\(\int x^m \sinh (a+b \log (c x^n)) \, dx\) [270]

Optimal result

Integrand size = 15, antiderivative size = 73 \[ \int x^m \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b n x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-b^2 n^2}+\frac {(1+m) x^{1+m} \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-b^2 n^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (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 = {5638} \[ \int x^m \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {(m+1) x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right )}{(-b n+m+1) (b n+m+1)}-\frac {b n x^{m+1} \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2} \]

[In]

Int[x^m*Sinh[a + b*Log[c*x^n]],x]

[Out]

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

Rule 5638

Int[((e_.)*(x_))^(m_.)*Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> Simp[(-(m + 1))*(e*x)^(
m + 1)*(Sinh[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)*(Cosh[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*} \text {integral}& = -\frac {b n x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-b^2 n^2}+\frac {(1+m) x^{1+m} \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m-b n) (1+m+b n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.74 \[ \int x^m \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{1+m} \left (-b n \cosh \left (a+b \log \left (c x^n\right )\right )+(1+m) \sinh \left (a+b \log \left (c x^n\right )\right )\right )}{(1+m-b n) (1+m+b n)} \]

[In]

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

[Out]

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

Maple [F]

\[\int x^{m} \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.34 \[ \int x^m \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \cosh \left (m \log \left (x\right )\right ) + b n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (m \log \left (x\right )\right ) - {\left ({\left (m + 1\right )} x \cosh \left (m \log \left (x\right )\right ) + {\left (m + 1\right )} x \sinh \left (m \log \left (x\right )\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{b^{2} n^{2} - m^{2} - 2 \, m - 1} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int x^m \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{b} x e^{\left (b \log \left (x^{n}\right ) + m \log \left (x\right ) + a\right )}}{2 \, {\left (b n + m + 1\right )}} + \frac {x e^{\left (-b \log \left (x^{n}\right ) + m \log \left (x\right ) - a\right )}}{2 \, {\left (b c^{b} n - c^{b} {\left (m + 1\right )}\right )}} \]

[In]

integrate(x^m*sinh(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))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (75) = 150\).

Time = 0.29 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.22 \[ \int x^m \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b c^{b} n x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {c^{b} m x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} - \frac {c^{b} x x^{b n} x^{m} e^{a}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )}} + \frac {b n x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} + \frac {m x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} + \frac {x x^{m} e^{\left (-a\right )}}{2 \, {\left (b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} c^{b} x^{b n}} \]

[In]

integrate(x^m*sinh(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))

Mupad [B] (verification not implemented)

Time = 1.45 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int x^m \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{2\,m+2\,b\,n+2}-\frac {x\,x^m\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (2\,m-2\,b\,n+2\right )} \]

[In]

int(x^m*sinh(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))