\(\int x^3 \cot ^2(d (a+b \log (c x^n))) \, dx\) [216]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F(-1)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 158 \[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(4 i-b d n) x^4}{4 b d n}+\frac {i x^4 \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i x^4 \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n} \] Output:

1/4*(4*I-b*d*n)*x^4/b/d/n+I*x^4*(1+exp(2*I*a*d)*(c*x^n)^(2*I*b*d))/b/d/n/( 
1-exp(2*I*a*d)*(c*x^n)^(2*I*b*d))-2*I*x^4*hypergeom([1, -2*I/b/d/n],[1-2*I 
/b/d/n],exp(2*I*a*d)*(c*x^n)^(2*I*b*d))/b/d/n
 

Mathematica [A] (verified)

Time = 3.57 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.11 \[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {x^4 \left (8 e^{2 i d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1-\frac {2 i}{b d n},2-\frac {2 i}{b d n},e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+(-2 i+b d n) \left (b d n+4 \cot \left (d \left (a+b \log \left (c x^n\right )\right )\right )+4 i \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )\right )\right )}{4 b d n (-2 i+b d n)} \] Input:

Integrate[x^3*Cot[d*(a + b*Log[c*x^n])]^2,x]
 

Output:

-1/4*(x^4*(8*E^((2*I)*d*(a + b*Log[c*x^n]))*Hypergeometric2F1[1, 1 - (2*I) 
/(b*d*n), 2 - (2*I)/(b*d*n), E^((2*I)*d*(a + b*Log[c*x^n]))] + (-2*I + b*d 
*n)*(b*d*n + 4*Cot[d*(a + b*Log[c*x^n])] + (4*I)*Hypergeometric2F1[1, (-2* 
I)/(b*d*n), 1 - (2*I)/(b*d*n), E^((2*I)*d*(a + b*Log[c*x^n]))])))/(b*d*n*( 
-2*I + b*d*n))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.34, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5009, 5007, 1004, 27, 959, 888}

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.

\(\displaystyle \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\)

\(\Big \downarrow \) 5009

\(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \int \left (c x^n\right )^{\frac {4}{n}-1} \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{n}\)

\(\Big \downarrow \) 5007

\(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1} \left (-i e^{2 i a d} \left (c x^n\right )^{2 i b d}-i\right )^2}{\left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^2}d\left (c x^n\right )}{n}\)

\(\Big \downarrow \) 1004

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 959

\(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {i \left (c x^n\right )^{4/n} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \left (\frac {8 e^{2 i a d} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1}}{1-e^{2 i a d} \left (c x^n\right )^{2 i b d}}d\left (c x^n\right )}{n}-\frac {1}{4} e^{2 i a d} (4+i b d n) \left (c x^n\right )^{4/n}\right )}{b d}\right )}{n}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {i \left (c x^n\right )^{4/n} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \left (2 e^{2 i a d} \left (c x^n\right )^{4/n} \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )-\frac {1}{4} e^{2 i a d} (4+i b d n) \left (c x^n\right )^{4/n}\right )}{b d}\right )}{n}\)

Input:

Int[x^3*Cot[d*(a + b*Log[c*x^n])]^2,x]
 

Output:

(x^4*((I*(c*x^n)^(4/n)*(1 + E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d)))/(b*d*(1 - 
E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))) - (I*(-1/4*(E^((2*I)*a*d)*(4 + I*b*d*n 
)*(c*x^n)^(4/n)) + 2*E^((2*I)*a*d)*(c*x^n)^(4/n)*Hypergeometric2F1[1, (-2* 
I)/(b*d*n), 1 - (2*I)/(b*d*n), E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d)]))/(b*d*E 
^((2*I)*a*d))))/(n*(c*x^n)^(4/n))
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 888
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 959
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p 
+ 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p 
 + 1) + 1))   Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, 
 n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

rule 1004
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) 
*((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1))   Int 
[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c 
*b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ 
n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && Lt 
Q[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
 

rule 5007
Int[Cot[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] 
:> Int[(e*x)^m*((-I - I*E^(2*I*a*d)*x^(2*I*b*d))/(1 - E^(2*I*a*d)*x^(2*I*b* 
d)))^p, x] /; FreeQ[{a, b, d, e, m, p}, x]
 

rule 5009
Int[Cot[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n))   Subst[Int[x 
^((m + 1)/n - 1)*Cot[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]

\[\int x^{3} {\cot \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]

Input:

int(x^3*cot(d*(a+b*ln(c*x^n)))^2,x)
 

Output:

int(x^3*cot(d*(a+b*ln(c*x^n)))^2,x)
 

Fricas [F]

\[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \cot \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \] Input:

integrate(x^3*cot(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")
 

Output:

integral(x^3*cot(b*d*log(c*x^n) + a*d)^2, x)
 

Sympy [F]

\[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x^{3} \cot ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \] Input:

integrate(x**3*cot(d*(a+b*ln(c*x**n)))**2,x)
 

Output:

Integral(x**3*cot(a*d + b*d*log(c*x**n))**2, x)
 

Maxima [F]

\[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \cot \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \] Input:

integrate(x^3*cot(d*(a+b*log(c*x^n)))^2,x, algorithm="maxima")
 

Output:

1/4*((b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2)*n*x^4*cos(2*b*d*l 
og(x^n) + 2*a*d)^2 + (b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2)*n 
*x^4*sin(2*b*d*log(x^n) + 2*a*d)^2 + b*d*n*x^4 - 2*(b*d*n*cos(2*b*d*log(c) 
) - 4*sin(2*b*d*log(c)))*x^4*cos(2*b*d*log(x^n) + 2*a*d) + 2*(b*d*n*sin(2* 
b*d*log(c)) + 4*cos(2*b*d*log(c)))*x^4*sin(2*b*d*log(x^n) + 2*a*d) - 16*(2 
*b^2*d^2*n^2*cos(2*b*d*log(c))*cos(2*b*d*log(x^n) + 2*a*d) - 2*b^2*d^2*n^2 
*sin(2*b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d) - b^2*d^2*n^2 - (b^2*d^2*co 
s(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*n^2*cos(2*b*d*log(x^n) + 
2*a*d)^2 - (b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*n^2 
*sin(2*b*d*log(x^n) + 2*a*d)^2)*integrate((x^3*cos(b*d*log(x^n) + a*d)*sin 
(b*d*log(c)) + x^3*cos(b*d*log(c))*sin(b*d*log(x^n) + a*d))/(2*b^2*d^2*n^2 
*cos(b*d*log(c))*cos(b*d*log(x^n) + a*d) - 2*b^2*d^2*n^2*sin(b*d*log(c))*s 
in(b*d*log(x^n) + a*d) + b^2*d^2*n^2 + (b^2*d^2*cos(b*d*log(c))^2 + b^2*d^ 
2*sin(b*d*log(c))^2)*n^2*cos(b*d*log(x^n) + a*d)^2 + (b^2*d^2*cos(b*d*log( 
c))^2 + b^2*d^2*sin(b*d*log(c))^2)*n^2*sin(b*d*log(x^n) + a*d)^2), x) + 16 
*(2*b^2*d^2*n^2*cos(2*b*d*log(c))*cos(2*b*d*log(x^n) + 2*a*d) - 2*b^2*d^2* 
n^2*sin(2*b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d) - b^2*d^2*n^2 - (b^2*d^2 
*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*n^2*cos(2*b*d*log(x^n) 
 + 2*a*d)^2 - (b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)* 
n^2*sin(2*b*d*log(x^n) + 2*a*d)^2)*integrate(-(x^3*cos(b*d*log(x^n) + a...
 

Giac [F(-1)]

Timed out. \[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \] Input:

integrate(x^3*cot(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x^3\,{\mathrm {cot}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \] Input:

int(x^3*cot(d*(a + b*log(c*x^n)))^2,x)
 

Output:

int(x^3*cot(d*(a + b*log(c*x^n)))^2, x)
 

Reduce [F]

\[ \int x^3 \cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\cot \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{2} x^{3}d x \] Input:

int(x^3*cot(d*(a+b*log(c*x^n)))^2,x)
                                                                                    
                                                                                    
 

Output:

int(cot(log(x**n*c)*b*d + a*d)**2*x**3,x)