\(\int F^{c (a+b x)} \tan (d+e x) \, dx\) [110]

Optimal result

Integrand size = 16, antiderivative size = 90 \[ \int F^{c (a+b x)} \tan (d+e x) \, dx=-\frac {i F^{c (a+b x)}}{b c \log (F)}+\frac {2 i F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int F^{c (a+b x)} \tan (d+e x) \, dx=\frac {i F^{c (a+b x)} \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )\right )}{b c \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4942, 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.

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

\(\Big \downarrow \) 4942

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

I*(-(F^(c*(a + b*x))/(b*c*Log[F])) + (2*F^(c*(a + b*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)) 
])/(b*c*Log[F]))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4942

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symb 
ol] :> Simp[I^n   Int[ExpandIntegrand[F^(c*(a + b*x))*((1 - E^(2*I*(d + e*x 
)))^n/(1 + E^(2*I*(d + e*x)))^n), x], x], x] /; FreeQ[{F, a, b, c, d, e}, x 
] && IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

2*(F^(b*c*x)*F^(a*c)*b*c*log(F)*sin(2*e*x + 2*d) - 2*F^(b*c*x)*F^(a*c)*e*c 
os(2*e*x + 2*d) - 2*F^(b*c*x)*F^(a*c)*e + 2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 
4*F^(a*c)*e^3 + (F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*cos(2*e*x + 2 
*d)^2 + (F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*sin(2*e*x + 2*d)^2 + 
2*(F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*cos(2*e*x + 2*d))*integrate 
((F^(b*c*x)*b*c*cos(4*e*x + 4*d)*log(F) + 2*F^(b*c*x)*b*c*cos(2*e*x + 2*d) 
*log(F) + F^(b*c*x)*b*c*log(F) + 2*F^(b*c*x)*e*sin(4*e*x + 4*d) + 4*F^(b*c 
*x)*e*sin(2*e*x + 2*d))/(b^2*c^2*log(F)^2 + (b^2*c^2*log(F)^2 + 4*e^2)*cos 
(4*e*x + 4*d)^2 + 4*(b^2*c^2*log(F)^2 + 4*e^2)*cos(2*e*x + 2*d)^2 + (b^2*c 
^2*log(F)^2 + 4*e^2)*sin(4*e*x + 4*d)^2 + 4*(b^2*c^2*log(F)^2 + 4*e^2)*sin 
(4*e*x + 4*d)*sin(2*e*x + 2*d) + 4*(b^2*c^2*log(F)^2 + 4*e^2)*sin(2*e*x + 
2*d)^2 + 4*e^2 + 2*(b^2*c^2*log(F)^2 + 4*e^2 + 2*(b^2*c^2*log(F)^2 + 4*e^2 
)*cos(2*e*x + 2*d))*cos(4*e*x + 4*d) + 4*(b^2*c^2*log(F)^2 + 4*e^2)*cos(2* 
e*x + 2*d)), x))/(b^2*c^2*log(F)^2 + (b^2*c^2*log(F)^2 + 4*e^2)*cos(2*e*x 
+ 2*d)^2 + (b^2*c^2*log(F)^2 + 4*e^2)*sin(2*e*x + 2*d)^2 + 4*e^2 + 2*(b^2* 
c^2*log(F)^2 + 4*e^2)*cos(2*e*x + 2*d))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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