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

Optimal. Leaf size=78 \[ \frac {i F^{a+b x}}{b e \log (F)}-\frac {2 i F^{a+b x} \, _2F_1\left (1,-\frac {i b \log (F)}{d};1-\frac {i b \log (F)}{d};e^{i (c+d x)}\right )}{b e \log (F)} \]

[Out]

I*F^(b*x+a)/b/e/ln(F)-2*I*F^(b*x+a)*hypergeom([1, -I*b*ln(F)/d],[1-I*b*ln(F)/d],exp(I*(d*x+c)))/b/e/ln(F)

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Rubi [A]  time = 0.12, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {4461, 4443, 2194, 2251} \[ \frac {i F^{a+b x}}{b e \log (F)}-\frac {2 i F^{a+b x} \, _2F_1\left (1,-\frac {i b \log (F)}{d};1-\frac {i b \log (F)}{d};e^{i (c+d x)}\right )}{b e \log (F)} \]

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 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])

Rule 4443

Int[Cot[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Dist[(-I)^n, Int[ExpandInteg
rand[(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 4461

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))*Cot[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]

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} \cot \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*}

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Mathematica [A]  time = 0.63, size = 66, normalized size = 0.85 \[ -\frac {i F^{a+b x} \left (-1+2 \, _2F_1\left (1,-\frac {i b \log (F)}{d};1-\frac {i b \log (F)}{d};\cos (c+d x)+i \sin (c+d x)\right )\right )}{b e \log (F)} \]

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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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {F^{b x + a} \sin \left (d x + c\right )}{e \cos \left (d x + c\right ) - e}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {F^{b x + a} \sin \left (d x + c\right )}{e \cos \left (d x + c\right ) - e}\,{d x} \]

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)

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {F^{b x +a} \sin \left (d x +c \right )}{e -e \cos \left (d x +c \right )}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{a+b\,x}\,\sin \left (c+d\,x\right )}{e-e\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {F^{a} F^{b x} \sin {\left (c + d x \right )}}{\cos {\left (c + d x \right )} - 1}\, dx}{e} \]

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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