\(\int F^{c (a+b x)} \csc ^2(d+e x) \sec ^3(d+e x) \, dx\) [127]

Optimal result

Integrand size = 26, antiderivative size = 346 \[ \int F^{c (a+b x)} \csc ^2(d+e x) \sec ^3(d+e x) \, dx=-\frac {2 i e^{5 i (d+e x)} F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )^2}+\frac {8 i e^{5 i (d+e x)} F^{c (a+b x)}}{e \left (1-e^{2 i (d+e x)}\right ) \left (1+e^{2 i (d+e x)}\right )^2}+\frac {e^{5 i (d+e x)} F^{c (a+b x)} (i e-b c \log (F))}{e^2 \left (1+e^{2 i (d+e x)}\right )}-\frac {2 b c e^{5 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (7-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right ) \log (F)}{e (5 e-i b c \log (F))}+\frac {e^{5 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (7-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \left (3 e^2+b^2 c^2 \log ^2(F)\right )}{e^2 (5 i e+b c \log (F))} \] Output:

-2*I*exp(5*I*(e*x+d))*F^(c*(b*x+a))/e/(1+exp(2*I*(e*x+d)))^2+8*I*exp(5*I*( 
e*x+d))*F^(c*(b*x+a))/e/(1-exp(2*I*(e*x+d)))/(1+exp(2*I*(e*x+d)))^2+exp(5* 
I*(e*x+d))*F^(c*(b*x+a))*(I*e-b*c*ln(F))/e^2/(1+exp(2*I*(e*x+d)))-2*b*c*ex 
p(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*e-I*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*I*e+b*c*ln(F))
 

Mathematica [A] (verified)

Time = 2.98 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} \csc ^2(d+e x) \sec ^3(d+e x) \, dx=-\frac {F^{c \left (a-\frac {b d}{e}\right )} \csc \left (\frac {1}{2} (d+e x)\right ) \sec \left (\frac {1}{2} (d+e x)\right ) \left (8 b c e e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},e^{2 i (d+e x)}\right ) \log (F) \sin (d+e x)+4 i e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \left (3 e^2+b^2 c^2 \log ^2(F)\right ) \sin (d+e x)+F^{\frac {b c (d+e x)}{e}} (e-i b c \log (F)) \sec ^2(d+e x) (e+3 e \cos (2 (d+e x))+b c \log (F) \sin (2 (d+e x)))\right )}{8 e^2 (e-i b c \log (F))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.01, 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.

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

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

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

Input:

Int[F^(c*(a + b*x))*Csc[d + e*x]^2*Sec[d + e*x]^3,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*I)*e 
 + b*c*Log[F]) - (8*E^((5*I)*d + (5*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric 
2F1[2, (5 - (I*b*c*Log[F])/e)/2, (7 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + 
e*x))])/((5*I)*e + 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*I)*e + 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*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

(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)*cos(e*x + d) + 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 - 10395*F^(a*c)*e^5)*F^(b*c*x)*sin(e*x + d) - 32*((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*lo 
g(F))*F^(b*c*x)*cos(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)*cos(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)*cos(e*x + d) - 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)*sin(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)*sin(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)*sin(e*x + d))*cos(10*e*x + 10*d) - 32*((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)*cos(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)*cos(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)*cos(e*x + d) - 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)*sin(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)*sin(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...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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