3.49 \(\int x^m \sin ^2(a+\frac{1}{4} \sqrt{-(1+m)^2} \log (c x^2)) \, dx\)

Optimal. Leaf size=106 \[ -\frac{e^{\frac{2 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{2}}}{8 (m+1)}-\frac{1}{4} e^{-\frac{2 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \log (x) \left (c x^2\right )^{\frac{1}{2} (-m-1)}+\frac{x^{m+1}}{2 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.14467, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {4493, 4489} \[ -\frac{e^{\frac{2 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{2}}}{8 (m+1)}-\frac{1}{4} e^{-\frac{2 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \log (x) \left (c x^2\right )^{\frac{1}{2} (-m-1)}+\frac{x^{m+1}}{2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/4]^2,x]

[Out]

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

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

Mathematica [F]  time = 0.3165, size = 0, normalized size = 0. \[ \int x^m \sin ^2\left (a+\frac{1}{4} \sqrt{-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/4]^2,x]

[Out]

Integrate[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/4]^2, x]

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Maple [F]  time = 0.061, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( \sin \left ( a+{\frac{\ln \left ( c{x}^{2} \right ) }{4}\sqrt{- \left ( 1+m \right ) ^{2}}} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sin(a+1/4*ln(c*x^2)*(-(1+m)^2)^(1/2))^2,x)

[Out]

int(x^m*sin(a+1/4*ln(c*x^2)*(-(1+m)^2)^(1/2))^2,x)

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Maxima [A]  time = 1.06182, size = 181, normalized size = 1.71 \begin{align*} -\frac{c^{m + 1} x^{2} x^{2 \, m} \cos \left (2 \, a\right ) - 4 \,{\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac{1}{2} \, m + \frac{1}{2}} x x^{m} + 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{1}{2} \, m} m +{\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac{1}{2} \, m}\right )} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/4*log(c*x^2)*(-(1+m)^2)^(1/2))^2,x, algorithm="maxima")

[Out]

-1/8*(c^(m + 1)*x^2*x^(2*m)*cos(2*a) - 4*(cos(2*a)^2 + sin(2*a)^2)*c^(1/2*m + 1/2)*x*x^m + 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^(1/2*m)*m + (co
s(2*a)^2 + sin(2*a)^2)*c^(1/2*m))*sqrt(c))

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Fricas [C]  time = 0.487449, size = 252, normalized size = 2.38 \begin{align*} -\frac{{\left (2 \,{\left (m + 1\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) - 2 \,{\left (m + 1\right )} \log \left (x\right ) + 4 i \, a\right )} \log \left (x\right ) - 4 \, e^{\left (-\frac{1}{2} \,{\left (m + 1\right )} \log \left (c\right ) -{\left (m + 1\right )} \log \left (x\right ) + 2 i \, a\right )} + 1\right )} e^{\left (\frac{1}{2} \,{\left (m + 1\right )} \log \left (c\right ) + 2 \,{\left (m + 1\right )} \log \left (x\right ) - 2 i \, a\right )}}{8 \,{\left (m + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/4*log(c*x^2)*(-(1+m)^2)^(1/2))^2,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*sin(a+1/4*ln(c*x**2)*(-(1+m)**2)**(1/2))**2,x)

[Out]

Integral(x**m*sin(a + sqrt(-m**2 - 2*m - 1)*log(c*x**2)/4)**2, x)

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Giac [C]  time = 1.89452, size = 473, normalized size = 4.46 \begin{align*} \frac{m^{2} x x^{m} e^{\left (\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) +{\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} - m x x^{m}{\left | m + 1 \right |} e^{\left (\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) +{\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + m^{2} x x^{m} e^{\left (-\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) -{\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + m x x^{m}{\left | m + 1 \right |} e^{\left (-\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) -{\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + 2 \,{\left (m + 1\right )}^{2} x x^{m} - 2 \, m^{2} x x^{m} + 2 \, m x x^{m} e^{\left (\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) +{\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} - x x^{m}{\left | m + 1 \right |} e^{\left (\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) +{\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + 2 \, m x x^{m} e^{\left (-\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) -{\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + x x^{m}{\left | m + 1 \right |} e^{\left (-\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) -{\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} - 4 \, m x x^{m} + x x^{m} e^{\left (\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) +{\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + x x^{m} e^{\left (-\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) -{\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} - 2 \, x x^{m}}{4 \,{\left ({\left (m + 1\right )}^{2} m - m^{3} +{\left (m + 1\right )}^{2} - 3 \, m^{2} - 3 \, m - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/4*log(c*x^2)*(-(1+m)^2)^(1/2))^2,x, algorithm="giac")

[Out]

1/4*(m^2*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 2*I*a) - m*x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*lo
g(c) + abs(m + 1)*log(x) - 2*I*a) + m^2*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 2*I*a) + m*x*x^m
*abs(m + 1)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 2*I*a) + 2*(m + 1)^2*x*x^m - 2*m^2*x*x^m + 2*m*x*x
^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 2*I*a) - x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*log(c) + abs(m +
1)*log(x) - 2*I*a) + 2*m*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 2*I*a) + x*x^m*abs(m + 1)*e^(-
1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 2*I*a) - 4*m*x*x^m + x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*l
og(x) - 2*I*a) + x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 2*I*a) - 2*x*x^m)/((m + 1)^2*m - m^3 +
(m + 1)^2 - 3*m^2 - 3*m - 1)