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

Optimal. Leaf size=82 \[ \frac{2 i F^{a+b x} \text{Hypergeometric2F1}\left (1,-\frac{i b \log (F)}{d},1-\frac{i b \log (F)}{d},-i e^{i (c+d x)}\right )}{b e \log (F)}-\frac{i F^{a+b x}}{b e \log (F)} \]

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

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Rubi [A]  time = 0.127621, antiderivative size = 82, normalized size of antiderivative = 1., 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 = {4459, 4442, 2194, 2251} \[ \frac{2 i F^{a+b x} \, _2F_1\left (1,-\frac{i b \log (F)}{d};1-\frac{i b \log (F)}{d};-i e^{i (c+d x)}\right )}{b e \log (F)}-\frac{i F^{a+b x}}{b e \log (F)} \]

Antiderivative was successfully verified.

[In]

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

Rule 4459

Int[Cos[(d_.) + (e_.)*(x_)]^(m_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)])^(n_
.), x_Symbol] :> Dist[g^n, Int[F^(c*(a + b*x))*Tan[(f*Pi)/(4*g) - d/2 - (e*x)/2]^m, x], x] /; FreeQ[{F, a, b,
c, d, e, f, g}, x] && EqQ[f^2 - g^2, 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} \cos (c+d x)}{e-e \sin (c+d x)} \, dx &=\frac{\int F^{a+b x} \tan \left (\frac{c}{2}+\frac{\pi }{4}+\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{\pi }{4}+\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{\pi }{4}+\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};-i e^{i (c+d x)}\right )}{b e \log (F)}\\ \end{align*}

Mathematica [A]  time = 2.57243, size = 64, normalized size = 0.78 \[ \frac{i F^{a+b x} \left (-1+2 \text{Hypergeometric2F1}\left (1,-\frac{i b \log (F)}{d},1-\frac{i b \log (F)}{d},-i e^{i (c+d x)}\right )\right )}{b e \log (F)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.175, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{bx+a}\cos \left ( dx+c \right ) }{e-e\sin \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, F^{b x} F^{a} b d \cos \left (d x + c\right ) \log \left (F\right ) + 2 \, F^{b x} F^{a} d^{2} \sin \left (d x + c\right ) -{\left (F^{a} b^{2} \log \left (F\right )^{2} + F^{a} d^{2}\right )} F^{b x} \cos \left (d x + c\right )^{2} -{\left (F^{a} b^{2} \log \left (F\right )^{2} + F^{a} d^{2}\right )} F^{b x} \sin \left (d x + c\right )^{2} +{\left (F^{a} b^{2} \log \left (F\right )^{2} - F^{a} d^{2}\right )} F^{b x} + \frac{{\left (F^{b x} d \cos \left (d x + c\right )^{2} - 2 \, F^{b x} b \cos \left (d x + c\right ) \log \left (F\right ) + F^{b x} d \sin \left (d x + c\right )^{2} - 2 \, F^{b x} d \sin \left (d x + c\right ) + F^{b x} d\right )}{\left ({\left (F^{a} b^{3} d \log \left (F\right )^{3} + F^{a} b d^{3} \log \left (F\right )\right )} e \cos \left (d x + c\right )^{2} +{\left (F^{a} b^{3} d \log \left (F\right )^{3} + F^{a} b d^{3} \log \left (F\right )\right )} e \sin \left (d x + c\right )^{2} - 2 \,{\left (F^{a} b^{3} d \log \left (F\right )^{3} + F^{a} b d^{3} \log \left (F\right )\right )} e \sin \left (d x + c\right ) +{\left (F^{a} b^{3} d \log \left (F\right )^{3} + F^{a} b d^{3} \log \left (F\right )\right )} e\right )}}{{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e \cos \left (d x + c\right )^{2} +{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e \sin \left (d x + c\right )^{2} - 2 \,{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e \sin \left (d x + c\right ) +{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e}}{{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e \cos \left (d x + c\right )^{2} +{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e \sin \left (d x + c\right )^{2} - 2 \,{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e \sin \left (d x + c\right ) +{\left (b^{3} \log \left (F\right )^{3} + b d^{2} \log \left (F\right )\right )} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*F^(b*x)*F^a*b*d*cos(d*x + c)*log(F) + 2*F^(b*x)*F^a*d^2*sin(d*x + c) - (F^a*b^2*log(F)^2 + F^a*d^2)*F^(b*x
)*cos(d*x + c)^2 - (F^a*b^2*log(F)^2 + F^a*d^2)*F^(b*x)*sin(d*x + c)^2 + (F^a*b^2*log(F)^2 - F^a*d^2)*F^(b*x)
+ 2*((F^a*b^3*d*log(F)^3 + F^a*b*d^3*log(F))*e*cos(d*x + c)^2 + (F^a*b^3*d*log(F)^3 + F^a*b*d^3*log(F))*e*sin(
d*x + c)^2 - 2*(F^a*b^3*d*log(F)^3 + F^a*b*d^3*log(F))*e*sin(d*x + c) + (F^a*b^3*d*log(F)^3 + F^a*b*d^3*log(F)
)*e)*integrate(-(2*F^(b*x)*b*cos(d*x + c)*log(F) - F^(b*x)*b*log(F)*sin(2*d*x + 2*c) + F^(b*x)*d*cos(2*d*x + 2
*c) + 2*F^(b*x)*d*sin(d*x + c) - F^(b*x)*d)/((b^2*log(F)^2 + d^2)*e*cos(2*d*x + 2*c)^2 + 4*(b^2*log(F)^2 + d^2
)*e*cos(d*x + c)^2 - 4*(b^2*log(F)^2 + d^2)*e*cos(d*x + c)*sin(2*d*x + 2*c) + (b^2*log(F)^2 + d^2)*e*sin(2*d*x
+ 2*c)^2 + 4*(b^2*log(F)^2 + d^2)*e*sin(d*x + c)^2 - 4*(b^2*log(F)^2 + d^2)*e*sin(d*x + c) + (b^2*log(F)^2 +
d^2)*e + 2*(2*(b^2*log(F)^2 + d^2)*e*sin(d*x + c) - (b^2*log(F)^2 + d^2)*e)*cos(2*d*x + 2*c)), x))/((b^3*log(F
)^3 + b*d^2*log(F))*e*cos(d*x + c)^2 + (b^3*log(F)^3 + b*d^2*log(F))*e*sin(d*x + c)^2 - 2*(b^3*log(F)^3 + b*d^
2*log(F))*e*sin(d*x + c) + (b^3*log(F)^3 + b*d^2*log(F))*e)

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

[Out]

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

[Out]

-Integral(F**a*F**(b*x)*cos(c + d*x)/(sin(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} \cos \left (d x + c\right )}{e \sin \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)*cos(d*x+c)/(e-e*sin(d*x+c)),x, algorithm="giac")

[Out]

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