Integrand size = 19, antiderivative size = 130 \[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 x^{1+m} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {2 i+2 i m-3 b n}{4 b n},-\frac {2 i+2 i m-7 b n}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+2 m+3 i b n} \] Output:
2*x^(1+m)*(1+exp(2*I*a)*(c*x^n)^(2*I*b))^(3/2)*hypergeom([3/2, -1/4*(2*I+2 *I*m-3*b*n)/b/n],[-1/4*(2*I+2*I*m-7*b*n)/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b)) *sec(a+b*ln(c*x^n))^(3/2)/(2+2*m+3*I*b*n)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(470\) vs. \(2(130)=260\).
Time = 7.53 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.62 \[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {\sqrt {2} x^{1+m-i b n} \left (-\left (\left (4+8 m+4 m^2+b^2 n^2\right ) x^{2 i b n} \sqrt {\frac {e^{i a} \left (c x^n\right )^{i b}}{1+e^{2 i a} \left (c x^n\right )^{2 i b}}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {i \left (1+m+\frac {3 i b n}{2}\right )}{2 b n},-\frac {2 i+2 i m-7 b n}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )\right )+(2+2 m+3 i b n) \left ((2+2 m+i b n) \sqrt {\frac {e^{i a} \left (c x^n\right )^{i b}}{1+e^{2 i a} \left (c x^n\right )^{2 i b}}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2 i+2 i m+b n}{4 b n},-\frac {2 i+2 i m-3 b n}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )-i \sqrt {2} x^{i b n} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} (b n \cos (b n \log (x))-2 (1+m) \sin (b n \log (x)))\right )\right )}{b n (-2 i-2 i m+3 b n) \left (-2 (1+m) \cos \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+b n \sin \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )} \] Input:
Integrate[x^m*Sec[a + b*Log[c*x^n]]^(3/2),x]
Output:
(Sqrt[2]*x^(1 + m - I*b*n)*(-((4 + 8*m + 4*m^2 + b^2*n^2)*x^((2*I)*b*n)*Sq rt[(E^(I*a)*(c*x^n)^(I*b))/(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Sqrt[1 + E ^((2*I)*a)*(c*x^n)^((2*I)*b)]*Hypergeometric2F1[1/2, ((-1/2*I)*(1 + m + (( 3*I)/2)*b*n))/(b*n), -1/4*(2*I + (2*I)*m - 7*b*n)/(b*n), -(E^((2*I)*a)*(c* x^n)^((2*I)*b))]) + (2 + 2*m + (3*I)*b*n)*((2 + 2*m + I*b*n)*Sqrt[(E^(I*a) *(c*x^n)^(I*b))/(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Sqrt[1 + E^((2*I)*a)* (c*x^n)^((2*I)*b)]*Hypergeometric2F1[1/2, -1/4*(2*I + (2*I)*m + b*n)/(b*n) , -1/4*(2*I + (2*I)*m - 3*b*n)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))] - I*Sqrt[2]*x^(I*b*n)*Sqrt[Sec[a + b*Log[c*x^n]]]*(b*n*Cos[b*n*Log[x]] - 2*( 1 + m)*Sin[b*n*Log[x]]))))/(b*n*(-2*I - (2*I)*m + 3*b*n)*(-2*(1 + m)*Cos[a - b*n*Log[x] + b*Log[c*x^n]] + b*n*Sin[a - b*n*Log[x] + b*Log[c*x^n]]))
Time = 0.34 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {5020, 5018, 888}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 5020 |
\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 5018 |
\(\displaystyle \frac {x^{m+1} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \left (c x^n\right )^{-\frac {m+1}{n}-\frac {3 i b}{2}} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \int \frac {\left (c x^n\right )^{\frac {3 i b}{2}+\frac {m+1}{n}-1}}{\left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^{3/2}}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 x^{m+1} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{4} \left (3-\frac {2 i (m+1)}{b n}\right ),-\frac {2 i m-7 b n+2 i}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 i b n+2 m+2}\) |
Input:
Int[x^m*Sec[a + b*Log[c*x^n]]^(3/2),x]
Output:
(2*x^(1 + m)*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Hypergeometric2F1[3 /2, (3 - ((2*I)*(1 + m))/(b*n))/4, -1/4*(2*I + (2*I)*m - 7*b*n)/(b*n), -(E ^((2*I)*a)*(c*x^n)^((2*I)*b))]*Sec[a + b*Log[c*x^n]]^(3/2))/(2 + 2*m + (3* I)*b*n)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[Sec[d*(a + b*Log[x])]^p*((1 + E^(2*I*a*d)*x^(2*I*b*d))^p/x^(I*b*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] /; F reeQ[{a, b, d, e, m, p}, x] && !IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ .), x_Symbol] :> Simp[(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])
\[\int x^{m} {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}d x\]
Input:
int(x^m*sec(a+b*ln(c*x^n))^(3/2),x)
Output:
int(x^m*sec(a+b*ln(c*x^n))^(3/2),x)
Exception generated. \[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^m*sec(a+b*log(c*x^n))^(3/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
Timed out. \[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x**m*sec(a+b*ln(c*x**n))**(3/2),x)
Output:
Timed out
\[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(x^m*sec(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")
Output:
integrate(x^m*sec(b*log(c*x^n) + a)^(3/2), x)
Timed out. \[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x^m*sec(a+b*log(c*x^n))^(3/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^m\,{\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{3/2} \,d x \] Input:
int(x^m*(1/cos(a + b*log(c*x^n)))^(3/2),x)
Output:
int(x^m*(1/cos(a + b*log(c*x^n)))^(3/2), x)
\[ \int x^m \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 x^{m} \sqrt {\sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}\, \sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) x -3 \left (\int x^{m} \sqrt {\sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}\, \sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) \tan \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )d x \right ) b n}{2 m +2} \] Input:
int(x^m*sec(a+b*log(c*x^n))^(3/2),x)
Output:
(2*x**m*sqrt(sec(log(x**n*c)*b + a))*sec(log(x**n*c)*b + a)*x - 3*int(x**m *sqrt(sec(log(x**n*c)*b + a))*sec(log(x**n*c)*b + a)*tan(log(x**n*c)*b + a ),x)*b*n)/(2*(m + 1))