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

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

[Out]

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

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Rubi [A]  time = 0.116375, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4460, 4442, 2194, 2251} \[ \frac{2 i F^{a+b x} \, _2F_1\left (1,-\frac{i b \log (F)}{d};1-\frac{i b \log (F)}{d};-e^{i (c+d x)}\right )}{b e \log (F)}-\frac{i F^{a+b x}}{b e \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4460

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

Rule 4442

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Dist[I^n, Int[ExpandIntegran
d[(F^(c*(a + b*x))*(1 - E^(2*I*(d + e*x)))^n)/(1 + E^(2*I*(d + e*x)))^n, x], x], x] /; FreeQ[{F, a, b, c, d, e
}, x] && IntegerQ[n]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx &=\frac{\int F^{a+b x} \tan \left (\frac{c}{2}+\frac{d x}{2}\right ) \, dx}{e}\\ &=\frac{i \int \left (-F^{a+b x}+\frac{2 F^{a+b x}}{1+e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}}\right ) \, dx}{e}\\ &=-\frac{i \int F^{a+b x} \, dx}{e}+\frac{(2 i) \int \frac{F^{a+b x}}{1+e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}} \, dx}{e}\\ &=-\frac{i F^{a+b x}}{b e \log (F)}+\frac{2 i F^{a+b x} \, _2F_1\left (1,-\frac{i b \log (F)}{d};1-\frac{i b \log (F)}{d};-e^{i (c+d x)}\right )}{b e \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.665764, size = 68, normalized size = 0.85 \[ \frac{i F^{a+b x} \left (-1+2 \text{Hypergeometric2F1}\left (1,-\frac{i b \log (F)}{d},1-\frac{i b \log (F)}{d},-\cos (c+d x)-i \sin (c+d x)\right )\right )}{b e \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.064, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{bx+a}\sin \left ( dx+c \right ) }{e+e\cos \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(F^(b*x+a)*sin(d*x+c)/(e+e*cos(d*x+c)),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^(b*x+a)*sin(d*x+c)/(e+e*cos(d*x+c)),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 x + a} \sin \left (d x + c\right )}{e \cos \left (d x + c\right ) + e}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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