\(\int F^{c (a+b x)} \cos ^2(d+e x) \, dx\) [12]

Optimal result

Integrand size = 18, antiderivative size = 128 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}+\frac {b c F^{c (a+b x)} \cos ^2(d+e x) \log (F)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)} \]

[Out]

2*e^2*F^(c*(b*x+a))/b/c/ln(F)/(4*e^2+b^2*c^2*ln(F)^2)+b*c*F^(c*(b*x+a))*cos(e*x+d)^2*ln(F)/(4*e^2+b^2*c^2*ln(F
)^2)+2*e*F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d)/(4*e^2+b^2*c^2*ln(F)^2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4520, 2225} \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {b c \log (F) \cos ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )} \]

[In]

Int[F^(c*(a + b*x))*Cos[d + e*x]^2,x]

[Out]

(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)

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 4520

Int[Cos[(d_.) + (e_.)*(x_)]^(m_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x
))*(Cos[d + e*x]^m/(e^2*m^2 + b^2*c^2*Log[F]^2)), x] + (Dist[(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] + 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]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*m^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
m, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {b c F^{c (a+b x)} \cos ^2(d+e x) \log (F)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {\left (2 e^2\right ) \int F^{c (a+b x)} \, dx}{4 e^2+b^2 c^2 \log ^2(F)} \\ & = \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 {b c F^{c (a+b x)} \cos ^2(d+e x) \log (F)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.66 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {F^{c (a+b x)} \left (4 e^2+b^2 c^2 \log ^2(F)+b^2 c^2 \cos (2 (d+e x)) \log ^2(F)+2 b c e \log (F) \sin (2 (d+e x))\right )}{8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \]

[In]

Integrate[F^(c*(a + b*x))*Cos[d + e*x]^2,x]

[Out]

(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)

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {F^{c \left (x b +a \right )} \left (b^{2} c^{2} \ln \left (F \right )^{2} \cos \left (2 e x +2 d \right )+b^{2} c^{2} \ln \left (F \right )^{2}+2 e \sin \left (2 e x +2 d \right ) b c \ln \left (F \right )+4 e^{2}\right )}{2 b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(94\)
risch	\(\frac {F^{c \left (x b +a \right )}}{2 b c \ln \left (F \right )}+\frac {F^{c \left (x b +a \right )} b c \ln \left (F \right ) \cos \left (2 e x +2 d \right )}{2 b^{2} c^{2} \ln \left (F \right )^{2}+8 e^{2}}+\frac {e \,F^{c \left (x b +a \right )} \sin \left (2 e x +2 d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(105\)
norman	\(\frac {\frac {\left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {\left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 e \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {4 e \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {2 \left (b^{2} c^{2} \ln \left (F \right )^{2}-2 e^{2}\right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\)	\(296\)

[In]

int(F^(c*(b*x+a))*cos(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.61 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {{\left (b^{2} c^{2} \cos \left (e x + d\right )^{2} \log \left (F\right )^{2} + 2 \, b c e \cos \left (e x + d\right ) \log \left (F\right ) \sin \left (e x + d\right ) + 2 \, e^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )} \]

[In]

integrate(F^(c*(b*x+a))*cos(e*x+d)^2,x, algorithm="fricas")

[Out]

(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))

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.08 (sec) , antiderivative size = 741, normalized size of antiderivative = 5.79 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\begin {cases} x \cos ^{2}{\left (d \right )} & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge e = 0 \\\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} + \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: F = 1 \\F^{a c} \left (\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} + \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e}\right ) & \text {for}\: b = 0 \\\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} + \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: c = 0 \\- \frac {F^{a c + b c x} x \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{4} + \frac {i F^{a c + b c x} x \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{2} + \frac {F^{a c + b c x} x \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{4} + \frac {i F^{a c + b c x} \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{2 b c \log {\left (F \right )}} + \frac {F^{a c + b c x} \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{b c \log {\left (F \right )}} & \text {for}\: e = - \frac {i b c \log {\left (F \right )}}{2} \\- \frac {F^{a c + b c x} x \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{4} + \frac {i F^{a c + b c x} x \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{2} + \frac {F^{a c + b c x} x \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{4} + \frac {F^{a c + b c x} \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{b c \log {\left (F \right )}} - \frac {3 i F^{a c + b c x} \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{2 b c \log {\left (F \right )}} & \text {for}\: e = \frac {i b c \log {\left (F \right )}}{2} \\\frac {F^{a c + b c x} b^{2} c^{2} \log {\left (F \right )}^{2} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} + \frac {2 F^{a c + b c x} b c e \log {\left (F \right )} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} + \frac {2 F^{a c + b c x} e^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} + \frac {2 F^{a c + b c x} e^{2} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(c*(b*x+a))*cos(e*x+d)**2,x)

[Out]

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 + I*F**(a*c + b*c*x)*si
n(I*b*c*x*log(F)/2 - d)*cos(I*b*c*x*log(F)/2 - d)/(2*b*c*log(F)) + F**(a*c + b*c*x)*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
))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (128) = 256\).

Time = 0.23 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.78 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {{\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} + 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} - 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x + 4 \, d\right ) - {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) - 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) + 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x + 4 \, d\right ) + 2 \, {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{2} + F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right )^{2} + 4 \, {\left (F^{a c} \cos \left (2 \, d\right )^{2} + F^{a c} \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )} F^{b c x}}{4 \, {\left (b^{3} c^{3} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{3} + b^{3} c^{3} \log \left (F\right )^{3} \sin \left (2 \, d\right )^{2} + 4 \, {\left (b c \cos \left (2 \, d\right )^{2} \log \left (F\right ) + b c \log \left (F\right ) \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )}} \]

[In]

integrate(F^(c*(b*x+a))*cos(e*x+d)^2,x, algorithm="maxima")

[Out]

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*lo
g(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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 915, normalized size of antiderivative = 7.15 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\text {Too large to display} \]

[In]

integrate(F^(c*(b*x+a))*cos(e*x+d)^2,x, algorithm="giac")

[Out]

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
*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))) + 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) - p
i*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(abs(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*log(abs(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*log(abs(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*pi*b*
c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F))) - 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*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F))))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)
))

Mupad [B] (verification not implemented)

Time = 28.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.77 \[ \int F^{c (a+b x)} \cos ^2(d+e x) \, dx=\frac {2\,F^{a\,c+b\,c\,x}\,e^2+F^{a\,c+b\,c\,x}\,b^2\,c^2\,{\cos \left (d+e\,x\right )}^2\,{\ln \left (F\right )}^2+2\,F^{a\,c+b\,c\,x}\,b\,c\,e\,\cos \left (d+e\,x\right )\,\sin \left (d+e\,x\right )\,\ln \left (F\right )}{b^3\,c^3\,{\ln \left (F\right )}^3+4\,b\,c\,e^2\,\ln \left (F\right )} \]

[In]

int(F^(c*(a + b*x))*cos(d + e*x)^2,x)

[Out]

(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))