\(\int F^{c (a+b x)} \sec ^4(d+e x) \, dx\) [17]

Optimal result

Integrand size = 18, antiderivative size = 143 \[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=-\frac {2 e^{2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) (2 i e-b c \log (F))}{3 e^2}-\frac {b c F^{c (a+b x)} \log (F) \sec ^2(d+e x)}{6 e^2}+\frac {F^{c (a+b x)} \sec ^2(d+e x) \tan (d+e x)}{3 e} \] Output:

-2/3*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)))*(2*I*e-b*c*ln(F))/e^2-1/6*b*c*F^(c*( 
b*x+a))*ln(F)*sec(e*x+d)^2/e^2+1/3*F^(c*(b*x+a))*sec(e*x+d)^2*tan(e*x+d)/e
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.78 \[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=\frac {F^{c (a+b x)} \left (4 e^{2 i (d+e x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) (-2 i e+b c \log (F))+\sec ^2(d+e x) (-b c \log (F)+2 e \tan (d+e x))\right )}{6 e^2} \] Input:

Integrate[F^(c*(a + b*x))*Sec[d + e*x]^4,x]
 

Output:

(F^(c*(a + b*x))*(4*E^((2*I)*(d + e*x))*Hypergeometric2F1[2, 1 - ((I/2)*b* 
c*Log[F])/e, 2 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + e*x))]*((-2*I)*e + b 
*c*Log[F]) + Sec[d + e*x]^2*(-(b*c*Log[F]) + 2*e*Tan[d + e*x])))/(6*e^2)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4948, 4951}

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.

\(\displaystyle \int \sec ^4(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 4948

\(\displaystyle \frac {1}{6} \left (\frac {b^2 c^2 \log ^2(F)}{e^2}+4\right ) \int F^{c (a+b x)} \sec ^2(d+e x)dx-\frac {b c \log (F) \sec ^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac {\tan (d+e x) \sec ^2(d+e x) F^{c (a+b x)}}{3 e}\)

\(\Big \downarrow \) 4951

\(\displaystyle \frac {2 e^{2 i (d+e x)} F^{c (a+b x)} \left (\frac {b^2 c^2 \log ^2(F)}{e^2}+4\right ) \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{3 (b c \log (F)+2 i e)}-\frac {b c \log (F) \sec ^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac {\tan (d+e x) \sec ^2(d+e x) F^{c (a+b x)}}{3 e}\)

Input:

Int[F^(c*(a + b*x))*Sec[d + e*x]^4,x]
 

Output:

(2*E^((2*I)*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 - ((I/2)*b*c 
*Log[F])/e, 2 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + e*x))]*(4 + (b^2*c^2* 
Log[F]^2)/e^2))/(3*((2*I)*e + b*c*Log[F])) - (b*c*F^(c*(a + b*x))*Log[F]*S 
ec[d + e*x]^2)/(6*e^2) + (F^(c*(a + b*x))*Sec[d + e*x]^2*Tan[d + e*x])/(3* 
e)
 

Defintions of rubi rules used

rule 4948

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_), x_Symbo 
l] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Sec[d + e*x]^(n - 2)/(e^2*(n - 1) 
*(n - 2))), x] + (Simp[F^(c*(a + b*x))*Sec[d + e*x]^(n - 1)*(Sin[d + e*x]/( 
e*(n - 1))), x] + Simp[(e^2*(n - 2)^2 + b^2*c^2*Log[F]^2)/(e^2*(n - 1)*(n - 
 2))   Int[F^(c*(a + b*x))*Sec[d + e*x]^(n - 2), x], x]) /; FreeQ[{F, a, b, 
 c, d, e}, x] && NeQ[b^2*c^2*Log[F]^2 + e^2*(n - 2)^2, 0] && GtQ[n, 1] && N 
eQ[n, 2]
 

rule 4951

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symb 
ol] :> Simp[2^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hy 
pergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(Log[F]/(2*e 
)), -E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \sec \left (e x +d \right )^{4}d x\]

Input:

int(F^(c*(b*x+a))*sec(e*x+d)^4,x)
 

Output:

int(F^(c*(b*x+a))*sec(e*x+d)^4,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sec \left (e x + d\right )^{4} \,d x } \] Input:

integrate(F^(c*(b*x+a))*sec(e*x+d)^4,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*sec(e*x + d)^4, x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=\int F^{c \left (a + b x\right )} \sec ^{4}{\left (d + e x \right )}\, dx \] Input:

integrate(F**(c*(b*x+a))*sec(e*x+d)**4,x)
 

Output:

Integral(F**(c*(a + b*x))*sec(d + e*x)**4, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sec \left (e x + d\right )^{4} \,d x } \] Input:

integrate(F^(c*(b*x+a))*sec(e*x+d)^4,x, algorithm="maxima")
 

Output:

16*(6*(F^(a*c)*b^5*c^5*log(F)^5 + 100*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 2304* 
F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(4*e*x + 4*d)^2 + 320*(F^(a*c)*b^3*c^ 
3*e^2*log(F)^3 + 64*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(2*e*x + 2*d)^2 + 
 6*(F^(a*c)*b^5*c^5*log(F)^5 + 100*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 2304*F^( 
a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(4*e*x + 4*d)^2 + 320*(F^(a*c)*b^3*c^3*e 
^2*log(F)^3 + 64*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(2*e*x + 2*d)^2 - 56 
0*(F^(a*c)*b^3*c^3*e^2*log(F)^3 - 32*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos 
(2*e*x + 2*d) + 40*(F^(a*c)*b^4*c^4*e*log(F)^4 - 104*F^(a*c)*b^2*c^2*e^3*l 
og(F)^2)*F^(b*c*x)*sin(2*e*x + 2*d) - 160*(F^(a*c)*b^3*c^3*e^2*log(F)^3 - 
20*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x) + ((F^(a*c)*b^5*c^5*log(F)^5 + 100*F^ 
(a*c)*b^3*c^3*e^2*log(F)^3 + 2304*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(4* 
e*x + 4*d) + 80*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 64*F^(a*c)*b*c*e^4*log(F)) 
*F^(b*c*x)*cos(2*e*x + 2*d) - 4*(F^(a*c)*b^4*c^4*e*log(F)^4 + 100*F^(a*c)* 
b^2*c^2*e^3*log(F)^2 + 2304*F^(a*c)*e^5)*F^(b*c*x)*sin(4*e*x + 4*d) + 8*(F 
^(a*c)*b^4*c^4*e*log(F)^4 + 40*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 1536*F^(a*c) 
*e^5)*F^(b*c*x)*sin(2*e*x + 2*d) - 160*(F^(a*c)*b^3*c^3*e^2*log(F)^3 - 20* 
F^(a*c)*b*c*e^4*log(F))*F^(b*c*x))*cos(8*e*x + 8*d) + 4*((F^(a*c)*b^5*c^5* 
log(F)^5 + 100*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 2304*F^(a*c)*b*c*e^4*log(F)) 
*F^(b*c*x)*cos(4*e*x + 4*d) + 80*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 64*F^(a*c 
)*b*c*e^4*log(F))*F^(b*c*x)*cos(2*e*x + 2*d) - 4*(F^(a*c)*b^4*c^4*e*log...
 

Giac [F]

\[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sec \left (e x + d\right )^{4} \,d x } \] Input:

integrate(F^(c*(b*x+a))*sec(e*x+d)^4,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*sec(e*x + d)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\cos \left (d+e\,x\right )}^4} \,d x \] Input:

int(F^(c*(a + b*x))/cos(d + e*x)^4,x)
 

Output:

int(F^(c*(a + b*x))/cos(d + e*x)^4, x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \sec ^4(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \sec \left (e x +d \right )^{4}d x \right ) \] Input:

int(F^(c*(b*x+a))*sec(e*x+d)^4,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*sec(d + e*x)**4,x)