\(\int F^{c (a+b x)} \sec ^2(d+e x) \tan (d+e x) \, dx\) [124]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int F^{c (a+b x)} \sec ^2(d+e x) \tan (d+e x) \, dx=\frac {2 e^{2 i (d+e x)} F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )^2}-\frac {2 b c e^{2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \log (F)}{e (2 i e+b c \log (F))} \] Output:

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

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.83 \[ \int F^{c (a+b x)} \sec ^2(d+e x) \tan (d+e x) \, dx=\frac {F^{-\frac {b c d}{e}} \left (-b^2 c^2 e^{\frac {(d+e x) (2 i e+b c \log (F))}{e}} F^{a c} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \log ^2(F)+b c F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \log (F) (2 i e+b c \log (F))+F^{c \left (a+b \left (\frac {d}{e}+x\right )\right )} (2 e-i b c \log (F)) \left (e \sec ^2(d+e x)-b c \log (F) \tan (d+e x)\right )\right )}{2 e^2 (2 e-i b c \log (F))} \] Input:

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

Output:

(-(b^2*c^2*E^(((d + e*x)*((2*I)*e + b*c*Log[F]))/e)*F^(a*c)*Hypergeometric 
2F1[1, 1 - ((I/2)*b*c*Log[F])/e, 2 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + 
e*x))]*Log[F]^2) + b*c*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))]*Log[F]* 
((2*I)*e + b*c*Log[F]) + F^(c*(a + b*(d/e + x)))*(2*e - I*b*c*Log[F])*(e*S 
ec[d + e*x]^2 - b*c*Log[F]*Tan[d + e*x]))/(2*e^2*F^((b*c*d)/e)*(2*e - I*b* 
c*Log[F]))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.36, 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.

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

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (\frac {8 i e^{2 i d+2 i e x} F^{a c+b c x}}{\left (1+e^{2 i (d+e x)}\right )^3}-\frac {4 i e^{2 i d+2 i e x} F^{a c+b c x}}{\left (1+e^{2 i (d+e x)}\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {8 e^{2 i d+2 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (2-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{2 e-i b c \log (F)}-\frac {4 e^{2 i d+2 i e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{2 e-i b c \log (F)}\)

Input:

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

Output:

(-4*E^((2*I)*d + (2*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, (2 - (I*b 
*c*Log[F])/e)/2, (4 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))])/(2*e - I 
*b*c*Log[F]) + (8*E^((2*I)*d + (2*I)*e*x)*F^(a*c + b*c*x)*Hypergeometric2F 
1[3, (2 - (I*b*c*Log[F])/e)/2, (4 - (I*b*c*Log[F])/e)/2, -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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

4*(6*(F^(a*c)*b^4*c^4*e*log(F)^4 + 52*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 576*F 
^(a*c)*e^5)*F^(b*c*x)*cos(4*e*x + 4*d)^2 - 6*(5*F^(a*c)*b^4*c^4*e*log(F)^4 
 + 164*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 576*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x 
 + 2*d)^2 + 6*(F^(a*c)*b^4*c^4*e*log(F)^4 + 52*F^(a*c)*b^2*c^2*e^3*log(F)^ 
2 + 576*F^(a*c)*e^5)*F^(b*c*x)*sin(4*e*x + 4*d)^2 - 6*(5*F^(a*c)*b^4*c^4*e 
*log(F)^4 + 164*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 576*F^(a*c)*e^5)*F^(b*c*x)* 
sin(2*e*x + 2*d)^2 + 2*(13*F^(a*c)*b^4*c^4*e*log(F)^4 - 956*F^(a*c)*b^2*c^ 
2*e^3*log(F)^2 + 576*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) - (F^(a*c)*b^ 
5*c^5*log(F)^5 - 428*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 576*F^(a*c)*b*c*e^4*lo 
g(F))*F^(b*c*x)*sin(2*e*x + 2*d) + 12*(F^(a*c)*b^4*c^4*e*log(F)^4 - 44*F^( 
a*c)*b^2*c^2*e^3*log(F)^2)*F^(b*c*x) + (2*(F^(a*c)*b^4*c^4*e*log(F)^4 + 52 
*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 576*F^(a*c)*e^5)*F^(b*c*x)*cos(4*e*x + 4*d 
) - 2*(5*F^(a*c)*b^4*c^4*e*log(F)^4 + 164*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 5 
76*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) + (F^(a*c)*b^5*c^5*log(F)^5 + 5 
2*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 576*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin 
(4*e*x + 4*d) - (F^(a*c)*b^5*c^5*log(F)^5 + 4*F^(a*c)*b^3*c^3*e^2*log(F)^3 
 - 1152*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) + 12*(F^(a*c)*b 
^4*c^4*e*log(F)^4 - 44*F^(a*c)*b^2*c^2*e^3*log(F)^2)*F^(b*c*x))*cos(6*e*x 
+ 6*d) - 2*(12*(F^(a*c)*b^4*c^4*e*log(F)^4 + 28*F^(a*c)*b^2*c^2*e^3*log(F) 
^2 - 288*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) + 3*(F^(a*c)*b^5*c^5*l...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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