3.3.74 \(\int \frac {1}{\sec ^{\frac {3}{2}}(a+b \log (c x^n))} \, dx\) [274]

Optimal. Leaf size=109 \[ \frac {2 x \, _2F_1\left (-\frac {3}{2},\frac {1}{4} \left (-3-\frac {2 i}{b n}\right );\frac {1}{4} \left (1-\frac {2 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-3 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[Out]

2*x*hypergeom([-3/2, -3/4-1/2*I/b/n],[1/4-1/2*I/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(2-3*I*b*n)/(1+exp(2*I*a)*(c
*x^n)^(2*I*b))^(3/2)/sec(a+b*ln(c*x^n))^(3/2)

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Rubi [A]
time = 0.05, 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 = {4599, 4603, 371} \begin {gather*} \frac {2 x \, _2F_1\left (-\frac {3}{2},\frac {1}{4} \left (-3-\frac {2 i}{b n}\right );\frac {1}{4} \left (1-\frac {2 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-3 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{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]]^(-3/2),x]

[Out]

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

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 {3}{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 {3}{2}}(a+b \log (x))} \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (x \left (c x^n\right )^{\frac {3 i b}{2}-\frac {1}{n}}\right ) \text {Subst}\left (\int x^{-1-\frac {3 i b}{2}+\frac {1}{n}} \left (1+e^{2 i a} x^{2 i b}\right )^{3/2} \, dx,x,c x^n\right )}{n \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ &=\frac {2 x \, _2F_1\left (-\frac {3}{2},\frac {1}{4} \left (-3-\frac {2 i}{b n}\right );\frac {1}{4} \left (1-\frac {2 i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-3 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ \end {align*}

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Mathematica [A]
time = 1.52, size = 168, normalized size = 1.54 \begin {gather*} \frac {2 x \left (3 b^2 n^2 \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\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 ) \sec ^2\left (a+b \log \left (c x^n\right )\right )+(2+i b n) \left (2+3 b n \tan \left (a+b \log \left (c x^n\right )\right )\right )\right )}{(2+3 i b n) (-2 i+b n) (2 i+3 b n) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {1}{\sec \left (a +b \ln \left (c \,x^{n}\right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(a+b*ln(c*x^n))^(3/2),x)

[Out]

int(1/sec(a+b*ln(c*x^n))^(3/2),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))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(b*log(c*x^n) + a)^(-3/2), 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))^(3/2),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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sec ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(a+b*ln(c*x**n))**(3/2),x)

[Out]

Integral(sec(a + b*log(c*x**n))**(-3/2), 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(1/sec(a+b*log(c*x^n))^(3/2),x, algorithm="giac")

[Out]

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

[Out]

int(1/(1/cos(a + b*log(c*x^n)))^(3/2), x)

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