\(\int (e x)^{-1+3 n} (a+b \sec (c+d x^n))^2 \, dx\) [77]

Optimal result

Integrand size = 24, antiderivative size = 390 \[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \arctan \left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {i b^2 x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d x^n\right )}\right )}{d^3 e n}-\frac {4 a b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.67, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4696, 4692, 3042, 4678, 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.

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

\(\Big \downarrow \) 4696

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 4678

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

rule 4692

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[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 4696

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

Maple [F]

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

Input:

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

Output:

int((e*x)^(-1+3*n)*(a+b*sec(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. 1032 vs. \(2 (365) = 730\).

Time = 0.15 (sec) , antiderivative size = 1032, normalized size of antiderivative = 2.65 \[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\frac {e^{3 n} \left (x^{3 n} \cos \left (x^{n} d +c \right ) a^{2} d^{3}+3 x^{3 n} \cos \left (x^{n} d +c \right ) a b \,d^{3}-12 \cos \left (x^{n} d +c \right ) \left (\int \frac {x^{3 n} \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{4}}{\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{4} x -2 \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} x +x}d x \right ) a b \,d^{3} n +24 \cos \left (x^{n} d +c \right ) \left (\int \frac {x^{3 n} \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{4} x -2 \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} x +x}d x \right ) a b \,d^{3} n +24 \cos \left (x^{n} d +c \right ) \left (\int \frac {x^{2 n} \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )}{\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{4} x -2 \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} x +x}d x \right ) a b \,d^{2} n -24 \cos \left (x^{n} d +c \right ) \left (\int \frac {x^{2 n} \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )}{\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{4} x -2 \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} x +x}d x \right ) b^{2} d^{2} n -6 \cos \left (x^{n} d +c \right ) \mathrm {log}\left (\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )-1\right ) a b +6 \cos \left (x^{n} d +c \right ) \mathrm {log}\left (\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )-1\right ) b^{2}+6 \cos \left (x^{n} d +c \right ) \mathrm {log}\left (\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )+1\right ) a b -6 \cos \left (x^{n} d +c \right ) \mathrm {log}\left (\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )+1\right ) b^{2}-3 x^{2 n} \sin \left (x^{n} d +c \right ) a b \,d^{2}+3 x^{2 n} \sin \left (x^{n} d +c \right ) b^{2} d^{2}-6 x^{n} a b d +6 x^{n} b^{2} d \right )}{3 \cos \left (x^{n} d +c \right ) d^{3} e n} \] Input:

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

Output:

(e**(3*n)*(x**(3*n)*cos(x**n*d + c)*a**2*d**3 + 3*x**(3*n)*cos(x**n*d + c) 
*a*b*d**3 - 12*cos(x**n*d + c)*int((x**(3*n)*tan((x**n*d + c)/2)**4)/(tan( 
(x**n*d + c)/2)**4*x - 2*tan((x**n*d + c)/2)**2*x + x),x)*a*b*d**3*n + 24* 
cos(x**n*d + c)*int((x**(3*n)*tan((x**n*d + c)/2)**2)/(tan((x**n*d + c)/2) 
**4*x - 2*tan((x**n*d + c)/2)**2*x + x),x)*a*b*d**3*n + 24*cos(x**n*d + c) 
*int((x**(2*n)*tan((x**n*d + c)/2))/(tan((x**n*d + c)/2)**4*x - 2*tan((x** 
n*d + c)/2)**2*x + x),x)*a*b*d**2*n - 24*cos(x**n*d + c)*int((x**(2*n)*tan 
((x**n*d + c)/2))/(tan((x**n*d + c)/2)**4*x - 2*tan((x**n*d + c)/2)**2*x + 
 x),x)*b**2*d**2*n - 6*cos(x**n*d + c)*log(tan((x**n*d + c)/2) - 1)*a*b + 
6*cos(x**n*d + c)*log(tan((x**n*d + c)/2) - 1)*b**2 + 6*cos(x**n*d + c)*lo 
g(tan((x**n*d + c)/2) + 1)*a*b - 6*cos(x**n*d + c)*log(tan((x**n*d + c)/2) 
 + 1)*b**2 - 3*x**(2*n)*sin(x**n*d + c)*a*b*d**2 + 3*x**(2*n)*sin(x**n*d + 
 c)*b**2*d**2 - 6*x**n*a*b*d + 6*x**n*b**2*d))/(3*cos(x**n*d + c)*d**3*e*n 
)