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

Optimal. Leaf size=87 \[ \frac{(e x)^n}{a e n}-\frac{2 b x^{-n} (e x)^n \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a+b}}\right )}{a d e n \sqrt{a-b} \sqrt{a+b}} \]

[Out]

(e*x)^n/(a*e*n) - (2*b*(e*x)^n*ArcTan[(Sqrt[a - b]*Tanh[(c + d*x^n)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a +
b]*d*e*n*x^n)

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Rubi [A]  time = 0.14618, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {5440, 5436, 3783, 2659, 208} \[ \frac{(e x)^n}{a e n}-\frac{2 b x^{-n} (e x)^n \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a+b}}\right )}{a d e n \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x)^n/(a*e*n) - (2*b*(e*x)^n*ArcTan[(Sqrt[a - b]*Tanh[(c + d*x^n)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a +
b]*d*e*n*x^n)

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 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 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(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 208

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

Rubi steps

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

Mathematica [A]  time = 0.146243, size = 80, normalized size = 0.92 \[ \frac{(e x)^n \left (\frac{2 b x^{-n} \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+c x^{-n}+d\right )}{a d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^n*(d + c/x^n + (2*b*ArcTan[((-a + b)*Tanh[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*x^n)))/(a*
d*e*n)

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Maple [C]  time = 0.121, size = 317, normalized size = 3.6 \begin{align*}{\frac{x}{an}{{\rm e}^{-{\frac{ \left ( -1+n \right ) \left ( i{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) \pi -i\pi \, \left ({\it csgn} \left ( iex \right ) \right ) ^{2}{\it csgn} \left ( ie \right ) -i\pi \, \left ({\it csgn} \left ( iex \right ) \right ) ^{2}{\it csgn} \left ( ix \right ) +i \left ({\it csgn} \left ( iex \right ) \right ) ^{3}\pi -2\,\ln \left ( e \right ) -2\,\ln \left ( x \right ) \right ) }{2}}}}}-2\,{\frac{b{{\rm e}^{-i/2\pi \,n{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) }}{{\rm e}^{i/2\pi \,n{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{i/2\pi \,n{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{-i/2\pi \,n \left ({\it csgn} \left ( iex \right ) \right ) ^{3}}}{{\rm e}^{i/2\pi \,{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) }}{{\rm e}^{-i/2\pi \,{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{-i/2\pi \,{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{i/2\pi \, \left ({\it csgn} \left ( iex \right ) \right ) ^{3}}}{e}^{n}{{\rm e}^{c}}}{aned\sqrt{{a}^{2}{{\rm e}^{2\,c}}-{{\rm e}^{2\,c}}{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a{{\rm e}^{2\,c+d{x}^{n}}}+2\,{{\rm e}^{c}}b}{\sqrt{{a}^{2}{{\rm e}^{2\,c}}-{{\rm e}^{2\,c}}{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.37508, size = 1381, normalized size = 15.87 \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+n)/(a+b*sech(c+d*x^n)),x, algorithm="fricas")

[Out]

[((a^2 - b^2)*d*cosh((n - 1)*log(e))*cosh(n*log(x)) + (a^2 - b^2)*d*cosh(n*log(x))*sinh((n - 1)*log(e)) - (sqr
t(-a^2 + b^2)*b*cosh((n - 1)*log(e)) + sqrt(-a^2 + b^2)*b*sinh((n - 1)*log(e)))*log((a*b + (b^2 + sqrt(-a^2 +
b^2)*b)*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + (a^2 - b^2 - sqrt(-a^2 + b^2)*b)*sinh(d*cosh(n*log(x))
+ d*sinh(n*log(x)) + c) + sqrt(-a^2 + b^2)*a)/(a*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + b)) + ((a^2
- b^2)*d*cosh((n - 1)*log(e)) + (a^2 - b^2)*d*sinh((n - 1)*log(e)))*sinh(n*log(x)))/((a^3 - a*b^2)*d*n), ((a^2
- b^2)*d*cosh((n - 1)*log(e))*cosh(n*log(x)) + (a^2 - b^2)*d*cosh(n*log(x))*sinh((n - 1)*log(e)) + 2*(sqrt(a^
2 - b^2)*b*cosh((n - 1)*log(e)) + sqrt(a^2 - b^2)*b*sinh((n - 1)*log(e)))*arctan(-(sqrt(a^2 - b^2)*a*cosh(d*co
sh(n*log(x)) + d*sinh(n*log(x)) + c) + sqrt(a^2 - b^2)*a*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + sqrt(
a^2 - b^2)*b)/(a^2 - b^2)) + ((a^2 - b^2)*d*cosh((n - 1)*log(e)) + (a^2 - b^2)*d*sinh((n - 1)*log(e)))*sinh(n*
log(x)))/((a^3 - a*b^2)*d*n)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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