\(\int F^{c (a+b x)} \csc (d+e x) \sec (d+e x) \, dx\) [112]

Optimal result

Integrand size = 22, antiderivative size = 83 \[ \int F^{c (a+b x)} \csc (d+e x) \sec (d+e x) \, dx=-\frac {4 e^{2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} \left (2-\frac {i b c \log (F)}{e}\right ),\frac {1}{4} \left (6-\frac {i b c \log (F)}{e}\right ),e^{4 i (d+e x)}\right )}{2 e-i b c \log (F)} \] Output:

-4*exp(2*I*(e*x+d))*F^(c*(b*x+a))*hypergeom([1, 1/2-1/4*I*b*c*ln(F)/e],[3/ 
2-1/4*I*b*c*ln(F)/e],exp(4*I*(e*x+d)))/(2*e-I*b*c*ln(F))
 

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33 \[ \int F^{c (a+b x)} \csc (d+e x) \sec (d+e x) \, dx=\frac {2 i F^{c (a+b x)} \left (\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )-\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right )\right )}{b c \log (F)} \] Input:

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

Output:

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

Rubi [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(228\) vs. \(2(83)=166\).

Time = 0.61 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.75, 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.

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

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

4*(F^(b*c*x)*F^(a*c)*b*c*log(F)*sin(2*e*x + 2*d) + 2*F^(b*c*x)*F^(a*c)*e*c 
os(2*e*x + 2*d) - (F^(b*c*x)*F^(a*c)*b*c*log(F)*sin(2*e*x + 2*d) + 2*F^(b* 
c*x)*F^(a*c)*e*cos(2*e*x + 2*d))*cos(4*e*x + 4*d) - 4*(F^(a*c)*b^2*c^2*e*l 
og(F)^2 + 4*F^(a*c)*e^3 + (F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*cos 
(4*e*x + 4*d)^2 + (F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*sin(4*e*x + 
 4*d)^2 - 2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*cos(4*e*x + 4*d)) 
*integrate((F^(b*c*x)*b*c*cos(2*e*x + 2*d)*log(F) - 2*F^(b*c*x)*e*sin(2*e* 
x + 2*d) + (F^(b*c*x)*b*c*cos(2*e*x + 2*d)*log(F) - 2*F^(b*c*x)*e*sin(2*e* 
x + 2*d))*cos(8*e*x + 8*d) - 2*(F^(b*c*x)*b*c*cos(2*e*x + 2*d)*log(F) - 2* 
F^(b*c*x)*e*sin(2*e*x + 2*d))*cos(4*e*x + 4*d) + (F^(b*c*x)*b*c*log(F)*sin 
(2*e*x + 2*d) + 2*F^(b*c*x)*e*cos(2*e*x + 2*d))*sin(8*e*x + 8*d) - 2*(F^(b 
*c*x)*b*c*log(F)*sin(2*e*x + 2*d) + 2*F^(b*c*x)*e*cos(2*e*x + 2*d))*sin(4* 
e*x + 4*d))/(b^2*c^2*log(F)^2 + (b^2*c^2*log(F)^2 + 4*e^2)*cos(8*e*x + 8*d 
)^2 + 4*(b^2*c^2*log(F)^2 + 4*e^2)*cos(4*e*x + 4*d)^2 + (b^2*c^2*log(F)^2 
+ 4*e^2)*sin(8*e*x + 8*d)^2 - 4*(b^2*c^2*log(F)^2 + 4*e^2)*sin(8*e*x + 8*d 
)*sin(4*e*x + 4*d) + 4*(b^2*c^2*log(F)^2 + 4*e^2)*sin(4*e*x + 4*d)^2 + 4*e 
^2 + 2*(b^2*c^2*log(F)^2 + 4*e^2 - 2*(b^2*c^2*log(F)^2 + 4*e^2)*cos(4*e*x 
+ 4*d))*cos(8*e*x + 8*d) - 4*(b^2*c^2*log(F)^2 + 4*e^2)*cos(4*e*x + 4*d)), 
 x) + (F^(b*c*x)*F^(a*c)*b*c*cos(2*e*x + 2*d)*log(F) - 2*F^(b*c*x)*F^(a*c) 
*e*sin(2*e*x + 2*d))*sin(4*e*x + 4*d))/(b^2*c^2*log(F)^2 + (b^2*c^2*log...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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