\(\int x^m \sec ^2(a+b \log (c x^n)) \, dx\) [279]

Optimal result

Integrand size = 17, antiderivative size = 102 \[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 i a} x^{1+m} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,-\frac {i (1+m)-2 b n}{2 b n},-\frac {i (1+m)-4 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+m+2 i b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4605, 4601, 371} \[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 i a} x^{m+1} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,-\frac {i (m+1)-2 b n}{2 b n},-\frac {i (m+1)-4 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{2 i b n+m+1} \]

[In]

Int[x^m*Sec[a + b*Log[c*x^n]]^2,x]

[Out]

(4*E^((2*I)*a)*x^(1 + m)*(c*x^n)^((2*I)*b)*Hypergeometric2F1[2, -1/2*(I*(1 + m) - 2*b*n)/(b*n), -1/2*(I*(1 + m
) - 4*b*n)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/(1 + m + (2*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 ^2(a+b \log (x)) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (4 e^{2 i a} x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int \frac {x^{-1+2 i b+\frac {1+m}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^2} \, dx,x,c x^n\right )}{n} \\ & = \frac {4 e^{2 i a} x^{1+m} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,-\frac {i (1+m)-2 b n}{2 b n},-\frac {i (1+m)-4 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+m+2 i b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 15.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.94 \[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {i x^{1+m} \left ((1+m+2 i b n) \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m)}{2 b n},1-\frac {i (1+m)}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )-e^{2 i a} (1+m) \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m+2 i b n)}{2 b n},-\frac {i (1+m+4 i b n)}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+i (1+m+2 i b n) \tan \left (a+b \log \left (c x^n\right )\right )\right )}{b n (1+m+2 i b n)} \]

[In]

Integrate[x^m*Sec[a + b*Log[c*x^n]]^2,x]

[Out]

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

Maple [F]

\[\int x^{m} {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]

[In]

int(x^m*sec(a+b*ln(c*x^n))^2,x)

[Out]

int(x^m*sec(a+b*ln(c*x^n))^2,x)

Fricas [F]

\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]

[In]

integrate(x^m*sec(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

integral(x^m*sec(b*log(c*x^n) + a)^2, x)

Sympy [F]

\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{m} \sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

[In]

integrate(x**m*sec(a+b*ln(c*x**n))**2,x)

[Out]

Integral(x**m*sec(a + b*log(c*x**n))**2, x)

Maxima [F]

\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]

[In]

integrate(x^m*sec(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

2*(x*x^m*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x*x^m*cos(2*b*log(c))*sin(2*b*log(x^n) + 2*a) - ((b^2*cos(2
*b*log(c))^2 + b^2*sin(2*b*log(c))^2 + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*m)*n^2*cos(2*b*log(x^n)
+ 2*a)^2 + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2 + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*m
)*n^2*sin(2*b*log(x^n) + 2*a)^2 + 2*(b^2*m*cos(2*b*log(c)) + b^2*cos(2*b*log(c)))*n^2*cos(2*b*log(x^n) + 2*a)
- 2*(b^2*m*sin(2*b*log(c)) + b^2*sin(2*b*log(c)))*n^2*sin(2*b*log(x^n) + 2*a) + (b^2*m + b^2)*n^2)*integrate((
x^m*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x^m*cos(2*b*log(c))*sin(2*b*log(x^n) + 2*a))/(2*b^2*n^2*cos(2*b*
log(c))*cos(2*b*log(x^n) + 2*a) - 2*b^2*n^2*sin(2*b*log(c))*sin(2*b*log(x^n) + 2*a) + (b^2*cos(2*b*log(c))^2 +
b^2*sin(2*b*log(c))^2)*n^2*cos(2*b*log(x^n) + 2*a)^2 + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*si
n(2*b*log(x^n) + 2*a)^2 + b^2*n^2), x))/(2*b*n*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) + (b*cos(2*b*log(c))^2
+ b*sin(2*b*log(c))^2)*n*cos(2*b*log(x^n) + 2*a)^2 - 2*b*n*sin(2*b*log(c))*sin(2*b*log(x^n) + 2*a) + (b*cos(2*
b*log(c))^2 + b*sin(2*b*log(c))^2)*n*sin(2*b*log(x^n) + 2*a)^2 + b*n)

Giac [F]

\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]

[In]

integrate(x^m*sec(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

integrate(x^m*sec(b*log(c*x^n) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {x^m}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]

[In]

int(x^m/cos(a + b*log(c*x^n))^2,x)

[Out]

int(x^m/cos(a + b*log(c*x^n))^2, x)