\(\int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx\) [121]

Optimal result

Integrand size = 26, antiderivative size = 327 \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=\frac {4 e^{5 i (d+e x)} F^{c (a+b x)}}{e \left (1-e^{2 i (d+e x)}\right )^2 \left (1+e^{2 i (d+e x)}\right )}-\frac {2 b c e^{5 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (7-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \log (F)}{e (5 i e+b c \log (F))}-\frac {e^{5 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (7-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right ) \left (3 e^2+b^2 c^2 \log ^2(F)\right )}{e^2 (5 e-i b c \log (F))}+\frac {e^{5 i (d+e x)} F^{c (a+b x)} \left (3 e+i b c \log (F)+e^{2 i (d+e x)} (e+i b c \log (F))\right )}{e^2 \left (1-e^{4 i (d+e x)}\right )} \] Output:

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

Mathematica [A] (verified)

Time = 2.64 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=-\frac {F^{c \left (a-\frac {b d}{e}\right )} \csc ^2(d+e x) \left (-8 i b c e e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \cos (d+e x) \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \log (F)+4 e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \cos (d+e x) \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right ) \left (3 e^2+b^2 c^2 \log ^2(F)\right )+F^{\frac {b c (d+e x)}{e}} \csc ^2(d+e x) (e-i b c \log (F)) (-e+3 e \cos (2 (d+e x))+b c \log (F) \sin (2 (d+e x)))\right )}{e^2 (e-i b c \log (F)) \left (\csc \left (\frac {1}{2} (d+e x)\right )-\sec \left (\frac {1}{2} (d+e x)\right )\right ) \left (\csc \left (\frac {1}{2} (d+e x)\right )+\sec \left (\frac {1}{2} (d+e x)\right )\right )} \] Input:

Integrate[F^(c*(a + b*x))*Csc[d + e*x]^3*Sec[d + e*x]^2,x]
 

Output:

-((F^(c*(a - (b*d)/e))*Csc[d + e*x]^2*((-8*I)*b*c*e*E^(((d + e*x)*(I*e + b 
*c*Log[F]))/e)*Cos[d + e*x]*Hypergeometric2F1[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] + 4*E^(((d + e*x 
)*(I*e + b*c*Log[F]))/e)*Cos[d + e*x]*Hypergeometric2F1[1, (e - I*b*c*Log[ 
F])/(2*e), 3/2 - ((I/2)*b*c*Log[F])/e, E^((2*I)*(d + e*x))]*(3*e^2 + b^2*c 
^2*Log[F]^2) + F^((b*c*(d + e*x))/e)*Csc[d + e*x]^2*(e - I*b*c*Log[F])*(-e 
 + 3*e*Cos[2*(d + e*x)] + b*c*Log[F]*Sin[2*(d + e*x)])))/(e^2*(e - I*b*c*L 
og[F])*(Csc[(d + e*x)/2] - Sec[(d + e*x)/2])*(Csc[(d + e*x)/2] + Sec[(d + 
e*x)/2])))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.07, 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.

\(\displaystyle \int \csc ^3(d+e x) \sec ^2(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (-\frac {12 i e^{5 i d+5 i e x} F^{a c+b c x}}{-1+e^{4 i (d+e x)}}+\frac {8 i e^{5 i d+5 i e x} F^{a c+b c x}}{\left (-1+e^{2 i (d+e x)}\right )^2}+\frac {4 i e^{5 i d+5 i e x} F^{a c+b c x}}{\left (1+e^{2 i (d+e x)}\right )^2}-\frac {8 i e^{5 i d+5 i e x} F^{a c+b c x}}{\left (-1+e^{2 i (d+e x)}\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {12 e^{5 i d+5 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{4} \left (9-\frac {i b c \log (F)}{e}\right ),e^{4 i (d+e x)}\right )}{5 e-i b c \log (F)}+\frac {4 e^{5 i d+5 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (7-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{5 e-i b c \log (F)}+\frac {8 e^{5 i d+5 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (7-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{5 e-i b c \log (F)}+\frac {8 e^{5 i d+5 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (7-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{5 e-i b c \log (F)}\)

Input:

Int[F^(c*(a + b*x))*Csc[d + e*x]^3*Sec[d + e*x]^2,x]
 

Output:

(12*E^((5*I)*d + (5*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (5 - (I*b 
*c*Log[F])/e)/4, (9 - (I*b*c*Log[F])/e)/4, E^((4*I)*(d + e*x))])/(5*e - I* 
b*c*Log[F]) + (4*E^((5*I)*d + (5*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1 
[2, (5 - (I*b*c*Log[F])/e)/2, (7 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x 
))])/(5*e - I*b*c*Log[F]) + (8*E^((5*I)*d + (5*I)*e*x)*F^(a*c + b*c*x)*Hyp 
ergeometric2F1[2, (5 - (I*b*c*Log[F])/e)/2, (7 - (I*b*c*Log[F])/e)/2, E^(( 
2*I)*(d + e*x))])/(5*e - I*b*c*Log[F]) + (8*E^((5*I)*d + (5*I)*e*x)*F^(a*c 
 + b*c*x)*Hypergeometric2F1[3, (5 - (I*b*c*Log[F])/e)/2, (7 - (I*b*c*Log[F 
])/e)/2, E^((2*I)*(d + e*x))])/(5*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]

\[\int F^{c \left (b x +a \right )} \csc \left (e x +d \right )^{3} \sec \left (e x +d \right )^{2}d x\]

Input:

int(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d)^2,x)
 

Output:

int(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d)^2,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right )^{3} \sec \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d)^2,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*csc(e*x + d)^3*sec(e*x + d)^2, x)
 

Sympy [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=\text {Timed out} \] Input:

integrate(F**(c*(b*x+a))*csc(e*x+d)**3*sec(e*x+d)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right )^{3} \sec \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d)^2,x, algorithm="maxima")
 

Output:

-(64*(5*F^(a*c)*b^4*c^4*e*log(F)^4 - 22*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 103 
95*F^(a*c)*e^5)*F^(b*c*x)*cos(e*x + d) - 1536*(3*F^(a*c)*b^3*c^3*e^2*log(F 
)^3 + 131*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(e*x + d) - 32*(5*(F^(a*c)* 
b^4*c^4*e*log(F)^4 + 130*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 3969*F^(a*c)*e^5)* 
F^(b*c*x)*cos(5*e*x + 5*d) + 2*(F^(a*c)*b^4*c^4*e*log(F)^4 + 46*F^(a*c)*b^ 
2*c^2*e^3*log(F)^2 - 2835*F^(a*c)*e^5)*F^(b*c*x)*cos(3*e*x + 3*d) + 2*(5*F 
^(a*c)*b^4*c^4*e*log(F)^4 - 22*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 10395*F^(a*c 
)*e^5)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^5*c^5*log(F)^5 + 130*F^(a*c)*b^ 
3*c^3*e^2*log(F)^3 + 3969*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(5*e*x + 5* 
d) - 24*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 81*F^(a*c)*b*c*e^4*log(F))*F^(b*c* 
x)*sin(3*e*x + 3*d) - 48*(3*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 131*F^(a*c)*b*c 
*e^4*log(F))*F^(b*c*x)*sin(e*x + d))*cos(10*e*x + 10*d) + 32*(5*(F^(a*c)*b 
^4*c^4*e*log(F)^4 + 130*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 3969*F^(a*c)*e^5)*F 
^(b*c*x)*cos(5*e*x + 5*d) + 2*(F^(a*c)*b^4*c^4*e*log(F)^4 + 46*F^(a*c)*b^2 
*c^2*e^3*log(F)^2 - 2835*F^(a*c)*e^5)*F^(b*c*x)*cos(3*e*x + 3*d) + 2*(5*F^ 
(a*c)*b^4*c^4*e*log(F)^4 - 22*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 10395*F^(a*c) 
*e^5)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^5*c^5*log(F)^5 + 130*F^(a*c)*b^3 
*c^3*e^2*log(F)^3 + 3969*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(5*e*x + 5*d 
) - 24*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 81*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x 
)*sin(3*e*x + 3*d) - 48*(3*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 131*F^(a*c)*b...
 

Giac [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \csc \left (e x + d\right )^{3} \sec \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d)^2,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*csc(e*x + d)^3*sec(e*x + d)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\cos \left (d+e\,x\right )}^2\,{\sin \left (d+e\,x\right )}^3} \,d x \] Input:

int(F^(c*(a + b*x))/(cos(d + e*x)^2*sin(d + e*x)^3),x)
 

Output:

int(F^(c*(a + b*x))/(cos(d + e*x)^2*sin(d + e*x)^3), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \csc ^3(d+e x) \sec ^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \csc \left (e x +d \right )^{3} \sec \left (e x +d \right )^{2}d x \right ) \] Input:

int(F^(c*(b*x+a))*csc(e*x+d)^3*sec(e*x+d)^2,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*csc(d + e*x)**3*sec(d + e*x)**2,x)