\(\int \frac {(e x)^{-1+3 n}}{(a+b \csc (c+d x^n))^2} \, dx\) [84]

Optimal result

Integrand size = 24, antiderivative size = 1417 \[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx =\text {Too large to display} \] Output:

1/3*(e*x)^(3*n)/a^2/e/n-4*I*b*(e*x)^(3*n)*polylog(3,I*a*exp(I*(c+d*x^n))/( 
b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^3/e/n/(x^(3*n))+2*b^2*(e*x)^(3 
*n)*ln(1+a*exp(I*(c+d*x^n))/(I*b-(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d^2/e/n/( 
x^(2*n))+2*b^2*(e*x)^(3*n)*ln(1+a*exp(I*(c+d*x^n))/(I*b+(a^2-b^2)^(1/2)))/ 
a^2/(a^2-b^2)/d^2/e/n/(x^(2*n))-I*b^3*(e*x)^(3*n)*ln(1-I*a*exp(I*(c+d*x^n) 
)/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d/e/n/(x^n)-2*I*b^2*(e*x)^(3* 
n)*polylog(2,-a*exp(I*(c+d*x^n))/(I*b-(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d^3/ 
e/n/(x^(3*n))+2*I*b*(e*x)^(3*n)*ln(1-I*a*exp(I*(c+d*x^n))/(b-(-a^2+b^2)^(1 
/2)))/a^2/(-a^2+b^2)^(1/2)/d/e/n/(x^n)+4*I*b*(e*x)^(3*n)*polylog(3,I*a*exp 
(I*(c+d*x^n))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^3/e/n/(x^(3*n)) 
-2*I*b^3*(e*x)^(3*n)*polylog(3,I*a*exp(I*(c+d*x^n))/(b-(-a^2+b^2)^(1/2)))/ 
a^2/(-a^2+b^2)^(3/2)/d^3/e/n/(x^(3*n))-2*I*b^2*(e*x)^(3*n)*polylog(2,-a*ex 
p(I*(c+d*x^n))/(I*b+(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d^3/e/n/(x^(3*n))-2*b^ 
3*(e*x)^(3*n)*polylog(2,I*a*exp(I*(c+d*x^n))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a 
^2+b^2)^(3/2)/d^2/e/n/(x^(2*n))+4*b*(e*x)^(3*n)*polylog(2,I*a*exp(I*(c+d*x 
^n))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2/e/n/(x^(2*n))+2*b^3*(e 
*x)^(3*n)*polylog(2,I*a*exp(I*(c+d*x^n))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b 
^2)^(3/2)/d^2/e/n/(x^(2*n))-4*b*(e*x)^(3*n)*polylog(2,I*a*exp(I*(c+d*x^n)) 
/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2/e/n/(x^(2*n))+2*I*b^3*(e*x 
)^(3*n)*polylog(3,I*a*exp(I*(c+d*x^n))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+...
 

Mathematica [F]

\[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx=\int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx \] Input:

Integrate[(e*x)^(-1 + 3*n)/(a + b*Csc[c + d*x^n])^2,x]
 

Output:

Integrate[(e*x)^(-1 + 3*n)/(a + b*Csc[c + d*x^n])^2, x]
 

Rubi [A] (verified)

Time = 2.39 (sec) , antiderivative size = 1152, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4697, 4693, 3042, 4679, 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.

\(\displaystyle \int \frac {(e x)^{3 n-1}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx\)

\(\Big \downarrow \) 4697

\(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int \frac {x^{3 n-1}}{\left (a+b \csc \left (d x^n+c\right )\right )^2}dx}{e}\)

\(\Big \downarrow \) 4693

\(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int \frac {x^{2 n}}{\left (a+b \csc \left (d x^n+c\right )\right )^2}dx^n}{e n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int \frac {x^{2 n}}{\left (a+b \csc \left (d x^n+c\right )\right )^2}dx^n}{e n}\)

\(\Big \downarrow \) 4679

\(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int \left (-\frac {2 b x^{2 n}}{a^2 \left (b+a \sin \left (d x^n+c\right )\right )}+\frac {x^{2 n}}{a^2}+\frac {b^2 x^{2 n}}{a^2 \left (b+a \sin \left (d x^n+c\right )\right )^2}\right )dx^n}{e n}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \left (\frac {2 b^2 \log \left (\frac {e^{i \left (d x^n+c\right )} a}{i b-\sqrt {a^2-b^2}}+1\right ) x^n}{a^2 \left (a^2-b^2\right ) d^2}+\frac {2 b^2 \log \left (\frac {e^{i \left (d x^n+c\right )} a}{i b+\sqrt {a^2-b^2}}+1\right ) x^n}{a^2 \left (a^2-b^2\right ) d^2}+\frac {4 b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right ) x^n}{a^2 \sqrt {b^2-a^2} d^2}-\frac {2 b^3 \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right ) x^n}{a^2 \left (b^2-a^2\right )^{3/2} d^2}-\frac {4 b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right ) x^n}{a^2 \sqrt {b^2-a^2} d^2}+\frac {2 b^3 \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right ) x^n}{a^2 \left (b^2-a^2\right )^{3/2} d^2}+\frac {2 i b \log \left (1-\frac {i a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right ) x^{2 n}}{a^2 \sqrt {b^2-a^2} d}-\frac {i b^3 \log \left (1-\frac {i a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right ) x^{2 n}}{a^2 \left (b^2-a^2\right )^{3/2} d}-\frac {2 i b \log \left (1-\frac {i a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right ) x^{2 n}}{a^2 \sqrt {b^2-a^2} d}+\frac {i b^3 \log \left (1-\frac {i a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right ) x^{2 n}}{a^2 \left (b^2-a^2\right )^{3/2} d}-\frac {i b^2 x^{2 n}}{a^2 \left (a^2-b^2\right ) d}-\frac {b^2 \cos \left (d x^n+c\right ) x^{2 n}}{a \left (a^2-b^2\right ) d \left (b+a \sin \left (d x^n+c\right )\right )}+\frac {x^{3 n}}{3 a^2}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{i b-\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right ) d^3}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{i b+\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right ) d^3}+\frac {4 i b \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 \sqrt {b^2-a^2} d^3}-\frac {2 i b^3 \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 \left (b^2-a^2\right )^{3/2} d^3}-\frac {4 i b \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a^2 \sqrt {b^2-a^2} d^3}+\frac {2 i b^3 \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a^2 \left (b^2-a^2\right )^{3/2} d^3}\right )}{e n}\)

Input:

Int[(e*x)^(-1 + 3*n)/(a + b*Csc[c + d*x^n])^2,x]
 

Output:

((e*x)^(3*n)*(((-I)*b^2*x^(2*n))/(a^2*(a^2 - b^2)*d) + x^(3*n)/(3*a^2) + ( 
2*b^2*x^n*Log[1 + (a*E^(I*(c + d*x^n)))/(I*b - Sqrt[a^2 - b^2])])/(a^2*(a^ 
2 - b^2)*d^2) + (2*b^2*x^n*Log[1 + (a*E^(I*(c + d*x^n)))/(I*b + Sqrt[a^2 - 
 b^2])])/(a^2*(a^2 - b^2)*d^2) - (I*b^3*x^(2*n)*Log[1 - (I*a*E^(I*(c + d*x 
^n)))/(b - Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + ((2*I)*b*x^(2* 
n)*Log[1 - (I*a*E^(I*(c + d*x^n)))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 
 + b^2]*d) + (I*b^3*x^(2*n)*Log[1 - (I*a*E^(I*(c + d*x^n)))/(b + Sqrt[-a^2 
 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) - ((2*I)*b*x^(2*n)*Log[1 - (I*a*E^(I 
*(c + d*x^n)))/(b + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) - ((2*I)* 
b^2*PolyLog[2, -((a*E^(I*(c + d*x^n)))/(I*b - Sqrt[a^2 - b^2]))])/(a^2*(a^ 
2 - b^2)*d^3) - ((2*I)*b^2*PolyLog[2, -((a*E^(I*(c + d*x^n)))/(I*b + Sqrt[ 
a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^3) - (2*b^3*x^n*PolyLog[2, (I*a*E^(I*(c 
+ d*x^n)))/(b - Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d^2) + (4*b*x^ 
n*PolyLog[2, (I*a*E^(I*(c + d*x^n)))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a 
^2 + b^2]*d^2) + (2*b^3*x^n*PolyLog[2, (I*a*E^(I*(c + d*x^n)))/(b + Sqrt[- 
a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d^2) - (4*b*x^n*PolyLog[2, (I*a*E^(I 
*(c + d*x^n)))/(b + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d^2) - ((2*I 
)*b^3*PolyLog[3, (I*a*E^(I*(c + d*x^n)))/(b - Sqrt[-a^2 + b^2])])/(a^2*(-a 
^2 + b^2)^(3/2)*d^3) + ((4*I)*b*PolyLog[3, (I*a*E^(I*(c + d*x^n)))/(b - Sq 
rt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d^3) + ((2*I)*b^3*PolyLog[3, (I...
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 4693

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csc[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

rule 4697

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_)*(x_))^(m_.), x 
_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m])   Int[x^m*( 
a + b*Csc[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]
 

Maple [F]

\[\int \frac {\left (e x \right )^{-1+3 n}}{{\left (a +b \csc \left (c +d \,x^{n}\right )\right )}^{2}}d x\]

Input:

int((e*x)^(-1+3*n)/(a+b*csc(c+d*x^n))^2,x)
 

Output:

int((e*x)^(-1+3*n)/(a+b*csc(c+d*x^n))^2,x)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3755 vs. \(2 (1291) = 2582\).

Time = 0.38 (sec) , antiderivative size = 3755, normalized size of antiderivative = 2.65 \[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((e*x)^(-1+3*n)/(a+b*csc(c+d*x^n))^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx=\int \frac {\left (e x\right )^{3 n - 1}}{\left (a + b \csc {\left (c + d x^{n} \right )}\right )^{2}}\, dx \] Input:

integrate((e*x)**(-1+3*n)/(a+b*csc(c+d*x**n))**2,x)
 

Output:

Integral((e*x)**(3*n - 1)/(a + b*csc(c + d*x**n))**2, x)
 

Maxima [F]

\[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx=\int { \frac {\left (e x\right )^{3 \, n - 1}}{{\left (b \csc \left (d x^{n} + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*x)^(-1+3*n)/(a+b*csc(c+d*x^n))^2,x, algorithm="maxima")
 

Output:

-1/3*(6*a*b^3*e^(3*n)*x^(2*n)*cos(d*x^n + c) - (a^4 - a^2*b^2)*d*e^(3*n)*x 
^(3*n)*cos(2*d*x^n + 2*c)^2 - 4*(a^2*b^2 - b^4)*d*e^(3*n)*x^(3*n)*cos(d*x^ 
n + c)^2 - (a^4 - a^2*b^2)*d*e^(3*n)*x^(3*n)*sin(2*d*x^n + 2*c)^2 - 4*(a^2 
*b^2 - b^4)*d*e^(3*n)*x^(3*n)*sin(d*x^n + c)^2 - 4*(a^3*b - a*b^3)*d*e^(3* 
n)*x^(3*n)*sin(d*x^n + c) - (a^4 - a^2*b^2)*d*e^(3*n)*x^(3*n) + 2*(3*a*b^3 
*e^(3*n)*x^(2*n)*cos(d*x^n + c) + 2*(a^3*b - a*b^3)*d*e^(3*n)*x^(3*n)*sin( 
d*x^n + c) + (a^4 - a^2*b^2)*d*e^(3*n)*x^(3*n))*cos(2*d*x^n + 2*c) - 3*((a 
^6 - a^4*b^2)*d*e*n*cos(2*d*x^n + 2*c)^2 + 4*(a^4*b^2 - a^2*b^4)*d*e*n*cos 
(d*x^n + c)^2 + 4*(a^5*b - a^3*b^3)*d*e*n*cos(d*x^n + c)*sin(2*d*x^n + 2*c 
) + (a^6 - a^4*b^2)*d*e*n*sin(2*d*x^n + 2*c)^2 + 4*(a^4*b^2 - a^2*b^4)*d*e 
*n*sin(d*x^n + c)^2 + 4*(a^5*b - a^3*b^3)*d*e*n*sin(d*x^n + c) + (a^6 - a^ 
4*b^2)*d*e*n - 2*(2*(a^5*b - a^3*b^3)*d*e*n*sin(d*x^n + c) + (a^6 - a^4*b^ 
2)*d*e*n)*cos(2*d*x^n + 2*c))*integrate(-2*(2*a^2*b^4*e^(3*n)*x^(2*n)*cos( 
2*c)*sin(2*d*x^n) + 2*a^2*b^4*e^(3*n)*x^(2*n)*cos(2*d*x^n)*sin(2*c) - 4*(a 
^3*b^3 - a*b^5)*e^(3*n)*x^(2*n)*cos(d*x^n)*cos(c) + 4*(a^3*b^3 - a*b^5)*e^ 
(3*n)*x^(2*n)*sin(d*x^n)*sin(c) - (2*a^3*b^3*e^(3*n)*x^(2*n)*cos(d*x^n + c 
) + (2*a^5*b - a^3*b^3)*d*e^(3*n)*x^(3*n)*sin(d*x^n + c))*cos(2*d*x^n + 2* 
c) + (2*(a^3*b^3 - a*b^5)*e^(3*n)*x^(2*n) + (2*a*b^5*e^(3*n)*x^(2*n)*cos(2 
*c) - (2*a^3*b^3 - a*b^5)*d*e^(3*n)*x^(3*n)*sin(2*c))*cos(2*d*x^n) + 2*((2 
*a^4*b^2 - 3*a^2*b^4 + b^6)*d*e^(3*n)*x^(3*n)*cos(c) + 2*(a^2*b^4 - b^6...
 

Giac [F]

\[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx=\int { \frac {\left (e x\right )^{3 \, n - 1}}{{\left (b \csc \left (d x^{n} + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*x)^(-1+3*n)/(a+b*csc(c+d*x^n))^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((e*x)^(3*n - 1)/(b*csc(d*x^n + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^{3\,n-1}}{{\left (a+\frac {b}{\sin \left (c+d\,x^n\right )}\right )}^2} \,d x \] Input:

int((e*x)^(3*n - 1)/(a + b/sin(c + d*x^n))^2,x)
 

Output:

int((e*x)^(3*n - 1)/(a + b/sin(c + d*x^n))^2, x)
 

Reduce [F]

\[ \int \frac {(e x)^{-1+3 n}}{\left (a+b \csc \left (c+d x^n\right )\right )^2} \, dx=\text {too large to display} \] Input:

int((e*x)^(-1+3*n)/(a+b*csc(c+d*x^n))^2,x)
 

Output:

(2*e**(3*n)*( - 72*sqrt( - a**2 + b**2)*atan((tan((x**n*d + c)/2)*b + a)/s 
qrt( - a**2 + b**2))*sin(x**n*d + c)*a**4 + 48*sqrt( - a**2 + b**2)*atan(( 
tan((x**n*d + c)/2)*b + a)/sqrt( - a**2 + b**2))*sin(x**n*d + c)*a**2*b**2 
 - 72*sqrt( - a**2 + b**2)*atan((tan((x**n*d + c)/2)*b + a)/sqrt( - a**2 + 
 b**2))*a**3*b + 48*sqrt( - a**2 + b**2)*atan((tan((x**n*d + c)/2)*b + a)/ 
sqrt( - a**2 + b**2))*a*b**3 + 6*x**(2*n)*cos(x**n*d + c)*a**4*b*d**2 - 6* 
x**(2*n)*cos(x**n*d + c)*a**2*b**3*d**2 - 36*x**n*cos(x**n*d + c)*a**5*d + 
 42*x**n*cos(x**n*d + c)*a**3*b**2*d - 6*x**n*cos(x**n*d + c)*a*b**4*d + x 
**(3*n)*sin(x**n*d + c)*a**4*b*d**3 - x**(3*n)*sin(x**n*d + c)*a**2*b**3*d 
**3 + x**(3*n)*a**3*b**2*d**3 - x**(3*n)*a*b**4*d**3 - 9*x**(2*n)*sin(x**n 
*d + c)*a**5*d**2 + 12*x**(2*n)*sin(x**n*d + c)*a**3*b**2*d**2 - 3*x**(2*n 
)*sin(x**n*d + c)*a*b**4*d**2 - 3*x**(2*n)*a**4*b*d**2 + 3*x**(2*n)*a**2*b 
**3*d**2 - 36*x**n*a**5*d + 60*x**n*a**3*b**2*d - 24*x**n*a*b**4*d + 12*in 
t(x**(3*n)/(tan((x**n*d + c)/2)**4*b**2*x + 4*tan((x**n*d + c)/2)**3*a*b*x 
 + 4*tan((x**n*d + c)/2)**2*a**2*x + 2*tan((x**n*d + c)/2)**2*b**2*x + 4*t 
an((x**n*d + c)/2)*a*b*x + b**2*x),x)*sin(x**n*d + c)*a**6*b*d**3*n - 18*i 
nt(x**(3*n)/(tan((x**n*d + c)/2)**4*b**2*x + 4*tan((x**n*d + c)/2)**3*a*b* 
x + 4*tan((x**n*d + c)/2)**2*a**2*x + 2*tan((x**n*d + c)/2)**2*b**2*x + 4* 
tan((x**n*d + c)/2)*a*b*x + b**2*x),x)*sin(x**n*d + c)*a**4*b**3*d**3*n + 
6*int(x**(3*n)/(tan((x**n*d + c)/2)**4*b**2*x + 4*tan((x**n*d + c)/2)**...