\(\int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx\) [149]

Optimal result

Integrand size = 34, antiderivative size = 117 \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=\frac {2 \sqrt {1-e^{2 i (d+e x)}} F^{c (a+b x)} \sqrt {g \cot (d+e x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {e-2 i b c \log (F)}{4 e},\frac {1}{4} \left (5-\frac {2 i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right ) \sqrt {f \sec (d+e x)}}{i e+2 b c \log (F)} \] Output:

2*(1-exp(2*I*(e*x+d)))^(1/2)*F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*hypergeom( 
[1/2, 1/4*(e-2*I*b*c*ln(F))/e],[5/4-1/2*I*b*c*ln(F)/e],exp(2*I*(e*x+d)))*( 
f*sec(e*x+d))^(1/2)/(I*e+2*b*c*ln(F))
 

Mathematica [A] (warning: unable to verify)

Time = 2.59 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04 \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=-\frac {4 F^{c (a+b x)} \sqrt {g \cot (d+e x)} (i+\cot (d+e x)) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{4}-\frac {i b c \log (F)}{2 e},\frac {5}{4}-\frac {i b c \log (F)}{2 e},\cos (2 (d+e x))+i \sin (2 (d+e x))\right ) \sqrt {f \sec (d+e x)} \sin ^2(d+e x)}{e-2 i b c \log (F)} \] Input:

Integrate[F^(c*(a + b*x))*Sqrt[g*Cot[d + e*x]]*Sqrt[f*Sec[d + e*x]],x]
 

Output:

(-4*F^(c*(a + b*x))*Sqrt[g*Cot[d + e*x]]*(I + Cot[d + e*x])*Hypergeometric 
2F1[1, 3/4 - ((I/2)*b*c*Log[F])/e, 5/4 - ((I/2)*b*c*Log[F])/e, Cos[2*(d + 
e*x)] + I*Sin[2*(d + e*x)]]*Sqrt[f*Sec[d + e*x]]*Sin[d + e*x]^2)/(e - (2*I 
)*b*c*Log[F])
 

Rubi [F]

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.

\(\displaystyle \int F^{c (a+b x)} \sqrt {f \sec (d+e x)} \sqrt {g \cot (d+e x)} \, dx\)

\(\Big \downarrow \) 7271

\(\displaystyle \frac {\sqrt {g \cot (d+e x)} \int F^{c (a+b x)} \sqrt {\cot (d+e x)} \sqrt {f \sec (d+e x)}dx}{\sqrt {\cot (d+e x)}}\)

\(\Big \downarrow \) 7271

\(\displaystyle \frac {\sqrt {f \sec (d+e x)} \sqrt {g \cot (d+e x)} \int F^{c (a+b x)} \sqrt {\cot (d+e x)} \sqrt {\sec (d+e x)}dx}{\sqrt {\cot (d+e x)} \sqrt {\sec (d+e x)}}\)

\(\Big \downarrow \) 7292

\(\displaystyle \frac {\sqrt {f \sec (d+e x)} \sqrt {g \cot (d+e x)} \int F^{a c+b x c} \sqrt {\cot (d+e x)} \sqrt {\sec (d+e x)}dx}{\sqrt {\cot (d+e x)} \sqrt {\sec (d+e x)}}\)

\(\Big \downarrow \) 7299

\(\displaystyle \frac {\sqrt {f \sec (d+e x)} \sqrt {g \cot (d+e x)} \int F^{a c+b x c} \sqrt {\cot (d+e x)} \sqrt {\sec (d+e x)}dx}{\sqrt {\cot (d+e x)} \sqrt {\sec (d+e x)}}\)

Input:

Int[F^(c*(a + b*x))*Sqrt[g*Cot[d + e*x]]*Sqrt[f*Sec[d + e*x]],x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \sqrt {g \cot \left (e x +d \right )}\, \sqrt {f \sec \left (e x +d \right )}d x\]

Input:

int(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*sec(e*x+d))^(1/2),x)
 

Output:

int(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*sec(e*x+d))^(1/2),x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=\int { \sqrt {g \cot \left (e x + d\right )} \sqrt {f \sec \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*sec(e*x+d))^(1/2),x, algor 
ithm="fricas")
 

Output:

integral(sqrt(g*cot(e*x + d))*sqrt(f*sec(e*x + d))*F^(b*c*x + a*c), x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=\int F^{c \left (a + b x\right )} \sqrt {f \sec {\left (d + e x \right )}} \sqrt {g \cot {\left (d + e x \right )}}\, dx \] Input:

integrate(F**(c*(b*x+a))*(g*cot(e*x+d))**(1/2)*(f*sec(e*x+d))**(1/2),x)
 

Output:

Integral(F**(c*(a + b*x))*sqrt(f*sec(d + e*x))*sqrt(g*cot(d + e*x)), x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=\int { \sqrt {g \cot \left (e x + d\right )} \sqrt {f \sec \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*sec(e*x+d))^(1/2),x, algor 
ithm="maxima")
 

Output:

integrate(sqrt(g*cot(e*x + d))*sqrt(f*sec(e*x + d))*F^((b*x + a)*c), x)
 

Giac [F]

\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=\int { \sqrt {g \cot \left (e x + d\right )} \sqrt {f \sec \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*sec(e*x+d))^(1/2),x, algor 
ithm="giac")
 

Output:

integrate(sqrt(g*cot(e*x + d))*sqrt(f*sec(e*x + d))*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=\int F^{c\,\left (a+b\,x\right )}\,\sqrt {g\,\mathrm {cot}\left (d+e\,x\right )}\,\sqrt {\frac {f}{\cos \left (d+e\,x\right )}} \,d x \] Input:

int(F^(c*(a + b*x))*(g*cot(d + e*x))^(1/2)*(f/cos(d + e*x))^(1/2),x)
 

Output:

int(F^(c*(a + b*x))*(g*cot(d + e*x))^(1/2)*(f/cos(d + e*x))^(1/2), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \sec (d+e x)} \, dx=\sqrt {g}\, f^{a c +\frac {1}{2}} \left (\int f^{b c x} \sqrt {\sec \left (e x +d \right )}\, \sqrt {\cot \left (e x +d \right )}d x \right ) \] Input:

int(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*sec(e*x+d))^(1/2),x)
 

Output:

sqrt(g)*f**((2*a*c + 1)/2)*int(f**(b*c*x)*sqrt(sec(d + e*x))*sqrt(cot(d + 
e*x)),x)