\(\int \frac {(e x)^m}{\sin ^{\frac {5}{2}}(d (a+b \log (c x^n)))} \, dx\) [78]

Optimal result

Integrand size = 23, antiderivative size = 150 \[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=\frac {2 (e x)^{1+m} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-\frac {2 i+2 i m-5 b d n}{4 b d n},-\frac {2 i+2 i m-9 b d n}{4 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (2+2 m+5 i b d n) \sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \] Output:

2*(e*x)^(1+m)*(1-exp(2*I*a*d)*(c*x^n)^(2*I*b*d))^(5/2)*hypergeom([5/2, -1/ 
4*(2*I+2*I*m-5*b*d*n)/b/d/n],[-1/4*(2*I+2*I*m-9*b*d*n)/b/d/n],exp(2*I*a*d) 
*(c*x^n)^(2*I*b*d))/e/(2+2*m+5*I*b*d*n)/sin(d*(a+b*ln(c*x^n)))^(5/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 2.10 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.43 \[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=\frac {2 x (e x)^m \left (-2-2 m-b d n \cot \left (d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )-\left (-1+e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right ) (2+2 m-i b d n) \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i+2 i m-3 b d n}{4 b d n},-\frac {2 i+2 i m-5 b d n}{4 b d n},e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+b d n \csc \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \csc \left (d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right ) \sin (b d n \log (x))\right )}{3 b^2 d^2 n^2 \sqrt {\sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )}} \] Input:

Integrate[(e*x)^m/Sin[d*(a + b*Log[c*x^n])]^(5/2),x]
 

Output:

(2*x*(e*x)^m*(-2 - 2*m - b*d*n*Cot[d*(a - b*n*Log[x] + b*Log[c*x^n])] - (- 
1 + E^((2*I)*d*(a + b*Log[c*x^n])))*(2 + 2*m - I*b*d*n)*Hypergeometric2F1[ 
1, -1/4*(2*I + (2*I)*m - 3*b*d*n)/(b*d*n), -1/4*(2*I + (2*I)*m - 5*b*d*n)/ 
(b*d*n), E^((2*I)*d*(a + b*Log[c*x^n]))] + b*d*n*Csc[d*(a + b*Log[c*x^n])] 
*Csc[d*(a - b*n*Log[x] + b*Log[c*x^n])]*Sin[b*d*n*Log[x]]))/(3*b^2*d^2*n^2 
*Sqrt[Sin[d*(a + b*Log[c*x^n])]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4996, 4994, 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.

\(\displaystyle \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx\)

\(\Big \downarrow \) 4996

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1}}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 4994

\(\displaystyle \frac {(e x)^{m+1} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{5/2} \left (c x^n\right )^{-\frac {m+1}{n}-\frac {5}{2} i b d} \int \frac {\left (c x^n\right )^{\frac {5 i b d}{2}+\frac {m+1}{n}-1}}{\left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{5/2}}d\left (c x^n\right )}{e n \sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {2 (e x)^{m+1} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i (m+1)}{b d n}\right ),-\frac {2 i m-9 b d n+2 i}{4 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (5 i b d n+2 m+2) \sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}\)

Input:

Int[(e*x)^m/Sin[d*(a + b*Log[c*x^n])]^(5/2),x]
 

Output:

(2*(e*x)^(1 + m)*(1 - E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))^(5/2)*Hypergeomet 
ric2F1[5/2, (5 - ((2*I)*(1 + m))/(b*d*n))/4, -1/4*(2*I + (2*I)*m - 9*b*d*n 
)/(b*d*n), E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d)])/(e*(2 + 2*m + (5*I)*b*d*n)* 
Sin[d*(a + b*Log[c*x^n])]^(5/2))
 

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 4994

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

rule 4996

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

\[\int \frac {\left (e x \right )^{m}}{{\sin \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{\frac {5}{2}}}d x\]

Input:

int((e*x)^m/sin(d*(a+b*ln(c*x^n)))^(5/2),x)
 

Output:

int((e*x)^m/sin(d*(a+b*ln(c*x^n)))^(5/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*x)^m/sin(d*(a+b*log(c*x^n)))^(5/2),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=\text {Timed out} \] Input:

integrate((e*x)**m/sin(d*(a+b*ln(c*x**n)))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=\int { \frac {\left (e x\right )^{m}}{\sin \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^m/sin(d*(a+b*log(c*x^n)))^(5/2),x, algorithm="maxima")
 

Output:

integrate((e*x)^m/sin((b*log(c*x^n) + a)*d)^(5/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=\text {Timed out} \] Input:

integrate((e*x)^m/sin(d*(a+b*log(c*x^n)))^(5/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=\int \frac {{\left (e\,x\right )}^m}{{\sin \left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^{5/2}} \,d x \] Input:

int((e*x)^m/sin(d*(a + b*log(c*x^n)))^(5/2),x)
 

Output:

int((e*x)^m/sin(d*(a + b*log(c*x^n)))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^m}{\sin ^{\frac {5}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )} \, dx=e^{m} \left (\int \frac {x^{m} \sqrt {\sin \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}}{{\sin \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{3}}d x \right ) \] Input:

int((e*x)^m/sin(d*(a+b*log(c*x^n)))^(5/2),x)
 

Output:

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