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

Optimal. Leaf size=149 \[ \frac{i b x^{-2 n} (e x)^{2 n} \text{PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac{i b x^{-2 n} (e x)^{2 n} \text{PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac{a (e x)^{2 n}}{2 e n}-\frac{2 i b x^{-n} (e x)^{2 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n} \]

[Out]

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

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Rubi [A]  time = 0.113521, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {14, 4208, 4204, 4181, 2279, 2391} \[ \frac{i b x^{-2 n} (e x)^{2 n} \text{PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac{i b x^{-2 n} (e x)^{2 n} \text{PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac{a (e x)^{2 n}}{2 e n}-\frac{2 i b x^{-n} (e x)^{2 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

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]]

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]

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 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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]

Rubi steps

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

Mathematica [A]  time = 0.53909, size = 188, normalized size = 1.26 \[ \frac{(e x)^{2 n} \cos \left (c+d x^n\right ) \left (a+b \sec \left (c+d x^n\right )\right ) \left (a+\frac{b x^{-2 n} \left (2 i \left (\text{PolyLog}\left (2,-i e^{-i \left (c+d x^n\right )}\right )-\text{PolyLog}\left (2,i e^{-i \left (c+d x^n\right )}\right )\right )+\left (-2 c-2 d x^n+\pi \right ) \left (\log \left (1-i e^{-i \left (c+d x^n\right )}\right )-\log \left (1+i e^{-i \left (c+d x^n\right )}\right )\right )-(\pi -2 c) \log \left (\cot \left (\frac{1}{4} \left (2 c+2 d x^n+\pi \right )\right )\right )\right )}{d^2}\right )}{2 e n \left (a \cos \left (c+d x^n\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(2*n)*Cos[c + d*x^n]*(a + (b*((-2*c + Pi - 2*d*x^n)*(Log[1 - I/E^(I*(c + d*x^n))] - Log[1 + I/E^(I*(c +
d*x^n))]) - (-2*c + Pi)*Log[Cot[(2*c + Pi + 2*d*x^n)/4]] + (2*I)*(PolyLog[2, (-I)/E^(I*(c + d*x^n))] - PolyLo
g[2, I/E^(I*(c + d*x^n))])))/(d^2*x^(2*n)))*(a + b*Sec[c + d*x^n]))/(2*e*n*(b + a*Cos[c + d*x^n]))

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Maple [C]  time = 0.346, size = 873, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

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*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)*csg
n(I*e*x)^2*Pi+I*csgn(I*e*x)^3*Pi-2*ln(x)-2*ln(e)))+I*b*(e^n)^2/e/n/d*(-1)^(-1/2*csgn(I*e)*csgn(I*e*x)^2)*(-1)^
(-1/2*csgn(I*x)*csgn(I*e*x)^2)*(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*(-exp(2*I*c))^(1/2)*x^n*ln(1+exp(I*x
^n*d)*(-exp(2*I*c))^(1/2))*exp(-I*Pi*n*csgn(I*e*x)^3)*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)*csgn(I*x)*csgn(I*e*x))*exp(1/2*I*Pi*csgn(I*e*x)^3)*exp(-I*c)-I*b*(e^n)^2
/e/n/d*(-1)^(-1/2*csgn(I*e)*csgn(I*e*x)^2)*(-1)^(-1/2*csgn(I*x)*csgn(I*e*x)^2)*(-1)^(1/2*csgn(I*e)*csgn(I*x)*c
sgn(I*e*x))*(-exp(2*I*c))^(1/2)*x^n*ln(1-exp(I*x^n*d)*(-exp(2*I*c))^(1/2))*exp(-I*Pi*n*csgn(I*e*x)^3)*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)*csgn(I*x)*csgn(I*e*x))*e
xp(1/2*I*Pi*csgn(I*e*x)^3)*exp(-I*c)+b*(e^n)^2/e/n/d^2*(-1)^(-1/2*csgn(I*e)*csgn(I*e*x)^2)*(-1)^(-1/2*csgn(I*x
)*csgn(I*e*x)^2)*(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*(-exp(2*I*c))^(1/2)*dilog(1+exp(I*x^n*d)*(-exp(2*I
*c))^(1/2))*exp(-I*Pi*n*csgn(I*e*x)^3)*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)*csgn(I*x)*csgn(I*e*x))*exp(1/2*I*Pi*csgn(I*e*x)^3)*exp(-I*c)-b*(e^n)^2/e/n/d^2*(-1)^(-1
/2*csgn(I*e)*csgn(I*e*x)^2)*(-1)^(-1/2*csgn(I*x)*csgn(I*e*x)^2)*(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*(-e
xp(2*I*c))^(1/2)*dilog(1-exp(I*x^n*d)*(-exp(2*I*c))^(1/2))*exp(-I*Pi*n*csgn(I*e*x)^3)*exp(I*Pi*n*csgn(I*e)*csg
n(I*e*x)^2)*exp(I*Pi*n*csgn(I*x)*csgn(I*e*x)^2)*exp(-I*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*exp(1/2*I*Pi*csgn
(I*e*x)^3)*exp(-I*c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.07783, size = 1170, normalized size = 7.85 \begin{align*} \frac{a d^{2} e^{2 \, n - 1} x^{2 \, n} - b c e^{2 \, n - 1} \log \left (\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right ) + i\right ) + b c e^{2 \, n - 1} \log \left (\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right ) + i\right ) - b c e^{2 \, n - 1} \log \left (-\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right ) + i\right ) + b c e^{2 \, n - 1} \log \left (-\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right ) + i\right ) - i \, b e^{2 \, n - 1}{\rm Li}_2\left (i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right )\right ) - i \, b e^{2 \, n - 1}{\rm Li}_2\left (i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right )\right ) + i \, b e^{2 \, n - 1}{\rm Li}_2\left (-i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right )\right ) + i \, b e^{2 \, n - 1}{\rm Li}_2\left (-i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right )\right ) +{\left (b d e^{2 \, n - 1} x^{n} + b c e^{2 \, n - 1}\right )} \log \left (i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right ) + 1\right ) -{\left (b d e^{2 \, n - 1} x^{n} + b c e^{2 \, n - 1}\right )} \log \left (i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right ) + 1\right ) +{\left (b d e^{2 \, n - 1} x^{n} + b c e^{2 \, n - 1}\right )} \log \left (-i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right ) + 1\right ) -{\left (b d e^{2 \, n - 1} x^{n} + b c e^{2 \, n - 1}\right )} \log \left (-i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right ) + 1\right )}{2 \, d^{2} n} \end{align*}

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]

1/2*(a*d^2*e^(2*n - 1)*x^(2*n) - b*c*e^(2*n - 1)*log(cos(d*x^n + c) + I*sin(d*x^n + c) + I) + b*c*e^(2*n - 1)*
log(cos(d*x^n + c) - I*sin(d*x^n + c) + I) - b*c*e^(2*n - 1)*log(-cos(d*x^n + c) + I*sin(d*x^n + c) + I) + b*c
*e^(2*n - 1)*log(-cos(d*x^n + c) - I*sin(d*x^n + c) + I) - I*b*e^(2*n - 1)*dilog(I*cos(d*x^n + c) + sin(d*x^n
+ c)) - I*b*e^(2*n - 1)*dilog(I*cos(d*x^n + c) - sin(d*x^n + c)) + I*b*e^(2*n - 1)*dilog(-I*cos(d*x^n + c) + s
in(d*x^n + c)) + I*b*e^(2*n - 1)*dilog(-I*cos(d*x^n + c) - sin(d*x^n + c)) + (b*d*e^(2*n - 1)*x^n + b*c*e^(2*n
- 1))*log(I*cos(d*x^n + c) + sin(d*x^n + c) + 1) - (b*d*e^(2*n - 1)*x^n + b*c*e^(2*n - 1))*log(I*cos(d*x^n +
c) - sin(d*x^n + c) + 1) + (b*d*e^(2*n - 1)*x^n + b*c*e^(2*n - 1))*log(-I*cos(d*x^n + c) + sin(d*x^n + c) + 1)
- (b*d*e^(2*n - 1)*x^n + b*c*e^(2*n - 1))*log(-I*cos(d*x^n + c) - sin(d*x^n + c) + 1))/(d^2*n)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x^{n} + c\right ) + a\right )} \left (e x\right )^{2 \, n - 1}\,{d x} \end{align*}

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