\(\int \frac {x^m}{\sec ^{\frac {3}{2}}(a+b \log (c x^n))} \, dx\) [285]

Optimal result

Integrand size = 19, antiderivative size = 130 \[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {2 i+2 i m+3 b n}{4 b n},-\frac {2 i+2 i m-b n}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+2 m-3 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \] Output:

2*x^(1+m)*hypergeom([-3/2, -1/4*(2*I+2*I*m+3*b*n)/b/n],[-1/4*(2*I+2*I*m-b* 
n)/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(2+2*m-3*I*b*n)/(1+exp(2*I*a)*(c*x^n) 
^(2*I*b))^(3/2)/sec(a+b*ln(c*x^n))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 1.77 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.55 \[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 x^{1+m} \left (3 b^2 n^2 \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i+2 i m-3 b n}{4 b n},-\frac {2 i+2 i m-5 b n}{4 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right ) \sec ^2\left (a+b \log \left (c x^n\right )\right )+(2+2 m+i b n) \left (2+2 m+3 b n \tan \left (a+b \log \left (c x^n\right )\right )\right )\right )}{(2+2 m+i b n) (2+2 m-3 i b n) (2+2 m+3 i b n) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \] Input:

Integrate[x^m/Sec[a + b*Log[c*x^n]]^(3/2),x]
 

Output:

(2*x^(1 + m)*(3*b^2*n^2*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))*Hypergeometric 
2F1[1, -1/4*(2*I + (2*I)*m - 3*b*n)/(b*n), -1/4*(2*I + (2*I)*m - 5*b*n)/(b 
*n), -E^((2*I)*(a + b*Log[c*x^n]))]*Sec[a + b*Log[c*x^n]]^2 + (2 + 2*m + I 
*b*n)*(2 + 2*m + 3*b*n*Tan[a + b*Log[c*x^n]])))/((2 + 2*m + I*b*n)*(2 + 2* 
m - (3*I)*b*n)*(2 + 2*m + (3*I)*b*n)*Sec[a + b*Log[c*x^n]]^(3/2))
 

Rubi [A] (verified)

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.

\(\displaystyle \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx\)

\(\Big \downarrow \) 5020

\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1}}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}d\left (c x^n\right )}{n}\)

\(\Big \downarrow \) 5018

\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}+\frac {3 i b}{2}} \int \left (c x^n\right )^{-\frac {3 i b}{2}+\frac {m+1}{n}-1} \left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^{3/2}d\left (c x^n\right )}{n \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {2 x^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4} \left (-\frac {2 i (m+1)}{b n}-3\right ),-\frac {2 i m-b n+2 i}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(-3 i b n+2 m+2) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}\)

Input:

Int[x^m/Sec[a + b*Log[c*x^n]]^(3/2),x]
 

Output:

(2*x^(1 + m)*Hypergeometric2F1[-3/2, (-3 - ((2*I)*(1 + m))/(b*n))/4, -1/4* 
(2*I + (2*I)*m - b*n)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/((2 + 2*m 
- (3*I)*b*n)*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Sec[a + b*Log[c*x^n 
]]^(3/2))
 

Defintions of rubi rules used

rule 888

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])
 

rule 5018

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]
 

rule 5020

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])
 

Maple [F]

\[\int \frac {x^{m}}{{\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}}d x\]

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)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Exception raised: TypeError} \] 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 (has polynomial part)
 

Sympy [F]

\[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {x^{m}}{\sec ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}}\, dx \] Input:

integrate(x**m/sec(a+b*ln(c*x**n))**(3/2),x)
 

Output:

Integral(x**m/sec(a + b*log(c*x**n))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {x^{m}}{\sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}} \,d x } \] 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)
 

Giac [F]

\[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {x^{m}}{\sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^m/sec(a+b*log(c*x^n))^(3/2),x, algorithm="giac")
 

Output:

integrate(x^m/sec(b*log(c*x^n) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {x^m}{{\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{3/2}} \,d x \] 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)
 

Reduce [F]

\[ \int \frac {x^m}{\sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {x^{m} \sqrt {\sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}}{{\sec \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}^{2}}d x \] Input:

int(x^m/sec(a+b*log(c*x^n))^(3/2),x)
 

Output:

int((x**m*sqrt(sec(log(x**n*c)*b + a)))/sec(log(x**n*c)*b + a)**2,x)