\(\int F^{c (a+b x)} \cos (d+e x) \cot (d+e x) \, dx\) [98]

Optimal result

Integrand size = 22, antiderivative size = 156 \[ \int F^{c (a+b x)} \cos (d+e x) \cot (d+e x) \, dx=\frac {e^{i (d+e x)} F^{c (a+b x)}}{2 (e-i b c \log (F))}-\frac {3 e^{-i (d+e x)} F^{c (a+b x)}}{2 (e+i b c \log (F))}+\frac {2 e^{-i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {-e-i b c \log (F)}{2 e},\frac {e-i b c \log (F)}{2 e},e^{2 i (d+e x)}\right )}{e+i b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 5.29 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.67 \[ \int F^{c (a+b x)} \cos (d+e x) \cot (d+e x) \, dx=F^{c (a+b x)} \left (\frac {\csc (d)}{b c \log (F)}-\frac {i \left (1+\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},\cos (d+e x)+i \sin (d+e x)\right ) (-1+\cos (d)+i \sin (d))\right )}{b c \log (F) (-1+\cos (d)+i \sin (d))}-\frac {i \left (1-\operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},-\cos (d+e x)-i \sin (d+e x)\right ) (1+\cos (d)+i \sin (d))\right )}{b c \log (F) (1+\cos (d)+i \sin (d))}+\frac {\cos (e x) (e \cos (d)-b c \log (F) \sin (d))}{e^2+b^2 c^2 \log ^2(F)}-\frac {(b c \cos (d) \log (F)+e \sin (d)) \sin (e x)}{e^2+b^2 c^2 \log ^2(F)}\right ) \] Input:

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

Output:

F^(c*(a + b*x))*(Csc[d]/(b*c*Log[F]) - (I*(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])) - (I*(1 - Hyper 
geometric2F1[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])) + (Cos[e*x]*(e*Cos[d] - b*c*Log[F]*Sin[d]))/(e^2 + b^2*c^2*Log[F]^2) 
 - ((b*c*Cos[d]*Log[F] + e*Sin[d])*Sin[e*x])/(e^2 + b^2*c^2*Log[F]^2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

int(F^(c*(b*x+a))*cos(e*x+d)*cot(e*x+d),x)
 

Output:

int(F^(c*(b*x+a))*cos(e*x+d)*cot(e*x+d),x)
 

Fricas [F]

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

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

Output:

integral(F^(b*c*x + a*c)*cos(e*x + d)*cot(e*x + d), x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a))*cos(e*x+d)*cot(e*x+d),x)
 

Output:

Integral(F**(c*(a + b*x))*cos(d + e*x)*cot(d + e*x), x)
 

Maxima [F]

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

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

Output:

1/2*(3*(3*F^(a*c)*b^2*c^2*e*log(F)^2 - 5*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + 
d) - (F^(a*c)*b^3*c^3*log(F)^3 - 23*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin( 
e*x + d) + ((F^(a*c)*b^2*c^2*e*log(F)^2 + 9*F^(a*c)*e^3)*F^(b*c*x)*cos(3*e 
*x + 3*d) - (F^(a*c)*b^2*c^2*e*log(F)^2 + 9*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x 
 + d) + (F^(a*c)*b^3*c^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*si 
n(3*e*x + 3*d) - (F^(a*c)*b^3*c^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*log(F))*F^( 
b*c*x)*sin(e*x + d))*cos(4*e*x + 4*d) - (4*(F^(a*c)*b^2*c^2*e*log(F)^2 + 9 
*F^(a*c)*e^3)*F^(b*c*x)*cos(2*e*x + 2*d) + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 9 
*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) + 3*(3*F^(a*c)*b^2*c^2 
*e*log(F)^2 - 5*F^(a*c)*e^3)*F^(b*c*x))*cos(3*e*x + 3*d) + 2*(2*(F^(a*c)*b 
^2*c^2*e*log(F)^2 + 9*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) - (F^(a*c)*b^3*c 
^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(e*x + d))*cos(2*e*x 
+ 2*d) - 8*((F^(a*c)*b^4*c^4*e*log(F)^4 + 10*F^(a*c)*b^2*c^2*e^3*log(F)^2 
+ 9*F^(a*c)*e^5)*cos(3*e*x + 3*d)^2 - 2*(F^(a*c)*b^4*c^4*e*log(F)^4 + 10*F 
^(a*c)*b^2*c^2*e^3*log(F)^2 + 9*F^(a*c)*e^5)*cos(3*e*x + 3*d)*cos(e*x + d) 
 + (F^(a*c)*b^4*c^4*e*log(F)^4 + 10*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 9*F^(a* 
c)*e^5)*cos(e*x + d)^2 + (F^(a*c)*b^4*c^4*e*log(F)^4 + 10*F^(a*c)*b^2*c^2* 
e^3*log(F)^2 + 9*F^(a*c)*e^5)*sin(3*e*x + 3*d)^2 - 2*(F^(a*c)*b^4*c^4*e*lo 
g(F)^4 + 10*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 9*F^(a*c)*e^5)*sin(3*e*x + 3*d) 
*sin(e*x + d) + (F^(a*c)*b^4*c^4*e*log(F)^4 + 10*F^(a*c)*b^2*c^2*e^3*lo...
 

Giac [F]

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

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

Output:

integrate(F^((b*x + a)*c)*cos(e*x + d)*cot(e*x + d), x)
 

Mupad [F(-1)]

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

int(F^(c*(a + b*x))*cos(d + e*x)*cot(d + e*x),x)
 

Output:

int(F^(c*(a + b*x))*cos(d + e*x)*cot(d + e*x), x)
 

Reduce [F]

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

int(F^(c*(b*x+a))*cos(e*x+d)*cot(e*x+d),x)
 

Output:

f**(a*c)*int(f**(b*c*x)*cos(d + e*x)*cot(d + e*x),x)