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

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

[Out]

(e*x)^n/(a*e*n) - (2*b*(e*x)^n*ArcTanh[(Sqrt[a - b]*Tan[(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 = 0.146948, 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 = {4208, 4204, 3783, 2659, 208} \[ \frac{(e x)^n}{a e n}-\frac{2 b x^{-n} (e x)^n \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a+b}}\right )}{a d e n \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 \sec \left (c+d x^n\right )} \, dx &=\frac{\left (x^{-n} (e x)^n\right ) \int \frac{x^{-1+n}}{a+b \sec \left (c+d x^n\right )} \, dx}{e}\\ &=\frac{\left (x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{1}{a+b \sec (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 \cos (c+d x)}{b}} \, dx,x,x^n\right )}{a e n}\\ &=\frac{(e x)^n}{a e n}-\frac{\left (2 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,\tan \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 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \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 = 0.260113, size = 80, normalized size = 0.92 \[ \frac{(e x)^n \left (\frac{2 b x^{-n} \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+c x^{-n}+d\right )}{a d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^n*(d + c/x^n + (2*b*ArcTanh[((-a + b)*Tan[(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 = 0.192, size = 314, 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{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}\pi -i{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}\pi +i \left ({\it csgn} \left ( iex \right ) \right ) ^{3}\pi -2\,\ln \left ( x \right ) -2\,\ln \left ( e \right ) \right ) }{2}}}}}+{\frac{2\,ib{e}^{n}{{\rm e}^{-{\frac{i}{2}} \left ( \pi \,n{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) -\pi \,n{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}-\pi \,n{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}+\pi \,n \left ({\it csgn} \left ( iex \right ) \right ) ^{3}-\pi \,{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) +\pi \,{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}+\pi \,{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}-\pi \, \left ({\it csgn} \left ( iex \right ) \right ) ^{3}-2\,c \right ) }}}{aned}\arctan \left ({\frac{2\,a{{\rm e}^{i \left ( d{x}^{n}+2\,c \right ) }}+2\,{{\rm e}^{ic}}b}{2}{\frac{1}{\sqrt{{a}^{2}{{\rm e}^{2\,ic}}-{{\rm e}^{2\,ic}}{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}{{\rm e}^{2\,ic}}-{{\rm e}^{2\,ic}}{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Timed out

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Fricas [A]  time = 1.88593, size = 633, normalized size = 7.28 \begin{align*} \left [\frac{2 \,{\left (a^{2} - b^{2}\right )} d e^{n - 1} x^{n} + \sqrt{a^{2} - b^{2}} b e^{n - 1} \log \left (\frac{2 \, a b \cos \left (d x^{n} + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{n} + c\right )^{2} + 2 \, a^{2} - b^{2} - 2 \,{\left (\sqrt{a^{2} - b^{2}} b \cos \left (d x^{n} + c\right ) + \sqrt{a^{2} - b^{2}} a\right )} \sin \left (d x^{n} + c\right )}{a^{2} \cos \left (d x^{n} + c\right )^{2} + 2 \, a b \cos \left (d x^{n} + c\right ) + b^{2}}\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d n}, \frac{{\left (a^{2} - b^{2}\right )} d e^{n - 1} x^{n} - \sqrt{-a^{2} + b^{2}} b e^{n - 1} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}} b \cos \left (d x^{n} + c\right ) + \sqrt{-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \sin \left (d x^{n} + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(a^2 - b^2)*d*e^(n - 1)*x^n + sqrt(a^2 - b^2)*b*e^(n - 1)*log((2*a*b*cos(d*x^n + c) - (a^2 - 2*b^2)*co
s(d*x^n + c)^2 + 2*a^2 - b^2 - 2*(sqrt(a^2 - b^2)*b*cos(d*x^n + c) + sqrt(a^2 - b^2)*a)*sin(d*x^n + c))/(a^2*c
os(d*x^n + c)^2 + 2*a*b*cos(d*x^n + c) + b^2)))/((a^3 - a*b^2)*d*n), ((a^2 - b^2)*d*e^(n - 1)*x^n - sqrt(-a^2
+ b^2)*b*e^(n - 1)*arctan(-(sqrt(-a^2 + b^2)*b*cos(d*x^n + c) + sqrt(-a^2 + b^2)*a)/((a^2 - b^2)*sin(d*x^n + c
))))/((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 \sec{\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*sec(c+d*x**n)),x)

[Out]

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

[Out]

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