\(\int F^{c (a+b x)} \csc ^2(d+e x) \sec ^2(d+e x) \, dx\) [120]

Optimal result

Integrand size = 26, antiderivative size = 82 \[ \int F^{c (a+b x)} \csc ^2(d+e x) \sec ^2(d+e x) \, dx=-\frac {16 e^{4 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{4} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {1}{4} \left (8-\frac {i b c \log (F)}{e}\right ),e^{4 i (d+e x)}\right )}{4 i e+b c \log (F)} \] Output:

-16*exp(4*I*(e*x+d))*F^(c*(b*x+a))*hypergeom([2, 1-1/4*I*b*c*ln(F)/e],[2-1 
/4*I*b*c*ln(F)/e],exp(4*I*(e*x+d)))/(4*I*e+b*c*ln(F))
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(215\) vs. \(2(82)=164\).

Time = 0.87 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.62 \[ \int F^{c (a+b x)} \csc ^2(d+e x) \sec ^2(d+e x) \, dx=\frac {2 F^{-\frac {b c d}{e}} \left (-b c e^{\frac {(d+e x) (4 i e+b c \log (F))}{e}} F^{a c} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c \log (F)}{4 e},2-\frac {i b c \log (F)}{4 e},e^{4 i (d+e x)}\right ) \log (F)-F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \cot (2 (d+e x)) (4 e-i b c \log (F))-i F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{4 e},1-\frac {i b c \log (F)}{4 e},e^{4 i (d+e x)}\right ) (4 e-i b c \log (F))\right )}{e (4 e-i b c \log (F))} \] Input:

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

Output:

(2*(-(b*c*E^(((d + e*x)*((4*I)*e + b*c*Log[F]))/e)*F^(a*c)*Hypergeometric2 
F1[1, 1 - ((I/4)*b*c*Log[F])/e, 2 - ((I/4)*b*c*Log[F])/e, E^((4*I)*(d + e* 
x))]*Log[F]) - F^(c*(a + b*(d/e + x)))*Cot[2*(d + e*x)]*(4*e - I*b*c*Log[F 
]) - I*F^(c*(a + b*(d/e + x)))*Hypergeometric2F1[1, ((-1/4*I)*b*c*Log[F])/ 
e, 1 - ((I/4)*b*c*Log[F])/e, E^((4*I)*(d + e*x))]*(4*e - I*b*c*Log[F])))/( 
e*F^((b*c*d)/e)*(4*e - I*b*c*Log[F]))
 

Rubi [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(470\) vs. \(2(82)=164\).

Time = 1.01 (sec) , antiderivative size = 470, normalized size of antiderivative = 5.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.

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

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {i b c \log (F)}{2 e},\frac {i b c \log (F)}{2 e}+1,-e^{-2 i (d+e x)}\right )}{b c \log (F)}-\frac {4 e^{4 i d+4 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (6-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{b c \log (F)+4 i e}-\frac {3 e^{4 i d+4 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,4-\frac {i b c \log (F)}{e},5-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{b c \log (F)+4 i e}-\frac {3 e^{4 i d+4 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,4-\frac {i b c \log (F)}{e},5-\frac {i b c \log (F)}{e},e^{i (d+e x)}\right )}{b c \log (F)+4 i e}-\frac {e^{4 i d+4 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,4-\frac {i b c \log (F)}{e},5-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{b c \log (F)+4 i e}-\frac {e^{4 i d+4 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,4-\frac {i b c \log (F)}{e},5-\frac {i b c \log (F)}{e},e^{i (d+e x)}\right )}{b c \log (F)+4 i e}\)

Input:

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

Output:

(-4*F^(a*c + b*c*x)*Hypergeometric2F1[2, ((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]) - (4*E^((4*I)*d + (4*I 
)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (4 - (I*b*c*Log[F])/e)/2, (6 - 
 (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))])/((4*I)*e + b*c*Log[F]) - (3*E 
^((4*I)*d + (4*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, 4 - (I*b*c*Log 
[F])/e, 5 - (I*b*c*Log[F])/e, -E^(I*(d + e*x))])/((4*I)*e + b*c*Log[F]) - 
(3*E^((4*I)*d + (4*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, 4 - (I*b*c 
*Log[F])/e, 5 - (I*b*c*Log[F])/e, E^(I*(d + e*x))])/((4*I)*e + b*c*Log[F]) 
 - (E^((4*I)*d + (4*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, 4 - (I*b* 
c*Log[F])/e, 5 - (I*b*c*Log[F])/e, -E^(I*(d + e*x))])/((4*I)*e + b*c*Log[F 
]) - (E^((4*I)*d + (4*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, 4 - (I* 
b*c*Log[F])/e, 5 - (I*b*c*Log[F])/e, E^(I*(d + e*x))])/((4*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

16*(96*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 64 
*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(4*e*x + 4*d)^2 + 2*(F^(a*c)*b^3*c^3 
*log(F)^3 + 64*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(4*e*x + 4*d)^2 - (F^( 
a*c)*b^3*c^3*log(F)^3 + 256*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(4*e*x + 
4*d) + 4*(5*F^(a*c)*b^2*c^2*e*log(F)^2 - 64*F^(a*c)*e^3)*F^(b*c*x)*sin(4*e 
*x + 4*d) + (96*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) - (F^(a*c)*b^3*c^3*log(F) 
^3 + 64*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(4*e*x + 4*d) + 4*(F^(a*c)*b^ 
2*c^2*e*log(F)^2 + 64*F^(a*c)*e^3)*F^(b*c*x)*sin(4*e*x + 4*d))*cos(8*e*x + 
 8*d) + 8*(F^(a*c)*b^5*c^5*e*log(F)^5 + 80*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 
1024*F^(a*c)*b*c*e^5*log(F) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 80*F^(a*c)*b^3 
*c^3*e^3*log(F)^3 + 1024*F^(a*c)*b*c*e^5*log(F))*cos(8*e*x + 8*d)^2 + 4*(F 
^(a*c)*b^5*c^5*e*log(F)^5 + 80*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 1024*F^(a*c) 
*b*c*e^5*log(F))*cos(4*e*x + 4*d)^2 + (F^(a*c)*b^5*c^5*e*log(F)^5 + 80*F^( 
a*c)*b^3*c^3*e^3*log(F)^3 + 1024*F^(a*c)*b*c*e^5*log(F))*sin(8*e*x + 8*d)^ 
2 - 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 80*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 1024 
*F^(a*c)*b*c*e^5*log(F))*sin(8*e*x + 8*d)*sin(4*e*x + 4*d) + 4*(F^(a*c)*b^ 
5*c^5*e*log(F)^5 + 80*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 1024*F^(a*c)*b*c*e^5* 
log(F))*sin(4*e*x + 4*d)^2 + 2*(F^(a*c)*b^5*c^5*e*log(F)^5 + 80*F^(a*c)*b^ 
3*c^3*e^3*log(F)^3 + 1024*F^(a*c)*b*c*e^5*log(F) - 2*(F^(a*c)*b^5*c^5*e*lo 
g(F)^5 + 80*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 1024*F^(a*c)*b*c*e^5*log(F))...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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