3.3.76 \(\int \frac {1}{\sec ^{\frac {5}{2}}(a+b \log (c x^n))} \, dx\) [276]

Optimal. Leaf size=110 \[ \frac {2 x \, _2F_1\left (-\frac {5}{2},\frac {1}{4} \left (-5-\frac {2 i}{b n}\right );-\frac {2 i+b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-5 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 110, 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 = {4599, 4603, 371} \begin {gather*} \frac {2 x \, _2F_1\left (-\frac {5}{2},\frac {1}{4} \left (-5-\frac {2 i}{b n}\right );-\frac {b n+2 i}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-5 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*Log[c*x^n]]^(-5/2),x]

[Out]

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

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 4599

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 4603

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(861\) vs. \(2(110)=220\).
time = 8.84, size = 861, normalized size = 7.83 \begin {gather*} \frac {30 i \sqrt {2} b^3 e^{-i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} n^3 x^{1-i b n} \sqrt {\frac {e^{i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{i b n}}{1+e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}}} \left ((2 i+b n) \left (1+e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}\right )+\left (-2 i-b n+e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} (-2 i+b n)\right ) \sqrt {1+e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}} \, _2F_1\left (\frac {1}{2},-\frac {2 i+b n}{4 b n};\frac {3}{4}-\frac {i}{2 b n};-e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}\right )\right )}{(-2 i+5 b n) (2 i+5 b n) \left (4+b^2 n^2\right ) \left (-2 i-b n+e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} (-2 i+b n)\right )}+\sqrt {\sec \left (a+b n \log (x)+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} \left (-\frac {x \cos (b n \log (x)) \left (12+55 b^2 n^2+12 \cos \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+65 b^2 n^2 \cos \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+4 b n \sin \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{4 (-2 i+5 b n) (2 i+5 b n) \left (-2 \cos \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+b n \sin \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}+\frac {x \sin (b n \log (x)) \left (-16 b n-4 b n \cos \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+12 \sin \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+65 b^2 n^2 \sin \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{4 (-2 i+5 b n) (2 i+5 b n) \left (-2 \cos \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+b n \sin \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}+\frac {x \sin (3 b n \log (x)) \left (5 b n \cos \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )-2 \sin \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{2 (-2 i+5 b n) (2 i+5 b n)}+\frac {x \cos (3 b n \log (x)) \left (2 \cos \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+5 b n \sin \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{2 (-2 i+5 b n) (2 i+5 b n)}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[a + b*Log[c*x^n]]^(-5/2),x]

[Out]

((30*I)*Sqrt[2]*b^3*n^3*x^(1 - I*b*n)*Sqrt[(E^(I*(a + b*(-(n*Log[x]) + Log[c*x^n])))*x^(I*b*n))/(1 + E^((2*I)*
(a + b*(-(n*Log[x]) + Log[c*x^n])))*x^((2*I)*b*n))]*((2*I + b*n)*(1 + E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n
])))*x^((2*I)*b*n)) + (-2*I - b*n + E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n])))*(-2*I + b*n))*Sqrt[1 + E^((2*
I)*(a + b*(-(n*Log[x]) + Log[c*x^n])))*x^((2*I)*b*n)]*Hypergeometric2F1[1/2, -1/4*(2*I + b*n)/(b*n), 3/4 - (I/
2)/(b*n), -(E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n])))*x^((2*I)*b*n))]))/(E^(I*(a + b*(-(n*Log[x]) + Log[c*x
^n])))*(-2*I + 5*b*n)*(2*I + 5*b*n)*(4 + b^2*n^2)*(-2*I - b*n + E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n])))*(
-2*I + b*n))) + Sqrt[Sec[a + b*n*Log[x] + b*(-(n*Log[x]) + Log[c*x^n])]]*(-1/4*(x*Cos[b*n*Log[x]]*(12 + 55*b^2
*n^2 + 12*Cos[2*(a + b*(-(n*Log[x]) + Log[c*x^n]))] + 65*b^2*n^2*Cos[2*(a + b*(-(n*Log[x]) + Log[c*x^n]))] + 4
*b*n*Sin[2*(a + b*(-(n*Log[x]) + Log[c*x^n]))]))/((-2*I + 5*b*n)*(2*I + 5*b*n)*(-2*Cos[a + b*(-(n*Log[x]) + Lo
g[c*x^n])] + b*n*Sin[a + b*(-(n*Log[x]) + Log[c*x^n])])) + (x*Sin[b*n*Log[x]]*(-16*b*n - 4*b*n*Cos[2*(a + b*(-
(n*Log[x]) + Log[c*x^n]))] + 12*Sin[2*(a + b*(-(n*Log[x]) + Log[c*x^n]))] + 65*b^2*n^2*Sin[2*(a + b*(-(n*Log[x
]) + Log[c*x^n]))]))/(4*(-2*I + 5*b*n)*(2*I + 5*b*n)*(-2*Cos[a + b*(-(n*Log[x]) + Log[c*x^n])] + b*n*Sin[a + b
*(-(n*Log[x]) + Log[c*x^n])])) + (x*Sin[3*b*n*Log[x]]*(5*b*n*Cos[3*(a + b*(-(n*Log[x]) + Log[c*x^n]))] - 2*Sin
[3*(a + b*(-(n*Log[x]) + Log[c*x^n]))]))/(2*(-2*I + 5*b*n)*(2*I + 5*b*n)) + (x*Cos[3*b*n*Log[x]]*(2*Cos[3*(a +
b*(-(n*Log[x]) + Log[c*x^n]))] + 5*b*n*Sin[3*(a + b*(-(n*Log[x]) + Log[c*x^n]))]))/(2*(-2*I + 5*b*n)*(2*I + 5
*b*n)))

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {1}{\sec \left (a +b \ln \left (c \,x^{n}\right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(a+b*ln(c*x^n))^(5/2),x)

[Out]

int(1/sec(a+b*ln(c*x^n))^(5/2),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(1/sec(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(b*log(c*x^n) + a)^(-5/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(a+b*ln(c*x**n))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3006 deep

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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(1/sec(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(b*log(c*x^n) + a)^(-5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/cos(a + b*log(c*x^n)))^(5/2),x)

[Out]

int(1/(1/cos(a + b*log(c*x^n)))^(5/2), x)

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