\(\int F^{c (a+b x)} \sin ^2(d+e x) \tan (d+e x) \, dx\) [108]

Optimal result

Integrand size = 24, antiderivative size = 181 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \tan (d+e x) \, dx=-\frac {i F^{c (a+b x)}}{b c \log (F)}+\frac {e^{2 i (d+e x)} F^{c (a+b x)}}{4 (2 e-i b c \log (F))}-\frac {7 e^{-2 i (d+e x)} F^{c (a+b x)}}{4 (2 e+i b c \log (F))}+\frac {2 e^{-2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (-2-\frac {i b c \log (F)}{e}\right ),-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{2 e+i b c \log (F)} \] Output:

-I*F^(c*(b*x+a))/b/c/ln(F)+exp(2*I*(e*x+d))*F^(c*(b*x+a))/(8*e-4*I*b*c*ln( 
F))-7/4*F^(c*(b*x+a))/exp(2*I*(e*x+d))/(I*b*c*ln(F)+2*e)+2*F^(c*(b*x+a))*h 
ypergeom([1, -1-1/2*I*b*c*ln(F)/e],[-1/2*I*b*c*ln(F)/e],-exp(2*I*(e*x+d))) 
/exp(2*I*(e*x+d))/(I*b*c*ln(F)+2*e)
 

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.39 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \tan (d+e x) \, dx=\frac {F^{-\frac {b c d}{e}} \left (2 i F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \left (4 e^2+b^2 c^2 \log ^2(F)\right )-b c \log (F) \left (2 e^{\frac {(d+e x) (2 i e+b c \log (F))}{e}} F^{a c} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) (2 e+i b c \log (F))+F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} (-2 e \cos (2 (d+e x))+b c \log (F) \sin (2 (d+e x)))\right )\right )}{2 \left (4 b c e^2 \log (F)+b^3 c^3 \log ^3(F)\right )} \] Input:

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

Output:

((2*I)*F^(c*(a + b*(d/e + x)))*Hypergeometric2F1[1, ((-1/2*I)*b*c*Log[F])/ 
e, 1 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + e*x))]*(4*e^2 + b^2*c^2*Log[F] 
^2) - b*c*Log[F]*(2*E^(((d + e*x)*((2*I)*e + b*c*Log[F]))/e)*F^(a*c)*Hyper 
geometric2F1[1, 1 - ((I/2)*b*c*Log[F])/e, 2 - ((I/2)*b*c*Log[F])/e, -E^((2 
*I)*(d + e*x))]*(2*e + I*b*c*Log[F]) + F^(c*(a + b*(d/e + x)))*(-2*e*Cos[2 
*(d + e*x)] + b*c*Log[F]*Sin[2*(d + e*x)])))/(2*F^((b*c*d)/e)*(4*b*c*e^2*L 
og[F] + b^3*c^3*Log[F]^3))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

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

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (\frac {7}{4} i e^{-2 i d-2 i e x} F^{a c+b c x}-i e^{2 i (d+e x)-2 i d-2 i e x} F^{a c+b c x}+\frac {1}{4} i e^{4 i (d+e x)-2 i d-2 i e x} F^{a c+b c x}-\frac {2 i e^{-2 i d-2 i e x} F^{a c+b c x}}{1+e^{2 i (d+e x)}}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((-I)*F^(a*c + b*c*x))/(b*c*Log[F]) + (E^((2*I)*d + x*((2*I)*e + b*c*Log[F 
]))*F^(a*c))/(4*(2*e - I*b*c*Log[F])) - (7*E^((-2*I)*d - x*((2*I)*e - b*c* 
Log[F]))*F^(a*c))/(4*(2*e + I*b*c*Log[F])) + (2*E^((-2*I)*d - (2*I)*e*x)*F 
^(a*c + b*c*x)*Hypergeometric2F1[1, (-2 - (I*b*c*Log[F])/e)/2, ((-1/2*I)*b 
*c*Log[F])/e, -E^((2*I)*(d + e*x))])/(2*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(cos(e*x + d)^2 - 1)*F^(b*c*x + a*c)*tan(e*x + d), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/4*(2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 16*F^(a*c)*e^3)*F^(b*c*x)*cos(4*e*x + 
 4*d)^2 - 14*(F^(a*c)*b^2*c^2*e*log(F)^2 + 16*F^(a*c)*e^3)*F^(b*c*x)*cos(2 
*e*x + 2*d)^2 + 2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 16*F^(a*c)*e^3)*F^(b*c*x)* 
sin(4*e*x + 4*d)^2 - 14*(F^(a*c)*b^2*c^2*e*log(F)^2 + 16*F^(a*c)*e^3)*F^(b 
*c*x)*sin(2*e*x + 2*d)^2 + 6*(3*F^(a*c)*b^2*c^2*e*log(F)^2 - 16*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 - 80*F^(a*c)*b*c 
*e^2*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) + (2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 
 16*F^(a*c)*e^3)*F^(b*c*x)*cos(4*e*x + 4*d) + 2*(F^(a*c)*b^2*c^2*e*log(F)^ 
2 + 16*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 
 + 16*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(4*e*x + 4*d) + (F^(a*c)*b^3*c^ 
3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(2*e*x + 2*d))*cos(6* 
e*x + 6*d) - 6*(2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 16*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 + 16*F^(a*c)*b*c*e^2*log(F))* 
F^(b*c*x)*sin(2*e*x + 2*d) - (3*F^(a*c)*b^2*c^2*e*log(F)^2 - 16*F^(a*c)*e^ 
3)*F^(b*c*x))*cos(4*e*x + 4*d) - 16*((F^(a*c)*b^4*c^4*e*log(F)^4 + 20*F^(a 
*c)*b^2*c^2*e^3*log(F)^2 + 64*F^(a*c)*e^5)*cos(4*e*x + 4*d)^2 + 2*(F^(a*c) 
*b^4*c^4*e*log(F)^4 + 20*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 64*F^(a*c)*e^5)*co 
s(4*e*x + 4*d)*cos(2*e*x + 2*d) + (F^(a*c)*b^4*c^4*e*log(F)^4 + 20*F^(a*c) 
*b^2*c^2*e^3*log(F)^2 + 64*F^(a*c)*e^5)*cos(2*e*x + 2*d)^2 + (F^(a*c)*b^4* 
c^4*e*log(F)^4 + 20*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 64*F^(a*c)*e^5)*sin(...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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