3.26 \(\int x^m \sin (a+\sqrt{-\frac{(1+m)^2}{n^2}} \log (c x^n)) \, dx\)

Optimal. Leaf size=133 \[ \frac{(m+1) x^{m+1} \log (x) e^{\frac{a n \sqrt{-\frac{(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{-\frac{m+1}{n}}}{2 n \sqrt{-\frac{(m+1)^2}{n^2}}}-\frac{x^{m+1} e^{\frac{a (m+1)}{n \sqrt{-\frac{(m+1)^2}{n^2}}}} \left (c x^n\right )^{\frac{m+1}{n}}}{4 n \sqrt{-\frac{(m+1)^2}{n^2}}} \]

[Out]

-(E^((a*(1 + m))/(Sqrt[-((1 + m)^2/n^2)]*n))*x^(1 + m)*(c*x^n)^((1 + m)/n))/(4*Sqrt[-((1 + m)^2/n^2)]*n) + (E^
((a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*(1 + m)*x^(1 + m)*Log[x])/(2*Sqrt[-((1 + m)^2/n^2)]*n*(c*x^n)^((1 + m)/
n))

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Rubi [A]  time = 0.277359, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {4493, 4489} \[ \frac{(m+1) x^{m+1} \log (x) e^{\frac{a n \sqrt{-\frac{(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{-\frac{m+1}{n}}}{2 n \sqrt{-\frac{(m+1)^2}{n^2}}}-\frac{x^{m+1} e^{\frac{a (m+1)}{n \sqrt{-\frac{(m+1)^2}{n^2}}}} \left (c x^n\right )^{\frac{m+1}{n}}}{4 n \sqrt{-\frac{(m+1)^2}{n^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sin[a + Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n]],x]

[Out]

-(E^((a*(1 + m))/(Sqrt[-((1 + m)^2/n^2)]*n))*x^(1 + m)*(c*x^n)^((1 + m)/n))/(4*Sqrt[-((1 + m)^2/n^2)]*n) + (E^
((a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*(1 + m)*x^(1 + m)*Log[x])/(2*Sqrt[-((1 + m)^2/n^2)]*n*(c*x^n)^((1 + m)/
n))

Rule 4493

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

Rule 4489

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(m + 1)^p/(2^p*b^p*d^p*p^p)
, Int[ExpandIntegrand[(e*x)^m*(E^((a*b*d^2*p)/(m + 1))/x^((m + 1)/p) - x^((m + 1)/p)/E^((a*b*d^2*p)/(m + 1)))^
p, x], x], x] /; FreeQ[{a, b, d, e, m}, x] && IGtQ[p, 0] && EqQ[b^2*d^2*p^2 + (m + 1)^2, 0]

Rubi steps

\begin{align*} \int x^m \sin \left (a+\sqrt{-\frac{(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx &=\frac{\left (x^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int x^{-1+\frac{1+m}{n}} \sin \left (a+\sqrt{-\frac{(1+m)^2}{n^2}} \log (x)\right ) \, dx,x,c x^n\right )}{n}\\ &=\frac{\left ((1+m) x^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int \left (\frac{e^{\frac{a \sqrt{-\frac{(1+m)^2}{n^2}} n}{1+m}}}{x}-e^{\frac{a (1+m)}{\sqrt{-\frac{(1+m)^2}{n^2}} n}} x^{-1+\frac{2 (1+m)}{n}}\right ) \, dx,x,c x^n\right )}{2 \sqrt{-\frac{(1+m)^2}{n^2}} n^2}\\ &=-\frac{e^{\frac{a (1+m)}{\sqrt{-\frac{(1+m)^2}{n^2}} n}} x^{1+m} \left (c x^n\right )^{\frac{1+m}{n}}}{4 \sqrt{-\frac{(1+m)^2}{n^2}} n}+\frac{e^{\frac{a \sqrt{-\frac{(1+m)^2}{n^2}} n}{1+m}} (1+m) x^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}} \log (x)}{2 \sqrt{-\frac{(1+m)^2}{n^2}} n}\\ \end{align*}

Mathematica [F]  time = 0.252815, size = 0, normalized size = 0. \[ \int x^m \sin \left (a+\sqrt{-\frac{(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^m*Sin[a + Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n]],x]

[Out]

Integrate[x^m*Sin[a + Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n]], x]

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}\sin \left ( a+\ln \left ( c{x}^{n} \right ) \sqrt{-{\frac{ \left ( 1+m \right ) ^{2}}{{n}^{2}}}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sin(a+ln(c*x^n)*(-(1+m)^2/n^2)^(1/2)),x)

[Out]

int(x^m*sin(a+ln(c*x^n)*(-(1+m)^2/n^2)^(1/2)),x)

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Maxima [A]  time = 1.19864, size = 111, normalized size = 0.83 \begin{align*} \frac{c^{\frac{2 \, m}{n} + \frac{2}{n}} x e^{\left (m \log \left (x\right ) + \frac{m \log \left (x^{n}\right )}{n} + \frac{\log \left (x^{n}\right )}{n}\right )} \sin \left (a\right ) + 2 \,{\left (m \sin \left (a\right ) + \sin \left (a\right )\right )} \log \left (x\right )}{4 \,{\left (c^{\frac{m}{n} + \frac{1}{n}} m + c^{\frac{m}{n} + \frac{1}{n}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+log(c*x^n)*(-(1+m)^2/n^2)^(1/2)),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 0.497432, size = 159, normalized size = 1.2 \begin{align*} \frac{{\left (i \, x^{2} x^{2 \, m} +{\left (-2 i \, m - 2 i\right )} e^{\left (\frac{2 \,{\left (i \, a n -{\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} \log \left (x\right )\right )} e^{\left (-\frac{i \, a n -{\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{4 \,{\left (m + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+log(c*x^n)*(-(1+m)^2/n^2)^(1/2)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sin{\left (a + \sqrt{- \frac{m^{2}}{n^{2}} - \frac{2 m}{n^{2}} - \frac{1}{n^{2}}} \log{\left (c x^{n} \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*sin(a+ln(c*x**n)*(-(1+m)**2/n**2)**(1/2)),x)

[Out]

Integral(x**m*sin(a + sqrt(-m**2/n**2 - 2*m/n**2 - 1/n**2)*log(c*x**n)), x)

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Giac [C]  time = 1.81321, size = 367, normalized size = 2.76 \begin{align*} \frac{-i \, m n^{2} x x^{m} e^{\left (i \, a - \frac{n{\left | m n + n \right |} \log \left (x\right ) +{\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + i \, m n^{2} x x^{m} e^{\left (-i \, a + \frac{n{\left | m n + n \right |} \log \left (x\right ) +{\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - i \, n^{2} x x^{m} e^{\left (i \, a - \frac{n{\left | m n + n \right |} \log \left (x\right ) +{\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - i \, n x x^{m}{\left | m n + n \right |} e^{\left (i \, a - \frac{n{\left | m n + n \right |} \log \left (x\right ) +{\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + i \, n^{2} x x^{m} e^{\left (-i \, a + \frac{n{\left | m n + n \right |} \log \left (x\right ) +{\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - i \, n x x^{m}{\left | m n + n \right |} e^{\left (-i \, a + \frac{n{\left | m n + n \right |} \log \left (x\right ) +{\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )}}{2 \,{\left (m^{2} n^{2} + 2 \, m n^{2} -{\left (m n + n\right )}^{2} + n^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+log(c*x^n)*(-(1+m)^2/n^2)^(1/2)),x, algorithm="giac")

[Out]

1/2*(-I*m*n^2*x*x^m*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + I*m*n^2*x*x^m*e^(-I*a + (n*a
bs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - I*n^2*x*x^m*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*lo
g(c))/n^2) - I*n*x*x^m*abs(m*n + n)*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + I*n^2*x*x^m*
e^(-I*a + (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - I*n*x*x^m*abs(m*n + n)*e^(-I*a + (n*abs(m*n + n
)*log(x) + abs(m*n + n)*log(c))/n^2))/(m^2*n^2 + 2*m*n^2 - (m*n + n)^2 + n^2)