\(\int (e x)^m \csc ^2(d (a+b \log (c x^n))) \, dx\) [321]

Optimal result

Integrand size = 21, antiderivative size = 119 \[ \int (e x)^m \csc ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {4 e^{2 i a d} (e x)^{1+m} \left (c x^n\right )^{2 i b d} \operatorname {Hypergeometric2F1}\left (2,-\frac {i (1+m)-2 b d n}{2 b d n},-\frac {i (1+m)-4 b d n}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (1+m+2 i b d n)} \] Output:

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

Mathematica [A] (verified)

Time = 13.02 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.89 \[ \int (e x)^m \csc ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {x (e x)^m \left ((1+m+2 i b d n) \cot \left (d \left (a+b \log \left (c x^n\right )\right )\right )+i (1+m+2 i b d n) \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m)}{2 b d n},1-\frac {i (1+m)}{2 b d n},e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+i e^{2 i a d} (1+m) \left (c x^n\right )^{2 i b d} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m+2 i b d n)}{2 b d n},-\frac {i (1+m+4 i b d n)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )\right )}{b d n (1+m+2 i b d n)} \] Input:

Integrate[(e*x)^m*Csc[d*(a + b*Log[c*x^n])]^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5021, 5017, 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.

\(\displaystyle \int (e x)^m \csc ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\)

\(\Big \downarrow \) 5021

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

\(\Big \downarrow \) 5017

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

\(\Big \downarrow \) 888

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

Input:

Int[(e*x)^m*Csc[d*(a + b*Log[c*x^n])]^2,x]
 

Output:

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

Defintions of rubi rules used

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 5017

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

rule 5021

Int[Csc[((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)*Csc[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]

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

Input:

int((e*x)^m*csc(d*(a+b*ln(c*x^n)))^2,x)
 

Output:

int((e*x)^m*csc(d*(a+b*ln(c*x^n)))^2,x)
 

Fricas [F]

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

integrate((e*x)^m*csc(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")
 

Output:

integral((e*x)^m*csc(b*d*log(c*x^n) + a*d)^2, x)
 

Sympy [F]

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

integrate((e*x)**m*csc(d*(a+b*ln(c*x**n)))**2,x)
 

Output:

Integral((e*x)**m*csc(a*d + b*d*log(c*x**n))**2, x)
 

Maxima [F]

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

integrate((e*x)^m*csc(d*(a+b*log(c*x^n)))^2,x, algorithm="maxima")
 

Output:

(2*e^m*x*x^m*cos(2*b*d*log(x^n) + 2*a*d)*sin(2*b*d*log(c)) + 2*e^m*x*x^m*c 
os(2*b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d) + (((b^2*d^2*cos(2*b*d*log(c) 
)^2 + b^2*d^2*sin(2*b*d*log(c))^2)*e^m*m + (b^2*d^2*cos(2*b*d*log(c))^2 + 
b^2*d^2*sin(2*b*d*log(c))^2)*e^m)*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)*e^m*m + (b^2*d^2* 
cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*e^m)*n^2*sin(2*b*d*log( 
x^n) + 2*a*d)^2 - 2*(b^2*d^2*e^m*m*cos(2*b*d*log(c)) + b^2*d^2*e^m*cos(2*b 
*d*log(c)))*n^2*cos(2*b*d*log(x^n) + 2*a*d) + 2*(b^2*d^2*e^m*m*sin(2*b*d*l 
og(c)) + b^2*d^2*e^m*sin(2*b*d*log(c)))*n^2*sin(2*b*d*log(x^n) + 2*a*d) + 
(b^2*d^2*e^m*m + b^2*d^2*e^m)*n^2)*integrate((x^m*cos(b*d*log(x^n) + a*d)* 
sin(b*d*log(c)) + x^m*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) 
)*sin(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*l 
og(c))^2 + b^2*d^2*sin(b*d*log(c))^2)*n^2*sin(b*d*log(x^n) + a*d)^2), x) - 
 (((b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*e^m*m + (b^ 
2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*e^m)*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)*e^m*m + (b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c) 
)^2)*e^m)*n^2*sin(2*b*d*log(x^n) + 2*a*d)^2 - 2*(b^2*d^2*e^m*m*cos(2*b*...
 

Giac [F]

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

integrate((e*x)^m*csc(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")
 

Output:

integrate((e*x)^m*csc((b*log(c*x^n) + a)*d)^2, x)
 

Mupad [F(-1)]

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

int((e*x)^m/sin(d*(a + b*log(c*x^n)))^2,x)
 

Output:

int((e*x)^m/sin(d*(a + b*log(c*x^n)))^2, x)
 

Reduce [F]

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

int((e*x)^m*csc(d*(a+b*log(c*x^n)))^2,x)
 

Output:

e**m*int(x**m*csc(log(x**n*c)*b*d + a*d)**2,x)