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

Optimal. Leaf size=80 \[ -\frac{2 e^{i (d+e x)} F^{c (a+b x)} \text{Hypergeometric2F1}\left (2,1-\frac{i b c \log (F)}{e},2-\frac{i b c \log (F)}{e},i e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \]

[Out]

(-2*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]))

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Rubi [A]  time = 0.0655502, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4456, 4450} \[ -\frac{2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac{i b c \log (F)}{e};2-\frac{i b c \log (F)}{e};i e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \]

Antiderivative was successfully verified.

[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 4456

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 + (e*x)/2 - (f*Pi)/(4*g)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] &&
EqQ[f^2 - g^2, 0] && ILtQ[n, 0]

Rule 4450

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))*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)))])/(I*e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && Int
egerQ[4*k] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx &=\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)} \, _2F_1\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]  time = 1.69873, size = 128, normalized size = 1.6 \[ \frac{2 F^{c (a+b x)} \left (-i \text{Hypergeometric2F1}\left (1,-\frac{i b c \log (F)}{e},1-\frac{i b c \log (F)}{e},-\sin (d+e x)+i \cos (d+e x)\right )+\frac{\sin \left (\frac{e x}{2}\right )}{\left (\sin \left (\frac{d}{2}\right )+\cos \left (\frac{d}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )}-\frac{1}{\cos (d)+i (\sin (d)+1)}\right )}{e f} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.119, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{f+f\sin \left ( ex+d \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{F^{b c x + a c}}{f \sin \left (e x + d\right ) + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{F^{a c} F^{b c x}}{\sin{\left (d + e x \right )} + 1}\, dx}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{f \sin \left (e x + d\right ) + f}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)