\(\int \frac {F^{c (a+b x)}}{(f+f \cos (d+e x))^2} \, dx\) [37]

Optimal result

Integrand size = 22, antiderivative size = 169 \[ \int \frac {F^{c (a+b x)}}{(f+f \cos (d+e x))^2} \, dx=-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right ) (i e-b c \log (F))}{3 e^2 f^2}-\frac {b c F^{c (a+b x)} \log (F) \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e f^2} \] Output:

-2/3*exp(I*(e*x+d))*F^(c*(b*x+a))*hypergeom([2, 1-I*b*c*ln(F)/e],[2-I*b*c* 
ln(F)/e],-exp(I*(e*x+d)))*(I*e-b*c*ln(F))/e^2/f^2-1/6*b*c*F^(c*(b*x+a))*ln 
(F)*sec(1/2*e*x+1/2*d)^2/e^2/f^2+1/6*F^(c*(b*x+a))*sec(1/2*e*x+1/2*d)^2*ta 
n(1/2*e*x+1/2*d)/e/f^2
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.86 \[ \int \frac {F^{c (a+b x)}}{(f+f \cos (d+e x))^2} \, dx=\frac {2 F^{c (a+b x)} \cos \left (\frac {1}{2} (d+e x)\right ) \left (-b c \cos \left (\frac {1}{2} (d+e x)\right ) \log (F)+4 e^{i (d+e x)} \cos ^3\left (\frac {1}{2} (d+e x)\right ) \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right ) (-i e+b c \log (F))+e \sin \left (\frac {1}{2} (d+e x)\right )\right )}{3 e^2 f^2 (1+\cos (d+e x))^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4957, 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.

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

\(\Big \downarrow \) 4957

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

\(\Big \downarrow \) 4948

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

\(\Big \downarrow \) 4951

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

Input:

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

Output:

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

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]
 

rule 4957

Int[(Cos[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.) 
*(x_))), x_Symbol] :> Simp[2^n*f^n   Int[F^(c*(a + b*x))*Cos[d/2 + e*(x/2)] 
^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[f - g, 0] && IL 
tQ[n, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {F^{c (a+b x)}}{(f+f \cos (d+e x))^2} \, dx=\frac {\int \frac {F^{a c + b c x}}{\cos ^{2}{\left (d + e x \right )} + 2 \cos {\left (d + e x \right )} + 1}\, dx}{f^{2}} \] Input:

integrate(F**(c*(b*x+a))/(f+f*cos(e*x+d))**2,x)
 

Output:

Integral(F**(a*c + b*c*x)/(cos(d + e*x)**2 + 2*cos(d + e*x) + 1), x)/f**2
 

Maxima [F]

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

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

Output:

4*(6*(F^(a*c)*b^5*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 144*F^( 
a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(2*e*x + 2*d)^2 + 80*(F^(a*c)*b^3*c^3*e^ 
2*log(F)^3 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(e*x + d)^2 + 6*(F^(a 
*c)*b^5*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 144*F^(a*c)*b*c*e 
^4*log(F))*F^(b*c*x)*sin(2*e*x + 2*d)^2 + 80*(F^(a*c)*b^3*c^3*e^2*log(F)^3 
 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(e*x + d)^2 - 140*(F^(a*c)*b^3* 
c^3*e^2*log(F)^3 - 8*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(e*x + d) + 20*( 
F^(a*c)*b^4*c^4*e*log(F)^4 - 26*F^(a*c)*b^2*c^2*e^3*log(F)^2)*F^(b*c*x)*si 
n(e*x + d) - 40*(F^(a*c)*b^3*c^3*e^2*log(F)^3 - 5*F^(a*c)*b*c*e^4*log(F))* 
F^(b*c*x) + ((F^(a*c)*b^5*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 
 144*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(2*e*x + 2*d) + 20*(F^(a*c)*b^3* 
c^3*e^2*log(F)^3 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(e*x + d) - 2*( 
F^(a*c)*b^4*c^4*e*log(F)^4 + 25*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 144*F^(a*c) 
*e^5)*F^(b*c*x)*sin(2*e*x + 2*d) + 4*(F^(a*c)*b^4*c^4*e*log(F)^4 + 10*F^(a 
*c)*b^2*c^2*e^3*log(F)^2 - 96*F^(a*c)*e^5)*F^(b*c*x)*sin(e*x + d) - 40*(F^ 
(a*c)*b^3*c^3*e^2*log(F)^3 - 5*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x))*cos(4*e* 
x + 4*d) + 4*((F^(a*c)*b^5*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 
+ 144*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(2*e*x + 2*d) + 20*(F^(a*c)*b^3 
*c^3*e^2*log(F)^3 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(e*x + d) - 2* 
(F^(a*c)*b^4*c^4*e*log(F)^4 + 25*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 144*F^(...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {F^{c (a+b x)}}{(f+f \cos (d+e x))^2} \, dx=\frac {f^{a c} \left (\int \frac {f^{b c x}}{\cos \left (e x +d \right )^{2}+2 \cos \left (e x +d \right )+1}d x \right )}{f^{2}} \] Input:

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

Output:

(f**(a*c)*int(f**(b*c*x)/(cos(d + e*x)**2 + 2*cos(d + e*x) + 1),x))/f**2