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

Optimal. Leaf size=79 \[ \frac{a^2 (e x)^n}{e n}+\frac{2 a b x^{-n} (e x)^n \tanh ^{-1}\left (\sin \left (c+d x^n\right )\right )}{d e n}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{d e n} \]

[Out]

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

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Rubi [A]  time = 0.0901147, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4208, 4204, 3773, 3770, 3767, 8} \[ \frac{a^2 (e x)^n}{e n}+\frac{2 a b x^{-n} (e x)^n \tanh ^{-1}\left (\sin \left (c+d x^n\right )\right )}{d e n}+\frac{b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(e*x)^n)/(e*n) + (2*a*b*(e*x)^n*ArcTanh[Sin[c + d*x^n]])/(d*e*n*x^n) + (b^2*(e*x)^n*Tan[c + d*x^n])/(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 3773

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.365918, size = 54, normalized size = 0.68 \[ \frac{x^{-n} (e x)^n \left (a^2 d x^n+2 a b \tanh ^{-1}\left (\sin \left (c+d x^n\right )\right )+b^2 \tan \left (c+d x^n\right )\right )}{d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.18, size = 276, normalized size = 3.5 \begin{align*}{\frac{{a}^{2}x}{n}{{\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 ( x \right ) -2\,\ln \left ( e \right ) \right ) }{2}}}}}+{\frac{2\,ix{b}^{2}}{dn{x}^{n} \left ( 1+{{\rm e}^{2\,i \left ( c+d{x}^{n} \right ) }} \right ) }{{\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 ( x \right ) -2\,\ln \left ( e \right ) \right ) }{2}}}}}-{\frac{4\,iab{e}^{n}\arctan \left ({{\rm e}^{i \left ( c+d{x}^{n} \right ) }} \right ){{\rm e}^{-{\frac{i}{2}}\pi \,{\it csgn} \left ( iex \right ) \left ( -1+n \right ) \left ( -{\it csgn} \left ( iex \right ) +{\it csgn} \left ( ix \right ) \right ) \left ( -{\it csgn} \left ( iex \right ) +{\it csgn} \left ( ie \right ) \right ) }}}{ned}} \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]

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*x*b^2*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)))/d/n/(x^n)/(1+e
xp(2*I*(c+d*x^n)))-4*I*a*b/n*e^n/e/d*arctan(exp(I*(c+d*x^n)))*exp(-1/2*I*Pi*csgn(I*e*x)*(-1+n)*(-csgn(I*e*x)+c
sgn(I*x))*(-csgn(I*e*x)+csgn(I*e)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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]

Exception raised: ValueError

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Fricas [A]  time = 1.77468, size = 271, normalized size = 3.43 \begin{align*} \frac{a^{2} d e^{n - 1} x^{n} \cos \left (d x^{n} + c\right ) + a b e^{n - 1} \cos \left (d x^{n} + c\right ) \log \left (\sin \left (d x^{n} + c\right ) + 1\right ) - a b e^{n - 1} \cos \left (d x^{n} + c\right ) \log \left (-\sin \left (d x^{n} + c\right ) + 1\right ) + b^{2} e^{n - 1} \sin \left (d x^{n} + c\right )}{d n \cos \left (d x^{n} + c\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]

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

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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

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