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

Optimal. Leaf size=128 \[ \frac {b c \log (F) \cos ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )} \]

[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]  time = 0.04, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4435, 2194} \[ \frac {b c \log (F) \cos ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )} \]

Antiderivative was successfully verified.

[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 2194

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 4435

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*} \int F^{c (a+b x)} \cos ^2(d+e x) \, dx &=\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]  time = 0.21, size = 85, normalized size = 0.66 \[ \frac {F^{c (a+b x)} \left (b^2 c^2 \log ^2(F) \cos (2 (d+e x))+b^2 c^2 \log ^2(F)+2 b c e \log (F) \sin (2 (d+e x))+4 e^2\right )}{2 b^3 c^3 \log ^3(F)+8 b c e^2 \log (F)} \]

Antiderivative was successfully verified.

[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)

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 78, normalized size = 0.61 \[ \frac {{\left (b^{2} c^{2} \cos \left (e x + d\right )^{2} \log \relax (F)^{2} + 2 \, b c e \cos \left (e x + d\right ) \log \relax (F) \sin \left (e x + d\right ) + 2 \, e^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \relax (F)^{3} + 4 \, b c e^{2} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [C]  time = 0.23, size = 933, normalized size = 7.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*x*e + 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*x*e + 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*x*e - 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*x*e - 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))) - 1/2*I*(-2*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*x*e + 2*I*d)/(4*I*pi*b*c*sgn(F) - 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*I*e) + 2*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*x*e - 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))) - 1/2*I*(-2*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*x*e - 2*I*d)/(4*I*pi*b*c*sgn(F) -
4*I*pi*b*c + 8*b*c*log(abs(F)) - 16*I*e) + 2*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*x*e + 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))) - 1/2*I*(-2*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))) + 2*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*l
og(abs(F)) + a*c*log(abs(F)))

________________________________________________________________________________________

maple [A]  time = 0.13, size = 153, normalized size = 1.20 \[ \frac {F^{c \left (b x +a \right )}}{2 b c \ln \relax (F )}+\frac {\frac {b c \ln \relax (F ) {\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )}}{4 e^{2}+b^{2} c^{2} \ln \relax (F )^{2}}+\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )} \tan \left (e x +d \right )}{4 e^{2}+b^{2} c^{2} \ln \relax (F )^{2}}-\frac {b c \ln \relax (F ) {\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )} \left (\tan ^{2}\left (e x +d \right )\right )}{4 e^{2}+b^{2} c^{2} \ln \relax (F )^{2}}}{2+2 \left (\tan ^{2}\left (e x +d \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b/c/ln(F)*F^(c*(b*x+a))+1/2*(1/(4*e^2+b^2*c^2*ln(F)^2)*b*c*ln(F)*exp(c*(b*x+a)*ln(F))+4/(4*e^2+b^2*c^2*ln(
F)^2)*e*exp(c*(b*x+a)*ln(F))*tan(e*x+d)-1/(4*e^2+b^2*c^2*ln(F)^2)*b*c*ln(F)*exp(c*(b*x+a)*ln(F))*tan(e*x+d)^2)
/(1+tan(e*x+d)^2)

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 356, normalized size = 2.78 \[ \frac {{\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \relax (F)^{2} + 2 \, F^{a c} b c e \log \relax (F) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \relax (F)^{2} - 2 \, F^{a c} b c e \log \relax (F) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x + 4 \, d\right ) - {\left (F^{a c} b^{2} c^{2} \log \relax (F)^{2} \sin \left (2 \, d\right ) - 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \relax (F)\right )} F^{b c x} \sin \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \log \relax (F)^{2} \sin \left (2 \, d\right ) + 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \relax (F)\right )} F^{b c x} \sin \left (2 \, e x + 4 \, d\right ) + 2 \, {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right )^{2} \log \relax (F)^{2} + F^{a c} b^{2} c^{2} \log \relax (F)^{2} \sin \left (2 \, d\right )^{2} + 4 \, {\left (F^{a c} \cos \left (2 \, d\right )^{2} + F^{a c} \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )} F^{b c x}}{4 \, {\left (b^{3} c^{3} \cos \left (2 \, d\right )^{2} \log \relax (F)^{3} + b^{3} c^{3} \log \relax (F)^{3} \sin \left (2 \, d\right )^{2} + 4 \, {\left (b c \cos \left (2 \, d\right )^{2} \log \relax (F) + b c \log \relax (F) \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 2.98, size = 98, normalized size = 0.77 \[ \frac {2\,F^{a\,c+b\,c\,x}\,e^2+F^{a\,c+b\,c\,x}\,b^2\,c^2\,{\cos \left (d+e\,x\right )}^2\,{\ln \relax (F)}^2+2\,F^{a\,c+b\,c\,x}\,b\,c\,e\,\cos \left (d+e\,x\right )\,\sin \left (d+e\,x\right )\,\ln \relax (F)}{b^3\,c^3\,{\ln \relax (F)}^3+4\,b\,c\,e^2\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 34.78, size = 627, normalized size = 4.90 \[ \begin {cases} \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 \\\tilde {\infty } e^{2} \left (e^{- \frac {2 i e}{b c}}\right )^{a c} \left (e^{- \frac {2 i e}{b c}}\right )^{b c x} \sin ^{2}{\left (d + e x \right )} + \tilde {\infty } e^{2} \left (e^{- \frac {2 i e}{b c}}\right )^{a c} \left (e^{- \frac {2 i e}{b c}}\right )^{b c x} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )} + \tilde {\infty } e^{2} \left (e^{- \frac {2 i e}{b c}}\right )^{a c} \left (e^{- \frac {2 i e}{b c}}\right )^{b c x} \cos ^{2}{\left (d + e x \right )} & \text {for}\: F = e^{- \frac {2 i e}{b c}} \\\tilde {\infty } e^{2} \left (e^{\frac {2 i e}{b c}}\right )^{a c} \left (e^{\frac {2 i e}{b c}}\right )^{b c x} \sin ^{2}{\left (d + e x \right )} + \tilde {\infty } e^{2} \left (e^{\frac {2 i e}{b c}}\right )^{a c} \left (e^{\frac {2 i e}{b c}}\right )^{b c x} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )} + \tilde {\infty } e^{2} \left (e^{\frac {2 i e}{b c}}\right )^{a c} \left (e^{\frac {2 i e}{b c}}\right )^{b c x} \cos ^{2}{\left (d + e x \right )} & \text {for}\: F = e^{\frac {2 i e}{b c}} \\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} F^{b c x} b^{2} c^{2} \log {\relax (F )}^{2} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\relax (F )}^{3} + 4 b c e^{2} \log {\relax (F )}} + \frac {2 F^{a c} F^{b c x} b c e \log {\relax (F )} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{b^{3} c^{3} \log {\relax (F )}^{3} + 4 b c e^{2} \log {\relax (F )}} + \frac {2 F^{a c} F^{b c x} e^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\relax (F )}^{3} + 4 b c e^{2} \log {\relax (F )}} + \frac {2 F^{a c} F^{b c x} e^{2} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\relax (F )}^{3} + 4 b c e^{2} \log {\relax (F )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((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)), (zoo*e**2*e
xp(-2*I*e/(b*c))**(a*c)*exp(-2*I*e/(b*c))**(b*c*x)*sin(d + e*x)**2 + zoo*e**2*exp(-2*I*e/(b*c))**(a*c)*exp(-2*
I*e/(b*c))**(b*c*x)*sin(d + e*x)*cos(d + e*x) + zoo*e**2*exp(-2*I*e/(b*c))**(a*c)*exp(-2*I*e/(b*c))**(b*c*x)*c
os(d + e*x)**2, Eq(F, exp(-2*I*e/(b*c)))), (zoo*e**2*exp(2*I*e/(b*c))**(a*c)*exp(2*I*e/(b*c))**(b*c*x)*sin(d +
e*x)**2 + zoo*e**2*exp(2*I*e/(b*c))**(a*c)*exp(2*I*e/(b*c))**(b*c*x)*sin(d + e*x)*cos(d + e*x) + zoo*e**2*exp
(2*I*e/(b*c))**(a*c)*exp(2*I*e/(b*c))**(b*c*x)*cos(d + e*x)**2, Eq(F, exp(2*I*e/(b*c)))), (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)*F**(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)*F**(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)*F**(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)*F**(b*c*x)*e**2*cos(d + e*x)**2/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)), Tr
ue))

________________________________________________________________________________________