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]

-1/4*exp(a*(1+m)/(-(1+m)^2)^(1/2))*x^(1+m)*(c*x^2)^(1/2+1/2*m)/(-(1+m)^2)^(1/2)+1/2*exp(a*(-(1+m)^2)^(1/2)/(1+
m))*(1+m)*x^(1+m)*(c*x^2)^(-1/2-1/2*m)*ln(x)/(-(1+m)^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]

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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fricas [C]  time = 0.44, size = 50, normalized size = 0.45 \[ \frac {{\left (i \, x^{2} x^{2 \, m} + {\left (-2 i \, m - 2 i\right )} e^{\left (-{\left (m + 1\right )} \log \relax (c) + 2 i \, a\right )} \log \relax (x)\right )} e^{\left (\frac {1}{2} \, {\left (m + 1\right )} \log \relax (c) - i \, a\right )}}{4 \, {\left (m + 1\right )}} \]

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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giac [C]  time = 0.79, size = 189, normalized size = 1.69 \[ -\frac {i \, m x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \relax (c) + {\left | m + 1 \right |} \log \relax (x) - i \, a\right )} - i \, x x^{m} {\left | m + 1 \right |} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \relax (c) + {\left | m + 1 \right |} \log \relax (x) - i \, a\right )} - i \, m x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \relax (c) - {\left | m + 1 \right |} \log \relax (x) + i \, a\right )} - i \, x x^{m} {\left | m + 1 \right |} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \relax (c) - {\left | m + 1 \right |} \log \relax (x) + i \, a\right )} + i \, x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \relax (c) + {\left | m + 1 \right |} \log \relax (x) - i \, a\right )} - i \, x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \relax (c) - {\left | m + 1 \right |} \log \relax (x) + i \, a\right )}}{2 \, {\left ({\left (m + 1\right )}^{2} - m^{2} - 2 \, m - 1\right )}} \]

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)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int x^{m} \sin \left (a +\frac {\ln \left (c \,x^{2}\right ) \sqrt {-\left (1+m \right )^{2}}}{2}\right )\, dx \]

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 = 0.35, size = 48, normalized size = 0.43 \[ \frac {c^{m + 1} x^{2} x^{2 \, m} \sin \relax (a) + 2 \, {\left (m \sin \relax (a) + \sin \relax (a)\right )} \log \relax (x)}{4 \, {\left (c^{\frac {1}{2} \, m} m + c^{\frac {1}{2} \, m}\right )} \sqrt {c}} \]

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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mupad [B]  time = 3.13, size = 139, normalized size = 1.24 \[ \frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,1{}\mathrm {i}}{2\,m+2-\sqrt {-{\left (m+1\right )}^2}\,2{}\mathrm {i}}-\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{2\,m+2+\sqrt {-{\left (m+1\right )}^2}\,2{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/c^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(-a*1i)/(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*1i)/(2*m - (-(m
+ 1)^2)^(1/2)*2i + 2) - (c^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(a*1i)*(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i
)/2)*1i)/(2*m + (-(m + 1)^2)^(1/2)*2i + 2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \sin {\left (a + \frac {\sqrt {- m^{2} - 2 m - 1} \log {\left (c x^{2} \right )}}{2} \right )}\, dx \]

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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