3.266 \(\int \sqrt {\sec (a+b \log (c x^n))} \, dx\)

Optimal. Leaf size=109 \[ \frac {2 x \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \, _2F_1\left (\frac {1}{2},\frac {1}{4} \left (1-\frac {2 i}{b n}\right );\frac {1}{4} \left (5-\frac {2 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}{2+i b n} \]

[Out]

2*x*hypergeom([1/2, 1/4-1/2*I/b/n],[5/4-1/2*I/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))*(1+exp(2*I*a)*(c*x^n)^(2*I*b))
^(1/2)*sec(a+b*ln(c*x^n))^(1/2)/(2+I*b*n)

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Rubi [A]  time = 0.07, antiderivative size = 109, 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 = {4503, 4507, 364} \[ \frac {2 x \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \, _2F_1\left (\frac {1}{2},\frac {1}{4} \left (1-\frac {2 i}{b n}\right );\frac {1}{4} \left (5-\frac {2 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}{2+i b n} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[a + b*Log[c*x^n]]],x]

[Out]

(2*x*Sqrt[1 + E^((2*I)*a)*(c*x^n)^((2*I)*b)]*Hypergeometric2F1[1/2, (1 - (2*I)/(b*n))/4, (5 - (2*I)/(b*n))/4,
-(E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Sqrt[Sec[a + b*Log[c*x^n]]])/(2 + I*b*n)

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 4503

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

Rule 4507

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

Rubi steps

\begin {align*} \int \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} \, dx &=\frac {\left (x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int x^{-1+\frac {1}{n}} \sqrt {\sec (a+b \log (x))} \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (x \left (c x^n\right )^{-\frac {i b}{2}-\frac {1}{n}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+\frac {i b}{2}+\frac {1}{n}}}{\sqrt {1+e^{2 i a} x^{2 i b}}} \, dx,x,c x^n\right )}{n}\\ &=\frac {2 x \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \, _2F_1\left (\frac {1}{2},\frac {1}{4} \left (1-\frac {2 i}{b n}\right );\frac {1}{4} \left (5-\frac {2 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}{2+i b n}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 99, normalized size = 0.91 \[ -\frac {2 i x \left (1+e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right ) \, _2F_1\left (1,\frac {3}{4}-\frac {i}{2 b n};\frac {5}{4}-\frac {i}{2 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right ) \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}{b n-2 i} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*I)*(1 + E^((2*I)*(a + b*Log[c*x^n])))*x*Hypergeometric2F1[1, 3/4 - (I/2)/(b*n), 5/4 - (I/2)/(b*n), -E^((2
*I)*(a + b*Log[c*x^n]))]*Sqrt[Sec[a + b*Log[c*x^n]]])/(-2*I + b*n)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*log(c*x^n))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*log(c*x^n))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \sqrt {\sec }\left (a +b \ln \left (c \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(a+b*ln(c*x^n))^(1/2),x)

[Out]

int(sec(a+b*ln(c*x^n))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*log(c*x^n))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*ln(c*x**n))**(1/2),x)

[Out]

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

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