {\left (m^{2} + 2 \, m + 1\right )} x \sinh \left (m \log \relax (x)\right )\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - 4 \, {\left ({\left (b m + b\right )} n x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \cosh \left (m \log \relax (x)\right ) + {\left (b m + b\right )} n x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (m \log \relax (x)\right )\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + {\left ({\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - {\left (4 \, b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} x\right )} \sinh \left (m \log \relax (x)\right )}{2 \, {\left (m^{3} - 4 \, {\left (b^{2} m + b^{2}\right )} n^{2} + 3 \, m^{2} + 3 \, m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*cosh(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

1/2*((m^2 + 2*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2*cosh(m*log(x)) - (4*b^2*n^2 - m^2 - 2*m - 1)*x*cosh(m
*log(x)) + ((m^2 + 2*m + 1)*x*cosh(m*log(x)) + (m^2 + 2*m + 1)*x*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) +
a)^2 - 4*((b*m + b)*n*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + (b*m + b)*n*x*cosh(b*n*log(x) + b*log
(c) + a)*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a) + ((m^2 + 2*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^
2 - (4*b^2*n^2 - m^2 - 2*m - 1)*x)*sinh(m*log(x)))/(m^3 - 4*(b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)

________________________________________________________________________________________

giac [B]  time = 0.22, size = 759, normalized size = 6.32 \[ \frac {b c^{2 \, b} m n x x^{2 \, b n} x^{m} e^{\left (2 \, a\right )}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )}} - \frac {c^{2 \, b} m^{2} x x^{2 \, b n} x^{m} e^{\left (2 \, a\right )}}{4 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )}} + \frac {b c^{2 \, b} n x x^{2 \, b n} x^{m} e^{\left (2 \, a\right )}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )}} + \frac {2 \, b^{2} n^{2} x x^{m}}{4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1} - \frac {c^{2 \, b} m x x^{2 \, b n} x^{m} e^{\left (2 \, a\right )}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )}} - \frac {c^{2 \, b} x x^{2 \, b n} x^{m} e^{\left (2 \, a\right )}}{4 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )}} - \frac {m^{2} x x^{m}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )}} - \frac {b m n x x^{m} e^{\left (-2 \, a\right )}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )} c^{2 \, b} x^{2 \, b n}} - \frac {m x x^{m}}{4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1} - \frac {m^{2} x x^{m} e^{\left (-2 \, a\right )}}{4 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )} c^{2 \, b} x^{2 \, b n}} - \frac {b n x x^{m} e^{\left (-2 \, a\right )}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )} c^{2 \, b} x^{2 \, b n}} - \frac {x x^{m}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )}} - \frac {m x x^{m} e^{\left (-2 \, a\right )}}{2 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )} c^{2 \, b} x^{2 \, b n}} - \frac {x x^{m} e^{\left (-2 \, a\right )}}{4 \, {\left (4 \, b^{2} m n^{2} + 4 \, b^{2} n^{2} - m^{3} - 3 \, m^{2} - 3 \, m - 1\right )} c^{2 \, b} x^{2 \, b n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*cosh(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

1/2*b*c^(2*b)*m*n*x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/4*c^(2*b)*m^2*
x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) + 1/2*b*c^(2*b)*n*x*x^(2*b*n)*x^m*e^
(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) + 2*b^2*n^2*x*x^m/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m
^2 - 3*m - 1) - 1/2*c^(2*b)*m*x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/4*
c^(2*b)*x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/2*m^2*x*x^m/(4*b^2*m*n^2
+ 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/2*b*m*n*x*x^m*e^(-2*a)/((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m
- 1)*c^(2*b)*x^(2*b*n)) - m*x*x^m/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/4*m^2*x*x^m*e^(-2*a)/
((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n)) - 1/2*b*n*x*x^m*e^(-2*a)/((4*b^2*m*n^2 +
4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n)) - 1/2*x*x^m/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*
m - 1) - 1/2*m*x*x^m*e^(-2*a)/((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n)) - 1/4*x*x^
m*e^(-2*a)/((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n))

________________________________________________________________________________________

maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\cosh ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*cosh(a+b*ln(c*x^n))^2,x)

[Out]

int(x^m*cosh(a+b*ln(c*x^n))^2,x)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 87, normalized size = 0.72 \[ \frac {c^{2 \, b} x e^{\left (2 \, b \log \left (x^{n}\right ) + m \log \relax (x) + 2 \, a\right )}}{4 \, {\left (2 \, b n + m + 1\right )}} - \frac {x e^{\left (-2 \, b \log \left (x^{n}\right ) + m \log \relax (x) - 2 \, a\right )}}{4 \, {\left (2 \, b c^{2 \, b} n - c^{2 \, b} {\left (m + 1\right )}\right )}} + \frac {x^{m + 1}}{2 \, {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*cosh(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

1/4*c^(2*b)*x*e^(2*b*log(x^n) + m*log(x) + 2*a)/(2*b*n + m + 1) - 1/4*x*e^(-2*b*log(x^n) + m*log(x) - 2*a)/(2*
b*c^(2*b)*n - c^(2*b)*(m + 1)) + 1/2*x^(m + 1)/(m + 1)

________________________________________________________________________________________

mupad [B]  time = 1.09, size = 73, normalized size = 0.61 \[ \frac {x\,x^m}{2\,m+2}+\frac {x\,x^m\,{\mathrm {e}}^{-2\,a}}{{\left (c\,x^n\right )}^{2\,b}\,\left (4\,m-8\,b\,n+4\right )}+\frac {x\,x^m\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}}{4\,m+8\,b\,n+4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*cosh(a + b*log(c*x^n))^2,x)

[Out]

(x*x^m)/(2*m + 2) + (x*x^m*exp(-2*a))/((c*x^n)^(2*b)*(4*m - 8*b*n + 4)) + (x*x^m*exp(2*a)*(c*x^n)^(2*b))/(4*m
+ 8*b*n + 4)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \log {\relax (x )} \cosh ^{2}{\relax (a )} & \text {for}\: b = 0 \wedge m = -1 \\\int x^{m} \cosh ^{2}{\left (a - \frac {m \log {\left (c x^{n} \right )}}{2 n} - \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {m + 1}{2 n} \\\int x^{m} \cosh ^{2}{\left (a + \frac {m \log {\left (c x^{n} \right )}}{2 n} + \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {m + 1}{2 n} \\\int \frac {\cosh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{x}\, dx & \text {for}\: m = -1 \\- \frac {2 b^{2} n^{2} x x^{m} \sinh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} + \frac {2 b^{2} n^{2} x x^{m} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} + \frac {2 b m n x x^{m} \sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} + \frac {2 b n x x^{m} \sinh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cosh {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} - \frac {m^{2} x x^{m} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} - \frac {2 m x x^{m} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} - \frac {x x^{m} \cosh ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} - m^{3} - 3 m^{2} - 3 m - 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*cosh(a+b*ln(c*x**n))**2,x)

[Out]

Piecewise((log(x)*cosh(a)**2, Eq(b, 0) & Eq(m, -1)), (Integral(x**m*cosh(a - m*log(c*x**n)/(2*n) - log(c*x**n)
/(2*n))**2, x), Eq(b, -(m + 1)/(2*n))), (Integral(x**m*cosh(a + m*log(c*x**n)/(2*n) + log(c*x**n)/(2*n))**2, x
), Eq(b, (m + 1)/(2*n))), (Integral(cosh(a + b*log(c*x**n))**2/x, x), Eq(m, -1)), (-2*b**2*n**2*x*x**m*sinh(a
+ b*n*log(x) + b*log(c))**2/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 3*m - 1) + 2*b**2*n**2*x*x**m*cosh(
a + b*n*log(x) + b*log(c))**2/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 3*m - 1) + 2*b*m*n*x*x**m*sinh(a
+ b*n*log(x) + b*log(c))*cosh(a + b*n*log(x) + b*log(c))/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 3*m -
1) + 2*b*n*x*x**m*sinh(a + b*n*log(x) + b*log(c))*cosh(a + b*n*log(x) + b*log(c))/(4*b**2*m*n**2 + 4*b**2*n**2
- m**3 - 3*m**2 - 3*m - 1) - m**2*x*x**m*cosh(a + b*n*log(x) + b*log(c))**2/(4*b**2*m*n**2 + 4*b**2*n**2 - m*
*3 - 3*m**2 - 3*m - 1) - 2*m*x*x**m*cosh(a + b*n*log(x) + b*log(c))**2/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3
*m**2 - 3*m - 1) - x*x**m*cosh(a + b*n*log(x) + b*log(c))**2/(4*b**2*m*n**2 + 4*b**2*n**2 - m**3 - 3*m**2 - 3*
m - 1), True))

________________________________________________________________________________________