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

Optimal. Leaf size=117 \[ -\frac{x^{m+1} e^{-\frac{2 a n \sqrt{-\frac{(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{\frac{m+1}{n}}}{8 (m+1)}-\frac{1}{4} x^{m+1} \log (x) e^{\frac{2 a n \sqrt{-\frac{(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{-\frac{m+1}{n}}+\frac{x^{m+1}}{2 (m+1)} \]

[Out]

x^(1 + m)/(2*(1 + m)) - (x^(1 + m)*(c*x^n)^((1 + m)/n))/(8*E^((2*a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*(1 + m))
- (E^((2*a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*x^(1 + m)*Log[x])/(4*(c*x^n)^((1 + m)/n))

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Rubi [A]  time = 0.15885, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {4493, 4489} \[ -\frac{x^{m+1} e^{-\frac{2 a n \sqrt{-\frac{(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{\frac{m+1}{n}}}{8 (m+1)}-\frac{1}{4} x^{m+1} \log (x) e^{\frac{2 a n \sqrt{-\frac{(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{-\frac{m+1}{n}}+\frac{x^{m+1}}{2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^(1 + m)/(2*(1 + m)) - (x^(1 + m)*(c*x^n)^((1 + m)/n))/(8*E^((2*a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*(1 + m))
- (E^((2*a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*x^(1 + m)*Log[x])/(4*(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 ^2\left (a+\frac{1}{2} \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 ^2\left (a+\frac{1}{2} \sqrt{-\frac{(1+m)^2}{n^2}} \log (x)\right ) \, dx,x,c x^n\right )}{n}\\ &=-\frac{\left (x^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int \left (\frac{e^{\frac{2 a \sqrt{-\frac{(1+m)^2}{n^2}} n}{1+m}}}{x}-2 x^{-1+\frac{1+m}{n}}+e^{-\frac{2 a \sqrt{-\frac{(1+m)^2}{n^2}} n}{1+m}} x^{-1+\frac{2 (1+m)}{n}}\right ) \, dx,x,c x^n\right )}{4 n}\\ &=\frac{x^{1+m}}{2 (1+m)}-\frac{e^{-\frac{2 a \sqrt{-\frac{(1+m)^2}{n^2}} n}{1+m}} x^{1+m} \left (c x^n\right )^{\frac{1+m}{n}}}{8 (1+m)}-\frac{1}{4} e^{\frac{2 a \sqrt{-\frac{(1+m)^2}{n^2}} n}{1+m}} x^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}} \log (x)\\ \end{align*}

Mathematica [F]  time = 0.356133, size = 0, normalized size = 0. \[ \int x^m \sin ^2\left (a+\frac{1}{2} \sqrt{-\frac{(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.086, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( \sin \left ( a+{\frac{\ln \left ( c{x}^{n} \right ) }{2}\sqrt{-{\frac{ \left ( 1+m \right ) ^{2}}{{n}^{2}}}}} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.37276, size = 234, normalized size = 2. \begin{align*} \frac{4 \,{\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac{m}{n} + \frac{1}{n}} x x^{m} - c^{\frac{2 \, m}{n} + \frac{2}{n}} x \cos \left (2 \, a\right ) e^{\left (m \log \left (x\right ) + \frac{m \log \left (x^{n}\right )}{n} + \frac{\log \left (x^{n}\right )}{n}\right )} - 2 \,{\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2} +{\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2}\right )} m\right )} \log \left (x\right )}{8 \,{\left ({\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac{m}{n} + \frac{1}{n}} m +{\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} 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+1/2*log(c*x^n)*(-(1+m)^2/n^2)^(1/2))^2,x, algorithm="maxima")

[Out]

1/8*(4*(cos(2*a)^2 + sin(2*a)^2)*c^(m/n + 1/n)*x*x^m - c^(2*m/n + 2/n)*x*cos(2*a)*e^(m*log(x) + m*log(x^n)/n +
log(x^n)/n) - 2*(cos(2*a)^3 + cos(2*a)*sin(2*a)^2 + (cos(2*a)^3 + cos(2*a)*sin(2*a)^2)*m)*log(x))/((cos(2*a)^
2 + sin(2*a)^2)*c^(m/n + 1/n)*m + (cos(2*a)^2 + sin(2*a)^2)*c^(m/n + 1/n))

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Fricas [C]  time = 0.495555, size = 316, normalized size = 2.7 \begin{align*} -\frac{{\left (2 \,{\left (m + 1\right )} e^{\left (-\frac{2 \,{\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n +{\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} \log \left (x\right ) - 4 \, e^{\left (-\frac{{\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n +{\left (m + 1\right )} \log \left (c\right )}{n}\right )} + 1\right )} e^{\left (\frac{2 \,{\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n +{\left (m + 1\right )} \log \left (c\right )\right )}}{n} + \frac{2 i \, a n -{\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{8 \,{\left (m + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*(m + 1)*e^(-2*((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n)*log(x) - 4*e^(-((m + 1)*n*log(x) - 2*I*
a*n + (m + 1)*log(c))/n) + 1)*e^(2*((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n + (2*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 ^{2}{\left (a + \frac{\sqrt{- \frac{m^{2}}{n^{2}} - \frac{2 m}{n^{2}} - \frac{1}{n^{2}}} \log{\left (c x^{n} \right )}}{2} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 2.53023, size = 672, normalized size = 5.74 \begin{align*} -\frac{m^{2} n^{2} x x^{m} e^{\left (2 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 )} + m^{2} n^{2} x x^{m} e^{\left (-2 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 \, m^{2} n^{2} x x^{m} + 2 \, m n^{2} x x^{m} e^{\left (2 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 )} + m n x x^{m}{\left | m n + n \right |} e^{\left (2 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 \, m n^{2} x x^{m} e^{\left (-2 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 )} - m n x x^{m}{\left | m n + n \right |} e^{\left (-2 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 )} - 4 \, m n^{2} x x^{m} + n^{2} x x^{m} e^{\left (2 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 )} + n x x^{m}{\left | m n + n \right |} e^{\left (2 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 )} + n^{2} x x^{m} e^{\left (-2 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 )} - n x x^{m}{\left | m n + n \right |} e^{\left (-2 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 n + n\right )}^{2} x x^{m} - 2 \, n^{2} x x^{m}}{4 \,{\left (m^{3} n^{2} + 3 \, m^{2} n^{2} -{\left (m n + n\right )}^{2} m + 3 \, 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+1/2*log(c*x^n)*(-(1+m)^2/n^2)^(1/2))^2,x, algorithm="giac")

[Out]

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