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

Optimal. Leaf size=103 \[ \frac{2 e^{i a} x^{m+1} \left (c x^n\right )^{i b} \text{Hypergeometric2F1}\left (1,-\frac{-b n+i m+i}{2 b n},-\frac{-3 b n+i (m+1)}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{i b n+m+1} \]

[Out]

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

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Rubi [A]  time = 0.0686398, antiderivative size = 99, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4509, 4505, 364} \[ \frac{2 e^{i a} x^{m+1} \left (c x^n\right )^{i b} \, _2F_1\left (1,\frac{1}{2} \left (1-\frac{i (m+1)}{b n}\right );-\frac{i (m+1)-3 b n}{2 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{i b n+m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sec[a + b*Log[c*x^n]],x]

[Out]

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

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])

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 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])

Rubi steps

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

Mathematica [A]  time = 0.212423, size = 94, normalized size = 0.91 \[ \frac{2 e^{i a} x^{m+1} \left (c x^n\right )^{i b} \text{Hypergeometric2F1}\left (1,\frac{1}{2}-\frac{i (m+1)}{2 b n},\frac{3}{2}-\frac{i (m+1)}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{i b n+m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(I*a)*x^(1 + m)*(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]))])/(1 + m + I*b*n)

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Maple [F]  time = 0.349, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}\sec \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sec{\left (a + b \log{\left (c x^{n} \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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