\(\int (e x)^{-1+3 n} (a+b \sec (c+d x^n)) \, dx\) [74]

Optimal result

Integrand size = 22, antiderivative size = 235 \[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right ) \, dx=\frac {a (e x)^{3 n}}{3 e n}-\frac {2 i b x^{-n} (e x)^{3 n} \arctan \left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 i b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {2 i b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {2 b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {2 b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,i e^{i \left (c+d x^n\right )}\right )}{d^3 e n} \] Output:

1/3*a*(e*x)^(3*n)/e/n-2*I*b*(e*x)^(3*n)*arctan(exp(I*(c+d*x^n)))/d/e/n/(x^ 
n)+2*I*b*(e*x)^(3*n)*polylog(2,-I*exp(I*(c+d*x^n)))/d^2/e/n/(x^(2*n))-2*I* 
b*(e*x)^(3*n)*polylog(2,I*exp(I*(c+d*x^n)))/d^2/e/n/(x^(2*n))-2*b*(e*x)^(3 
*n)*polylog(3,-I*exp(I*(c+d*x^n)))/d^3/e/n/(x^(3*n))+2*b*(e*x)^(3*n)*polyl 
og(3,I*exp(I*(c+d*x^n)))/d^3/e/n/(x^(3*n))
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2010, 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.

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a (e x)^{3 n}}{3 e n}-\frac {2 i b x^{-n} (e x)^{3 n} \arctan \left (e^{i \left (c+d x^n\right )}\right )}{d e n}-\frac {2 b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,-i e^{i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {2 b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,i e^{i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {2 i b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,-i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}-\frac {2 i b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}\)

Input:

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

Output:

(a*(e*x)^(3*n))/(3*e*n) - ((2*I)*b*(e*x)^(3*n)*ArcTan[E^(I*(c + d*x^n))])/ 
(d*e*n*x^n) + ((2*I)*b*(e*x)^(3*n)*PolyLog[2, (-I)*E^(I*(c + d*x^n))])/(d^ 
2*e*n*x^(2*n)) - ((2*I)*b*(e*x)^(3*n)*PolyLog[2, I*E^(I*(c + d*x^n))])/(d^ 
2*e*n*x^(2*n)) - (2*b*(e*x)^(3*n)*PolyLog[3, (-I)*E^(I*(c + d*x^n))])/(d^3 
*e*n*x^(3*n)) + (2*b*(e*x)^(3*n)*PolyLog[3, I*E^(I*(c + d*x^n))])/(d^3*e*n 
*x^(3*n))
 

Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [F]

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

Input:

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

Output:

int((e*x)^(-1+3*n)*(a+b*sec(c+d*x^n)),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. 655 vs. \(2 (215) = 430\).

Time = 0.13 (sec) , antiderivative size = 655, normalized size of antiderivative = 2.79 \[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*(2*a*d^3*e^(3*n - 1)*x^(3*n) - 6*I*b*d*e^(3*n - 1)*x^n*dilog(I*cos(d*x 
^n + c) + sin(d*x^n + c)) - 6*I*b*d*e^(3*n - 1)*x^n*dilog(I*cos(d*x^n + c) 
 - sin(d*x^n + c)) + 6*I*b*d*e^(3*n - 1)*x^n*dilog(-I*cos(d*x^n + c) + sin 
(d*x^n + c)) + 6*I*b*d*e^(3*n - 1)*x^n*dilog(-I*cos(d*x^n + c) - sin(d*x^n 
 + c)) + 3*b*c^2*e^(3*n - 1)*log(cos(d*x^n + c) + I*sin(d*x^n + c) + I) - 
3*b*c^2*e^(3*n - 1)*log(cos(d*x^n + c) - I*sin(d*x^n + c) + I) + 3*b*c^2*e 
^(3*n - 1)*log(-cos(d*x^n + c) + I*sin(d*x^n + c) + I) - 3*b*c^2*e^(3*n - 
1)*log(-cos(d*x^n + c) - I*sin(d*x^n + c) + I) - 6*b*e^(3*n - 1)*polylog(3 
, I*cos(d*x^n + c) + sin(d*x^n + c)) + 6*b*e^(3*n - 1)*polylog(3, I*cos(d* 
x^n + c) - sin(d*x^n + c)) - 6*b*e^(3*n - 1)*polylog(3, -I*cos(d*x^n + c) 
+ sin(d*x^n + c)) + 6*b*e^(3*n - 1)*polylog(3, -I*cos(d*x^n + c) - sin(d*x 
^n + c)) + 3*(b*d^2*e^(3*n - 1)*x^(2*n) - b*c^2*e^(3*n - 1))*log(I*cos(d*x 
^n + c) + sin(d*x^n + c) + 1) - 3*(b*d^2*e^(3*n - 1)*x^(2*n) - b*c^2*e^(3* 
n - 1))*log(I*cos(d*x^n + c) - sin(d*x^n + c) + 1) + 3*(b*d^2*e^(3*n - 1)* 
x^(2*n) - b*c^2*e^(3*n - 1))*log(-I*cos(d*x^n + c) + sin(d*x^n + c) + 1) - 
 3*(b*d^2*e^(3*n - 1)*x^(2*n) - b*c^2*e^(3*n - 1))*log(-I*cos(d*x^n + c) - 
 sin(d*x^n + c) + 1))/(d^3*n)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right ) \, dx=\frac {e^{3 n} \left (x^{3 n} a +x^{3 n} b -6 \left (\int \frac {x^{3 n} \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} x -x}d x \right ) b n \right )}{3 e n} \] Input:

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

Output:

(e**(3*n)*(x**(3*n)*a + x**(3*n)*b - 6*int((x**(3*n)*tan((x**n*d + c)/2)** 
2)/(tan((x**n*d + c)/2)**2*x - x),x)*b*n))/(3*e*n)