\(\int \frac {\cot ^2(d (a+b \log (c x^n)))}{x^3} \, dx\) [222]

Optimal result

Integrand size = 19, antiderivative size = 155 \[ \int \frac {\cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^3} \, dx=\frac {1+\frac {2 i}{b d n}}{2 x^2}+\frac {i \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{b d n},1+\frac {i}{b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4594, 4592, 516, 470, 371} \[ \int \frac {\cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^3} \, dx=-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{b d n},1+\frac {i}{b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2}+\frac {i \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}+\frac {1+\frac {2 i}{b d n}}{2 x^2} \]

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

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] - Dist[(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 516

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] + Dist[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] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 4592

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 4594

Int[Cot[((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)*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])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c x^n\right )^{2/n} \text {Subst}\left (\int x^{-1-\frac {2}{n}} \cot ^2(d (a+b \log (x))) \, dx,x,c x^n\right )}{n x^2} \\ & = \frac {\left (c x^n\right )^{2/n} \text {Subst}\left (\int \frac {x^{-1-\frac {2}{n}} \left (-i-i e^{2 i a d} x^{2 i b d}\right )^2}{\left (1-e^{2 i a d} x^{2 i b d}\right )^2} \, dx,x,c x^n\right )}{n x^2} \\ & = \frac {i \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {\left (i e^{-2 i a d} \left (c x^n\right )^{2/n}\right ) \text {Subst}\left (\int \frac {x^{-1-\frac {2}{n}} \left (-\frac {2 e^{2 i a d} (2+i b d n)}{n}-\frac {2 e^{4 i a d} (2-i b d n) x^{2 i b d}}{n}\right )}{1-e^{2 i a d} x^{2 i b d}} \, dx,x,c x^n\right )}{2 b d n x^2} \\ & = \frac {1+\frac {2 i}{b d n}}{2 x^2}+\frac {i \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}+\frac {\left (4 i \left (c x^n\right )^{2/n}\right ) \text {Subst}\left (\int \frac {x^{-1-\frac {2}{n}}}{1-e^{2 i a d} x^{2 i b d}} \, dx,x,c x^n\right )}{b d n^2 x^2} \\ & = \frac {1+\frac {2 i}{b d n}}{2 x^2}+\frac {i \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{b d n},1+\frac {i}{b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.00 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.13 \[ \int \frac {\cot ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^3} \, dx=\frac {2 e^{2 i d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1+\frac {i}{b d n},2+\frac {i}{b d n},e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+(i+b d n) \left (b d n-2 \cot \left (d \left (a+b \log \left (c x^n\right )\right )\right )-2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{b d n},1+\frac {i}{b d n},e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{2 b d n (i+b d n) x^2} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-1/2*((b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2)*n*cos(2*b*d*log(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*sin(2*b*d*log(x^n) + 2*a*d)^2 + b*d*n - 2*(b*d*n*cos(2*b*d*log(c)) + 2*sin
(2*b*d*log(c)))*cos(2*b*d*log(x^n) + 2*a*d) - 4*(2*b^2*d^2*n^2*x^2*cos(2*b*d*log(c))*cos(2*b*d*log(x^n) + 2*a*
d) - 2*b^2*d^2*n^2*x^2*sin(2*b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d) - b^2*d^2*n^2*x^2 - (b^2*d^2*cos(2*b*d*lo
g(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*n^2*x^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*x^2*sin(2*b*d*log(x^n) + 2*a*d)^2)*integrate((cos(b*d*log(x^n) + a*d)*sin(b*d
*log(c)) + cos(b*d*log(c))*sin(b*d*log(x^n) + a*d))/(2*b^2*d^2*n^2*x^3*cos(b*d*log(c))*cos(b*d*log(x^n) + a*d)
- 2*b^2*d^2*n^2*x^3*sin(b*d*log(c))*sin(b*d*log(x^n) + a*d) + b^2*d^2*n^2*x^3 + (b^2*d^2*cos(b*d*log(c))^2 +
b^2*d^2*sin(b*d*log(c))^2)*n^2*x^3*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*lo
g(c))^2)*n^2*x^3*sin(b*d*log(x^n) + a*d)^2), x) + 4*(2*b^2*d^2*n^2*x^2*cos(2*b*d*log(c))*cos(2*b*d*log(x^n) +
2*a*d) - 2*b^2*d^2*n^2*x^2*sin(2*b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d) - b^2*d^2*n^2*x^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*x^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*x^2*sin(2*b*d*log(x^n) + 2*a*d)^2)*integrate(-(cos(b*d*log(x^n) + a*d)*si
n(b*d*log(c)) + cos(b*d*log(c))*sin(b*d*log(x^n) + a*d))/(2*b^2*d^2*n^2*x^3*cos(b*d*log(c))*cos(b*d*log(x^n) +
a*d) - 2*b^2*d^2*n^2*x^3*sin(b*d*log(c))*sin(b*d*log(x^n) + a*d) - b^2*d^2*n^2*x^3 - (b^2*d^2*cos(b*d*log(c))
^2 + b^2*d^2*sin(b*d*log(c))^2)*n^2*x^3*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*x^3*sin(b*d*log(x^n) + a*d)^2), x) + 2*(b*d*n*sin(2*b*d*log(c)) - 2*cos(2*b*d*log(c)))*sin(2
*b*d*log(x^n) + 2*a*d))/(2*b*d*n*x^2*cos(2*b*d*log(c))*cos(2*b*d*log(x^n) + 2*a*d) - 2*b*d*n*x^2*sin(2*b*d*log
(c))*sin(2*b*d*log(x^n) + 2*a*d) - (b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2)*n*x^2*cos(2*b*d*log(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^2*sin(2*b*d*log(x^n) + 2*a*d)^2 - b*d*n
*x^2)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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