\(\int F^{c (a+b x)} \csc ^3(d+e x) \, dx\) [7]

Optimal result

Integrand size = 18, antiderivative size = 137 \[ \int F^{c (a+b x)} \csc ^3(d+e x) \, dx=-\frac {F^{c (a+b x)} \cot (d+e x) \csc (d+e x)}{2 e}-\frac {b c F^{c (a+b x)} \csc (d+e x) \log (F)}{2 e^2}-\frac {e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right ) (e+i b c \log (F))}{e^2} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(334\) vs. \(2(137)=274\).

Time = 3.31 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.44 \[ \int F^{c (a+b x)} \csc ^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (-e \csc ^2\left (\frac {1}{2} (d+e x)\right )-4 b c \csc (d) \log (F)+\csc (d) \left (\frac {4 e^2}{b c \log (F)}+4 b c \log (F)\right )+e \sec ^2\left (\frac {1}{2} (d+e x)\right )-\frac {4 i \left (e^2+b^2 c^2 \log ^2(F)\right ) \left (1+\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},\cos (d+e x)+i \sin (d+e x)\right ) (-1+\cos (d)+i \sin (d))\right )}{b c \log (F) (-1+\cos (d)+i \sin (d))}-\frac {4 i \left (e^2+b^2 c^2 \log ^2(F)\right ) \left (1-\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},-\cos (d+e x)-i \sin (d+e x)\right ) (1+\cos (d)+i \sin (d))\right )}{b c \log (F) (1+\cos (d)+i \sin (d))}+2 b c \csc \left (\frac {d}{2}\right ) \csc \left (\frac {1}{2} (d+e x)\right ) \log (F) \sin \left (\frac {e x}{2}\right )-2 b c \log (F) \sec \left (\frac {d}{2}\right ) \sec \left (\frac {1}{2} (d+e x)\right ) \sin \left (\frac {e x}{2}\right )\right )}{8 e^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4949, 4953}

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.

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

\(\Big \downarrow \) 4949

\(\displaystyle \frac {1}{2} \left (\frac {b^2 c^2 \log ^2(F)}{e^2}+1\right ) \int F^{c (a+b x)} \csc (d+e x)dx-\frac {b c \log (F) \csc (d+e x) F^{c (a+b x)}}{2 e^2}-\frac {\cot (d+e x) \csc (d+e x) F^{c (a+b x)}}{2 e}\)

\(\Big \downarrow \) 4953

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 4949

Int[Csc[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbo 
l] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Csc[d + e*x]^(n - 2)/(e^2*(n - 1) 
*(n - 2))), x] + (-Simp[F^(c*(a + b*x))*Csc[d + e*x]^(n - 1)*(Cos[d + e*x]/ 
(e*(n - 1))), x] + Simp[(e^2*(n - 2)^2 + b^2*c^2*Log[F]^2)/(e^2*(n - 1)*(n 
- 2))   Int[F^(c*(a + b*x))*Csc[d + e*x]^(n - 2), x], x]) /; FreeQ[{F, a, b 
, c, d, e}, x] && NeQ[b^2*c^2*Log[F]^2 + e^2*(n - 2)^2, 0] && GtQ[n, 1] && 
NeQ[n, 2]
 

rule 4953

Int[Csc[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symb 
ol] :> Simp[(-2*I)^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F] 
))*Hypergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(Log[F] 
/(2*e)), E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ 
[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

8*(48*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F)*sin(e*x + d) - 6*(F^(a*c)*b^2*c^2*e 
*log(F)^2 - 15*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) - (48*F^(b*c*x)*F^(a*c) 
*b*c*e^2*log(F)*sin(e*x + d) - 3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 25*F^(a*c)* 
e^3)*F^(b*c*x)*cos(3*e*x + 3*d) - 6*(F^(a*c)*b^2*c^2*e*log(F)^2 - 15*F^(a* 
c)*e^3)*F^(b*c*x)*cos(e*x + d) - (F^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)*b* 
c*e^2*log(F))*F^(b*c*x)*sin(3*e*x + 3*d))*cos(6*e*x + 6*d) + 3*(48*F^(b*c* 
x)*F^(a*c)*b*c*e^2*log(F)*sin(e*x + d) - 3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 2 
5*F^(a*c)*e^3)*F^(b*c*x)*cos(3*e*x + 3*d) - 6*(F^(a*c)*b^2*c^2*e*log(F)^2 
- 15*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) - (F^(a*c)*b^3*c^3*log(F)^3 + 25* 
F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(3*e*x + 3*d))*cos(4*e*x + 4*d) + 3*( 
3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 25*F^(a*c)*e^3)*F^(b*c*x)*cos(2*e*x + 2*d) 
 - (F^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(2* 
e*x + 2*d) - (F^(a*c)*b^2*c^2*e*log(F)^2 + 25*F^(a*c)*e^3)*F^(b*c*x))*cos( 
3*e*x + 3*d) - 18*(8*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F)*sin(e*x + d) - (F^(a 
*c)*b^2*c^2*e*log(F)^2 - 15*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d))*cos(2*e*x 
 + 2*d) - 6*(F^(a*c)*b^5*c^5*e*log(F)^5*sin(d) + F^(a*c)*b^4*c^4*e^2*cos(d 
)*log(F)^4 + 34*F^(a*c)*b^3*c^3*e^3*log(F)^3*sin(d) + 34*F^(a*c)*b^2*c^2*e 
^4*cos(d)*log(F)^2 + 225*F^(a*c)*b*c*e^5*log(F)*sin(d) + 225*F^(a*c)*e^6*c 
os(d) + (F^(a*c)*b^5*c^5*e*log(F)^5*sin(d) + F^(a*c)*b^4*c^4*e^2*cos(d)*lo 
g(F)^4 + 34*F^(a*c)*b^3*c^3*e^3*log(F)^3*sin(d) + 34*F^(a*c)*b^2*c^2*e^...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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