3.250 \(\int \sec ^3(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=85 \[ \frac {8 e^{3 i a} x \left (c x^n\right )^{3 i b} \, _2F_1\left (3,\frac {1}{2} \left (3-\frac {i}{b n}\right );\frac {1}{2} \left (5-\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+3 i b n} \]

[Out]

8*exp(3*I*a)*x*(c*x^n)^(3*I*b)*hypergeom([3, 3/2-1/2*I/b/n],[5/2-1/2*I/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(1+3*
I*b*n)

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Rubi [A]  time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4503, 4505, 364} \[ \frac {8 e^{3 i a} x \left (c x^n\right )^{3 i b} \, _2F_1\left (3,\frac {1}{2} \left (3-\frac {i}{b n}\right );\frac {1}{2} \left (5-\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+3 i b n} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*Log[c*x^n]]^3,x]

[Out]

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

Rubi steps

\begin {align*} \int \sec ^3\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}} \sec ^3(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (8 e^{3 i a} x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+3 i b+\frac {1}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^3} \, dx,x,c x^n\right )}{n}\\ &=\frac {8 e^{3 i a} x \left (c x^n\right )^{3 i b} \, _2F_1\left (3,\frac {1}{2} \left (3-\frac {i}{b n}\right );\frac {1}{2} \left (5-\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+3 i b n}\\ \end {align*}

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Mathematica [A]  time = 4.41, size = 120, normalized size = 1.41 \[ \frac {x \left (\left (b n \tan \left (a+b \log \left (c x^n\right )\right )-1\right ) \sec \left (a+b \log \left (c x^n\right )\right )+2 e^{i a} (1-i b n) \left (c x^n\right )^{i b} \, _2F_1\left (1,\frac {1}{2}-\frac {i}{2 b n};\frac {3}{2}-\frac {i}{2 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{2 b^2 n^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[a + b*Log[c*x^n]]^3,x]

[Out]

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

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sec \left (b \log \left (c x^{n}\right ) + a\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

integral(sec(b*log(c*x^n) + a)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec \left (b \log \left (c x^{n}\right ) + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

integrate(sec(b*log(c*x^n) + a)^3, x)

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \sec ^{3}\left (a +b \ln \left (c \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(a + b*log(c*x^n))^3,x)

[Out]

int(1/cos(a + b*log(c*x^n))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(a + b*log(c*x**n))**3, x)

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