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

Optimal. Leaf size=157 \[ -\frac{2 b \left (2 a^2-b^2\right ) x^{-n} (e x)^n \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a+b}}\right )}{a^2 d e n (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a d e n \left (a^2-b^2\right ) \left (a+b \sec \left (c+d x^n\right )\right )}+\frac{(e x)^n}{a^2 e n} \]

[Out]

(e*x)^n/(a^2*e*n) - (2*b*(2*a^2 - b^2)*(e*x)^n*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x^n)/2])/Sqrt[a + b]])/(a^2*(a
- b)^(3/2)*(a + b)^(3/2)*d*e*n*x^n) + (b^2*(e*x)^n*Tan[c + d*x^n])/(a*(a^2 - b^2)*d*e*n*x^n*(a + b*Sec[c + d*x
^n]))

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Rubi [A]  time = 0.286647, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {4208, 4204, 3785, 3919, 3831, 2659, 208} \[ -\frac{2 b \left (2 a^2-b^2\right ) x^{-n} (e x)^n \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a+b}}\right )}{a^2 d e n (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a d e n \left (a^2-b^2\right ) \left (a+b \sec \left (c+d x^n\right )\right )}+\frac{(e x)^n}{a^2 e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x)^n/(a^2*e*n) - (2*b*(2*a^2 - b^2)*(e*x)^n*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x^n)/2])/Sqrt[a + b]])/(a^2*(a
- b)^(3/2)*(a + b)^(3/2)*d*e*n*x^n) + (b^2*(e*x)^n*Tan[c + d*x^n])/(a*(a^2 - b^2)*d*e*n*x^n*(a + b*Sec[c + d*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 3785

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, 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}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx &=\frac{\left (x^{-n} (e x)^n\right ) \int \frac{x^{-1+n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx}{e}\\ &=\frac{\left (x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b \sec (c+d x))^2} \, dx,x,x^n\right )}{e n}\\ &=\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \sec \left (c+d x^n\right )\right )}-\frac{\left (x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{-a^2+b^2+a b \sec (c+d x)}{a+b \sec (c+d x)} \, dx,x,x^n\right )}{a \left (a^2-b^2\right ) e n}\\ &=\frac{(e x)^n}{a^2 e n}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \sec \left (c+d x^n\right )\right )}+\frac{\left (\left (-a^2 b+b \left (-a^2+b^2\right )\right ) x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx,x,x^n\right )}{a^2 \left (a^2-b^2\right ) e n}\\ &=\frac{(e x)^n}{a^2 e n}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \sec \left (c+d x^n\right )\right )}+\frac{\left (\left (-a^2 b+b \left (-a^2+b^2\right )\right ) 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^2 b \left (a^2-b^2\right ) e n}\\ &=\frac{(e x)^n}{a^2 e n}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \sec \left (c+d x^n\right )\right )}+\frac{\left (2 \left (-a^2 b+b \left (-a^2+b^2\right )\right ) 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^2 b \left (a^2-b^2\right ) d e n}\\ &=\frac{(e x)^n}{a^2 e n}-\frac{2 b \left (2 a^2-b^2\right ) 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^2 (a-b)^{3/2} (a+b)^{3/2} d e n}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \sec \left (c+d x^n\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.775071, size = 191, normalized size = 1.22 \[ \frac{x^{-n} (e x)^n \left (\sqrt{a^2-b^2} \left (b \left (\left (a^2-b^2\right ) \left (c+d x^n\right )+a b \sin \left (c+d x^n\right )\right )+a \left (a^2-b^2\right ) \left (c+d x^n\right ) \cos \left (c+d x^n\right )\right )-2 b \left (b^2-2 a^2\right ) \left (a \cos \left (c+d x^n\right )+b\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a^2-b^2}}\right )\right )}{a^2 d e n (a-b) (a+b) \sqrt{a^2-b^2} \left (a \cos \left (c+d x^n\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.257, size = 706, normalized size = 4.5 \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+n)/(a+b*sec(c+d*x^n))^2,x)

[Out]

1/a^2/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^2*x/a^2/(a^2-b^2)/d/n/(x^n)/(exp(2*I*(c+d*x^n))*a+2*b*e
xp(I*(c+d*x^n))+a)*((-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*(-1)^(-1/2*csgn(I*e)*csgn(I*e*x)^2)*(-1)^(-1/2*
csgn(I*x)*csgn(I*e*x)^2)/x/e*x^n*e^n*b*exp(1/2*I*Pi*csgn(I*e*x)^3)*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(I*x^n*d)*exp(I*c)+(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*(-1)^(-1/2*csgn(I*e)*csgn(I*e*x)^2)*(-
1)^(-1/2*csgn(I*x)*csgn(I*e*x)^2)/x/e*x^n*e^n*a*exp(-1/2*I*Pi*csgn(I*e*x)*(n*csgn(I*e*x)^2-n*csgn(I*e)*csgn(I*
e*x)-n*csgn(I*x)*csgn(I*e*x)+n*csgn(I*e)*csgn(I*x)-csgn(I*e*x)^2)))+2*I*b/a^2*(-2*a^2+b^2)/(-a^2+b^2)/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)*csgn(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)*c
sgn(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))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.01166, size = 1347, normalized size = 8.58 \begin{align*} \left [\frac{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d e^{n - 1} x^{n} \cos \left (d x^{n} + c\right ) + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d e^{n - 1} x^{n} + 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} e^{n - 1} \sin \left (d x^{n} + c\right ) +{\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt{a^{2} - b^{2}} e^{n - 1} \cos \left (d x^{n} + c\right ) +{\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt{a^{2} - b^{2}} e^{n - 1}\right )} \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 ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d n \cos \left (d x^{n} + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d n\right )}}, \frac{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d e^{n - 1} x^{n} \cos \left (d x^{n} + c\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d e^{n - 1} x^{n} +{\left (a^{3} b^{2} - a b^{4}\right )} e^{n - 1} \sin \left (d x^{n} + c\right ) -{\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt{-a^{2} + b^{2}} e^{n - 1} \cos \left (d x^{n} + c\right ) +{\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt{-a^{2} + b^{2}} e^{n - 1}\right )} \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^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d n \cos \left (d x^{n} + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\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))^2,x, algorithm="fricas")

[Out]

[1/2*(2*(a^5 - 2*a^3*b^2 + a*b^4)*d*e^(n - 1)*x^n*cos(d*x^n + c) + 2*(a^4*b - 2*a^2*b^3 + b^5)*d*e^(n - 1)*x^n
+ 2*(a^3*b^2 - a*b^4)*e^(n - 1)*sin(d*x^n + c) + ((2*a^3*b - a*b^3)*sqrt(a^2 - b^2)*e^(n - 1)*cos(d*x^n + c)
+ (2*a^2*b^2 - b^4)*sqrt(a^2 - b^2)*e^(n - 1))*log((2*a*b*cos(d*x^n + c) - (a^2 - 2*b^2)*cos(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*cos(d*x^n + c)^2 + 2
*a*b*cos(d*x^n + c) + b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*n*cos(d*x^n + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d
*n), ((a^5 - 2*a^3*b^2 + a*b^4)*d*e^(n - 1)*x^n*cos(d*x^n + c) + (a^4*b - 2*a^2*b^3 + b^5)*d*e^(n - 1)*x^n + (
a^3*b^2 - a*b^4)*e^(n - 1)*sin(d*x^n + c) - ((2*a^3*b - a*b^3)*sqrt(-a^2 + b^2)*e^(n - 1)*cos(d*x^n + c) + (2*
a^2*b^2 - b^4)*sqrt(-a^2 + b^2)*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^7 - 2*a^5*b^2 + a^3*b^4)*d*n*cos(d*x^n + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d
*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+n)/(a+b*sec(c+d*x**n))**2,x)

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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