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

Optimal. Leaf size=156 \[ \frac{2 b \left (2 a^2-b^2\right ) x^{-n} (e x)^n \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d e n \left (a^2-b^2\right )^{3/2}}-\frac{b^2 x^{-n} (e x)^n \cot \left (c+d x^n\right )}{a d e n \left (a^2-b^2\right ) \left (a+b \csc \left (c+d x^n\right )\right )}+\frac{(e x)^n}{a^2 e n} \]

[Out]

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

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Rubi [A]  time = 0.285968, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {4209, 4205, 3785, 3919, 3831, 2660, 618, 206} \[ \frac{2 b \left (2 a^2-b^2\right ) x^{-n} (e x)^n \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d e n \left (a^2-b^2\right )^{3/2}}-\frac{b^2 x^{-n} (e x)^n \cot \left (c+d x^n\right )}{a d e n \left (a^2-b^2\right ) \left (a+b \csc \left (c+d x^n\right )\right )}+\frac{(e x)^n}{a^2 e n} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^(-1 + n)/(a + b*Csc[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[(a + b*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2
- b^2)^(3/2)*d*e*n*x^n) - (b^2*(e*x)^n*Cot[c + d*x^n])/(a*(a^2 - b^2)*d*e*n*x^n*(a + b*Csc[c + d*x^n]))

Rule 4209

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_)*(x_))^(m_.), x_Symbol] :> Dist[(e^IntPart[m]*(e*x
)^FracPart[m])/x^FracPart[m], Int[x^m*(a + b*Csc[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]

Rule 4205

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Csc[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 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(e x)^{-1+n}}{\left (a+b \csc \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 \csc \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 \csc (c+d x))^2} \, dx,x,x^n\right )}{e n}\\ &=-\frac{b^2 x^{-n} (e x)^n \cot \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \csc \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 \csc (c+d x)}{a+b \csc (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 \cot \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \csc \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{\csc (c+d x)}{a+b \csc (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 \cot \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \csc \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 \sin (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 \cot \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \csc \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{2 a x}{b}+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{b^2 x^{-n} (e x)^n \cot \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \csc \left (c+d x^n\right )\right )}-\frac{\left (4 \left (-a^2 b+b \left (-a^2+b^2\right )\right ) x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (1-\frac{a^2}{b^2}\right )-x^2} \, dx,x,\frac{2 a}{b}+2 \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{b \left (\frac{a}{b}+\tan \left (\frac{1}{2} \left (c+d x^n\right )\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2} d e n}-\frac{b^2 x^{-n} (e x)^n \cot \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (a+b \csc \left (c+d x^n\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.959075, size = 176, normalized size = 1.13 \[ \frac{x^{-n} (e x)^n \left (\sqrt{b^2-a^2} \left (\left (a^2-b^2\right ) \left (c+d x^n\right ) \left (a+b \csc \left (c+d x^n\right )\right )-a b^2 \cot \left (c+d x^n\right )\right )+2 b \left (b^2-2 a^2\right ) \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{b^2-a^2}}\right ) \left (a+b \csc \left (c+d x^n\right )\right )\right )}{a^2 d e n (a-b) (a+b) \sqrt{b^2-a^2} \left (a+b \csc \left (c+d x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \text{hanged} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*csc(c+d*x^n))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.609418, size = 1353, normalized size = 8.67 \begin{align*} \left [\frac{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d e^{n - 1} x^{n} \sin \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} \cos \left (d x^{n} + c\right ) +{\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt{a^{2} - b^{2}} e^{n - 1} \sin \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{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{n} + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}} a \cos \left (d x^{n} + c\right ) + a^{2} + b^{2} + 2 \,{\left (\sqrt{a^{2} - b^{2}} b \cos \left (d x^{n} + c\right ) + a b\right )} \sin \left (d x^{n} + c\right )}{a^{2} \cos \left (d x^{n} + c\right )^{2} - 2 \, a b \sin \left (d x^{n} + c\right ) - a^{2} - b^{2}}\right )}{2 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d n \sin \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} \sin \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} \cos \left (d x^{n} + c\right ) +{\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt{-a^{2} + b^{2}} e^{n - 1} \sin \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 \sin \left (d x^{n} + c\right ) + \sqrt{-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \cos \left (d x^{n} + c\right )}\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d n \sin \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*csc(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*sin(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)*cos(d*x^n + c) + ((2*a^3*b - a*b^3)*sqrt(a^2 - b^2)*e^(n - 1)*sin(d*x^n + c)
+ (2*a^2*b^2 - b^4)*sqrt(a^2 - b^2)*e^(n - 1))*log(((a^2 - 2*b^2)*cos(d*x^n + c)^2 + 2*sqrt(a^2 - b^2)*a*cos(d
*x^n + c) + a^2 + b^2 + 2*(sqrt(a^2 - b^2)*b*cos(d*x^n + c) + a*b)*sin(d*x^n + c))/(a^2*cos(d*x^n + c)^2 - 2*a
*b*sin(d*x^n + c) - a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*n*sin(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*sin(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)*cos(d*x^n + c) + ((2*a^3*b - a*b^3)*sqrt(-a^2 + b^2)*e^(n - 1)*sin(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*sin(d*x^n + c) + sqrt(-a^2 + b^2)*a
)/((a^2 - b^2)*cos(d*x^n + c))))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*n*sin(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*csc(c+d*x**n))**2,x)

[Out]

Timed out

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

[Out]

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