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

Optimal. Leaf size=328 \[ \frac {b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (-\frac {a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 e n \sqrt {b^2-a^2}}-\frac {b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (-\frac {a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 e n \sqrt {b^2-a^2}}+\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d e n \sqrt {b^2-a^2}}-\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d e n \sqrt {b^2-a^2}}+\frac {(e x)^{2 n}}{2 a e n} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.60, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4208, 4204, 4191, 3321, 2264, 2190, 2279, 2391} \[ \frac {b x^{-2 n} (e x)^{2 n} \text {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 e n \sqrt {b^2-a^2}}-\frac {b x^{-2 n} (e x)^{2 n} \text {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d^2 e n \sqrt {b^2-a^2}}+\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d e n \sqrt {b^2-a^2}}-\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d e n \sqrt {b^2-a^2}}+\frac {(e x)^{2 n}}{2 a e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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 4208

Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[(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]

Rubi steps

\begin {align*} \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx &=\frac {\left (x^{-2 n} (e x)^{2 n}\right ) \int \frac {x^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx}{e}\\ &=\frac {\left (x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \frac {x}{a+b \sec (c+d x)} \, dx,x,x^n\right )}{e n}\\ &=\frac {\left (x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \left (\frac {x}{a}-\frac {b x}{a (b+a \cos (c+d x))}\right ) \, dx,x,x^n\right )}{e n}\\ &=\frac {(e x)^{2 n}}{2 a e n}-\frac {\left (b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \frac {x}{b+a \cos (c+d x)} \, dx,x,x^n\right )}{a e n}\\ &=\frac {(e x)^{2 n}}{2 a e n}-\frac {\left (2 b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{a+2 b e^{i (c+d x)}+a e^{2 i (c+d x)}} \, dx,x,x^n\right )}{a e n}\\ &=\frac {(e x)^{2 n}}{2 a e n}-\frac {\left (2 b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 b-2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^n\right )}{\sqrt {-a^2+b^2} e n}+\frac {\left (2 b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 b+2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^n\right )}{\sqrt {-a^2+b^2} e n}\\ &=\frac {(e x)^{2 n}}{2 a e n}+\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {\left (i b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^n\right )}{a \sqrt {-a^2+b^2} d e n}+\frac {\left (i b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^n\right )}{a \sqrt {-a^2+b^2} d e n}\\ &=\frac {(e x)^{2 n}}{2 a e n}+\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {\left (b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 a x}{2 b-2 \sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{a \sqrt {-a^2+b^2} d^2 e n}+\frac {\left (b x^{-2 n} (e x)^{2 n}\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 a x}{2 b+2 \sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{a \sqrt {-a^2+b^2} d^2 e n}\\ &=\frac {(e x)^{2 n}}{2 a e n}+\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}+\frac {b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (-\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n}-\frac {b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (-\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.95, size = 861, normalized size = 2.62 \[ \frac {(e x)^{2 n} \left (b+a \cos \left (d x^n+c\right )\right ) \left (1-\frac {2 b x^{-2 n} \left (2 \left (d x^n+c\right ) \tanh ^{-1}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )-2 \left (c+\cos ^{-1}\left (-\frac {b}{a}\right )\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )+\left (\cos ^{-1}\left (-\frac {b}{a}\right )-2 i \tanh ^{-1}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )+2 i \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{-\frac {1}{2} i \left (d x^n+c\right )}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cos \left (d x^n+c\right )}}\right )+\left (\cos ^{-1}\left (-\frac {b}{a}\right )+2 i \left (\tanh ^{-1}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )-\tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{\frac {1}{2} i \left (d x^n+c\right )}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cos \left (d x^n+c\right )}}\right )-\left (\cos ^{-1}\left (-\frac {b}{a}\right )-2 i \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (a-b-i \sqrt {a^2-b^2}\right ) \left (i \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )+1\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )\right )}\right )-\left (\cos ^{-1}\left (-\frac {b}{a}\right )+2 i \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (-i a+i b+\sqrt {a^2-b^2}\right ) \left (\tan \left (\frac {1}{2} \left (d x^n+c\right )\right )+i\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )\right )}\right )+i \left (\text {Li}_2\left (\frac {\left (b-i \sqrt {a^2-b^2}\right ) \left (a+b-\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )\right )}\right )-\text {Li}_2\left (\frac {\left (b+i \sqrt {a^2-b^2}\right ) \left (a+b-\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (d x^n+c\right )\right )\right )}\right )\right )\right )}{\sqrt {a^2-b^2} d^2}\right ) \sec \left (d x^n+c\right )}{2 a e n \left (a+b \sec \left (d x^n+c\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(2*n)*(b + a*Cos[c + d*x^n])*(1 - (2*b*(2*(c + d*x^n)*ArcTanh[((a + b)*Cot[(c + d*x^n)/2])/Sqrt[a^2 - b
^2]] - 2*(c + ArcCos[-(b/a)])*ArcTanh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + (ArcCos[-(b/a)] - (2*I)*
ArcTanh[((a + b)*Cot[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + (2*I)*ArcTanh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 -
b^2]])*Log[Sqrt[a^2 - b^2]/(Sqrt[2]*Sqrt[a]*E^((I/2)*(c + d*x^n))*Sqrt[b + a*Cos[c + d*x^n]])] + (ArcCos[-(b/a
)] + (2*I)*(ArcTanh[((a + b)*Cot[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] - ArcTanh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[
a^2 - b^2]]))*Log[(Sqrt[a^2 - b^2]*E^((I/2)*(c + d*x^n)))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cos[c + d*x^n]])] - (Arc
Cos[-(b/a)] - (2*I)*ArcTanh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*(a - b - I*Sqrt[a^2 -
b^2])*(1 + I*Tan[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))] - (ArcCos[-(b/a)] + (2*I)*
ArcTanh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*((-I)*a + I*b + Sqrt[a^2 - b^2])*(I + Tan[
(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))] + I*(PolyLog[2, ((b - I*Sqrt[a^2 - b^2])*(a
+ b - Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))] - PolyLog[2, ((b
+ I*Sqrt[a^2 - b^2])*(a + b - Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^n
)/2]))])))/(Sqrt[a^2 - b^2]*d^2*x^(2*n)))*Sec[c + d*x^n])/(2*a*e*n*(a + b*Sec[c + d*x^n]))

________________________________________________________________________________________

fricas [B]  time = 0.71, size = 1291, normalized size = 3.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^n + c) + 2*I*a*sin(d*x^n + c) + 2*a*sqrt(-(
a^2 - b^2)/a^2) + 2*b) - 2*I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^n + c) - 2*I*a*sin(d*x^n
+ c) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) + 2*I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos(d*x^n +
c) + 2*I*a*sin(d*x^n + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 2*I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*l
og(-2*a*cos(d*x^n + c) - 2*I*a*sin(d*x^n + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 2*(a^2 - b^2)*d^2*e^(2*n -
1)*x^(2*n) + 2*a*b*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2) + b)*cos(d*x^n
+ c) + (2*I*a*sqrt(-(a^2 - b^2)/a^2) + 2*I*b)*sin(d*x^n + c) + 2*a)/a + 1) + 2*a*b*e^(2*n - 1)*sqrt(-(a^2 - b^
2)/a^2)*dilog(-1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2) + b)*cos(d*x^n + c) + (-2*I*a*sqrt(-(a^2 - b^2)/a^2) - 2*I*b)*
sin(d*x^n + c) + 2*a)/a + 1) - 2*a*b*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*dilog(1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2)
- b)*cos(d*x^n + c) - (2*I*a*sqrt(-(a^2 - b^2)/a^2) - 2*I*b)*sin(d*x^n + c) - 2*a)/a + 1) - 2*a*b*e^(2*n - 1)
*sqrt(-(a^2 - b^2)/a^2)*dilog(1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2) - b)*cos(d*x^n + c) - (-2*I*a*sqrt(-(a^2 - b^2)
/a^2) + 2*I*b)*sin(d*x^n + c) - 2*a)/a + 1) + (2*I*a*b*d*e^(2*n - 1)*x^n*sqrt(-(a^2 - b^2)/a^2) + 2*I*a*b*c*e^
(2*n - 1)*sqrt(-(a^2 - b^2)/a^2))*log(1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2) + b)*cos(d*x^n + c) + (2*I*a*sqrt(-(a^2
- b^2)/a^2) + 2*I*b)*sin(d*x^n + c) + 2*a)/a) + (-2*I*a*b*d*e^(2*n - 1)*x^n*sqrt(-(a^2 - b^2)/a^2) - 2*I*a*b*
c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2))*log(1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2) + b)*cos(d*x^n + c) + (-2*I*a*sqrt(
-(a^2 - b^2)/a^2) - 2*I*b)*sin(d*x^n + c) + 2*a)/a) + (2*I*a*b*d*e^(2*n - 1)*x^n*sqrt(-(a^2 - b^2)/a^2) + 2*I*
a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2))*log(-1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2) - b)*cos(d*x^n + c) - (2*I*a*s
qrt(-(a^2 - b^2)/a^2) - 2*I*b)*sin(d*x^n + c) - 2*a)/a) + (-2*I*a*b*d*e^(2*n - 1)*x^n*sqrt(-(a^2 - b^2)/a^2) -
2*I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2))*log(-1/2*(2*(a*sqrt(-(a^2 - b^2)/a^2) - b)*cos(d*x^n + c) - (-2
*I*a*sqrt(-(a^2 - b^2)/a^2) + 2*I*b)*sin(d*x^n + c) - 2*a)/a))/((a^3 - a*b^2)*d^2*n)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x\right )^{2 \, n - 1}}{b \sec \left (d x^{n} + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.52, size = 1308, normalized size = 3.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^{2\,n-1}}{a+\frac {b}{\cos \left (c+d\,x^n\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x\right )^{2 n - 1}}{a + b \sec {\left (c + d x^{n} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________