\(\int F^{c (a+b x)} \tan ^2(d+e x) \, dx\) [116]

Optimal result

Integrand size = 18, antiderivative size = 112 \[ \int F^{c (a+b x)} \tan ^2(d+e x) \, dx=\frac {2 i F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )}-\frac {2 i F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{e}-\frac {F^{c (a+b x)}}{b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95 \[ \int F^{c (a+b x)} \tan ^2(d+e x) \, dx=F^{c (a+b x)} \left (\frac {2 i}{e+e e^{2 i d}}-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{e}-\frac {1}{b c \log (F)}+\frac {\sec (d) \sec (d+e x) \sin (e x)}{e}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.

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

\(\Big \downarrow \) 4942

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-(F^(c*(a + b*x))/(b*c*Log[F])) + (4*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]) - (4*F^(c*(a + b*x))*Hypergeometric2F1[2, ((-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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-((F^(a*c)*b^4*c^4*log(F)^4 + 20*F^(a*c)*b^2*c^2*e^2*log(F)^2 + 64*F^(a*c) 
*e^4)*F^(b*c*x)*cos(4*e*x + 4*d)^2 - 4*(F^(a*c)*b^4*c^4*log(F)^4 + 12*F^(a 
*c)*b^2*c^2*e^2*log(F)^2 - 64*F^(a*c)*e^4)*F^(b*c*x)*cos(2*e*x + 2*d)^2 + 
(F^(a*c)*b^4*c^4*log(F)^4 + 20*F^(a*c)*b^2*c^2*e^2*log(F)^2 + 64*F^(a*c)*e 
^4)*F^(b*c*x)*sin(4*e*x + 4*d)^2 - 4*(F^(a*c)*b^4*c^4*log(F)^4 + 12*F^(a*c 
)*b^2*c^2*e^2*log(F)^2 - 64*F^(a*c)*e^4)*F^(b*c*x)*sin(2*e*x + 2*d)^2 - 16 
*(11*F^(a*c)*b^2*c^2*e^2*log(F)^2 - 16*F^(a*c)*e^4)*F^(b*c*x)*cos(2*e*x + 
2*d) + 8*(5*F^(a*c)*b^3*c^3*e*log(F)^3 - 16*F^(a*c)*b*c*e^3*log(F))*F^(b*c 
*x)*sin(2*e*x + 2*d) + (F^(a*c)*b^4*c^4*log(F)^4 - 76*F^(a*c)*b^2*c^2*e^2* 
log(F)^2 + 64*F^(a*c)*e^4)*F^(b*c*x) + 2*(8*(F^(a*c)*b^2*c^2*e^2*log(F)^2 
+ 16*F^(a*c)*e^4)*F^(b*c*x)*cos(2*e*x + 2*d) + 4*(F^(a*c)*b^3*c^3*e*log(F) 
^3 + 16*F^(a*c)*b*c*e^3*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) + (F^(a*c)*b^4* 
c^4*log(F)^4 - 28*F^(a*c)*b^2*c^2*e^2*log(F)^2 + 64*F^(a*c)*e^4)*F^(b*c*x) 
)*cos(4*e*x + 4*d) + 16*(F^(a*c)*b^6*c^6*e*log(F)^6 + 20*F^(a*c)*b^4*c^4*e 
^3*log(F)^4 + 64*F^(a*c)*b^2*c^2*e^5*log(F)^2 + (F^(a*c)*b^6*c^6*e*log(F)^ 
6 + 20*F^(a*c)*b^4*c^4*e^3*log(F)^4 + 64*F^(a*c)*b^2*c^2*e^5*log(F)^2)*cos 
(4*e*x + 4*d)^2 + 4*(F^(a*c)*b^6*c^6*e*log(F)^6 + 20*F^(a*c)*b^4*c^4*e^3*l 
og(F)^4 + 64*F^(a*c)*b^2*c^2*e^5*log(F)^2)*cos(2*e*x + 2*d)^2 + (F^(a*c)*b 
^6*c^6*e*log(F)^6 + 20*F^(a*c)*b^4*c^4*e^3*log(F)^4 + 64*F^(a*c)*b^2*c^2*e 
^5*log(F)^2)*sin(4*e*x + 4*d)^2 + 4*(F^(a*c)*b^6*c^6*e*log(F)^6 + 20*F^...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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