\(\int F^{c (a+b x)} \sin (d+e x) \tan ^2(d+e x) \, dx\) [115]

Optimal result

Integrand size = 24, antiderivative size = 207 \[ \int F^{c (a+b x)} \sin (d+e x) \tan ^2(d+e x) \, dx=-\frac {2 e^{-i (d+e x)} F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )}-\frac {2 b c e^{-i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {-e-i b c \log (F)}{2 e},\frac {e-i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \log (F)}{e (i e-b c \log (F))}+\frac {e^{i (d+e x)} F^{c (a+b x)}}{2 (e-i b c \log (F))}+\frac {5 e^{-i (d+e x)} F^{c (a+b x)}}{2 (e+i b c \log (F))} \] Output:

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.19 \[ \int F^{c (a+b x)} \sin (d+e x) \tan ^2(d+e x) \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (4 i b c e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \cos (d+e x) \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \log (F) (e+i b c \log (F))+F^{\frac {b c (d+e x)}{e}} \left (3 e^2+e^2 \cos (2 (d+e x))+2 b^2 c^2 \log ^2(F)-b c e \log (F) \sin (2 (d+e x))\right )\right )}{2 e (e-i b c \log (F)) (e+i b c \log (F)) \left (\cos \left (\frac {1}{2} (d+e x)\right )-\sin \left (\frac {1}{2} (d+e x)\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )} \] Input:

Integrate[F^(c*(a + b*x))*Sin[d + e*x]*Tan[d + e*x]^2,x]
 

Output:

(F^(c*(a - (b*d)/e))*((4*I)*b*c*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*Lo 
g[F])/e, -E^((2*I)*(d + e*x))]*Log[F]*(e + I*b*c*Log[F]) + F^((b*c*(d + e* 
x))/e)*(3*e^2 + e^2*Cos[2*(d + e*x)] + 2*b^2*c^2*Log[F]^2 - b*c*e*Log[F]*S 
in[2*(d + e*x)])))/(2*e*(e - I*b*c*Log[F])*(e + I*b*c*Log[F])*(Cos[(d + e* 
x)/2] - Sin[(d + e*x)/2])*(Cos[(d + e*x)/2] + Sin[(d + e*x)/2]))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.26, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

\(\displaystyle \int \sin (d+e x) \tan ^2(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (-\frac {5}{2} i e^{-i d-i e x} F^{a c+b c x}+\frac {1}{2} i e^{2 i (d+e x)-i d-i e x} F^{a c+b c x}+\frac {6 i e^{-i d-i e x} F^{a c+b c x}}{1+e^{2 i (d+e x)}}-\frac {4 i e^{-i d-i e x} F^{a c+b c x}}{\left (1+e^{2 i (d+e x)}\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {6 e^{-i d-i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,-\frac {e+i b c \log (F)}{2 e},\frac {e-i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{e+i b c \log (F)}+\frac {4 e^{-i d-i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,-\frac {e+i b c \log (F)}{2 e},\frac {e-i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{e+i b c \log (F)}+\frac {F^{a c} e^{x (b c \log (F)+i e)+i d}}{2 (e-i b c \log (F))}+\frac {5 F^{a c} e^{-x (-b c \log (F)+i e)-i d}}{2 (e+i b c \log (F))}\)

Input:

Int[F^(c*(a + b*x))*Sin[d + e*x]*Tan[d + e*x]^2,x]
 

Output:

(E^(I*d + x*(I*e + b*c*Log[F]))*F^(a*c))/(2*(e - I*b*c*Log[F])) + (5*E^((- 
I)*d - x*(I*e - b*c*Log[F]))*F^(a*c))/(2*(e + I*b*c*Log[F])) - (6*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]) + (4* 
E^((-I)*d - I*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, -1/2*(e + I*b*c*Lo 
g[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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(F^(c*(b*x+a))*sin(e*x+d)*tan(e*x+d)^2,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*sin(e*x + d)*tan(e*x + d)^2, x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a))*sin(e*x+d)*tan(e*x+d)**2,x)
 

Output:

Integral(F**(c*(a + b*x))*sin(d + e*x)*tan(d + e*x)**2, x)
 

Maxima [F]

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

integrate(F^(c*(b*x+a))*sin(e*x+d)*tan(e*x+d)^2,x, algorithm="maxima")
 

Output:

1/2*(3*(11*F^(a*c)*b^4*c^4*e*log(F)^4 - 234*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 
 75*F^(a*c)*e^5)*F^(b*c*x)*cos(e*x + d) - (F^(a*c)*b^5*c^5*log(F)^5 - 254* 
F^(a*c)*b^3*c^3*e^2*log(F)^3 + 705*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(e 
*x + d) + ((F^(a*c)*b^4*c^4*e*log(F)^4 + 34*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 
 225*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 + 34*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 225*F^(a*c)*e^5)*F^(b*c*x)*cos(3*e* 
x + 3*d) + (F^(a*c)*b^4*c^4*e*log(F)^4 + 34*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 
 225*F^(a*c)*e^5)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^5*c^5*log(F)^5 + 34* 
F^(a*c)*b^3*c^3*e^2*log(F)^3 + 225*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(5 
*e*x + 5*d) + 2*(F^(a*c)*b^5*c^5*log(F)^5 + 34*F^(a*c)*b^3*c^3*e^2*log(F)^ 
3 + 225*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(3*e*x + 3*d) + (F^(a*c)*b^5* 
c^5*log(F)^5 + 34*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 225*F^(a*c)*b*c*e^4*log(F 
))*F^(b*c*x)*sin(e*x + d))*cos(6*e*x + 6*d) + (7*(F^(a*c)*b^4*c^4*e*log(F) 
^4 + 34*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 225*F^(a*c)*e^5)*F^(b*c*x)*cos(4*e* 
x + 4*d) - (25*F^(a*c)*b^4*c^4*e*log(F)^4 + 562*F^(a*c)*b^2*c^2*e^3*log(F) 
^2 - 1575*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) + 3*(F^(a*c)*b^5*c^5*log 
(F)^5 + 34*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 225*F^(a*c)*b*c*e^4*log(F))*F^(b 
*c*x)*sin(4*e*x + 4*d) - 3*(F^(a*c)*b^5*c^5*log(F)^5 + 2*F^(a*c)*b^3*c^3*e 
^2*log(F)^3 - 575*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) + 3*( 
11*F^(a*c)*b^4*c^4*e*log(F)^4 - 234*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 75*F...
 

Giac [F]

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

integrate(F^(c*(b*x+a))*sin(e*x+d)*tan(e*x+d)^2,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*sin(e*x + d)*tan(e*x + d)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

f**(a*c)*int(f**(b*c*x)*sin(d + e*x)*tan(d + e*x)**2,x)