Integrand size = 15, antiderivative size = 109 \[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 x \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{4} \left (3-\frac {2 i}{b n}\right ),\frac {1}{4} \left (7-\frac {2 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+3 i b n} \] Output:
2*x*(1+exp(2*I*a)*(c*x^n)^(2*I*b))^(3/2)*hypergeom([3/2, 3/4-1/2*I/b/n],[7 /4-1/2*I/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))*sec(a+b*ln(c*x^n))^(3/2)/(2+3*I *b*n)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(415\) vs. \(2(109)=218\).
Time = 4.34 (sec) , antiderivative size = 415, normalized size of antiderivative = 3.81 \[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {\sqrt {2} x^{1-i b n} \left (-\left (\left (4+b^2 n^2\right ) x^{2 i b n} \sqrt {\frac {e^{i a} \left (c x^n\right )^{i b}}{1+e^{2 i a} \left (c x^n\right )^{2 i b}}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4}-\frac {i}{2 b n},\frac {7}{4}-\frac {i}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )\right )+(-2 i+3 b n) \left ((2 i-b n) \sqrt {\frac {e^{i a} \left (c x^n\right )^{i b}}{1+e^{2 i a} \left (c x^n\right )^{2 i b}}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2 i+b n}{4 b n},\frac {3}{4}-\frac {i}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )+\sqrt {2} x^{i b n} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} (b n \cos (b n \log (x))-2 \sin (b n \log (x)))\right )\right )}{b n (-2 i+3 b n) \left (-2 \cos \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+b n \sin \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )} \] Input:
Integrate[Sec[a + b*Log[c*x^n]]^(3/2),x]
Output:
(Sqrt[2]*x^(1 - I*b*n)*(-((4 + b^2*n^2)*x^((2*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, 3/4 - (I/2)/(b*n), 7/4 - (I/2)/(b*n), - (E^((2*I)*a)*(c*x^n)^((2*I)*b))]) + (-2*I + 3*b*n)*((2*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 + b*n)/(b*n), 3/4 - (I/2)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))] + Sqrt[2]*x^(I*b*n)*Sqrt[S ec[a + b*Log[c*x^n]]]*(b*n*Cos[b*n*Log[x]] - 2*Sin[b*n*Log[x]]))))/(b*n*(- 2*I + 3*b*n)*(-2*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 = 0.31 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5014, 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.
\(\displaystyle \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 5014 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 5018 |
\(\displaystyle \frac {x \left (c x^n\right )^{-\frac {1}{n}-\frac {3 i b}{2}} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \int \frac {\left (c x^n\right )^{\frac {3 i b}{2}+\frac {1}{n}-1}}{\left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^{3/2}}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 x \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{4} \left (3-\frac {2 i}{b n}\right ),\frac {1}{4} \left (7-\frac {2 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+3 i b n}\) |
Input:
Int[Sec[a + b*Log[c*x^n]]^(3/2),x]
Output:
(2*x*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Hypergeometric2F1[3/2, (3 - (2*I)/(b*n))/4, (7 - (2*I)/(b*n))/4, -(E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Se c[a + b*Log[c*x^n]]^(3/2))/(2 + (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[Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Si mp[x/(n*(c*x^n)^(1/n)) Subst[Int[x^(1/n - 1)*Sec[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
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]
\[\int {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}d x\]
Input:
int(sec(a+b*ln(c*x^n))^(3/2),x)
Output:
int(sec(a+b*ln(c*x^n))^(3/2),x)
Exception generated. \[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(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)
\[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \sec ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \] Input:
integrate(sec(a+b*ln(c*x**n))**(3/2),x)
Output:
Integral(sec(a + b*log(c*x**n))**(3/2), x)
\[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(sec(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")
Output:
integrate(sec(b*log(c*x^n) + a)^(3/2), x)
Timed out. \[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(sec(a+b*log(c*x^n))^(3/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int {\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{3/2} \,d x \] Input:
int((1/cos(a + b*log(c*x^n)))^(3/2),x)
Output:
int((1/cos(a + b*log(c*x^n)))^(3/2), x)
\[ \int \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\sqrt {\sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}\, \sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) x -\frac {3 \left (\int \sqrt {\sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}\, \sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) \tan \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )d x \right ) b n}{2} \] Input:
int(sec(a+b*log(c*x^n))^(3/2),x)
Output:
(2*sqrt(sec(log(x**n*c)*b + a))*sec(log(x**n*c)*b + a)*x - 3*int(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