\(\int F^{c (a+b x)} \sin (d+e x) \tan (d+e x) \, dx\) [109]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int F^{c (a+b x)} \sin (d+e x) \tan (d+e x) \, dx=-\frac {3 e^{-i (d+e x)} F^{c (a+b x)}}{2 (i e-b c \log (F))}+\frac {2 e^{-i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {-e-i b c \log (F)}{2 e},\frac {e-i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{i e-b c \log (F)}-\frac {e^{i (d+e x)} F^{c (a+b x)}}{2 (i e+b c \log (F))} \] Output:

-3/2*F^(c*(b*x+a))/exp(I*(e*x+d))/(I*e-b*c*ln(F))+2*F^(c*(b*x+a))*hypergeo 
m([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)))/e 
xp(I*(e*x+d))/(I*e-b*c*ln(F))-exp(I*(e*x+d))*F^(c*(b*x+a))/(2*I*e+2*b*c*ln 
(F))
 

Mathematica [A] (verified)

Time = 4.52 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.82 \[ \int F^{c (a+b x)} \sin (d+e x) \tan (d+e x) \, dx=\frac {2 e^{i (d+e x)} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{i e+b c \log (F)}-\frac {F^{c (a+b x)} (b c \cos (d+e x) \log (F)+e \sin (d+e x))}{e^2+b^2 c^2 \log ^2(F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 179, 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.

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

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(F^(b*c*x + a*c)*sin(e*x + d)*tan(e*x + d), x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a))*sin(e*x+d)*tan(e*x+d),x)
 

Output:

Integral(F**(c*(a + b*x))*sin(d + e*x)*tan(d + e*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

integrate(F^((b*x + a)*c)*sin(e*x + d)*tan(e*x + d), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

f**(a*c)*int(f**(b*c*x)*sin(d + e*x)*tan(d + e*x),x)