\(\int F^{c (a+b x)} \sqrt {f \sec (d+e x)} \sqrt {g \tan (d+e x)} \, dx\) [148]

Optimal result

Integrand size = 34, antiderivative size = 158 \[ \int F^{c (a+b x)} \sqrt {f \sec (d+e x)} \sqrt {g \tan (d+e x)} \, dx=\frac {i \left (e^{2 i (d+e x)}\right )^{\frac {1}{4} \left (-1+\frac {2 i b c \log (F)}{e}\right )} \left (1-e^{2 i (d+e x)}\right ) \left (1+e^{2 i (d+e x)}\right ) F^{c (a+b x)} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4} \left (3+\frac {2 i b c \log (F)}{e}\right ),1,\frac {5}{2},1-e^{2 i (d+e x)},\frac {1}{2} \left (1-e^{2 i (d+e x)}\right )\right ) \sqrt {f \sec (d+e x)} \sqrt {g \tan (d+e x)}}{6 e} \] Output:

1/6*I*exp(2*I*(e*x+d))^(-1/4+1/2*I*b*c*ln(F)/e)*(1-exp(2*I*(e*x+d)))*(1+ex 
p(2*I*(e*x+d)))*F^(c*(b*x+a))*AppellF1(3/2,3/4+1/2*I*b*c*ln(F)/e,1,5/2,1-e 
xp(2*I*(e*x+d)),1/2-1/2*exp(2*I*(e*x+d)))*(f*sec(e*x+d))^(1/2)*(g*tan(e*x+ 
d))^(1/2)/e
 

Mathematica [F]

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

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

Output:

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

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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7299

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

Input:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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