\(\int x^m \cos ^3(a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log (c x^n)) \, dx\) [108]

Optimal result

Integrand size = 33, antiderivative size = 226 \[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {8 x^{1+m} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)}-\frac {4 x^{1+m} \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)}+\frac {4 \sqrt {-\frac {(1+m)^2}{n^2}} n x^{1+m} \sin \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)^2}-\frac {6 \sqrt {-\frac {(1+m)^2}{n^2}} n x^{1+m} \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \sin \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)^2} \] Output:

8*x^(1+m)*cos(a+1/2*(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))/(5+5*m)-4*x^(1+m)*cos( 
a+1/2*(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))^3/(5+5*m)+4/5*(-(1+m)^2/n^2)^(1/2)*n 
*x^(1+m)*sin(a+1/2*(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))/(1+m)^2-6/5*(-(1+m)^2/n 
^2)^(1/2)*n*x^(1+m)*cos(a+1/2*(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))^2*sin(a+1/2* 
(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))/(1+m)^2
 

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.70 \[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x^{1+m} \left (10 (1+m) \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )-2 (1+m) \cos \left (3 a+\frac {3}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )+\sqrt {-\frac {(1+m)^2}{n^2}} n \left (5 \sin \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )-3 \sin \left (3 a+\frac {3}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )\right )\right )}{10 (1+m)^2} \] Input:

Integrate[x^m*Cos[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]^3,x]
 

Output:

(x^(1 + m)*(10*(1 + m)*Cos[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2] - 2* 
(1 + m)*Cos[3*a + (3*Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2] + Sqrt[-((1 + m 
)^2/n^2)]*n*(5*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2] - 3*Sin[3*a 
+ (3*Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2])))/(10*(1 + m)^2)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {4991, 4989}

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.

\(\displaystyle \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 4991

\(\displaystyle \frac {6}{5} \int x^m \cos \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )dx-\frac {4 x^{m+1} \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)}-\frac {6 n \sqrt {-\frac {(m+1)^2}{n^2}} x^{m+1} \sin \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right ) \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)^2}\)

\(\Big \downarrow \) 4989

\(\displaystyle -\frac {4 x^{m+1} \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)}-\frac {6 n \sqrt {-\frac {(m+1)^2}{n^2}} x^{m+1} \sin \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right ) \cos ^2\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)^2}+\frac {6}{5} \left (\frac {2 n \sqrt {-\frac {(m+1)^2}{n^2}} x^{m+1} \sin \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{3 (m+1)^2}+\frac {4 x^{m+1} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{3 (m+1)}\right )\)

Input:

Int[x^m*Cos[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]^3,x]
 

Output:

(-4*x^(1 + m)*Cos[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]^3)/(5*(1 + m) 
) - (6*Sqrt[-((1 + m)^2/n^2)]*n*x^(1 + m)*Cos[a + (Sqrt[-((1 + m)^2/n^2)]* 
Log[c*x^n])/2]^2*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2])/(5*(1 + m 
)^2) + (6*((4*x^(1 + m)*Cos[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2])/(3 
*(1 + m)) + (2*Sqrt[-((1 + m)^2/n^2)]*n*x^(1 + m)*Sin[a + (Sqrt[-((1 + m)^ 
2/n^2)]*Log[c*x^n])/2])/(3*(1 + m)^2)))/5
 

Defintions of rubi rules used

rule 4989

Int[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]*((e_.)*(x_))^(m_.), x_ 
Symbol] :> Simp[(m + 1)*(e*x)^(m + 1)*(Cos[d*(a + b*Log[c*x^n])]/(b^2*d^2*e 
*n^2 + e*(m + 1)^2)), x] + Simp[b*d*n*(e*x)^(m + 1)*(Sin[d*(a + b*Log[c*x^n 
])]/(b^2*d^2*e*n^2 + e*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x] & 
& NeQ[b^2*d^2*n^2 + (m + 1)^2, 0]
 

rule 4991

Int[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_)*((e_.)*(x_))^(m_. 
), x_Symbol] :> Simp[(m + 1)*(e*x)^(m + 1)*(Cos[d*(a + b*Log[c*x^n])]^p/(b^ 
2*d^2*e*n^2*p^2 + e*(m + 1)^2)), x] + (Simp[b*d*n*p*(e*x)^(m + 1)*Sin[d*(a 
+ b*Log[c*x^n])]*(Cos[d*(a + b*Log[c*x^n])]^(p - 1)/(b^2*d^2*e*n^2*p^2 + e* 
(m + 1)^2)), x] + Simp[b^2*d^2*n^2*p*((p - 1)/(b^2*d^2*n^2*p^2 + (m + 1)^2) 
)   Int[(e*x)^m*Cos[d*(a + b*Log[c*x^n])]^(p - 2), x], x]) /; FreeQ[{a, b, 
c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 + (m + 1)^2, 0]
 

Maple [A] (verified)

Time = 125.53 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(-\frac {x^{1+m} \left (\left (1+m \right ) \cos \left (3 a +\frac {3 \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \ln \left (c \,x^{n}\right )}{2}\right )+\frac {3 \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, n \sin \left (3 a +\frac {3 \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \ln \left (c \,x^{n}\right )}{2}\right )}{2}+\left (1+m \right ) \cos \left (\sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \ln \left (c \,x^{n}\right )+2 a \right )+\sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, n \sin \left (\sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \ln \left (c \,x^{n}\right )+2 a \right )+\left (-5 m -5\right ) \cos \left (a +\sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \ln \left (\sqrt {c \,x^{n}}\right )\right )-\frac {5 \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \sin \left (a +\sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \ln \left (\sqrt {c \,x^{n}}\right )\right ) n}{2}\right )}{5 \left (1+m \right )^{2}}\)	\(218\)

Input:

int(x^m*cos(a+1/2*(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))^3,x,method=_RETURNVERBOS 
E)
 

Output:

-1/5*x^(1+m)*((1+m)*cos(3*a+3/2*(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))+3/2*(-(1+m 
)^2/n^2)^(1/2)*n*sin(3*a+3/2*(-(1+m)^2/n^2)^(1/2)*ln(c*x^n))+(1+m)*cos((-( 
1+m)^2/n^2)^(1/2)*ln(c*x^n)+2*a)+(-(1+m)^2/n^2)^(1/2)*n*sin((-(1+m)^2/n^2) 
^(1/2)*ln(c*x^n)+2*a)+(-5*m-5)*cos(a+(-(1+m)^2/n^2)^(1/2)*ln((c*x^n)^(1/2) 
))-5/2*(-(1+m)^2/n^2)^(1/2)*sin(a+(-(1+m)^2/n^2)^(1/2)*ln((c*x^n)^(1/2)))* 
n)/(1+m)^2
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.57 \[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {{\left (5 \, e^{\left (-\frac {{\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )}{n}\right )} + 15 \, e^{\left (-\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} - 5 \, e^{\left (-\frac {3 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} + 1\right )} e^{\left (\frac {5 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{2 \, n} + \frac {2 i \, a n - {\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{20 \, {\left (m + 1\right )}} \] Input:

integrate(x^m*cos(a+1/2*(-(1+m)^2/n^2)^(1/2)*log(c*x^n))^3,x, algorithm="f 
ricas")
 

Output:

1/20*(5*e^(-((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n) + 15*e^(-2*(( 
m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n) - 5*e^(-3*((m + 1)*n*log(x) 
 - 2*I*a*n + (m + 1)*log(c))/n) + 1)*e^(5/2*((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]

\[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int x^{m} \cos ^{3}{\left (a + \frac {\sqrt {- \frac {m^{2}}{n^{2}} - \frac {2 m}{n^{2}} - \frac {1}{n^{2}}} \log {\left (c x^{n} \right )}}{2} \right )}\, dx \] Input:

integrate(x**m*cos(a+1/2*(-(1+m)**2/n**2)**(1/2)*ln(c*x**n))**3,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.86 \[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {{\left (c^{\frac {3 \, m}{n} + \frac {3}{n}} x \cos \left (3 \, a\right ) e^{\left (m \log \left (x\right ) + \frac {3 \, m \log \left (x^{n}\right )}{n} + \frac {3 \, \log \left (x^{n}\right )}{n}\right )} + 5 \, c^{\frac {2 \, m}{n} + \frac {2}{n}} x \cos \left (a\right ) e^{\left (m \log \left (x\right ) + \frac {2 \, m \log \left (x^{n}\right )}{n} + \frac {2 \, \log \left (x^{n}\right )}{n}\right )} + 15 \, c^{\frac {m}{n} + \frac {1}{n}} x \cos \left (a\right ) e^{\left (m \log \left (x\right ) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} - 5 \, x x^{m} \cos \left (3 \, a\right )\right )} e^{\left (-\frac {3 \, m \log \left (x^{n}\right )}{2 \, n} - \frac {3 \, \log \left (x^{n}\right )}{2 \, n}\right )}}{20 \, {\left (c^{\frac {3 \, m}{2 \, n} + \frac {3}{2 \, n}} m + c^{\frac {3 \, m}{2 \, n} + \frac {3}{2 \, n}}\right )}} \] Input:

integrate(x^m*cos(a+1/2*(-(1+m)^2/n^2)^(1/2)*log(c*x^n))^3,x, algorithm="m 
axima")
 

Output:

1/20*(c^(3*m/n + 3/n)*x*cos(3*a)*e^(m*log(x) + 3*m*log(x^n)/n + 3*log(x^n) 
/n) + 5*c^(2*m/n + 2/n)*x*cos(a)*e^(m*log(x) + 2*m*log(x^n)/n + 2*log(x^n) 
/n) + 15*c^(m/n + 1/n)*x*cos(a)*e^(m*log(x) + m*log(x^n)/n + log(x^n)/n) - 
 5*x*x^m*cos(3*a))*e^(-3/2*m*log(x^n)/n - 3/2*log(x^n)/n)/(c^(3/2*m/n + 3/ 
2/n)*m + c^(3/2*m/n + 3/2/n))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.58 (sec) , antiderivative size = 1870, normalized size of antiderivative = 8.27 \[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \] Input:

integrate(x^m*cos(a+1/2*(-(1+m)^2/n^2)^(1/2)*log(c*x^n))^3,x, algorithm="g 
iac")
 

Output:

1/4*(8*m^3*n^4*x*x^m*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)* 
log(c))/n^2) + 24*m^3*n^4*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs( 
m*n + n)*log(c))/n^2) + 24*m^3*n^4*x*x^m*e^(-I*a + 1/2*(n*abs(m*n + n)*log 
(x) + abs(m*n + n)*log(c))/n^2) + 8*m^3*n^4*x*x^m*e^(-3*I*a + 3/2*(n*abs(m 
*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*m^2*n^4*x*x^m*e^(3*I*a - 3 
/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 12*m^2*n^3*x*x^m*a 
bs(m*n + n)*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n 
^2) + 72*m^2*n^4*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)* 
log(c))/n^2) + 12*m^2*n^3*x*x^m*abs(m*n + n)*e^(I*a - 1/2*(n*abs(m*n + n)* 
log(x) + abs(m*n + n)*log(c))/n^2) + 72*m^2*n^4*x*x^m*e^(-I*a + 1/2*(n*abs 
(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 12*m^2*n^3*x*x^m*abs(m*n + 
n)*e^(-I*a + 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*m 
^2*n^4*x*x^m*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c)) 
/n^2) - 12*m^2*n^3*x*x^m*abs(m*n + n)*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log( 
x) + abs(m*n + n)*log(c))/n^2) - 2*(m*n + n)^2*m*n^2*x*x^m*e^(3*I*a - 3/2* 
(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*m*n^4*x*x^m*e^(3*I 
*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*m*n^3*x*x 
^m*abs(m*n + n)*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c 
))/n^2) - 54*(m*n + n)^2*m*n^2*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + 
 abs(m*n + n)*log(c))/n^2) + 72*m*n^4*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n...
 

Mupad [B] (verification not implemented)

Time = 22.37 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.23 \[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}{2}}}\,\left (2\,m+2+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,1{}\mathrm {i}\right )}{4\,{\left (m+1\right )}^2}+\frac {x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}{2}}\,\left (2\,m+2-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,1{}\mathrm {i}\right )}{4\,{\left (m+1\right )}^2}-\frac {x\,x^m\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,3{}\mathrm {i}}{2}}}\,\left (2\,m+2+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,3{}\mathrm {i}\right )}{20\,{\left (m+1\right )}^2}-\frac {x\,x^m\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,3{}\mathrm {i}}{2}}\,\left (2\,m+2-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,3{}\mathrm {i}\right )}{20\,{\left (m+1\right )}^2} \] Input:

int(x^m*cos(a + (log(c*x^n)*(-(m + 1)^2/n^2)^(1/2))/2)^3,x)
 

Output:

(x*x^m*exp(-a*1i)/(c*x^n)^(((- (2*m)/n^2 - 1/n^2 - m^2/n^2)^(1/2)*1i)/2)*( 
2*m + n*(-(m + 1)^2/n^2)^(1/2)*1i + 2))/(4*(m + 1)^2) + (x*x^m*exp(a*1i)*( 
c*x^n)^(((- (2*m)/n^2 - 1/n^2 - m^2/n^2)^(1/2)*1i)/2)*(2*m - n*(-(m + 1)^2 
/n^2)^(1/2)*1i + 2))/(4*(m + 1)^2) - (x*x^m*exp(-a*3i)/(c*x^n)^(((- (2*m)/ 
n^2 - 1/n^2 - m^2/n^2)^(1/2)*3i)/2)*(2*m + n*(-(m + 1)^2/n^2)^(1/2)*3i + 2 
))/(20*(m + 1)^2) - (x*x^m*exp(a*3i)*(c*x^n)^(((- (2*m)/n^2 - 1/n^2 - m^2/ 
n^2)^(1/2)*3i)/2)*(2*m - n*(-(m + 1)^2/n^2)^(1/2)*3i + 2))/(20*(m + 1)^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.67 \[ \int x^m \cos ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {2 x^{m} x \left (-10 \cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right ) {\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right )}^{2}+22 \cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right )-15 {\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right )}^{3}+21 \sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+2 a n}{2 n}\right )\right )}{65 m +65} \] Input:

int(x^m*cos(a+1/2*(-(1+m)^2/n^2)^(1/2)*log(c*x^n))^3,x)
 

Output:

(2*x**m*x*( - 10*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 + 22*cos((log(x**n*c)*m + log( 
x**n*c) + 2*a*n)/(2*n)) - 15*sin((log(x**n*c)*m + log(x**n*c) + 2*a*n)/(2* 
n))**3 + 21*sin((log(x**n*c)*m + log(x**n*c) + 2*a*n)/(2*n))))/(65*(m + 1) 
)