3.3.41 \(\int \frac {\sec (a+b \log (c x^n))}{x^2} \, dx\) [241]

Optimal. Leaf size=87 \[ -\frac {2 e^{i a} \left (c x^n\right )^{i b} \, _2F_1\left (1,\frac {1}{2} \left (1+\frac {i}{b n}\right );\frac {1}{2} \left (3+\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(1-i b n) x} \]

[Out]

-2*exp(I*a)*(c*x^n)^(I*b)*hypergeom([1, 1/2+1/2*I/b/n],[3/2+1/2*I/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(1-I*b*n)/
x

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Rubi [A]
time = 0.04, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4605, 4601, 371} \begin {gather*} -\frac {2 e^{i a} \left (c x^n\right )^{i b} \, _2F_1\left (1,\frac {1}{2} \left (1+\frac {i}{b n}\right );\frac {1}{2} \left (3+\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{x (1-i b n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*Log[c*x^n]]/x^2,x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 4601

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 4605

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

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Mathematica [A]
time = 0.73, size = 85, normalized size = 0.98 \begin {gather*} \frac {2 e^{i a} \left (c x^n\right )^{i b} \, _2F_1\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 )}{(-1+i b n) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(I*a)*(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
]))])/((-1 + I*b*n)*x)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*cos(a + b*log(c*x^n))),x)

[Out]

int(1/(x^2*cos(a + b*log(c*x^n))), x)

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