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

Optimal. Leaf size=112 \[ \frac{(m+1) e^{\frac{a \sqrt{-(m+1)^2}}{m+1}} x^{m+1} \log (x) \left (c x^2\right )^{\frac{1}{2} (-m-1)}}{2 \sqrt{-(m+1)^2}}-\frac{e^{\frac{a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{2}}}{4 \sqrt{-(m+1)^2}} \]

[Out]

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

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Rubi [A]  time = 0.194204, antiderivative size = 112, 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} \[ \frac{(m+1) e^{\frac{a \sqrt{-(m+1)^2}}{m+1}} x^{m+1} \log (x) \left (c x^2\right )^{\frac{1}{2} (-m-1)}}{2 \sqrt{-(m+1)^2}}-\frac{e^{\frac{a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{2}}}{4 \sqrt{-(m+1)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [F]  time = 0.230448, size = 0, normalized size = 0. \[ \int x^m \sin \left (a+\frac{1}{2} \sqrt{-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.064, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}\sin \left ( a+{\frac{\ln \left ( c{x}^{2} \right ) }{2}\sqrt{- \left ( 1+m \right ) ^{2}}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06175, size = 65, normalized size = 0.58 \begin{align*} \frac{c^{m + 1} x^{2} x^{2 \, m} \sin \left (a\right ) + 2 \,{\left (m \sin \left (a\right ) + \sin \left (a\right )\right )} \log \left (x\right )}{4 \,{\left (c^{\frac{1}{2} \, m} m + 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+1/2*log(c*x^2)*(-(1+m)^2)^(1/2)),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 0.486322, size = 149, normalized size = 1.33 \begin{align*} \frac{{\left (i \, x^{2} x^{2 \, m} +{\left (-2 i \, m - 2 i\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) + 2 i \, a\right )} \log \left (x\right )\right )} e^{\left (\frac{1}{2} \,{\left (m + 1\right )} \log \left (c\right ) - i \, a\right )}}{4 \,{\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^2)*(-(1+m)^2)^(1/2)),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [C]  time = 1.4687, size = 255, normalized size = 2.28 \begin{align*} -\frac{i \, 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 ) - i \, a\right )} - i \, 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 ) - i \, a\right )} - i \, 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 ) + i \, a\right )} - i \, 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 ) + i \, a\right )} + i \, x x^{m} e^{\left (\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) +{\left | m + 1 \right |} \log \left (x\right ) - i \, a\right )} - i \, x x^{m} e^{\left (-\frac{1}{2} \,{\left | m + 1 \right |} \log \left (c\right ) -{\left | m + 1 \right |} \log \left (x\right ) + i \, a\right )}}{2 \,{\left ({\left (m + 1\right )}^{2} - m^{2} - 2 \, 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^2)*(-(1+m)^2)^(1/2)),x, algorithm="giac")

[Out]

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