Integrand size = 18, antiderivative size = 100 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}+\frac {F^{c (a+b x)} \cos (d+e x) (b c \cos (d+e x) \log (F)+2 e \sin (d+e x))}{4 e^2+b^2 c^2 \log ^2(F)} \] Output:
2*e^2*F^(c*(b*x+a))/b/c/ln(F)/(4*e^2+b^2*c^2*ln(F)^2)+F^(c*(b*x+a))*cos(e* x+d)*(b*c*cos(e*x+d)*ln(F)+2*e*sin(e*x+d))/(4*e^2+b^2*c^2*ln(F)^2)
Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.85 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {F^{c (a+b x)} \left (4 e^2+b^2 c^2 \log ^2(F)+b^2 c^2 \cos (2 (d+e x)) \log ^2(F)+2 b c e \log (F) \sin (2 (d+e x))\right )}{8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \] Input:
Integrate[F^(c*(a + b*x))*Cos[d + e*x]^2,x]
Output:
(F^(c*(a + b*x))*(4*e^2 + b^2*c^2*Log[F]^2 + b^2*c^2*Cos[2*(d + e*x)]*Log[ F]^2 + 2*b*c*e*Log[F]*Sin[2*(d + e*x)]))/(8*b*c*e^2*Log[F] + 2*b^3*c^3*Log [F]^3)
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4935, 2624}
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.
| \(\displaystyle \int \cos ^2(d+e x) F^{c (a+b x)} \, dx\) |
| \(\Big \downarrow \) 4935 |
| \(\displaystyle \frac {2 e^2 \int F^{c (a+b x)}dx}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {b c \log (F) \cos ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}\) |
| \(\Big \downarrow \) 2624 |
| \(\displaystyle \frac {b c \log (F) \cos ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )}\) |
Input:
Int[F^(c*(a + b*x))*Cos[d + e*x]^2,x]
Output:
(2*e^2*F^(c*(a + b*x)))/(b*c*Log[F]*(4*e^2 + b^2*c^2*Log[F]^2)) + (b*c*F^( c*(a + b*x))*Cos[d + e*x]^2*Log[F])/(4*e^2 + b^2*c^2*Log[F]^2) + (2*e*F^(c *(a + b*x))*Cos[d + e*x]*Sin[d + e*x])/(4*e^2 + b^2*c^2*Log[F]^2)
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Int[Cos[(d_.) + (e_.)*(x_)]^(m_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbo l] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Cos[d + e*x]^m/(e^2*m^2 + b^2*c^2*Lo g[F]^2)), x] + (Simp[e*m*F^(c*(a + b*x))*Sin[d + e*x]*(Cos[d + e*x]^(m - 1) /(e^2*m^2 + b^2*c^2*Log[F]^2)), x] + Simp[(m*(m - 1)*e^2)/(e^2*m^2 + b^2*c^ 2*Log[F]^2) Int[F^(c*(a + b*x))*Cos[d + e*x]^(m - 2), x], x]) /; FreeQ[{F , a, b, c, d, e}, x] && NeQ[e^2*m^2 + b^2*c^2*Log[F]^2, 0] && GtQ[m, 1]
Time = 0.41 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94
| method | result | size |
| parallelrisch | \(\frac {F^{c \left (b x +a \right )} \left (b^{2} c^{2} \ln \left (F \right )^{2} \cos \left (2 e x +2 d \right )+b^{2} c^{2} \ln \left (F \right )^{2}+2 e \sin \left (2 e x +2 d \right ) b c \ln \left (F \right )+4 e^{2}\right )}{2 b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\) | \(94\) |
| risch | \(\frac {F^{c \left (b x +a \right )}}{2 b c \ln \left (F \right )}+\frac {F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (2 e x +2 d \right )}{2 b^{2} c^{2} \ln \left (F \right )^{2}+8 e^{2}}+\frac {e \,F^{c \left (b x +a \right )} \sin \left (2 e x +2 d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(105\) |
| orering | \(\frac {\left (3 b^{2} c^{2} \ln \left (F \right )^{2}+4 e^{2}\right ) F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {3 \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (e x +d \right )^{2}-2 F^{c \left (b x +a \right )} \cos \left (e x +d \right ) e \sin \left (e x +d \right )\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \cos \left (e x +d \right )^{2}-4 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (e x +d \right ) e \sin \left (e x +d \right )+2 F^{c \left (b x +a \right )} e^{2} \sin \left (e x +d \right )^{2}-2 F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{2} e^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\) | \(266\) |
| norman | \(\frac {\frac {\left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {\left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {2 \left (b^{2} c^{2} \ln \left (F \right )^{2}-2 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\) | \(296\) |
Input:
int(F^(c*(b*x+a))*cos(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
1/2*F^(c*(b*x+a))*(b^2*c^2*ln(F)^2*cos(2*e*x+2*d)+b^2*c^2*ln(F)^2+2*e*sin( 2*e*x+2*d)*b*c*ln(F)+4*e^2)/b/c/ln(F)/(4*e^2+b^2*c^2*ln(F)^2)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {{\left (b^{2} c^{2} \cos \left (e x + d\right )^{2} \log \left (F\right )^{2} + 2 \, b c e \cos \left (e x + d\right ) \log \left (F\right ) \sin \left (e x + d\right ) + 2 \, e^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)^2,x, algorithm="fricas")
Output:
(b^2*c^2*cos(e*x + d)^2*log(F)^2 + 2*b*c*e*cos(e*x + d)*log(F)*sin(e*x + d ) + 2*e^2)*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3 + 4*b*c*e^2*log(F))
Result contains complex when optimal does not.
Time = 1.20 (sec) , antiderivative size = 743, normalized size of antiderivative = 7.43 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx =\text {Too large to display} \] Input:
integrate(F**(c*(b*x+a))*cos(e*x+d)**2,x)
Output:
Piecewise((x*cos(d)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)), (x*sin (d + e*x)**2/2 + x*cos(d + e*x)**2/2 + sin(d + e*x)*cos(d + e*x)/(2*e), Eq (F, 1)), (F**(a*c)*(x*sin(d + e*x)**2/2 + x*cos(d + e*x)**2/2 + sin(d + e* x)*cos(d + e*x)/(2*e)), Eq(b, 0)), (x*sin(d + e*x)**2/2 + x*cos(d + e*x)** 2/2 + sin(d + e*x)*cos(d + e*x)/(2*e), Eq(c, 0)), (-F**(a*c + b*c*x)*x*sin (I*b*c*x*log(F)/2 - d)**2/4 + I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/2 - d)*cos(I*b*c*x*log(F)/2 - d)/2 + F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F)/2 - d)**2/4 + F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/2 - d)**2/(b*c*log(F)) - 3* I*F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/2 - d)*cos(I*b*c*x*log(F)/2 - d)/(2* b*c*log(F)), Eq(e, -I*b*c*log(F)/2)), (-F**(a*c + b*c*x)*x*sin(I*b*c*x*log (F)/2 + d)**2/4 + I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/2 + d)*cos(I*b*c *x*log(F)/2 + d)/2 + F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F)/2 + d)**2/4 + F **(a*c + b*c*x)*sin(I*b*c*x*log(F)/2 + d)**2/(b*c*log(F)) - 3*I*F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/2 + d)*cos(I*b*c*x*log(F)/2 + d)/(2*b*c*log(F)), Eq(e, I*b*c*log(F)/2)), (F**(a*c + b*c*x)*b**2*c**2*log(F)**2*cos(d + e*x )**2/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)) + 2*F**(a*c + b*c*x)*b*c*e* log(F)*sin(d + e*x)*cos(d + e*x)/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)) + 2*F**(a*c + b*c*x)*e**2*sin(d + e*x)**2/(b**3*c**3*log(F)**3 + 4*b*c*e* *2*log(F)) + 2*F**(a*c + b*c*x)*e**2*cos(d + e*x)**2/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)), True))
Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (100) = 200\).
Time = 0.05 (sec) , antiderivative size = 356, normalized size of antiderivative = 3.56 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {{\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} + 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} - 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x + 4 \, d\right ) - {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) - 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) + 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x + 4 \, d\right ) + 2 \, {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{2} + F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right )^{2} + 4 \, {\left (F^{a c} \cos \left (2 \, d\right )^{2} + F^{a c} \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )} F^{b c x}}{4 \, {\left (b^{3} c^{3} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{3} + b^{3} c^{3} \log \left (F\right )^{3} \sin \left (2 \, d\right )^{2} + 4 \, {\left (b c \cos \left (2 \, d\right )^{2} \log \left (F\right ) + b c \log \left (F\right ) \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )}} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)^2,x, algorithm="maxima")
Output:
1/4*((F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 + 2*F^(a*c)*b*c*e*log(F)*sin(2*d)) *F^(b*c*x)*cos(2*e*x) + (F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 - 2*F^(a*c)*b*c *e*log(F)*sin(2*d))*F^(b*c*x)*cos(2*e*x + 4*d) - (F^(a*c)*b^2*c^2*log(F)^2 *sin(2*d) - 2*F^(a*c)*b*c*e*cos(2*d)*log(F))*F^(b*c*x)*sin(2*e*x) + (F^(a* c)*b^2*c^2*log(F)^2*sin(2*d) + 2*F^(a*c)*b*c*e*cos(2*d)*log(F))*F^(b*c*x)* sin(2*e*x + 4*d) + 2*(F^(a*c)*b^2*c^2*cos(2*d)^2*log(F)^2 + F^(a*c)*b^2*c^ 2*log(F)^2*sin(2*d)^2 + 4*(F^(a*c)*cos(2*d)^2 + F^(a*c)*sin(2*d)^2)*e^2)*F ^(b*c*x))/(b^3*c^3*cos(2*d)^2*log(F)^3 + b^3*c^3*log(F)^3*sin(2*d)^2 + 4*( b*c*cos(2*d)^2*log(F) + b*c*log(F)*sin(2*d)^2)*e^2)
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 915, normalized size of antiderivative = 9.15 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)^2,x, algorithm="giac")
Output:
1/2*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/ 2*pi*a*c + 2*e*x + 2*d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn (F) - pi*b*c + 4*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 4*e)*sin(1/2*pi*b*c*x*s gn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + 2*e*x + 2*d)/(4*b^ 2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 4*e)^2))*e^(b*c*x*log(abs( F)) + a*c*log(abs(F))) + 1/2*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c - 2*e*x - 2*d)*log(abs(F))/(4*b^2*c^2*lo g(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 4*e)^2) + (pi*b*c*sgn(F) - pi*b*c - 4*e)*sin(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi *a*c - 2*e*x - 2*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 4 *e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + (2*b*c*cos(-1/2*pi*b*c*x *sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)*log(abs(F))/(4*b^ 2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2) - (pi*b*c*sgn(F) - pi*b* c)*sin(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a* c)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2))*e^(b*c*x*log(ab s(F)) + a*c*log(abs(F))) + I*(I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c + 2*I*e*x + 2*I*d)/(4*I*pi*b*c*sgn(F) - 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*I*e) - I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c - 2*I*e*x - 2*I*d)/( -4*I*pi*b*c*sgn(F) + 4*I*pi*b*c + 8*b*c*log(abs(F)) - 16*I*e))*e^(b*c*x...
Time = 0.00 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {2\,F^{a\,c+b\,c\,x}\,e^2+F^{a\,c+b\,c\,x}\,b^2\,c^2\,{\cos \left (d+e\,x\right )}^2\,{\ln \left (F\right )}^2+2\,F^{a\,c+b\,c\,x}\,b\,c\,e\,\cos \left (d+e\,x\right )\,\sin \left (d+e\,x\right )\,\ln \left (F\right )}{b^3\,c^3\,{\ln \left (F\right )}^3+4\,b\,c\,e^2\,\ln \left (F\right )} \] Input:
int(F^(c*(a + b*x))*cos(d + e*x)^2,x)
Output:
(2*F^(a*c + b*c*x)*e^2 + F^(a*c + b*c*x)*b^2*c^2*cos(d + e*x)^2*log(F)^2 + 2*F^(a*c + b*c*x)*b*c*e*cos(d + e*x)*sin(d + e*x)*log(F))/(b^3*c^3*log(F) ^3 + 4*b*c*e^2*log(F))
\[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \cos \left (e x +d \right )^{2}d x \right ) \] Input:
int(F^(c*(b*x+a))*cos(e*x+d)^2,x)
Output:
f**(a*c)*int(f**(b*c*x)*cos(d + e*x)**2,x)