Integrand size = 26, antiderivative size = 82 \[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=\frac {i F^{a+b x}}{b e \log (F)}-\frac {2 i F^{a+b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},i e^{i (c+d x)}\right )}{b e \log (F)} \] 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],I*exp(I*(d*x+c)))/b/e/ln(F)
Time = 3.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=-\frac {i F^{a+b x} \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},i e^{i (c+d x)}\right )\right )}{b e \log (F)} \] Input:
Integrate[(F^(a + b*x)*Cos[c + d*x])/(e + e*Sin[c + d*x]),x]
Output:
((-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])
Time = 0.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {4962, 4943, 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.
\(\displaystyle \int \frac {F^{a+b x} \cos (c+d x)}{e \sin (c+d x)+e} \, dx\) |
\(\Big \downarrow \) 4962 |
\(\displaystyle \frac {\int F^{a+b x} \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{e}\) |
\(\Big \downarrow \) 4943 |
\(\displaystyle -\frac {i \int \left (\frac {2 F^{a+b x}}{1-e^{\frac {1}{2} i (2 c+2 d x+\pi )}}-F^{a+b x}\right )dx}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {F^{a+b x}}{b \log (F)}+\frac {2 F^{a+b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},e^{\frac {1}{2} i (2 c+2 d x+\pi )}\right )}{b \log (F)}\right )}{e}\) |
Input:
Int[(F^(a + b*x)*Cos[c + d*x])/(e + e*Sin[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/2)*(2*c + Pi + 2*d*x))])/(b*Log[ F])))/e
Int[Cot[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), 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]
Int[Cos[(d_.) + (e_.)*(x_)]^(m_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)])^(n_.), x_Symbol] :> Simp[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]
\[\int \frac {F^{b x +a} \cos \left (d x +c \right )}{e +e \sin \left (d x +c \right )}d x\]
Input:
int(F^(b*x+a)*cos(d*x+c)/(e+e*sin(d*x+c)),x)
Output:
int(F^(b*x+a)*cos(d*x+c)/(e+e*sin(d*x+c)),x)
\[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=\int { \frac {F^{b x + a} \cos \left (d x + c\right )}{e \sin \left (d x + c\right ) + e} \,d x } \] Input:
integrate(F^(b*x+a)*cos(d*x+c)/(e+e*sin(d*x+c)),x, algorithm="fricas")
Output:
integral(F^(b*x + a)*cos(d*x + c)/(e*sin(d*x + c) + e), x)
\[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=\frac {\int \frac {F^{a + b x} \cos {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{e} \] Input:
integrate(F**(b*x+a)*cos(d*x+c)/(e+e*sin(d*x+c)),x)
Output:
Integral(F**(a + b*x)*cos(c + d*x)/(sin(c + d*x) + 1), x)/e
\[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=\int { \frac {F^{b x + a} \cos \left (d x + c\right )}{e \sin \left (d x + c\right ) + e} \,d x } \] Input:
integrate(F^(b*x+a)*cos(d*x+c)/(e+e*sin(d*x+c)),x, algorithm="maxima")
Output:
-(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*l og(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)
\[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=\int { \frac {F^{b x + a} \cos \left (d x + c\right )}{e \sin \left (d x + c\right ) + e} \,d x } \] Input:
integrate(F^(b*x+a)*cos(d*x+c)/(e+e*sin(d*x+c)),x, algorithm="giac")
Output:
integrate(F^(b*x + a)*cos(d*x + c)/(e*sin(d*x + c) + e), x)
Timed out. \[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=\int \frac {F^{a+b\,x}\,\cos \left (c+d\,x\right )}{e+e\,\sin \left (c+d\,x\right )} \,d x \] Input:
int((F^(a + b*x)*cos(c + d*x))/(e + e*sin(c + d*x)),x)
Output:
int((F^(a + b*x)*cos(c + d*x))/(e + e*sin(c + d*x)), x)
\[ \int \frac {F^{a+b x} \cos (c+d x)}{e+e \sin (c+d x)} \, dx=\frac {f^{a} \left (\int \frac {f^{b x} \cos \left (d x +c \right )}{\sin \left (d x +c \right )+1}d x \right )}{e} \] Input:
int(F^(b*x+a)*cos(d*x+c)/(e+e*sin(d*x+c)),x)
Output:
(f**a*int((f**(b*x)*cos(c + d*x))/(sin(c + d*x) + 1),x))/e