3.243 \(\int x^2 \sec ^2(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=87 \[ \frac{4 e^{2 i a} x^3 \left (c x^n\right )^{2 i b} \text{Hypergeometric2F1}\left (2,\frac{1}{2} \left (2-\frac{3 i}{b n}\right ),\frac{1}{2} \left (4-\frac{3 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3+2 i b n} \]

[Out]

(4*E^((2*I)*a)*x^3*(c*x^n)^((2*I)*b)*Hypergeometric2F1[2, (2 - (3*I)/(b*n))/2, (4 - (3*I)/(b*n))/2, -(E^((2*I)
*a)*(c*x^n)^((2*I)*b))])/(3 + (2*I)*b*n)

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Rubi [A]  time = 0.075054, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {4509, 4505, 364} \[ \frac{4 e^{2 i a} x^3 \left (c x^n\right )^{2 i b} \, _2F_1\left (2,\frac{1}{2} \left (2-\frac{3 i}{b n}\right );\frac{1}{2} \left (4-\frac{3 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3+2 i b n} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sec[a + b*Log[c*x^n]]^2,x]

[Out]

(4*E^((2*I)*a)*x^3*(c*x^n)^((2*I)*b)*Hypergeometric2F1[2, (2 - (3*I)/(b*n))/2, (4 - (3*I)/(b*n))/2, -(E^((2*I)
*a)*(c*x^n)^((2*I)*b))])/(3 + (2*I)*b*n)

Rule 4509

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Sec[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a
, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

Rule 4505

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[2^p*E^(I*a*d*p), Int[((e*x)
^m*x^(I*b*d*p))/(1 + E^(2*I*a*d)*x^(2*I*b*d))^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{\left (x^3 \left (c x^n\right )^{-3/n}\right ) \operatorname{Subst}\left (\int x^{-1+\frac{3}{n}} \sec ^2(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac{\left (4 e^{2 i a} x^3 \left (c x^n\right )^{-3/n}\right ) \operatorname{Subst}\left (\int \frac{x^{-1+2 i b+\frac{3}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^2} \, dx,x,c x^n\right )}{n}\\ &=\frac{4 e^{2 i a} x^3 \left (c x^n\right )^{2 i b} \, _2F_1\left (2,\frac{1}{2} \left (2-\frac{3 i}{b n}\right );\frac{1}{2} \left (4-\frac{3 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3+2 i b n}\\ \end{align*}

Mathematica [A]  time = 5.53073, size = 160, normalized size = 1.84 \[ \frac{x^3 \left (3 e^{2 i a} \left (c x^n\right )^{2 i b} \text{Hypergeometric2F1}\left (1,1-\frac{3 i}{2 b n},2-\frac{3 i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(2 b n-3 i) \left (\tan \left (a+b \log \left (c x^n\right )\right )-i \text{Hypergeometric2F1}\left (1,-\frac{3 i}{2 b n},1-\frac{3 i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )\right )\right )}{b n (2 b n-3 i)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*Sec[a + b*Log[c*x^n]]^2,x]

[Out]

(x^3*(3*E^((2*I)*a)*(c*x^n)^((2*I)*b)*Hypergeometric2F1[1, 1 - ((3*I)/2)/(b*n), 2 - ((3*I)/2)/(b*n), -E^((2*I)
*(a + b*Log[c*x^n]))] + (-3*I + 2*b*n)*((-I)*Hypergeometric2F1[1, ((-3*I)/2)/(b*n), 1 - ((3*I)/2)/(b*n), -E^((
2*I)*(a + b*Log[c*x^n]))] + Tan[a + b*Log[c*x^n]])))/(b*n*(-3*I + 2*b*n))

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Maple [F]  time = 1.485, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( \sec \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sec(a+b*ln(c*x^n))^2,x)

[Out]

int(x^2*sec(a+b*ln(c*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(x^2*sec(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sec(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

integral(x^2*sec(b*log(c*x^n) + a)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sec ^{2}{\left (a + b \log{\left (c x^{n} \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*sec(a+b*ln(c*x**n))**2,x)

[Out]

Integral(x**2*sec(a + b*log(c*x**n))**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sec(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

integrate(x^2*sec(b*log(c*x^n) + a)^2, x)