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

Optimal. Leaf size=87 \[ -\frac {4 e^{2 i a} \left (c x^n\right )^{2 i b} \, _2F_1\left (2,\frac {1}{2} \left (2+\frac {i}{b n}\right );\frac {1}{2} \left (4+\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{x (1-2 i b n)} \]

[Out]

-4*exp(2*I*a)*(c*x^n)^(2*I*b)*hypergeom([2, 1+1/2*I/b/n],[2+1/2*I/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(1-2*I*b*n
)/x

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, 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} \[ -\frac {4 e^{2 i a} \left (c x^n\right )^{2 i b} \, _2F_1\left (2,\frac {1}{2} \left (2+\frac {i}{b n}\right );\frac {1}{2} \left (4+\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{x (1-2 i b n)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*Log[c*x^n]]^2/x^2,x]

[Out]

(-4*E^((2*I)*a)*(c*x^n)^((2*I)*b)*Hypergeometric2F1[2, (2 + I/(b*n))/2, (4 + I/(b*n))/2, -(E^((2*I)*a)*(c*x^n)
^((2*I)*b))])/((1 - (2*I)*b*n)*x)

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])

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 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])

Rubi steps

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

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Mathematica [A]  time = 3.74, size = 160, normalized size = 1.84 \[ \frac {(1-2 i b n) \left (\, _2F_1\left (1,\frac {i}{2 b n};1+\frac {i}{2 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+i \tan \left (a+b \log \left (c x^n\right )\right )\right )-e^{2 i a} \left (c x^n\right )^{2 i b} \, _2F_1\left (1,1+\frac {i}{2 b n};2+\frac {i}{2 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{b n x (2 b n+i)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{2}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{2}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.33, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*cos(a + b*log(c*x^n))^2),x)

[Out]

int(1/(x^2*cos(a + b*log(c*x^n))^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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