\(\int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx\) [114]

Optimal result

Integrand size = 24, antiderivative size = 280 \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=\frac {2 e^{4 i (d+e x)} F^{c (a+b x)}}{e \left (1-e^{2 i (d+e x)}\right )^2}+\frac {2 e^{4 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (6-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{4 e-i b c \log (F)}+\frac {e^{4 i (d+e x)} F^{c (a+b x)} (2 e+i b c \log (F))}{e^2 \left (1-e^{2 i (d+e x)}\right )}-\frac {e^{4 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (6-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right ) \left (2+\frac {b^2 c^2 \log ^2(F)}{e^2}\right )}{4 e-i b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 5.32 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.58 \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=\frac {F^{-\frac {b c d}{e}} \left (-2 e^2 F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) (2 e-i b c \log (F))+F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right ) \left (4 e^3-2 i b c e^2 \log (F)+2 b^2 c^2 e \log ^2(F)-i b^3 c^3 \log ^3(F)\right )-b c \log (F) \left (2 i e^2 e^{\frac {(d+e x) (2 i e+b c \log (F))}{e}} F^{a c} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )+F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} (2 i e+b c \log (F)) \left (e \csc ^2(d+e x)+b c \cot (d+e x) \log (F)\right )+i e^{\frac {(d+e x) (2 i e+b c \log (F))}{e}} F^{a c} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right ) \left (2 e^2+b^2 c^2 \log ^2(F)\right )\right )\right )}{2 b c e^2 \log (F) (2 i e+b c \log (F))} \] Input:

Integrate[F^(c*(a + b*x))*Csc[d + e*x]^3*Sec[d + e*x],x]
 

Output:

(-2*e^2*F^(c*(a + b*(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))]*(2*e - I*b*c*Log[F]) + 
 F^(c*(a + b*(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))]*(4*e^3 - (2*I)*b*c*e^2*Log[F] 
+ 2*b^2*c^2*e*Log[F]^2 - I*b^3*c^3*Log[F]^3) - b*c*Log[F]*((2*I)*e^2*E^((( 
d + e*x)*((2*I)*e + b*c*Log[F]))/e)*F^(a*c)*Hypergeometric2F1[1, 1 - ((I/2 
)*b*c*Log[F])/e, 2 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + e*x))] + F^(c*(a 
 + b*(d/e + x)))*((2*I)*e + b*c*Log[F])*(e*Csc[d + e*x]^2 + b*c*Cot[d + e* 
x]*Log[F]) + I*E^(((d + e*x)*((2*I)*e + b*c*Log[F]))/e)*F^(a*c)*Hypergeome 
tric2F1[1, 1 - ((I/2)*b*c*Log[F])/e, 2 - ((I/2)*b*c*Log[F])/e, E^((2*I)*(d 
 + e*x))]*(2*e^2 + b^2*c^2*Log[F]^2)))/(2*b*c*e^2*F^((b*c*d)/e)*Log[F]*((2 
*I)*e + b*c*Log[F]))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.78, 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.

\(\displaystyle \int \csc ^3(d+e x) \sec (d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (-\frac {4 i e^{4 i d+4 i e x} F^{a c+b c x}}{-1+e^{4 i (d+e x)}}+\frac {4 i e^{4 i d+4 i e x} F^{a c+b c x}}{\left (-1+e^{2 i (d+e x)}\right )^2}-\frac {8 i e^{4 i d+4 i e x} F^{a c+b c x}}{\left (-1+e^{2 i (d+e x)}\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 i F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {i b c \log (F)}{4 e},\frac {i b c \log (F)}{4 e}+1,e^{-4 i (d+e x)}\right )}{b c \log (F)}+\frac {4 i F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {i b c \log (F)}{2 e},\frac {i b c \log (F)}{2 e}+1,e^{-2 i (d+e x)}\right )}{b c \log (F)}+\frac {8 e^{4 i d+4 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (6-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{4 e-i b c \log (F)}\)

Input:

Int[F^(c*(a + b*x))*Csc[d + e*x]^3*Sec[d + e*x],x]
 

Output:

((-4*I)*F^(a*c + b*c*x)*Hypergeometric2F1[1, ((I/4)*b*c*Log[F])/e, 1 + ((I 
/4)*b*c*Log[F])/e, E^((-4*I)*(d + e*x))])/(b*c*Log[F]) + ((4*I)*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]) + (8*E^((4*I)*d + (4*I)*e*x)*F^(a*c + 
 b*c*x)*Hypergeometric2F1[3, (4 - (I*b*c*Log[F])/e)/2, (6 - (I*b*c*Log[F]) 
/e)/2, E^((2*I)*(d + e*x))])/(4*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]

\[\int F^{c \left (b x +a \right )} \csc \left (e x +d \right )^{3} \sec \left (e x +d \right )d x\]

Input:

int(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d),x)
 

Output:

int(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d),x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right )^{3} \sec \left (e x + d\right ) \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d),x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*csc(e*x + d)^3*sec(e*x + d), x)
 

Sympy [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=\text {Timed out} \] Input:

integrate(F**(c*(b*x+a))*csc(e*x+d)**3*sec(e*x+d),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right )^{3} \sec \left (e x + d\right ) \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d),x, algorithm="maxima")
 

Output:

(128*(F^(a*c)*b^4*c^4*e*log(F)^4 + 40*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 1536* 
F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d)^2 + 128*(F^(a*c)*b^4*c^4*e*log(F)^ 
4 + 40*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 1536*F^(a*c)*e^5)*F^(b*c*x)*sin(2*e* 
x + 2*d)^2 + 192*(F^(a*c)*b^4*c^4*e*log(F)^4 - 56*F^(a*c)*b^2*c^2*e^3*log( 
F)^2 - 512*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) + 384*(11*F^(a*c)*b^3*c 
^3*e^2*log(F)^3 + 256*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) - 
 128*(F^(a*c)*b^4*c^4*e*log(F)^4 - 32*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 768*F 
^(a*c)*e^5)*F^(b*c*x) + 16*(4*(F^(a*c)*b^4*c^4*e*log(F)^4 + 100*F^(a*c)*b^ 
2*c^2*e^3*log(F)^2 + 2304*F^(a*c)*e^5)*F^(b*c*x)*cos(4*e*x + 4*d) + 4*(F^( 
a*c)*b^4*c^4*e*log(F)^4 + 40*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 1536*F^(a*c)*e 
^5)*F^(b*c*x)*cos(2*e*x + 2*d) + (F^(a*c)*b^5*c^5*log(F)^5 + 100*F^(a*c)*b 
^3*c^3*e^2*log(F)^3 + 2304*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(4*e*x + 4 
*d) - 40*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 64*F^(a*c)*b*c*e^4*log(F))*F^(b*c 
*x)*sin(2*e*x + 2*d) + 8*(F^(a*c)*b^4*c^4*e*log(F)^4 - 32*F^(a*c)*b^2*c^2* 
e^3*log(F)^2 - 768*F^(a*c)*e^5)*F^(b*c*x))*cos(8*e*x + 8*d) - 32*(4*(F^(a* 
c)*b^4*c^4*e*log(F)^4 + 100*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 2304*F^(a*c)*e^ 
5)*F^(b*c*x)*cos(4*e*x + 4*d) + 4*(F^(a*c)*b^4*c^4*e*log(F)^4 + 40*F^(a*c) 
*b^2*c^2*e^3*log(F)^2 - 1536*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) + (F^ 
(a*c)*b^5*c^5*log(F)^5 + 100*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 2304*F^(a*c)*b 
*c*e^4*log(F))*F^(b*c*x)*sin(4*e*x + 4*d) - 40*(F^(a*c)*b^3*c^3*e^2*log...
 

Giac [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right )^{3} \sec \left (e x + d\right ) \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d),x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*csc(e*x + d)^3*sec(e*x + d), x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{\cos \left (d+e\,x\right )\,{\sin \left (d+e\,x\right )}^3} \,d x \] Input:

int(F^(c*(a + b*x))/(cos(d + e*x)*sin(d + e*x)^3),x)
 

Output:

int(F^(c*(a + b*x))/(cos(d + e*x)*sin(d + e*x)^3), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec (d+e x) \, dx=f^{a c} \left (\int f^{b c x} \csc \left (e x +d \right )^{3} \sec \left (e x +d \right )d x \right ) \] Input:

int(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d),x)
 

Output:

f**(a*c)*int(f**(b*c*x)*csc(d + e*x)**3*sec(d + e*x),x)