\(\int \frac {(e x)^{-1+n}}{(a+b \sec (c+d x^n))^2} \, dx\) [81]

Optimal result

Integrand size = 22, antiderivative size = 157 \[ \int \frac {(e x)^{-1+n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\frac {(e x)^n}{a^2 e n}-\frac {2 b \left (2 a^2-b^2\right ) x^{-n} (e x)^n \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d e n}+\frac {b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \sec \left (c+d x^n\right )\right )} \] Output:

(e*x)^n/a^2/e/n-2*b*(2*a^2-b^2)*(e*x)^n*arctanh((a-b)^(1/2)*tan(1/2*c+1/2* 
d*x^n)/(a+b)^(1/2))/a^2/(a-b)^(3/2)/(a+b)^(3/2)/d/e/n/(x^n)+b^2*(e*x)^n*ta 
n(c+d*x^n)/a/(a^2-b^2)/d/e/n/(x^n)/(a+b*sec(c+d*x^n))
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.22 \[ \int \frac {(e x)^{-1+n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\frac {x^{-n} (e x)^n \left (-2 b \left (-2 a^2+b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right ) \left (b+a \cos \left (c+d x^n\right )\right )+\sqrt {a^2-b^2} \left (a \left (a^2-b^2\right ) \left (c+d x^n\right ) \cos \left (c+d x^n\right )+b \left (\left (a^2-b^2\right ) \left (c+d x^n\right )+a b \sin \left (c+d x^n\right )\right )\right )\right )}{a^2 (a-b) (a+b) \sqrt {a^2-b^2} d e n \left (b+a \cos \left (c+d x^n\right )\right )} \] Input:

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

Output:

((e*x)^n*(-2*b*(-2*a^2 + b^2)*ArcTanh[((-a + b)*Tan[(c + d*x^n)/2])/Sqrt[a 
^2 - b^2]]*(b + a*Cos[c + d*x^n]) + Sqrt[a^2 - b^2]*(a*(a^2 - b^2)*(c + d* 
x^n)*Cos[c + d*x^n] + b*((a^2 - b^2)*(c + d*x^n) + a*b*Sin[c + d*x^n]))))/ 
(a^2*(a - b)*(a + b)*Sqrt[a^2 - b^2]*d*e*n*x^n*(b + a*Cos[c + d*x^n]))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {4696, 4692, 3042, 4272, 25, 3042, 4407, 3042, 4318, 3042, 3138, 221}

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.

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

\(\Big \downarrow \) 4696

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4272

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {x^{-n} (e x)^n \left (\frac {\frac {\left (a^2-b^2\right ) x^n}{a}-\frac {2 b \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan \left (c+d x^n\right )}{a d \left (a^2-b^2\right ) \left (a+b \sec \left (c+d x^n\right )\right )}\right )}{e n}\)

Input:

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

Output:

((e*x)^n*((((a^2 - b^2)*x^n)/a - (2*b*(2*a^2 - b^2)*ArcTanh[(Sqrt[a - b]*T 
an[(c + d*x^n)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/(a*(a^2 - 
b^2)) + (b^2*Tan[c + d*x^n])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x^n]))))/(e 
*n*x^n)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 4272

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[ 
c + d*x]*((a + b*Csc[c + d*x])^(n + 1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Sim 
p[1/(a*(n + 1)*(a^2 - b^2))   Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^2 - 
b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x 
], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ 
erQ[2*n]
 

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4407

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.70 (sec) , antiderivative size = 614, normalized size of antiderivative = 3.91

method	result	size
risch	\(\frac {x \,{\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right ) \pi +i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi +i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi -i \operatorname {csgn}\left (i e x \right )^{3} \pi +2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}}}{a^{2} n}+\frac {2 i b^{2} e^{n} \left (-1\right )^{-\frac {\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} \left (-1\right )^{-\frac {\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} \left (-1\right )^{\frac {\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )}{2}} \left (b \,{\mathrm e}^{\frac {i \left (-\pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+\pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+\pi n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-\pi n \operatorname {csgn}\left (i e x \right )^{3}+\pi \operatorname {csgn}\left (i e x \right )^{3}+2 d \,x^{n}+2 c \right )}{2}}+{\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i e x \right ) \left (-\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) n +\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right ) n +n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )-n \operatorname {csgn}\left (i e x \right )^{2}+\operatorname {csgn}\left (i e x \right )^{2}\right )}{2}} a \right )}{a^{2} \left (a^{2}-b^{2}\right ) d n \left (a \,{\mathrm e}^{2 i \left (c +d \,x^{n}\right )}+2 b \,{\mathrm e}^{i \left (c +d \,x^{n}\right )}+a \right ) e}+\frac {2 i \arctan \left (\frac {2 a \,{\mathrm e}^{i \left (d \,x^{n}+2 c \right )}+2 \,{\mathrm e}^{i c} b}{2 \sqrt {a^{2} {\mathrm e}^{2 i c}-{\mathrm e}^{2 i c} b^{2}}}\right ) e^{n} \left (-2 a^{2}+b^{2}\right ) b \,{\mathrm e}^{\frac {i \left (-\pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+\pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+\pi n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-\pi n \operatorname {csgn}\left (i e x \right )^{3}+\pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )-\pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}+\pi \operatorname {csgn}\left (i e x \right )^{3}+2 c \right )}{2}}}{\sqrt {a^{2} {\mathrm e}^{2 i c}-{\mathrm e}^{2 i c} b^{2}}\, d e n \left (-a^{2}+b^{2}\right ) a^{2}}\)	\(614\)

Input:

int((e*x)^(-1+n)/(a+b*sec(c+d*x^n))^2,x,method=_RETURNVERBOSE)
 

Output:

1/a^2/n*x*exp(1/2*(-1+n)*(-I*csgn(I*e)*csgn(I*x)*csgn(I*e*x)*Pi+I*csgn(I*e 
)*csgn(I*e*x)^2*Pi+I*csgn(I*x)*csgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*Pi+2*ln(x) 
+2*ln(e)))+2*I*b^2/a^2/(a^2-b^2)/d/n/(a*exp(2*I*(c+d*x^n))+2*b*exp(I*(c+d* 
x^n))+a)*e^n*(-1)^(-1/2*csgn(I*x)*csgn(I*e*x)^2)*(-1)^(-1/2*csgn(I*e)*csgn 
(I*e*x)^2)*(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*(b*exp(1/2*I*(-Pi*n* 
csgn(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*n*csgn(I*e)*csgn(I*e*x)^2+Pi*n*csgn(I*x 
)*csgn(I*e*x)^2-Pi*n*csgn(I*e*x)^3+Pi*csgn(I*e*x)^3+2*d*x^n+2*c))+exp(1/2* 
I*Pi*csgn(I*e*x)*(-csgn(I*e)*csgn(I*x)*n+csgn(I*e)*csgn(I*e*x)*n+n*csgn(I* 
x)*csgn(I*e*x)-n*csgn(I*e*x)^2+csgn(I*e*x)^2))*a)/e+2*I*arctan(1/2*(2*a*ex 
p(I*(d*x^n+2*c))+2*exp(I*c)*b)/(a^2*exp(2*I*c)-exp(2*I*c)*b^2)^(1/2))/(a^2 
*exp(2*I*c)-exp(2*I*c)*b^2)^(1/2)/d/e*e^n/n/(-a^2+b^2)*(-2*a^2+b^2)/a^2*b* 
exp(1/2*I*(-Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*n*csgn(I*e)*csgn(I*e*x 
)^2+Pi*n*csgn(I*x)*csgn(I*e*x)^2-Pi*n*csgn(I*e*x)^3+Pi*csgn(I*e)*csgn(I*x) 
*csgn(I*e*x)-Pi*csgn(I*e)*csgn(I*e*x)^2-Pi*csgn(I*x)*csgn(I*e*x)^2+Pi*csgn 
(I*e*x)^3+2*c))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 628, normalized size of antiderivative = 4.00 \[ \int \frac {(e x)^{-1+n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\left [\frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d e^{n - 1} x^{n} \cos \left (d x^{n} + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d e^{n - 1} x^{n} + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} e^{n - 1} \sin \left (d x^{n} + c\right ) + {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {a^{2} - b^{2}} e^{n - 1} \cos \left (d x^{n} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}} e^{n - 1}\right )} \log \left (\frac {2 \, a b \cos \left (d x^{n} + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{n} + c\right )^{2} + 2 \, a^{2} - b^{2} - 2 \, {\left (\sqrt {a^{2} - b^{2}} b \cos \left (d x^{n} + c\right ) + \sqrt {a^{2} - b^{2}} a\right )} \sin \left (d x^{n} + c\right )}{a^{2} \cos \left (d x^{n} + c\right )^{2} + 2 \, a b \cos \left (d x^{n} + c\right ) + b^{2}}\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d n \cos \left (d x^{n} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d n\right )}}, \frac {{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d e^{n - 1} x^{n} \cos \left (d x^{n} + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d e^{n - 1} x^{n} + {\left (a^{3} b^{2} - a b^{4}\right )} e^{n - 1} \sin \left (d x^{n} + c\right ) - {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {-a^{2} + b^{2}} e^{n - 1} \cos \left (d x^{n} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}} e^{n - 1}\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} b \cos \left (d x^{n} + c\right ) + \sqrt {-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \sin \left (d x^{n} + c\right )}\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d n \cos \left (d x^{n} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d n}\right ] \] Input:

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

Output:

[1/2*(2*(a^5 - 2*a^3*b^2 + a*b^4)*d*e^(n - 1)*x^n*cos(d*x^n + c) + 2*(a^4* 
b - 2*a^2*b^3 + b^5)*d*e^(n - 1)*x^n + 2*(a^3*b^2 - a*b^4)*e^(n - 1)*sin(d 
*x^n + c) + ((2*a^3*b - a*b^3)*sqrt(a^2 - b^2)*e^(n - 1)*cos(d*x^n + c) + 
(2*a^2*b^2 - b^4)*sqrt(a^2 - b^2)*e^(n - 1))*log((2*a*b*cos(d*x^n + c) - ( 
a^2 - 2*b^2)*cos(d*x^n + c)^2 + 2*a^2 - b^2 - 2*(sqrt(a^2 - b^2)*b*cos(d*x 
^n + c) + sqrt(a^2 - b^2)*a)*sin(d*x^n + c))/(a^2*cos(d*x^n + c)^2 + 2*a*b 
*cos(d*x^n + c) + b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*n*cos(d*x^n + c) + 
 (a^6*b - 2*a^4*b^3 + a^2*b^5)*d*n), ((a^5 - 2*a^3*b^2 + a*b^4)*d*e^(n - 1 
)*x^n*cos(d*x^n + c) + (a^4*b - 2*a^2*b^3 + b^5)*d*e^(n - 1)*x^n + (a^3*b^ 
2 - a*b^4)*e^(n - 1)*sin(d*x^n + c) - ((2*a^3*b - a*b^3)*sqrt(-a^2 + b^2)* 
e^(n - 1)*cos(d*x^n + c) + (2*a^2*b^2 - b^4)*sqrt(-a^2 + b^2)*e^(n - 1))*a 
rctan(-(sqrt(-a^2 + b^2)*b*cos(d*x^n + c) + sqrt(-a^2 + b^2)*a)/((a^2 - b^ 
2)*sin(d*x^n + c))))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*n*cos(d*x^n + c) + (a^ 
6*b - 2*a^4*b^3 + a^2*b^5)*d*n)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

((a^4 - a^2*b^2)*d*e^n*x^n*cos(2*d*x^n + 2*c)^2 + 4*(a^2*b^2 - b^4)*d*e^n* 
x^n*cos(d*x^n + c)^2 + (a^4 - a^2*b^2)*d*e^n*x^n*sin(2*d*x^n + 2*c)^2 + 2* 
a*b^3*e^n*sin(d*x^n + c) + 4*(a^2*b^2 - b^4)*d*e^n*x^n*sin(d*x^n + c)^2 + 
4*(a^3*b - a*b^3)*d*e^n*x^n*cos(d*x^n + c) + (a^4 - a^2*b^2)*d*e^n*x^n - 2 
*(a*b^3*e^n*sin(d*x^n + c) - 2*(a^3*b - a*b^3)*d*e^n*x^n*cos(d*x^n + c) - 
(a^4 - a^2*b^2)*d*e^n*x^n)*cos(2*d*x^n + 2*c) - 2*((2*a^8*b - 3*a^6*b^3 + 
a^4*b^5)*d*e^(n + 1)*n*cos(2*d*x^n + 2*c)^2*cos(c) + 4*(2*a^6*b^3 - 3*a^4* 
b^5 + a^2*b^7)*d*e^(n + 1)*n*cos(d*x^n + c)^2*cos(c) + (2*a^8*b - 3*a^6*b^ 
3 + a^4*b^5)*d*e^(n + 1)*n*cos(c)*sin(2*d*x^n + 2*c)^2 + 4*(2*a^7*b^2 - 3* 
a^5*b^4 + a^3*b^6)*d*e^(n + 1)*n*cos(c)*sin(2*d*x^n + 2*c)*sin(d*x^n + c) 
+ 4*(2*a^6*b^3 - 3*a^4*b^5 + a^2*b^7)*d*e^(n + 1)*n*cos(c)*sin(d*x^n + c)^ 
2 + 4*(2*a^7*b^2 - 3*a^5*b^4 + a^3*b^6)*d*e^(n + 1)*n*cos(d*x^n + c)*cos(c 
) + (2*a^8*b - 3*a^6*b^3 + a^4*b^5)*d*e^(n + 1)*n*cos(c) + 2*(2*(2*a^7*b^2 
 - 3*a^5*b^4 + a^3*b^6)*d*e^(n + 1)*n*cos(d*x^n + c)*cos(c) + (2*a^8*b - 3 
*a^6*b^3 + a^4*b^5)*d*e^(n + 1)*n*cos(c))*cos(2*d*x^n + 2*c))*integrate((a 
^3*x^n*cos(2*d*x^n + 2*c)*cos(d*x^n) + a^3*x^n*sin(2*d*x^n + 2*c)*sin(d*x^ 
n) + 2*(a^2*b - b^3)*x^n*cos(d*x^n)^2*cos(c) + 2*(a^2*b - b^3)*x^n*cos(c)* 
sin(d*x^n)^2 + (a^3 - a*b^2)*x^n*cos(d*x^n) - (a*b^2*x^n*cos(d*x^n)*cos(2* 
c) + a*b^2*x^n*sin(d*x^n)*sin(2*c))*cos(2*d*x^n) - (a*b^2*x^n*cos(2*c)*sin 
(d*x^n) - a*b^2*x^n*cos(d*x^n)*sin(2*c))*sin(2*d*x^n))/(a^8*e*x*cos(2*d...
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 19.71 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.94 \[ \int \frac {(e x)^{-1+n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\frac {\frac {b^2\,x\,{\left (e\,x\right )}^{n-1}\,2{}\mathrm {i}}{a\,d\,n\,x^n\,\left (a^2-b^2\right )}+\frac {b^3\,x\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x^n\,1{}\mathrm {i}}\,{\left (e\,x\right )}^{n-1}\,2{}\mathrm {i}}{a^2\,d\,n\,x^n\,\left (a^2-b^2\right )}}{a+a\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x^n\,2{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x^n\,1{}\mathrm {i}}}+\frac {x\,{\left (e\,x\right )}^{n-1}}{a^2\,n}+\frac {b\,x\,\ln \left (-2\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,\left (b^3\,x\,{\left (e\,x\right )}^{n-1}-2\,a^2\,b\,x\,{\left (e\,x\right )}^{n-1}\right )-\frac {b\,x\,\left (a^4-a^2\,b^2\right )\,\left (a+b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\right )\,{\left (e\,x\right )}^{n-1}\,\left (2\,a^2-b^2\right )\,2{}\mathrm {i}}{a^2\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,{\left (e\,x\right )}^{n-1}\,\left (2\,a^2-b^2\right )}{a^2\,d\,n\,x^n\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {b\,x\,\ln \left (-2\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,\left (b^3\,x\,{\left (e\,x\right )}^{n-1}-2\,a^2\,b\,x\,{\left (e\,x\right )}^{n-1}\right )+\frac {b\,x\,\left (a^4-a^2\,b^2\right )\,\left (a+b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\right )\,{\left (e\,x\right )}^{n-1}\,\left (2\,a^2-b^2\right )\,2{}\mathrm {i}}{a^2\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,{\left (e\,x\right )}^{n-1}\,\left (2\,a^2-b^2\right )}{a^2\,d\,n\,x^n\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:

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

Output:

((b^2*x*(e*x)^(n - 1)*2i)/(a*d*n*x^n*(a^2 - b^2)) + (b^3*x*exp(c*1i + d*x^ 
n*1i)*(e*x)^(n - 1)*2i)/(a^2*d*n*x^n*(a^2 - b^2)))/(a + a*exp(c*2i + d*x^n 
*2i) + 2*b*exp(c*1i + d*x^n*1i)) + (x*(e*x)^(n - 1))/(a^2*n) + (b*x*log(- 
2*exp(c*1i)*exp(d*x^n*1i)*(b^3*x*(e*x)^(n - 1) - 2*a^2*b*x*(e*x)^(n - 1)) 
- (b*x*(a^4 - a^2*b^2)*(a + b*exp(c*1i)*exp(d*x^n*1i))*(e*x)^(n - 1)*(2*a^ 
2 - b^2)*2i)/(a^2*(a + b)^(3/2)*(a - b)^(3/2)))*(e*x)^(n - 1)*(2*a^2 - b^2 
))/(a^2*d*n*x^n*(a + b)^(3/2)*(a - b)^(3/2)) - (b*x*log((b*x*(a^4 - a^2*b^ 
2)*(a + b*exp(c*1i)*exp(d*x^n*1i))*(e*x)^(n - 1)*(2*a^2 - b^2)*2i)/(a^2*(a 
 + b)^(3/2)*(a - b)^(3/2)) - 2*exp(c*1i)*exp(d*x^n*1i)*(b^3*x*(e*x)^(n - 1 
) - 2*a^2*b*x*(e*x)^(n - 1)))*(e*x)^(n - 1)*(2*a^2 - b^2))/(a^2*d*n*x^n*(a 
 + b)^(3/2)*(a - b)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.78 \[ \int \frac {(e x)^{-1+n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\frac {e^{n} \left (-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (x^{n} d +c \right ) a^{3} b +2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (x^{n} d +c \right ) a \,b^{3}-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2} b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{4}+x^{n} \cos \left (x^{n} d +c \right ) a^{5} d -2 x^{n} \cos \left (x^{n} d +c \right ) a^{3} b^{2} d +x^{n} \cos \left (x^{n} d +c \right ) a \,b^{4} d +x^{n} a^{4} b d -2 x^{n} a^{2} b^{3} d +x^{n} b^{5} d +\sin \left (x^{n} d +c \right ) a^{3} b^{2}-\sin \left (x^{n} d +c \right ) a \,b^{4}\right )}{a^{2} d e n \left (\cos \left (x^{n} d +c \right ) a^{5}-2 \cos \left (x^{n} d +c \right ) a^{3} b^{2}+\cos \left (x^{n} d +c \right ) a \,b^{4}+a^{4} b -2 a^{2} b^{3}+b^{5}\right )} \] Input:

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

Output:

(e**n*( - 4*sqrt( - a**2 + b**2)*atan((tan((x**n*d + c)/2)*a - tan((x**n*d 
 + c)/2)*b)/sqrt( - a**2 + b**2))*cos(x**n*d + c)*a**3*b + 2*sqrt( - a**2 
+ b**2)*atan((tan((x**n*d + c)/2)*a - tan((x**n*d + c)/2)*b)/sqrt( - a**2 
+ b**2))*cos(x**n*d + c)*a*b**3 - 4*sqrt( - a**2 + b**2)*atan((tan((x**n*d 
 + c)/2)*a - tan((x**n*d + c)/2)*b)/sqrt( - a**2 + b**2))*a**2*b**2 + 2*sq 
rt( - a**2 + b**2)*atan((tan((x**n*d + c)/2)*a - tan((x**n*d + c)/2)*b)/sq 
rt( - a**2 + b**2))*b**4 + x**n*cos(x**n*d + c)*a**5*d - 2*x**n*cos(x**n*d 
 + c)*a**3*b**2*d + x**n*cos(x**n*d + c)*a*b**4*d + x**n*a**4*b*d - 2*x**n 
*a**2*b**3*d + x**n*b**5*d + sin(x**n*d + c)*a**3*b**2 - sin(x**n*d + c)*a 
*b**4))/(a**2*d*e*n*(cos(x**n*d + c)*a**5 - 2*cos(x**n*d + c)*a**3*b**2 + 
cos(x**n*d + c)*a*b**4 + a**4*b - 2*a**2*b**3 + b**5))