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

Optimal. Leaf size=73 \[ \frac {(m+1) x^{m+1} \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2}-\frac {b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2} \]

[Out]

(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*ln(c*x^n))/((1+m)^2-b^2*n^2)

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {5528} \[ \frac {(m+1) x^{m+1} \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2}-\frac {b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Cosh[a + b*Log[c*x^n]],x]

[Out]

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

Rule 5528

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]

Rubi steps

\begin {align*} \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}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 54, normalized size = 0.74 \[ \frac {x^{m+1} \left ((m+1) \cosh \left (a+b \log \left (c x^n\right )\right )-b n \sinh \left (a+b \log \left (c x^n\right )\right )\right )}{(-b n+m+1) (b n+m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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))

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fricas [A]  time = 0.47, size = 99, normalized size = 1.36 \[ -\frac {{\left (m + 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \cosh \left (m \log \relax (x)\right ) + {\left (m + 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (m \log \relax (x)\right ) - {\left (b n x \cosh \left (m \log \relax (x)\right ) + b n x \sinh \left (m \log \relax (x)\right )\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{b^{2} n^{2} - m^{2} - 2 \, m - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + (m + 1)*x*cosh(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)

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giac [B]  time = 0.16, size = 235, normalized size = 3.22 \[ \frac {b c^{b} n x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {c^{b} m x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {c^{b} x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {b n x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} - \frac {m x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} - \frac {x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int x^{m} \cosh \left (a +b \ln \left (c \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 64, normalized size = 0.88 \[ \frac {c^{b} x e^{\left (b \log \left (x^{n}\right ) + m \log \relax (x) + a\right )}}{2 \, {\left (b n + m + 1\right )}} - \frac {x e^{\left (-b \log \left (x^{n}\right ) + m \log \relax (x) - a\right )}}{2 \, {\left (b c^{b} n - c^{b} {\left (m + 1\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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mupad [B]  time = 1.05, size = 55, normalized size = 0.75 \[ \frac {x\,x^m\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (2\,m-2\,b\,n+2\right )}+\frac {x\,x^m\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{2\,m+2\,b\,n+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*cosh(a + b*log(c*x^n)),x)

[Out]

(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*n*log(x) + b*log(c))/(b**2*n**2 - m**2 - 2*m - 1) - m*x*x**m*cosh(a + b*n*log(x) + b*log(c))/(
b**2*n**2 - m**2 - 2*m - 1) - x*x**m*cosh(a + b*n*log(x) + b*log(c))/(b**2*n**2 - m**2 - 2*m - 1), True))

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