\(\int \frac {F^{c (a+b x)}}{(f+f \sin (d+e x))^2} \, dx\) [31]

Optimal result

Integrand size = 22, antiderivative size = 184 \[ \int \frac {F^{c (a+b x)}}{(f+f \sin (d+e x))^2} \, dx=-\frac {F^{c (a+b x)} \cot \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right ) \csc ^2\left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )}{6 e f^2}-\frac {b c F^{c (a+b x)} \csc ^2\left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right ) \log (F)}{6 e^2 f^2}-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},i e^{i (d+e x)}\right ) (e+i b c \log (F))}{3 e^2 f^2} \] Output:

-1/6*F^(c*(b*x+a))*cot(1/2*d+1/4*Pi+1/2*e*x)*csc(1/2*d+1/4*Pi+1/2*e*x)^2/e 
/f^2-1/6*b*c*F^(c*(b*x+a))*csc(1/2*d+1/4*Pi+1/2*e*x)^2*ln(F)/e^2/f^2-2/3*e 
xp(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],I*exp(I*(e*x+d)))*(e+I*b*c*ln(F))/e^2/f^2
 

Mathematica [A] (verified)

Time = 3.04 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.30 \[ \int \frac {F^{c (a+b x)}}{(f+f \sin (d+e x))^2} \, dx=\frac {F^{c (a+b x)} \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right ) \left (2 e^2 \sin \left (\frac {1}{2} (d+e x)\right )-e (e+b c \log (F)) \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )+2 \left (e^2+b^2 c^2 \log ^2(F)\right ) \sin \left (\frac {1}{2} (d+e x)\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^2-(1-i) \left (1-(1-i) \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},i \cos (d+e x)-\sin (d+e x)\right )\right ) \left (e^2+b^2 c^2 \log ^2(F)\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^3\right )}{3 e^3 f^2 (1+\sin (d+e x))^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 197, 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 = {4956, 4949, 4952}

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.

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

\(\Big \downarrow \) 4956

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

\(\Big \downarrow \) 4949

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

\(\Big \downarrow \) 4952

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

Input:

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

Output:

((-2*F^(c*(a + b*x))*Cot[d/2 + Pi/4 + (e*x)/2]*Csc[d/2 + Pi/4 + (e*x)/2]^2 
)/(3*e) - (2*b*c*F^(c*(a + b*x))*Csc[d/2 + Pi/4 + (e*x)/2]^2*Log[F])/(3*e^ 
2) - (8*E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 - (I*b*c*Lo 
g[F])/e, 2 - (I*b*c*Log[F])/e, I*E^(I*(d + e*x))]*(1 + (b^2*c^2*Log[F]^2)/ 
e^2))/(3*(e - I*b*c*Log[F])))/(4*f^2)
 

Defintions of rubi rules used

rule 4949

Int[Csc[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbo 
l] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Csc[d + e*x]^(n - 2)/(e^2*(n - 1) 
*(n - 2))), x] + (-Simp[F^(c*(a + b*x))*Csc[d + e*x]^(n - 1)*(Cos[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))*Csc[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] && 
NeQ[n, 2]
 

rule 4952

Int[Csc[(d_.) + Pi*(k_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_ 
))), x_Symbol] :> Simp[(-2*I)^n*E^(I*k*n*Pi)*E^(I*n*(d + e*x))*(F^(c*(a + b 
*x))/(I*e*n + b*c*Log[F]))*Hypergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 
 1 + n/2 - I*b*c*(Log[F]/(2*e)), E^(2*I*k*Pi)*E^(2*I*(d + e*x))], x] /; Fre 
eQ[{F, a, b, c, d, e}, x] && IntegerQ[4*k] && IntegerQ[n]
 

rule 4956

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)] 
)^(n_.), x_Symbol] :> Simp[2^n*f^n   Int[F^(c*(a + b*x))*Cos[d/2 - f*(Pi/(4 
*g)) + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[ 
f^2 - g^2, 0] && ILtQ[n, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(F^(c*(b*x+a))/(f+f*sin(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 - 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)*cos(e*x + d) - 
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)*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) + 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 
)*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*lo 
g(F)^2 + 144*F^(a*c)*e^5)*F^(b*c*x)*sin(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)*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*(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)*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) + (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^...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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