\(\int x^m \sec (a+b \log (c x^n)) \, dx\) [280]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 15, antiderivative size = 103 \[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 e^{i a} x^{1+m} \left (c x^n\right )^{i b} \operatorname {Hypergeometric2F1}\left (1,-\frac {i+i m-b n}{2 b n},-\frac {i (1+m)-3 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+m+i b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4605, 4601, 371} \[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 e^{i a} x^{m+1} \left (c x^n\right )^{i b} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (1-\frac {i (m+1)}{b n}\right ),-\frac {i (m+1)-3 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{i b n+m+1} \]

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

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 4601

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 4605

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*} \text {integral}& = \frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {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 ) \text {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} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 1.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 e^{i a} x^{1+m} \left (c x^n\right )^{i b} \operatorname {Hypergeometric2F1}\left (1,\frac {-i-i m+b n}{2 b n},-\frac {i (1+m+3 i b n)}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{1+m+i b n} \]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right ) \,d x } \]

[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)

Sympy [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{m} \sec {\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right ) \,d x } \]

[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)

Giac [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right ) \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {x^m}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )} \,d x \]

[In]

int(x^m/cos(a + b*log(c*x^n)),x)

[Out]

int(x^m/cos(a + b*log(c*x^n)), x)