3.252 \(\int \frac{\sec ^3(a+b \log (c x^n))}{x^2} \, dx\)

Optimal. Leaf size=87 \[ -\frac{8 e^{3 i a} \left (c x^n\right )^{3 i b} \text{Hypergeometric2F1}\left (3,\frac{1}{2} \left (3+\frac{i}{b n}\right ),\frac{1}{2} \left (5+\frac{i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{x (1-3 i b n)} \]

[Out]

(-8*E^((3*I)*a)*(c*x^n)^((3*I)*b)*Hypergeometric2F1[3, (3 + I/(b*n))/2, (5 + I/(b*n))/2, -(E^((2*I)*a)*(c*x^n)
^((2*I)*b))])/((1 - (3*I)*b*n)*x)

________________________________________________________________________________________

Rubi [A]  time = 0.0745732, 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} \[ -\frac{8 e^{3 i a} \left (c x^n\right )^{3 i b} \, _2F_1\left (3,\frac{1}{2} \left (3+\frac{i}{b n}\right );\frac{1}{2} \left (5+\frac{i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{x (1-3 i b n)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*Log[c*x^n]]^3/x^2,x]

[Out]

(-8*E^((3*I)*a)*(c*x^n)^((3*I)*b)*Hypergeometric2F1[3, (3 + I/(b*n))/2, (5 + I/(b*n))/2, -(E^((2*I)*a)*(c*x^n)
^((2*I)*b))])/((1 - (3*I)*b*n)*x)

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 \frac{\sec ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx &=\frac{\left (c x^n\right )^{\frac{1}{n}} \operatorname{Subst}\left (\int x^{-1-\frac{1}{n}} \sec ^3(a+b \log (x)) \, dx,x,c x^n\right )}{n x}\\ &=\frac{\left (8 e^{3 i a} \left (c x^n\right )^{\frac{1}{n}}\right ) \operatorname{Subst}\left (\int \frac{x^{-1+3 i b-\frac{1}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^3} \, dx,x,c x^n\right )}{n x}\\ &=-\frac{8 e^{3 i a} \left (c x^n\right )^{3 i b} \, _2F_1\left (3,\frac{1}{2} \left (3+\frac{i}{b n}\right );\frac{1}{2} \left (5+\frac{i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(1-3 i b n) x}\\ \end{align*}

Mathematica [A]  time = 4.6288, size = 123, normalized size = 1.41 \[ \frac{\left (b n \tan \left (a+b \log \left (c x^n\right )\right )+1\right ) \sec \left (a+b \log \left (c x^n\right )\right )-2 i e^{i a} (b n-i) \left (c x^n\right )^{i b} \text{Hypergeometric2F1}\left (1,\frac{1}{2}+\frac{i}{2 b n},\frac{3}{2}+\frac{i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{2 b^2 n^2 x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[a + b*Log[c*x^n]]^3/x^2,x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 2.144, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \sec \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{3}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(a+b*ln(c*x^n))^3/x^2,x)

[Out]

int(sec(a+b*ln(c*x^n))^3/x^2,x)

________________________________________________________________________________________

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(sec(a+b*log(c*x^n))^3/x^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sec \left (b \log \left (c x^{n}\right ) + a\right )^{3}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*log(c*x^n))^3/x^2,x, algorithm="fricas")

[Out]

integral(sec(b*log(c*x^n) + a)^3/x^2, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (a + b \log{\left (c x^{n} \right )} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*ln(c*x**n))**3/x**2,x)

[Out]

Integral(sec(a + b*log(c*x**n))**3/x**2, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (b \log \left (c x^{n}\right ) + a\right )^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*log(c*x^n))^3/x^2,x, algorithm="giac")

[Out]

integrate(sec(b*log(c*x^n) + a)^3/x^2, x)