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

Optimal. Leaf size=184 \[ -\frac{2 e^{i (d+e x)} F^{c (a+b x)} (e+i b c \log (F)) \text{Hypergeometric2F1}\left (2,1-\frac{i b c \log (F)}{e},2-\frac{i b c \log (F)}{e},i e^{i (d+e x)}\right )}{3 e^2 f^2}-\frac{b c \log (F) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}-\frac{\cot \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \]

[Out]

-(F^(c*(a + b*x))*Cot[d/2 + Pi/4 + (e*x)/2]*Csc[d/2 + Pi/4 + (e*x)/2]^2)/(6*e*f^2) - (b*c*F^(c*(a + b*x))*Csc[
d/2 + Pi/4 + (e*x)/2]^2*Log[F])/(6*e^2*f^2) - (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))]*(e + I*b*c*Log[F]))/(3*e^2*f^2)

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Rubi [A]  time = 0.0997605, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4456, 4448, 4450} \[ -\frac{2 e^{i (d+e x)} F^{c (a+b x)} (e+i b c \log (F)) \, _2F_1\left (2,1-\frac{i b c \log (F)}{e};2-\frac{i b c \log (F)}{e};i e^{i (d+e x)}\right )}{3 e^2 f^2}-\frac{b c \log (F) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}-\frac{\cot \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(F^(c*(a + b*x))*Cot[d/2 + Pi/4 + (e*x)/2]*Csc[d/2 + Pi/4 + (e*x)/2]^2)/(6*e*f^2) - (b*c*F^(c*(a + b*x))*Csc[
d/2 + Pi/4 + (e*x)/2]^2*Log[F])/(6*e^2*f^2) - (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))]*(e + I*b*c*Log[F]))/(3*e^2*f^2)

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 4448

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Sec[d + e*x]^(n - 2))/(e^2*(n - 1)*(n - 2)), x] + (Dist[(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] + Simp[(F^(c*(a + b*x))*Sec[d + e*x]^(n - 1)*Sin[d +
e*x])/(e*(n - 1)), 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 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))^2} \, dx &=\frac{\int F^{c (a+b x)} \sec ^4\left (\frac{d}{2}-\frac{\pi }{4}+\frac{e x}{2}\right ) \, dx}{4 f^2}\\ &=-\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{\left (1+\frac{b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \sec ^2\left (\frac{d}{2}-\frac{\pi }{4}+\frac{e x}{2}\right ) \, dx}{6 f^2}\\ &=-\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)} \, _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 ) (e+i b c \log (F))}{3 e^2 f^2}\\ \end{align*}

Mathematica [A]  time = 3.23028, size = 240, normalized size = 1.3 \[ \frac{F^{c (a+b x)} \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right ) \left ((-1+i) \left (b^2 c^2 \log ^2(F)+e^2\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )^3 \left (1-(1-i) \text{Hypergeometric2F1}\left (1,-\frac{i b c \log (F)}{e},1-\frac{i b c \log (F)}{e},-\sin (d+e x)+i \cos (d+e x)\right )\right )+2 \sin \left (\frac{1}{2} (d+e x)\right ) \left (b^2 c^2 \log ^2(F)+e^2\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )^2-e (b c \log (F)+e) \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )+2 e^2 \sin \left (\frac{1}{2} (d+e x)\right )\right )}{3 e^3 f^2 (\sin (d+e x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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] - Si
n[d + e*x]])*(e^2 + b^2*c^2*Log[F]^2)*(Cos[(d + e*x)/2] + Sin[(d + e*x)/2])^3))/(3*e^3*f^2*(1 + Sin[d + e*x])^
2)

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Maple [F]  time = 0.373, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{ \left ( f+f\sin \left ( ex+d \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int(F^(c*(b*x+a))/(f+f*sin(e*x+d))^2,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))^2,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^{2} \cos \left (e x + d\right )^{2} - 2 \, f^{2} \sin \left (e x + d\right ) - 2 \, f^{2}}, 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))^2,x, algorithm="fricas")

[Out]

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)

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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 ^{2}{\left (d + e x \right )} + 2 \sin{\left (d + e x \right )} + 1}\, dx}{f^{2}} \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))**2,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (f \sin \left (e x + d\right ) + f\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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