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

Optimal. Leaf size=217 \[ \frac {a (e x)^{3 n}}{3 e n}+\frac {2 i b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{d x^n+c}\right )}{d^3 e n}-\frac {2 i b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{d x^n+c}\right )}{d^3 e n}-\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{d x^n+c}\right )}{d^2 e n}+\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{d x^n+c}\right )}{d^2 e n}+\frac {2 b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{c+d x^n}\right )}{d e n} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 217, normalized size of antiderivative = 1.00, 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, 5440, 5436, 4180, 2531, 2282, 6589} \[ \frac {2 i b x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (3,-i e^{c+d x^n}\right )}{d^3 e n}-\frac {2 i b x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (3,i e^{c+d x^n}\right )}{d^3 e n}-\frac {2 i b x^{-2 n} (e x)^{3 n} \text {PolyLog}\left (2,-i e^{c+d x^n}\right )}{d^2 e n}+\frac {2 i b x^{-2 n} (e x)^{3 n} \text {PolyLog}\left (2,i e^{c+d x^n}\right )}{d^2 e n}+\frac {a (e x)^{3 n}}{3 e n}+\frac {2 b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{c+d x^n}\right )}{d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(e*x)^(3*n))/(3*e*n) + (2*b*(e*x)^(3*n)*ArcTan[E^(c + d*x^n)])/(d*e*n*x^n) - ((2*I)*b*(e*x)^(3*n)*PolyLog[2
, (-I)*E^(c + d*x^n)])/(d^2*e*n*x^(2*n)) + ((2*I)*b*(e*x)^(3*n)*PolyLog[2, I*E^(c + d*x^n)])/(d^2*e*n*x^(2*n))
+ ((2*I)*b*(e*x)^(3*n)*PolyLog[3, (-I)*E^(c + d*x^n)])/(d^3*e*n*x^(3*n)) - ((2*I)*b*(e*x)^(3*n)*PolyLog[3, I*
E^(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 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 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 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5436

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sech[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 5440

Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[(e^IntPart[m]*(e*
x)^FracPart[m])/x^FracPart[m], Int[x^m*(a + b*Sech[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, 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 \text {sech}\left (c+d x^n\right )\right ) \, dx &=\int \left (a (e x)^{-1+3 n}+b (e x)^{-1+3 n} \text {sech}\left (c+d x^n\right )\right ) \, dx\\ &=\frac {a (e x)^{3 n}}{3 e n}+b \int (e x)^{-1+3 n} \text {sech}\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} \text {sech}\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 \text {sech}(c+d x) \, dx,x,x^n\right )}{e n}\\ &=\frac {a (e x)^{3 n}}{3 e n}+\frac {2 b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{c+d x^n}\right )}{d e n}-\frac {\left (2 i b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{c+d x}\right ) \, dx,x,x^n\right )}{d e n}+\frac {\left (2 i b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{c+d x}\right ) \, dx,x,x^n\right )}{d e n}\\ &=\frac {a (e x)^{3 n}}{3 e n}+\frac {2 b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{c+d x^n}\right )}{d e n}-\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{c+d x^n}\right )}{d^2 e n}+\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{c+d 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^{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^{c+d x}\right ) \, dx,x,x^n\right )}{d^2 e n}\\ &=\frac {a (e x)^{3 n}}{3 e n}+\frac {2 b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{c+d x^n}\right )}{d e n}-\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{c+d x^n}\right )}{d^2 e n}+\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{c+d x^n}\right )}{d^2 e n}+\frac {\left (2 i b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{c+d x^n}\right )}{d^3 e n}-\frac {\left (2 i b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{c+d x^n}\right )}{d^3 e n}\\ &=\frac {a (e x)^{3 n}}{3 e n}+\frac {2 b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{c+d x^n}\right )}{d e n}-\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{c+d x^n}\right )}{d^2 e n}+\frac {2 i b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{c+d x^n}\right )}{d^2 e n}+\frac {2 i b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{c+d x^n}\right )}{d^3 e n}-\frac {2 i b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{c+d x^n}\right )}{d^3 e n}\\ \end {align*}

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Mathematica [F]  time = 10.53, size = 0, normalized size = 0.00 \[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [C]  time = 0.49, size = 1072, normalized size = 4.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*d^3*cosh((3*n - 1)*log(e))*cosh(n*log(x))^3 + a*d^3*cosh(n*log(x))^3*sinh((3*n - 1)*log(e)) + (a*d^3*co
sh((3*n - 1)*log(e)) + a*d^3*sinh((3*n - 1)*log(e)))*sinh(n*log(x))^3 + 3*(a*d^3*cosh((3*n - 1)*log(e))*cosh(n
*log(x)) + a*d^3*cosh(n*log(x))*sinh((3*n - 1)*log(e)))*sinh(n*log(x))^2 + (6*I*b*d*cosh((3*n - 1)*log(e))*cos
h(n*log(x)) + 6*I*b*d*cosh(n*log(x))*sinh((3*n - 1)*log(e)) + (6*I*b*d*cosh((3*n - 1)*log(e)) + 6*I*b*d*sinh((
3*n - 1)*log(e)))*sinh(n*log(x)))*dilog(I*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + I*sinh(d*cosh(n*log(
x)) + d*sinh(n*log(x)) + c)) + (-6*I*b*d*cosh((3*n - 1)*log(e))*cosh(n*log(x)) - 6*I*b*d*cosh(n*log(x))*sinh((
3*n - 1)*log(e)) + (-6*I*b*d*cosh((3*n - 1)*log(e)) - 6*I*b*d*sinh((3*n - 1)*log(e)))*sinh(n*log(x)))*dilog(-I
*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) - I*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c)) + (3*I*b*c^2
*cosh((3*n - 1)*log(e)) + 3*I*b*c^2*sinh((3*n - 1)*log(e)))*log(cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c)
+ sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + I) + (-3*I*b*c^2*cosh((3*n - 1)*log(e)) - 3*I*b*c^2*sinh((3*
n - 1)*log(e)))*log(cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) +
c) - I) + (-3*I*b*d^2*cosh((3*n - 1)*log(e))*cosh(n*log(x))^2 + 3*I*b*c^2*cosh((3*n - 1)*log(e)) + (-3*I*b*d^
2*cosh((3*n - 1)*log(e)) - 3*I*b*d^2*sinh((3*n - 1)*log(e)))*sinh(n*log(x))^2 + (-3*I*b*d^2*cosh(n*log(x))^2 +
3*I*b*c^2)*sinh((3*n - 1)*log(e)) + (-6*I*b*d^2*cosh((3*n - 1)*log(e))*cosh(n*log(x)) - 6*I*b*d^2*cosh(n*log(
x))*sinh((3*n - 1)*log(e)))*sinh(n*log(x)))*log(I*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + I*sinh(d*cos
h(n*log(x)) + d*sinh(n*log(x)) + c) + 1) + (3*I*b*d^2*cosh((3*n - 1)*log(e))*cosh(n*log(x))^2 - 3*I*b*c^2*cosh
((3*n - 1)*log(e)) + (3*I*b*d^2*cosh((3*n - 1)*log(e)) + 3*I*b*d^2*sinh((3*n - 1)*log(e)))*sinh(n*log(x))^2 +
(3*I*b*d^2*cosh(n*log(x))^2 - 3*I*b*c^2)*sinh((3*n - 1)*log(e)) + (6*I*b*d^2*cosh((3*n - 1)*log(e))*cosh(n*log
(x)) + 6*I*b*d^2*cosh(n*log(x))*sinh((3*n - 1)*log(e)))*sinh(n*log(x)))*log(-I*cosh(d*cosh(n*log(x)) + d*sinh(
n*log(x)) + c) - I*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + 1) + (-6*I*b*cosh((3*n - 1)*log(e)) - 6*I*b
*sinh((3*n - 1)*log(e)))*polylog(3, I*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + I*sinh(d*cosh(n*log(x))
+ d*sinh(n*log(x)) + c)) + (6*I*b*cosh((3*n - 1)*log(e)) + 6*I*b*sinh((3*n - 1)*log(e)))*polylog(3, -I*cosh(d*
cosh(n*log(x)) + d*sinh(n*log(x)) + c) - I*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c)) + 3*(a*d^3*cosh((3*n
- 1)*log(e))*cosh(n*log(x))^2 + a*d^3*cosh(n*log(x))^2*sinh((3*n - 1)*log(e)))*sinh(n*log(x)))/(d^3*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {sech}\left (d x^{n} + c\right ) + a\right )} \left (e x\right )^{3 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.16, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{3 n -1} \left (a +b \,\mathrm {sech}\left (c +d \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, b \int \frac {\left (e x\right )^{3 \, n - 1}}{e^{\left (d x^{n} + c\right )} + e^{\left (-d x^{n} - c\right )}}\,{d x} + \frac {\left (e x\right )^{3 \, n} a}{3 \, e n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x^n\right )}\right )\,{\left (e\,x\right )}^{3\,n-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{3 n - 1} \left (a + b \operatorname {sech}{\left (c + d x^{n} \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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