\(\int F^{c (a+b x)} \csc ^2(d+e x) \sec (d+e x) \, dx\) [113]

Optimal result

Integrand size = 24, antiderivative size = 219 \[ \int F^{c (a+b x)} \csc ^2(d+e x) \sec (d+e x) \, dx=\frac {2 i e^{3 i (d+e x)} F^{c (a+b x)}}{e \left (1-e^{2 i (d+e x)}\right )}-\frac {2 b c e^{3 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right ) \log (F)}{e (3 e-i b c \log (F))}-\frac {2 e^{3 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{3 i e+b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.97 \[ \int F^{c (a+b x)} \csc ^2(d+e x) \sec (d+e x) \, dx=-\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (2 i e e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )+2 b c e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right ) \log (F)+F^{\frac {b c (d+e x)}{e}} \csc (d+e x) (e-i b c \log (F))\right )}{e (e-i b c \log (F))} \] Input:

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

Output:

-((F^(c*(a - (b*d)/e))*((2*I)*e*E^(((d + e*x)*(I*e + b*c*Log[F]))/e)*Hyper 
geometric2F1[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*b*c*E^(((d + e*x)*(I*e + b*c*Log[F]))/e)*Hypergeomet 
ric2F1[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)*Csc[d + e*x]*(e - I*b*c*Log[F])) 
)/(e*(e - I*b*c*Log[F])))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.79, 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.

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

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

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

Input:

Int[F^(c*(a + b*x))*Csc[d + e*x]^2*Sec[d + e*x],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*I)*e 
+ b*c*Log[F]) - (4*E^((3*I)*d + (3*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2 
F1[2, (3 - (I*b*c*Log[F])/e)/2, (5 - (I*b*c*Log[F])/e)/2, E^((2*I)*(d + e* 
x))])/((3*I)*e + 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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