\(\int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx\) [137]

Optimal result

Integrand size = 22, antiderivative size = 80 \[ \int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx=-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},i e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \]

[Out]

-2*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],I*exp(I*(e*x+d)))/f/(e-I*b*c*
ln(F))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4541, 4535} \[ \int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx=-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},i e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \]

[In]

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

[Out]

(-2*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, I*E^(I*(d
+ e*x))])/(f*(e - I*b*c*Log[F]))

Rule 4535

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + Pi*(k_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[2^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] /; FreeQ[{F, a, b, c, d, e}, x] && Int
egerQ[4*k] && IntegerQ[n]

Rule 4541

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)])^(n_.), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int F^{c (a+b x)} \sec ^2\left (\frac {d}{2}-\frac {\pi }{4}+\frac {e x}{2}\right ) \, dx}{2 f} \\ & = -\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 )}{f (e-i b c \log (F))} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.87 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.60 \[ \int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx=\frac {2 F^{c (a+b x)} \left (-i \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},i \cos (d+e x)-\sin (d+e x)\right )-\frac {1}{\cos (d)+i (1+\sin (d))}+\frac {\sin \left (\frac {e x}{2}\right )}{\left (\cos \left (\frac {d}{2}\right )+\sin \left (\frac {d}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )}\right )}{e f} \]

[In]

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

[Out]

(2*F^(c*(a + b*x))*((-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]] - (Cos[d] + I*(1 + Sin[d]))^(-1) + Sin[(e*x)/2]/((Cos[d/2] + Sin[d/2])*(Cos[(d + e*x)/2] + Sin[(d +
e*x)/2]))))/(e*f)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

2*(6*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(
e*x + d)^2 + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(e*x + d)^2 + (5*F^(a*c)*b^2
*c^2*e*log(F)^2 - 4*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F
))*F^(b*c*x)*sin(e*x + d) - (6*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) + (F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)
*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(e*x + d))*cos(2*
e*x + 2*d) - 2*((F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f*cos
(2*e*x + 2*d)^2 + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f
*cos(e*x + d)^2 + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f
*cos(e*x + d)*sin(2*e*x + 2*d) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*
e^5*log(F))*f*sin(2*e*x + 2*d)^2 + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*
b*c*e^5*log(F))*f*sin(e*x + d)^2 + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*
b*c*e^5*log(F))*f*sin(e*x + d) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*
e^5*log(F))*f - 2*(2*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*
f*sin(e*x + d) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f)*c
os(2*e*x + 2*d))*integrate((3*F^(b*c*x)*b*c*e*cos(3*e*x + 3*d)*log(F) - 9*F^(b*c*x)*b*c*e*cos(e*x + d)*log(F)
- 9*F^(b*c*x)*b*c*e*log(F)*sin(2*e*x + 2*d) - 3*(b^2*c^2*log(F)^2 - 2*e^2)*F^(b*c*x)*cos(2*e*x + 2*d) - (b^2*c
^2*log(F)^2 - 2*e^2)*F^(b*c*x)*sin(3*e*x + 3*d) + 3*(b^2*c^2*log(F)^2 - 2*e^2)*F^(b*c*x)*sin(e*x + d) + (b^2*c
^2*log(F)^2 - 2*e^2)*F^(b*c*x))/((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(3*e*x + 3*d)^2 + 9*
(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(2*e*x + 2*d)^2 + 9*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2
*log(F)^2 + 4*e^4)*f*cos(e*x + d)^2 + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(3*e*x + 3*d)^2
+ 18*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d)*sin(2*e*x + 2*d) + 9*(b^4*c^4*log(F)^
4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(2*e*x + 2*d)^2 + 9*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^
4)*f*sin(e*x + d)^2 + 6*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d) + (b^4*c^4*log(F)^4
+ 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f - 6*((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d) +
(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(2*e*x + 2*d))*cos(3*e*x + 3*d) - 6*(3*(b^4*c^4*log(
F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f)
*cos(2*e*x + 2*d) + 2*(3*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(2*e*x + 2*d) - 3*(b^4*c^4*l
og(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d) - (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)
*f)*sin(3*e*x + 3*d)), x) + ((F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(e*x + d) - (F
^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*F^(b*c*x)*sin(e*x + d) + 2*(F^(a*c)*b^2*c^2*e*log(F)^2 - 2*F^(a*c)*
e^3)*F^(b*c*x))*sin(2*e*x + 2*d))/((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(2*e*x + 2*d)^2 +
4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d)^2 + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*l
og(F)^2 + 4*e^4)*f*cos(e*x + d)*sin(2*e*x + 2*d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(2
*e*x + 2*d)^2 + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d)^2 + 4*(b^4*c^4*log(F)^4 +
5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f - 2*(2
*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)
^2 + 4*e^4)*f)*cos(2*e*x + 2*d))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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