\(\int F^{c (a+b x)} \sec (d+e x) \tan (d+e x) \, dx\) [117]

Optimal result

Integrand size = 22, antiderivative size = 132 \[ \int F^{c (a+b x)} \sec (d+e x) \tan (d+e x) \, dx=\frac {2 e^{i (d+e x)} F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )}-\frac {2 b c e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \log (F)}{e (i e+b c \log (F))} \] Output:

2*exp(I*(e*x+d))*F^(c*(b*x+a))/e/(1+exp(2*I*(e*x+d)))-2*b*c*exp(I*(e*x+d)) 
*F^(c*(b*x+a))*hypergeom([1, 1/2*(e-I*b*c*ln(F))/e],[3/2-1/2*I*b*c*ln(F)/e 
],-exp(2*I*(e*x+d)))*ln(F)/e/(I*e+b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int F^{c (a+b x)} \sec (d+e x) \tan (d+e x) \, dx=-\frac {2 b c e^{i (d+e x)} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \log (F)}{e (i e+b c \log (F))}+\frac {F^{a c+b c x} \sec (d+e x)}{e} \] Input:

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

Output:

(-2*b*c*E^(I*(d + e*x))*F^(a*c + b*c*x)*Hypergeometric2F1[1, 1/2 - ((I/2)* 
b*c*Log[F])/e, 3/2 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + e*x))]*Log[F])/( 
e*(I*e + b*c*Log[F])) + (F^(a*c + b*c*x)*Sec[d + e*x])/e
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.33, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4974, 2009}

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.

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

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (\frac {4 i e^{i d+i e x} F^{a c+b c x}}{\left (1+e^{2 i (d+e x)}\right )^2}-\frac {2 i e^{i d+i e x} F^{a c+b c x}}{1+e^{2 i (d+e x)}}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-2*E^(I*d + I*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (e - I*b*c*Log[F] 
)/(2*e), (3 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))])/(e - I*b*c*Log[F 
]) + (4*E^(I*d + I*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, (e - I*b*c*Lo 
g[F])/(2*e), (3 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))])/(e - I*b*c*L 
og[F])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4974

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[( 
d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^(c*(a + b*x)), 
 G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IGtQ[ 
m, 0] && IGtQ[n, 0] && TrigQ[G] && TrigQ[H]
 

Maple [F]

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

Input:

int(F^(c*(b*x+a))*sec(e*x+d)*tan(e*x+d),x)
 

Output:

int(F^(c*(b*x+a))*sec(e*x+d)*tan(e*x+d),x)
 

Fricas [F]

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

integrate(F^(c*(b*x+a))*sec(e*x+d)*tan(e*x+d),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(F**(c*(b*x+a))*sec(e*x+d)*tan(e*x+d),x)
 

Output:

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

Maxima [F]

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

integrate(F^(c*(b*x+a))*sec(e*x+d)*tan(e*x+d),x, algorithm="maxima")
 

Output:

-2*((7*F^(a*c)*b^2*c^2*e*log(F)^2 - 9*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) 
+ (F^(a*c)*b^3*c^3*log(F)^3 - 15*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(e*x 
 + d) - ((F^(a*c)*b^2*c^2*e*log(F)^2 + 9*F^(a*c)*e^3)*F^(b*c*x)*cos(3*e*x 
+ 3*d) - (7*F^(a*c)*b^2*c^2*e*log(F)^2 - 9*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x 
+ d) + (F^(a*c)*b^3*c^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin 
(3*e*x + 3*d) - (F^(a*c)*b^3*c^3*log(F)^3 - 15*F^(a*c)*b*c*e^2*log(F))*F^( 
b*c*x)*sin(e*x + d))*cos(4*e*x + 4*d) - (2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 9 
*F^(a*c)*e^3)*F^(b*c*x)*cos(2*e*x + 2*d) - 2*(F^(a*c)*b^3*c^3*log(F)^3 + 9 
*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) + (F^(a*c)*b^2*c^2*e*l 
og(F)^2 + 9*F^(a*c)*e^3)*F^(b*c*x))*cos(3*e*x + 3*d) + 2*((7*F^(a*c)*b^2*c 
^2*e*log(F)^2 - 9*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^3*c^3*l 
og(F)^3 - 15*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(e*x + d))*cos(2*e*x + 2 
*d) + 8*(F^(a*c)*b^5*c^5*e*log(F)^5 + 10*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 9* 
F^(a*c)*b*c*e^5*log(F) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 10*F^(a*c)*b^3*c^3* 
e^3*log(F)^3 + 9*F^(a*c)*b*c*e^5*log(F))*cos(4*e*x + 4*d)^2 + 4*(F^(a*c)*b 
^5*c^5*e*log(F)^5 + 10*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 9*F^(a*c)*b*c*e^5*lo 
g(F))*cos(2*e*x + 2*d)^2 + (F^(a*c)*b^5*c^5*e*log(F)^5 + 10*F^(a*c)*b^3*c^ 
3*e^3*log(F)^3 + 9*F^(a*c)*b*c*e^5*log(F))*sin(4*e*x + 4*d)^2 + 4*(F^(a*c) 
*b^5*c^5*e*log(F)^5 + 10*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 9*F^(a*c)*b*c*e^5* 
log(F))*sin(4*e*x + 4*d)*sin(2*e*x + 2*d) + 4*(F^(a*c)*b^5*c^5*e*log(F)...
 

Giac [F]

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

integrate(F^(c*(b*x+a))*sec(e*x+d)*tan(e*x+d),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(F^(c*(b*x+a))*sec(e*x+d)*tan(e*x+d),x)
 

Output:

f**(a*c)*int(f**(b*c*x)*sec(d + e*x)*tan(d + e*x),x)