\(\int F^{c (a+b x)} \csc (d+e x) \sec ^3(d+e x) \, dx\) [126]

Optimal result

Integrand size = 24, antiderivative size = 276 \[ \int F^{c (a+b x)} \csc (d+e x) \sec ^3(d+e x) \, dx=-\frac {2 e^{4 i (d+e x)} F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )^2}-\frac {2 e^{4 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (6-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{4 e-i b c \log (F)}-\frac {e^{4 i (d+e x)} F^{c (a+b x)} (2 e+i b c \log (F))}{e^2 \left (1+e^{2 i (d+e x)}\right )}+\frac {e^{4 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (6-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \left (2+\frac {b^2 c^2 \log ^2(F)}{e^2}\right )}{4 e-i b c \log (F)} \] 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 = 5.75 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.61 \[ \int F^{c (a+b x)} \csc (d+e x) \sec ^3(d+e x) \, dx=\frac {F^{-\frac {b c d}{e}} \left (-2 i e^2 F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right ) (2 e-i b c \log (F))+F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \left (4 i e^3+2 b c e^2 \log (F)+2 i b^2 c^2 e \log ^2(F)+b^3 c^3 \log ^3(F)\right )-b c \log (F) \left (2 e^2 e^{\frac {(d+e x) (2 i e+b c \log (F))}{e}} F^{a c} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right )+e^{\frac {(d+e x) (2 i e+b c \log (F))}{e}} F^{a c} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \left (2 e^2+b^2 c^2 \log ^2(F)\right )-F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} (2 e-i b c \log (F)) \left (e \sec ^2(d+e x)-b c \log (F) \tan (d+e x)\right )\right )\right )}{2 b c e^2 \log (F) (2 e-i b c \log (F))} \] Input:

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

Output:

((-2*I)*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*I)*e^3 + 2*b*c*e^2*Log 
[F] + (2*I)*b^2*c^2*e*Log[F]^2 + b^3*c^3*Log[F]^3) - b*c*Log[F]*(2*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))] + 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))]*(2*e^2 + b 
^2*c^2*Log[F]^2) - F^(c*(a + b*(d/e + x)))*(2*e - I*b*c*Log[F])*(e*Sec[d + 
 e*x]^2 - b*c*Log[F]*Tan[d + e*x])))/(2*b*c*e^2*F^((b*c*d)/e)*Log[F]*(2*e 
- I*b*c*Log[F]))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.80, 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.

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

\(\Big \downarrow \) 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]*Sec[d + e*x]^3,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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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