\(\int x^m \cos ^2(a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log (c x^n)) \, dx\) [106]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {4582, 4578} \[ \int x^m \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x^{m+1} e^{-\frac {2 a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{\frac {m+1}{n}}}{8 (m+1)}+\frac {1}{4} x^{m+1} \log (x) e^{\frac {2 a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{-\frac {m+1}{n}}+\frac {x^{m+1}}{2 (m+1)} \]

[In]

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

[Out]

x^(1 + m)/(2*(1 + m)) + (x^(1 + m)*(c*x^n)^((1 + m)/n))/(8*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))*x^(1 + m)*Log[x])/(4*(c*x^n)^((1 + m)/n))

Rule 4578

Int[Cos[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2^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 4582

Int[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Cos[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 {\left (x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log (x)\right ) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int \left (\frac {e^{\frac {2 a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}}}{x}+2 x^{-1+\frac {1+m}{n}}+e^{-\frac {2 a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} x^{-1+\frac {2 (1+m)}{n}}\right ) \, dx,x,c x^n\right )}{4 n} \\ & = \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) \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int x^m \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {{\left (2 \, {\left (m + 1\right )} e^{\left (-\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} \log \left (x\right ) + 4 \, e^{\left (-\frac {{\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )}{n}\right )} + 1\right )} e^{\left (\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n} + \frac {2 i \, a n - {\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{8 \, {\left (m + 1\right )}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

Integral(x**m*cos(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)

none

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.47 \[ \int x^m \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {4 \, {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}} x x^{m} + c^{\frac {2 \, m}{n} + \frac {2}{n}} x \cos \left (2 \, a\right ) e^{\left (m \log \left (x\right ) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} + 2 \, {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2} + {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2}\right )} m\right )} \log \left (x\right )}{8 \, {\left ({\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}} m + {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}}\right )}} \]

[In]

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

[Out]

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*cos(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 = 4.23 (sec) , antiderivative size = 498, normalized size of antiderivative = 4.26 \[ \int x^m \cos ^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 )}} \]

[In]

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

[Out]

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) + a
bs(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)*log(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 = 28.38 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.22 \[ \int x^m \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m}{2\,m+2}+\frac {x\,x^m\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}}{4\,m+4-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,4{}\mathrm {i}}+\frac {x\,x^m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}{4\,m+4+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,4{}\mathrm {i}} \]

[In]

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

[Out]

(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)