\(\int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx\) [119]

Optimal result

Integrand size = 24, antiderivative size = 215 \[ \int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx=-\frac {2 e^{3 i (d+e x)} F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )}-\frac {2 e^{3 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{3 e-i b c \log (F)}+\frac {2 b c e^{3 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \log (F)}{e (3 i e+b c \log (F))} \] Output:

-2*exp(3*I*(e*x+d))*F^(c*(b*x+a))/e/(1+exp(2*I*(e*x+d)))-2*exp(3*I*(e*x+d) 
)*F^(c*(b*x+a))*hypergeom([1, 3/2-1/2*I*b*c*ln(F)/e],[5/2-1/2*I*b*c*ln(F)/ 
e],exp(2*I*(e*x+d)))/(3*e-I*b*c*ln(F))+2*b*c*exp(3*I*(e*x+d))*F^(c*(b*x+a) 
)*hypergeom([1, 3/2-1/2*I*b*c*ln(F)/e],[5/2-1/2*I*b*c*ln(F)/e],-exp(2*I*(e 
*x+d)))*ln(F)/e/(3*I*e+b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.98 \[ \int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (-2 e e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right )+2 i b c e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \log (F)+F^{\frac {b c (d+e x)}{e}} (e-i b c \log (F)) \sec (d+e x)\right )}{e (e-i b c \log (F))} \] Input:

Integrate[F^(c*(a + b*x))*Csc[d + e*x]*Sec[d + e*x]^2,x]
 

Output:

(F^(c*(a - (b*d)/e))*(-2*e*E^(((d + e*x)*(I*e + b*c*Log[F]))/e)*Hypergeome 
tric2F1[1, (e - I*b*c*Log[F])/(2*e), 3/2 - ((I/2)*b*c*Log[F])/e, E^((2*I)* 
(d + e*x))] + (2*I)*b*c*E^(((d + e*x)*(I*e + b*c*Log[F]))/e)*Hypergeometri 
c2F1[1, (e - I*b*c*Log[F])/(2*e), 3/2 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d 
 + e*x))]*Log[F] + F^((b*c*(d + e*x))/e)*(e - I*b*c*Log[F])*Sec[d + e*x])) 
/(e*(e - I*b*c*Log[F]))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

\(\displaystyle \int \csc (d+e x) \sec ^2(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 e^{3 i d+3 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{4} \left (7-\frac {i b c \log (F)}{e}\right ),e^{4 i (d+e x)}\right )}{3 e-i b c \log (F)}-\frac {4 e^{3 i d+3 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{3 e-i b c \log (F)}\)

Input:

Int[F^(c*(a + b*x))*Csc[d + e*x]*Sec[d + e*x]^2,x]
 

Output:

(-4*E^((3*I)*d + (3*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (3 - (I*b 
*c*Log[F])/e)/4, (7 - (I*b*c*Log[F])/e)/4, E^((4*I)*(d + e*x))])/(3*e - I* 
b*c*Log[F]) - (4*E^((3*I)*d + (3*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1 
[2, (3 - (I*b*c*Log[F])/e)/2, (5 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x 
))])/(3*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]

\[\int F^{c \left (b x +a \right )} \csc \left (e x +d \right ) \sec \left (e x +d \right )^{2}d x\]

Input:

int(F^(c*(b*x+a))*csc(e*x+d)*sec(e*x+d)^2,x)
 

Output:

int(F^(c*(b*x+a))*csc(e*x+d)*sec(e*x+d)^2,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right ) \sec \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)*sec(e*x+d)^2,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*csc(e*x + d)*sec(e*x + d)^2, x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx=\int F^{c \left (a + b x\right )} \csc {\left (d + e x \right )} \sec ^{2}{\left (d + e x \right )}\, dx \] Input:

integrate(F**(c*(b*x+a))*csc(e*x+d)*sec(e*x+d)**2,x)
 

Output:

Integral(F**(c*(a + b*x))*csc(d + e*x)*sec(d + e*x)**2, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right ) \sec \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)*sec(e*x+d)^2,x, algorithm="maxima")
 

Output:

(128*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F)*sin(e*x + d) - 16*(F^(a*c)*b^2*c^2*e 
*log(F)^2 - 15*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) - 8*(16*F^(b*c*x)*F^(a* 
c)*b*c*e^2*log(F)*sin(e*x + d) + 3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 25*F^(a*c 
)*e^3)*F^(b*c*x)*cos(3*e*x + 3*d) - 2*(F^(a*c)*b^2*c^2*e*log(F)^2 - 15*F^( 
a*c)*e^3)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)* 
b*c*e^2*log(F))*F^(b*c*x)*sin(3*e*x + 3*d))*cos(6*e*x + 6*d) - 8*(16*F^(b* 
c*x)*F^(a*c)*b*c*e^2*log(F)*sin(e*x + d) + 3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 
 25*F^(a*c)*e^3)*F^(b*c*x)*cos(3*e*x + 3*d) - 2*(F^(a*c)*b^2*c^2*e*log(F)^ 
2 - 15*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^3*c^3*log(F)^3 + 2 
5*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(3*e*x + 3*d))*cos(4*e*x + 4*d) + 8 
*(3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 25*F^(a*c)*e^3)*F^(b*c*x)*cos(2*e*x + 2* 
d) - (F^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin( 
2*e*x + 2*d) + 3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 25*F^(a*c)*e^3)*F^(b*c*x))* 
cos(3*e*x + 3*d) + 16*(8*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F)*sin(e*x + d) - ( 
F^(a*c)*b^2*c^2*e*log(F)^2 - 15*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d))*cos(2 
*e*x + 2*d) - (b^4*c^4*log(F)^4 + 34*b^2*c^2*e^2*log(F)^2 + 225*e^4 + (b^4 
*c^4*log(F)^4 + 34*b^2*c^2*e^2*log(F)^2 + 225*e^4)*cos(6*e*x + 6*d)^2 + (b 
^4*c^4*log(F)^4 + 34*b^2*c^2*e^2*log(F)^2 + 225*e^4)*cos(4*e*x + 4*d)^2 + 
(b^4*c^4*log(F)^4 + 34*b^2*c^2*e^2*log(F)^2 + 225*e^4)*cos(2*e*x + 2*d)^2 
+ (b^4*c^4*log(F)^4 + 34*b^2*c^2*e^2*log(F)^2 + 225*e^4)*sin(6*e*x + 6*...
 

Giac [F]

\[ \int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right ) \sec \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)*sec(e*x+d)^2,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*csc(e*x + d)*sec(e*x + d)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \csc (d+e x) \sec ^2(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\cos \left (d+e\,x\right )}^2\,\sin \left (d+e\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

int(F^(c*(b*x+a))*csc(e*x+d)*sec(e*x+d)^2,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*csc(d + e*x)*sec(d + e*x)**2,x)