\(\int F^{c (a+b x)} \sec (d+e x) \tan ^2(d+e x) \, dx\) [123]

Optimal result

Integrand size = 24, antiderivative size = 196 \[ \int F^{c (a+b x)} \sec (d+e x) \tan ^2(d+e x) \, dx=\frac {2 i e^{i (d+e x)} F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )^2}-\frac {e^{i (d+e x)} F^{c (a+b x)} (i e+b c \log (F))}{e^2 \left (1+e^{2 i (d+e x)}\right )}-\frac {e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) (e-b c \log (F)) (e+b c \log (F))}{e^2 (i e+b c \log (F))} \] Output:

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

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.86 \[ \int F^{c (a+b x)} \sec (d+e x) \tan ^2(d+e x) \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (2 i e^{\frac {(d+e x) (i e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \left (e^2-b^2 c^2 \log ^2(F)\right )+F^{\frac {b c (d+e x)}{e}} (e-i b c \log (F)) \sec (d+e x) (-b c \log (F)+e \tan (d+e x))\right )}{2 e^2 (e-i b c \log (F))} \] Input:

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

Output:

(F^(c*(a - (b*d)/e))*((2*I)*E^(((d + e*x)*(I*e + b*c*Log[F]))/e)*Hypergeom 
etric2F1[1, (e - I*b*c*Log[F])/(2*e), 3/2 - ((I/2)*b*c*Log[F])/e, -E^((2*I 
)*(d + e*x))]*(e^2 - b^2*c^2*Log[F]^2) + F^((b*c*(d + e*x))/e)*(e - I*b*c* 
Log[F])*Sec[d + e*x]*(-(b*c*Log[F]) + e*Tan[d + e*x])))/(2*e^2*(e - I*b*c* 
Log[F]))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.35, 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.

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

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (\frac {2 e^{i d+i e x} F^{a c+b c x}}{-1-e^{2 i (d+e x)}}+\frac {8 e^{i d+i e x} F^{a c+b c x}}{\left (1+e^{2 i (d+e x)}\right )^2}-\frac {8 e^{i d+i e x} F^{a c+b c x}}{\left (1+e^{2 i (d+e x)}\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 e^{i d+i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{b c \log (F)+i e}+\frac {8 e^{i d+i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{b c \log (F)+i e}-\frac {8 e^{i d+i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{b c \log (F)+i e}\)

Input:

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

Output:

(-2*E^(I*d + I*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (e - I*b*c*Log[F] 
)/(2*e), (3 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))])/(I*e + b*c*Log[F 
]) + (8*E^(I*d + I*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, (e - I*b*c*Lo 
g[F])/(2*e), (3 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))])/(I*e + b*c*L 
og[F]) - (8*E^(I*d + I*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[3, (e - I*b* 
c*Log[F])/(2*e), (3 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))])/(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-2*((F^(a*c)*b^5*c^5*log(F)^5 - 158*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 417*F^( 
a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(e*x + d) - (23*F^(a*c)*b^4*c^4*e*log(F) 
^4 - 418*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 135*F^(a*c)*e^5)*F^(b*c*x)*sin(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)*cos(5*e*x + 5*d) - 2*(F^(a*c)*b^5*c^5*l 
og(F)^5 + 14*F^(a*c)*b^3*c^3*e^2*log(F)^3 - 275*F^(a*c)*b*c*e^4*log(F))*F^ 
(b*c*x)*cos(3*e*x + 3*d) + (F^(a*c)*b^5*c^5*log(F)^5 - 158*F^(a*c)*b^3*c^3 
*e^2*log(F)^3 + 417*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(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)*sin(5*e*x + 5*d) + 12*(F^(a*c)*b^4*c^4*e*log(F)^4 + 24*F^(a*c) 
*b^2*c^2*e^3*log(F)^2 - 25*F^(a*c)*e^5)*F^(b*c*x)*sin(3*e*x + 3*d) - (23*F 
^(a*c)*b^4*c^4*e*log(F)^4 - 418*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 135*F^(a*c) 
*e^5)*F^(b*c*x)*sin(e*x + d))*cos(6*e*x + 6*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)*cos(4*e*x + 4*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)*cos(2*e*x + 2*d) + 3*(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)*sin(4*e*x + 4*d) + 3*(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)*sin(2*e*x + 2*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)...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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