\(\int x^m \sin ^2(a+\frac {1}{4} \sqrt {-(1+m)^2} \log (c x^2)) \, dx\) [49]

Optimal result

Integrand size = 30, antiderivative size = 106 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {x^{1+m}}{2 (1+m)}-\frac {e^{\frac {2 a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1+m}{2}}}{8 (1+m)}-\frac {1}{4} e^{-\frac {2 a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)} \log (x) \]

[Out]

1/2*x^(1+m)/(1+m)-1/8*exp(2*a*(1+m)/(-(1+m)^2)^(1/2))*x^(1+m)*(c*x^2)^(1/2+1/2*m)/(1+m)-1/4*x^(1+m)*(c*x^2)^(-
1/2-1/2*m)*ln(x)/exp(2*a*(1+m)/(-(1+m)^2)^(1/2))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {4581, 4577} \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {e^{\frac {2 a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac {m+1}{2}}}{8 (m+1)}-\frac {1}{4} e^{-\frac {2 a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \log (x) \left (c x^2\right )^{\frac {1}{2} (-m-1)}+\frac {x^{m+1}}{2 (m+1)} \]

[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 4577

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 4581

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*} \text {integral}& = \frac {1}{2} \left (x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \text {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 ) \text {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]

\[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]

[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]

Maple [F]

\[\int x^{m} {\sin \left (a +\frac {\ln \left (c \,x^{2}\right ) \sqrt {-\left (1+m \right )^{2}}}{4}\right )}^{2}d x\]

[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)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {{\left (2 \, {\left (m + 1\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) - 2 \, {\left (m + 1\right )} \log \left (x\right ) + 4 i \, a\right )} \log \left (x\right ) - 4 \, e^{\left (-\frac {1}{2} \, {\left (m + 1\right )} \log \left (c\right ) - {\left (m + 1\right )} \log \left (x\right ) + 2 i \, a\right )} + 1\right )} e^{\left (\frac {1}{2} \, {\left (m + 1\right )} \log \left (c\right ) + 2 \, {\left (m + 1\right )} \log \left (x\right ) - 2 i \, a\right )}}{8 \, {\left (m + 1\right )}} \]

[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)

Sympy [F]

\[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^{m} \sin ^{2}{\left (a + \frac {\sqrt {- m^{2} - 2 m - 1} \log {\left (c x^{2} \right )}}{4} \right )}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.26 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {c^{m + 1} x^{2} x^{2 \, m} \cos \left (2 \, a\right ) - 4 \, {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {1}{2} \, m + \frac {1}{2}} x x^{m} + 2 \, {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2} + {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2}\right )} m\right )} \log \left (x\right )}{8 \, {\left ({\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {1}{2} \, m} m + {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {1}{2} \, m}\right )} \sqrt {c}} \]

[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))

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.40 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.30 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {m^{2} x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} - m x x^{m} {\left | m + 1 \right |} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + m^{2} x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + m x x^{m} {\left | m + 1 \right |} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + 2 \, {\left (m + 1\right )}^{2} x x^{m} - 2 \, m^{2} x x^{m} + 2 \, m x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} - x x^{m} {\left | m + 1 \right |} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + 2 \, m x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + x x^{m} {\left | m + 1 \right |} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} - 4 \, m x x^{m} + x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} - 2 \, x x^{m}}{4 \, {\left ({\left (m + 1\right )}^{2} m - m^{3} + {\left (m + 1\right )}^{2} - 3 \, m^{2} - 3 \, m - 1\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 27.99 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.41 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {x\,x^m}{2\,m+2}-\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,x\,x^m\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}}{4\,m+4-\sqrt {-{\left (m+1\right )}^2}\,4{}\mathrm {i}}-\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,x\,x^m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}{4\,m+4+\sqrt {-{\left (m+1\right )}^2}\,4{}\mathrm {i}} \]

[In]

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

[Out]

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