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

Optimal result

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

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4996, 4992, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 4996

\(\displaystyle \frac {1}{2} x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \int \left (c x^2\right )^{\frac {m-1}{2}} \sin ^2\left (a+\frac {1}{4} \sqrt {-(m+1)^2} \log \left (c x^2\right )\right )d\left (c x^2\right )\)

\(\Big \downarrow \) 4992

\(\displaystyle -\frac {1}{8} x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \int \left (-2 \left (c x^2\right )^{\frac {m-1}{2}}+e^{\frac {2 a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^m+\frac {e^{-\frac {2 a (m+1)}{\sqrt {-(m+1)^2}}}}{c x^2}\right )d\left (c x^2\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{8} x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \left (-\frac {e^{\frac {2 a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^{m+1}}{m+1}-e^{-\frac {2 a (m+1)}{\sqrt {-(m+1)^2}}} \log \left (c x^2\right )+\frac {4 \left (c x^2\right )^{\frac {m+1}{2}}}{m+1}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4992

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] 
:> Simp[(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 4996

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ 
.), x_Symbol] :> Simp[(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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {{\left (2 \, {\left (m + 1\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) - 2 \, {\left (m + 1\right )} \log \left (x\right ) + 4 i \, a\right )} \log \left (x\right ) - 4 \, e^{\left (-\frac {1}{2} \, {\left (m + 1\right )} \log \left (c\right ) - {\left (m + 1\right )} \log \left (x\right ) + 2 i \, a\right )} + 1\right )} e^{\left (\frac {1}{2} \, {\left (m + 1\right )} \log \left (c\right ) + 2 \, {\left (m + 1\right )} \log \left (x\right ) - 2 i \, a\right )}}{8 \, {\left (m + 1\right )}} \] Input:

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

Output:

-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]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.26 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {c^{m + 1} x^{2} x^{2 \, m} \cos \left (2 \, a\right ) - 4 \, {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {1}{2} \, m + \frac {1}{2}} x x^{m} + 2 \, {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2} + {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2}\right )} m\right )} \log \left (x\right )}{8 \, {\left ({\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {1}{2} \, m} m + {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {1}{2} \, m}\right )} \sqrt {c}} \] Input:

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

Output:

-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 + (cos( 
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 = 1.25 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.30 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {m^{2} x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} - m x x^{m} {\left | m + 1 \right |} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + m^{2} x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + m x x^{m} {\left | m + 1 \right |} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + 2 \, {\left (m + 1\right )}^{2} x x^{m} - 2 \, m^{2} x x^{m} + 2 \, m x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} - x x^{m} {\left | m + 1 \right |} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + 2 \, m x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} + x x^{m} {\left | m + 1 \right |} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} - 4 \, m x x^{m} + x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - 2 i \, a\right )} + x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + 2 i \, a\right )} - 2 \, x x^{m}}{4 \, {\left ({\left (m + 1\right )}^{2} m - m^{3} + {\left (m + 1\right )}^{2} - 3 \, m^{2} - 3 \, m - 1\right )}} \] Input:

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

Output:

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)*log(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*a 
bs(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)*log(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 = 20.50 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.41 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {x\,x^m}{2\,m+2}-\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,x\,x^m\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}}{4\,m+4-\sqrt {-{\left (m+1\right )}^2}\,4{}\mathrm {i}}-\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,x\,x^m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}{4\,m+4+\sqrt {-{\left (m+1\right )}^2}\,4{}\mathrm {i}} \] Input:

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

Output:

(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)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.75 \[ \int x^m \sin ^2\left (a+\frac {1}{4} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {x^{m} x \left (-2 \cos \left (\frac {\mathrm {log}\left (c \,x^{2}\right ) m}{4}+\frac {\mathrm {log}\left (c \,x^{2}\right )}{4}+a \right ) \sin \left (\frac {\mathrm {log}\left (c \,x^{2}\right ) m}{4}+\frac {\mathrm {log}\left (c \,x^{2}\right )}{4}+a \right )+2 {\sin \left (\frac {\mathrm {log}\left (c \,x^{2}\right ) m}{4}+\frac {\mathrm {log}\left (c \,x^{2}\right )}{4}+a \right )}^{2}+1\right )}{4 m +4} \] Input:

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

Output:

(x**m*x*( - 2*cos((log(c*x**2)*m + log(c*x**2) + 4*a)/4)*sin((log(c*x**2)* 
m + log(c*x**2) + 4*a)/4) + 2*sin((log(c*x**2)*m + log(c*x**2) + 4*a)/4)** 
2 + 1))/(4*(m + 1))