\(\int F^{c (a+b x)} \tan ^3(d+e x) \, dx\) [122]

Optimal result

Integrand size = 18, antiderivative size = 171 \[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=-\frac {2 F^{c (a+b x)}}{e \left (1+e^{2 i (d+e x)}\right )^2}+\frac {i F^{c (a+b x)}}{b c \log (F)}+\frac {F^{c (a+b x)} (2 e-i b c \log (F))}{e^2 \left (1+e^{2 i (d+e x)}\right )}-i F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \left (\frac {2}{b c \log (F)}-\frac {b c \log (F)}{e^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.02 \[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (2 i e^2-i b^2 c^2 \log ^2(F)-2 i \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-\cos (2 (d+e x))-i \sin (2 (d+e x))\right ) \left (2 e^2-b^2 c^2 \log ^2(F)\right )+b c e \log (F) \sec ^2(d+e x)-b^2 c^2 \log ^2(F) \sec (d) \sec (d+e x) \sin (e x)-b^2 c^2 \log ^2(F) \tan (d)\right )}{2 b c e^2 \log (F)} \] Input:

Integrate[F^(c*(a + b*x))*Tan[d + e*x]^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.27, 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.

\(\displaystyle \int \tan ^3(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 4942

\(\displaystyle -i \int \left (\frac {6 F^{c (a+b x)}}{1+e^{2 i (d+e x)}}-\frac {12 F^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^2}+\frac {8 F^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^3}-F^{c (a+b x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -i \left (\frac {6 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{b c \log (F)}-\frac {12 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{b c \log (F)}+\frac {8 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{b c \log (F)}-\frac {F^{c (a+b x)}}{b c \log (F)}\right )\)

Input:

Int[F^(c*(a + b*x))*Tan[d + e*x]^3,x]
 

Output:

(-I)*(-(F^(c*(a + b*x))/(b*c*Log[F])) + (6*F^(c*(a + b*x))*Hypergeometric2 
F1[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]) - (12*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]) 
 + (8*F^(c*(a + b*x))*Hypergeometric2F1[3, ((-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]

\[\int F^{c \left (b x +a \right )} \tan \left (e x +d \right )^{3}d x\]

Input:

int(F^(c*(b*x+a))*tan(e*x+d)^3,x)
 

Output:

int(F^(c*(b*x+a))*tan(e*x+d)^3,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \tan \left (e x + d\right )^{3} \,d x } \] Input:

integrate(F^(c*(b*x+a))*tan(e*x+d)^3,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*tan(e*x + d)^3, x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=\int F^{c \left (a + b x\right )} \tan ^{3}{\left (d + e x \right )}\, dx \] Input:

integrate(F**(c*(b*x+a))*tan(e*x+d)**3,x)
 

Output:

Integral(F**(c*(a + b*x))*tan(d + e*x)**3, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \tan \left (e x + d\right )^{3} \,d x } \] Input:

integrate(F^(c*(b*x+a))*tan(e*x+d)^3,x, algorithm="maxima")
 

Output:

2*(18*(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 - 54*(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)^2 + 18*(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 - 54*(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)*si 
n(2*e*x + 2*d)^2 + 18*(3*F^(a*c)*b^4*c^4*e*log(F)^4 - 212*F^(a*c)*b^2*c^2* 
e^3*log(F)^2 + 640*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) - 3*(F^(a*c)*b^ 
5*c^5*log(F)^5 - 268*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 1216*F^(a*c)*b*c*e^4*l 
og(F))*F^(b*c*x)*sin(2*e*x + 2*d) + 24*(F^(a*c)*b^4*c^4*e*log(F)^4 - 46*F^ 
(a*c)*b^2*c^2*e^3*log(F)^2 + 88*F^(a*c)*e^5)*F^(b*c*x) + 3*(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) - 6*(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) + (F^(a*c)*b^5 
*c^5*log(F)^5 + 52*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) + 36*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 36*F^( 
a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(2*e*x + 2*d) + 8*(F^(a*c)*b^4*c^4*e*log 
(F)^4 - 46*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 88*F^(a*c)*e^5)*F^(b*c*x))*cos(6 
*e*x + 6*d) - 3*(12*(F^(a*c)*b^4*c^4*e*log(F)^4 + 16*F^(a*c)*b^2*c^2*e^3*l 
og(F)^2 - 720*F^(a*c)*e^5)*F^(b*c*x)*cos(2*e*x + 2*d) + 3*(F^(a*c)*b^5*...
 

Giac [F]

\[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \tan \left (e x + d\right )^{3} \,d x } \] Input:

integrate(F^(c*(b*x+a))*tan(e*x+d)^3,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*tan(e*x + d)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\mathrm {tan}\left (d+e\,x\right )}^3 \,d x \] Input:

int(F^(c*(a + b*x))*tan(d + e*x)^3,x)
 

Output:

int(F^(c*(a + b*x))*tan(d + e*x)^3, x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \tan ^3(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \tan \left (e x +d \right )^{3}d x \right ) \] Input:

int(F^(c*(b*x+a))*tan(e*x+d)^3,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*tan(d + e*x)**3,x)