\(\int x^m \sin ^2(a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log (c x^n)) \, dx\) [33]

Optimal result

Integrand size = 33, antiderivative size = 117 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x^{1+m}}{2 (1+m)}-\frac {e^{-\frac {2 a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} x^{1+m} \left (c x^n\right )^{\frac {1+m}{n}}}{8 (1+m)}-\frac {1}{4} e^{\frac {2 a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \log (x) \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 4996

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

\(\Big \downarrow \) 4992

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=-\frac {{\left (2 \, {\left (m + 1\right )} e^{\left (-\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} \log \left (x\right ) - 4 \, e^{\left (-\frac {{\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )}{n}\right )} + 1\right )} e^{\left (\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n} + \frac {2 i \, a n - {\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{8 \, {\left (m + 1\right )}} \] Input:

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

Output:

-1/8*(2*(m + 1)*e^(-2*((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n)*log 
(x) - 4*e^(-((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n) + 1)*e^(2*((m 
 + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n + (2*I*a*n - (m + 1)*log(c))/ 
n)/(m + 1)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.48 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {4 \, {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}} x x^{m} - c^{\frac {2 \, m}{n} + \frac {2}{n}} x \cos \left (2 \, a\right ) e^{\left (m \log \left (x\right ) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} - 2 \, {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2} + {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2}\right )} m\right )} \log \left (x\right )}{8 \, {\left ({\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}} m + {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}}\right )}} \] Input:

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

Output:

1/8*(4*(cos(2*a)^2 + sin(2*a)^2)*c^(m/n + 1/n)*x*x^m - c^(2*m/n + 2/n)*x*c 
os(2*a)*e^(m*log(x) + m*log(x^n)/n + log(x^n)/n) - 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^(m/n + 1/n)*m + (cos(2*a)^2 + sin(2*a)^2)*c^(m/n + 1/n))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.08 (sec) , antiderivative size = 498, normalized size of antiderivative = 4.26 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=-\frac {m^{2} n^{2} x x^{m} e^{\left (2 i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + m^{2} n^{2} x x^{m} e^{\left (-2 i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - 2 \, m^{2} n^{2} x x^{m} + 2 \, m n^{2} x x^{m} e^{\left (2 i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + m n x x^{m} {\left | m n + n \right |} e^{\left (2 i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + 2 \, m n^{2} x x^{m} e^{\left (-2 i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - m n x x^{m} {\left | m n + n \right |} e^{\left (-2 i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - 4 \, m n^{2} x x^{m} + n^{2} x x^{m} e^{\left (2 i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + n x x^{m} {\left | m n + n \right |} e^{\left (2 i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + n^{2} x x^{m} e^{\left (-2 i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - n x x^{m} {\left | m n + n \right |} e^{\left (-2 i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + 2 \, {\left (m n + n\right )}^{2} x x^{m} - 2 \, n^{2} x x^{m}}{4 \, {\left (m^{3} n^{2} + 3 \, m^{2} n^{2} - {\left (m n + n\right )}^{2} m + 3 \, m n^{2} - {\left (m n + n\right )}^{2} + n^{2}\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 21.49 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.24 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m}{2\,m+2}-\frac {x\,x^m\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}}{4\,m+4-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,4{}\mathrm {i}}-\frac {x\,x^m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}{4\,m+4+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,4{}\mathrm {i}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x^{m} x \left (-2 \cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right ) \sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right )+2 {\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right )}^{2}+1\right )}{4 m +4} \] Input:

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

Output:

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