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

Optimal. Leaf size=102 \[ \frac {8 e^{3 i a} x^{m+1} \left (c x^n\right )^{3 i b} \, _2F_1\left (3,-\frac {i (m+1)-3 b n}{2 b n};-\frac {i (m+1)-5 b n}{2 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3 i b n+m+1} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4509, 4505, 364} \[ \frac {8 e^{3 i a} x^{m+1} \left (c x^n\right )^{3 i b} \, _2F_1\left (3,-\frac {i (m+1)-3 b n}{2 b n};-\frac {i (m+1)-5 b n}{2 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3 i b n+m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sec[a + b*Log[c*x^n]]^3,x]

[Out]

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

Rule 4509

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

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Mathematica [A]  time = 5.60, size = 134, normalized size = 1.31 \[ \frac {x^{m+1} \left (-2 \sec \left (a+b \log \left (c x^n\right )\right ) \left (-b n \tan \left (a+b \log \left (c x^n\right )\right )+m+1\right )+4 e^{i a} (-i b n+m+1) \left (c x^n\right )^{i b} \, _2F_1\left (1,\frac {1}{2}-\frac {i (m+1)}{2 b n};\frac {3}{2}-\frac {i (m+1)}{2 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{4 b^2 n^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m*Sec[a + b*Log[c*x^n]]^3,x]

[Out]

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

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fricas [F]  time = 1.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m*sec(b*log(c*x^n) + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m*sec(b*log(c*x^n) + a)^3, x)

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maple [F]  time = 1.75, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\sec ^{3}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^m*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(x^m*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 \[ \int \frac {x^m}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/cos(a + b*log(c*x^n))^3,x)

[Out]

int(x^m/cos(a + b*log(c*x^n))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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