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

Optimal. Leaf size=235 \[ -\frac{2 b x^{-3 n} (e x)^{3 n} \text{PolyLog}\left (3,-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac{2 b x^{-3 n} (e x)^{3 n} \text{PolyLog}\left (3,i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac{2 i b x^{-2 n} (e x)^{3 n} \text{PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac{2 i b x^{-2 n} (e x)^{3 n} \text{PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac{a (e x)^{3 n}}{3 e n}-\frac{2 i b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n} \]

[Out]

(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)*P
olyLog[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)*P
olyLog[3, I*E^(I*(c + d*x^n))])/(d^3*e*n*x^(3*n))

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Rubi [A]  time = 0.18986, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {14, 4208, 4204, 4181, 2531, 2282, 6589} \[ -\frac{2 b x^{-3 n} (e x)^{3 n} \text{PolyLog}\left (3,-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac{2 b x^{-3 n} (e x)^{3 n} \text{PolyLog}\left (3,i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac{2 i b x^{-2 n} (e x)^{3 n} \text{PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac{2 i b x^{-2 n} (e x)^{3 n} \text{PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac{a (e x)^{3 n}}{3 e n}-\frac{2 i b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)*P
olyLog[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)*P
olyLog[3, I*E^(I*(c + d*x^n))])/(d^3*e*n*x^(3*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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right ) \, dx &=\int \left (a (e x)^{-1+3 n}+b (e x)^{-1+3 n} \sec \left (c+d x^n\right )\right ) \, dx\\ &=\frac{a (e x)^{3 n}}{3 e n}+b \int (e x)^{-1+3 n} \sec \left (c+d x^n\right ) \, dx\\ &=\frac{a (e x)^{3 n}}{3 e n}+\frac{\left (b x^{-3 n} (e x)^{3 n}\right ) \int x^{-1+3 n} \sec \left (c+d x^n\right ) \, dx}{e}\\ &=\frac{a (e x)^{3 n}}{3 e n}+\frac{\left (b x^{-3 n} (e x)^{3 n}\right ) \operatorname{Subst}\left (\int x^2 \sec (c+d x) \, dx,x,x^n\right )}{e n}\\ &=\frac{a (e x)^{3 n}}{3 e n}-\frac{2 i b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}-\frac{\left (2 b x^{-3 n} (e x)^{3 n}\right ) \operatorname{Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d e n}+\frac{\left (2 b x^{-3 n} (e x)^{3 n}\right ) \operatorname{Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d e n}\\ &=\frac{a (e x)^{3 n}}{3 e n}-\frac{2 i b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac{2 i b x^{-2 n} (e x)^{3 n} \text{Li}_2\left (-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} \text{Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac{\left (2 i b x^{-3 n} (e x)^{3 n}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}+\frac{\left (2 i b x^{-3 n} (e x)^{3 n}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}\\ &=\frac{a (e x)^{3 n}}{3 e n}-\frac{2 i b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac{2 i b x^{-2 n} (e x)^{3 n} \text{Li}_2\left (-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} \text{Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac{\left (2 b x^{-3 n} (e x)^{3 n}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac{\left (2 b x^{-3 n} (e x)^{3 n}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{d^3 e n}\\ &=\frac{a (e x)^{3 n}}{3 e n}-\frac{2 i b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac{2 i b x^{-2 n} (e x)^{3 n} \text{Li}_2\left (-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} \text{Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac{2 b x^{-3 n} (e x)^{3 n} \text{Li}_3\left (-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac{2 b x^{-3 n} (e x)^{3 n} \text{Li}_3\left (i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}\\ \end{align*}

Mathematica [F]  time = 1.17997, size = 0, normalized size = 0. \[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.378, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{-1+3\,n} \left ( a+b\sec \left ( c+d{x}^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.19279, size = 1640, normalized size = 6.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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) + s
in(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(c
os(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)

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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+3*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 )^{3 \, n - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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