\(\int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx\) [67]

Optimal result

Integrand size = 26, antiderivative size = 80 \[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=-\frac {i F^{a+b x}}{b e \log (F)}+\frac {2 i F^{a+b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},-e^{i (c+d x)}\right )}{b e \log (F)} \] Output:

-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)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=\frac {i F^{a+b x} \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},-\cos (c+d x)-i \sin (c+d x)\right )\right )}{b e \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {4963, 4942, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 4963

\(\displaystyle \frac {\int F^{a+b x} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{e}\)

\(\Big \downarrow \) 4942

\(\displaystyle \frac {i \int \left (\frac {2 F^{a+b x}}{1+e^{i (c+d x)}}-F^{a+b x}\right )dx}{e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i \left (-\frac {F^{a+b x}}{b \log (F)}+\frac {2 F^{a+b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},-e^{i (c+d x)}\right )}{b \log (F)}\right )}{e}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4942

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symb 
ol] :> Simp[I^n   Int[ExpandIntegrand[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 4963

Int[(Cos[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.) 
*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :> Simp[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]
 

Maple [F]

\[\int \frac {F^{b x +a} \sin \left (d x +c \right )}{e +e \cos \left (d x +c \right )}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=\int { \frac {F^{b x + a} \sin \left (d x + c\right )}{e \cos \left (d x + c\right ) + e} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=\frac {\int \frac {F^{a + b x} \sin {\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{e} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=\int { \frac {F^{b x + a} \sin \left (d x + c\right )}{e \cos \left (d x + c\right ) + e} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=\int { \frac {F^{b x + a} \sin \left (d x + c\right )}{e \cos \left (d x + c\right ) + e} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=\int \frac {F^{a+b\,x}\,\sin \left (c+d\,x\right )}{e+e\,\cos \left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {F^{a+b x} \sin (c+d x)}{e+e \cos (c+d x)} \, dx=\frac {f^{a} \left (\int \frac {f^{b x} \sin \left (d x +c \right )}{\cos \left (d x +c \right )+1}d x \right )}{e} \] Input:

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

Output:

(f**a*int((f**(b*x)*sin(c + d*x))/(cos(c + d*x) + 1),x))/e