3.1.53 \(\int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx\) [53]

Optimal. Leaf size=82 \[ \frac {i F^{a+b x}}{b e \log (F)}-\frac {2 i F^{a+b x} \, _2F_1\left (1,-\frac {i b \log (F)}{d};1-\frac {i b \log (F)}{d};i e^{i (c+d x)}\right )}{b e \log (F)} \]

[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],I*exp(I*(d*x+c)))/b/e/ln(F)

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Rubi [A]
time = 0.09, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 = {4547, 4527, 2225, 2283} \begin {gather*} \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 {gather*}

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 2225

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 2283

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

Rule 4527

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 4547

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]

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*}

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Mathematica [A]
time = 1.61, size = 64, normalized size = 0.78 \begin {gather*} -\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};i e^{i (c+d x)}\right )\right )}{b e \log (F)} \end {gather*}

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 = 0.24, size = 0, normalized size = 0.00 \[\int \frac {F^{b x +a} \cos \left (d x +c \right )}{e +e \sin \left (d x +c \right )}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {F^{a} F^{b x} \cos {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{e} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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