\(\int (e x)^m \sec ^p(d (a+b \log (c x^n))) \, dx\) [286]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {Hypergeometric2F1}\left (p,-\frac {i+i m-b d n p}{2 b d n},\frac {1}{2} \left (2-\frac {i (1+m)}{b d n}+p\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (1+m+i b d n p)} \] Output:

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

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {2^p x (e x)^m \left (\frac {e^{i a d} \left (c x^n\right )^{i b d}}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {Hypergeometric2F1}\left (p,-\frac {i (1+m+i b d n p)}{2 b d n},\frac {1}{2} \left (2-\frac {i (1+m)}{b d n}+p\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+m+i b d n p} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.22, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5020, 5018, 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.

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

\(\Big \downarrow \) 5020

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

\(\Big \downarrow \) 5018

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

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 5018

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

rule 5020

Int[((e_.)*(x_))^(m_.)*Sec[((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)*Sec[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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((e*x)**m*sec(d*(a+b*ln(c*x**n)))**p,x)
 

Output:

Integral((e*x)**m*sec(a*d + b*d*log(c*x**n))**p, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\left (e\,x\right )}^m\,{\left (\frac {1}{\cos \left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}\right )}^p \,d x \] Input:

int((e*x)^m*(1/cos(d*(a + b*log(c*x^n))))^p,x)
 

Output:

int((e*x)^m*(1/cos(d*(a + b*log(c*x^n))))^p, x)
 

Reduce [F]

\[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {e^{m} \left (x^{m} {\sec \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{p} x -\left (\int x^{m} {\sec \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{p} \tan \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )d x \right ) b d n p \right )}{m +1} \] Input:

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

Output:

(e**m*(x**m*sec(log(x**n*c)*b*d + a*d)**p*x - int(x**m*sec(log(x**n*c)*b*d 
 + a*d)**p*tan(log(x**n*c)*b*d + a*d),x)*b*d*n*p))/(m + 1)