\(\int F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)} \, dx\) [133]

Optimal result

Integrand size = 34, antiderivative size = 44 \[ \int F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)} \, dx=\frac {F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)}}{b c \log (F)} \] Output:

F^(c*(b*x+a))*(f*cos(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2)/b/c/ln(F)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)} \, dx=\frac {F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)}}{b c \log (F)} \] Input:

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

Output:

(F^(c*(a + b*x))*Sqrt[f*Cos[d + e*x]]*Sqrt[g*Sec[d + e*x]])/(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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7299

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

Input:

Int[F^(c*(a + b*x))*Sqrt[f*Cos[d + e*x]]*Sqrt[g*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 [A] (verified)

Time = 0.56 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93

method	result	size
orering	\(\frac {F^{c \left (b x +a \right )} \sqrt {f \cos \left (e x +d \right )}\, \sqrt {g \sec \left (e x +d \right )}}{b c \ln \left (F \right )}\)	\(41\)
risch	\(\frac {\sqrt {2}\, \sqrt {f \cos \left (e x +d \right )}\, \sqrt {\frac {g \,{\mathrm e}^{i \left (e x +d \right )}}{1+{\mathrm e}^{2 i \left (e x +d \right )}}}\, F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}\)	\(60\)

Input:

int(F^(c*(b*x+a))*(f*cos(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2),x,method=_RETU 
RNVERBOSE)
 

Output:

F^(c*(b*x+a))*(f*cos(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2)/b/c/ln(F)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)} \, dx=\frac {\sqrt {f \cos \left (e x + d\right )} F^{b c x + a c} \sqrt {\frac {g}{\cos \left (e x + d\right )}}}{b c \log \left (F\right )} \] Input:

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

Output:

sqrt(f*cos(e*x + d))*F^(b*c*x + a*c)*sqrt(g/cos(e*x + d))/(b*c*log(F))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.64 \[ \int F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)} \, dx=\frac {F^{b c x} F^{a c} \sqrt {f} \sqrt {g}}{b c \log \left (F\right )} \] Input:

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

Output:

F^(b*c*x)*F^(a*c)*sqrt(f)*sqrt(g)/(b*c*log(F))
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 16.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int F^{c (a+b x)} \sqrt {f \cos (d+e x)} \sqrt {g \sec (d+e x)} \, dx=\frac {F^{c\,\left (a+b\,x\right )}\,\sqrt {f\,\cos \left (d+e\,x\right )}\,\sqrt {\frac {g}{\cos \left (d+e\,x\right )}}}{b\,c\,\ln \left (F\right )} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(g)*f**((2*a*c + 1)/2)*(2*f**(b*c*x)*sqrt(sec(d + e*x))*sqrt(cos(d + 
e*x)) + int((f**(b*c*x)*sqrt(sec(d + e*x))*sqrt(cos(d + e*x))*sin(d + e*x) 
)/cos(d + e*x),x)*e - int(f**(b*c*x)*sqrt(sec(d + e*x))*sqrt(cos(d + e*x)) 
*tan(d + e*x),x)*e))/(2*log(f)*b*c)