3.11 \(\int F^{c (a+b x)} \cos ^3(d+e x) \, dx\)
Optimal. Leaf size=199 \[ \frac{6 e^3 \sin (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}+\frac{b c \log (F) \cos ^3(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}+\frac{6 b c e^2 \log (F) \cos (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}+\frac{3 e \sin (d+e x) \cos ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2} \]
[Out]
(b*c*F^(c*(a + b*x))*Cos[d + e*x]^3*Log[F])/(9*e^2 + b^2*c^2*Log[F]^2) + (6*b*c*e^2*F^(c*(a + b*x))*Cos[d + e*
x]*Log[F])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) + (3*e*F^(c*(a + b*x))*Cos[d + e*x]^2*Sin[d +
e*x])/(9*e^2 + b^2*c^2*Log[F]^2) + (6*e^3*F^(c*(a + b*x))*Sin[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4
*c^4*Log[F]^4)
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Rubi [A] time = 0.0528869, antiderivative size = 199, normalized size of antiderivative = 1.,
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, 4433} \[ \frac{6 e^3 \sin (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}+\frac{b c \log (F) \cos ^3(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}+\frac{6 b c e^2 \log (F) \cos (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}+\frac{3 e \sin (d+e x) \cos ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2} \]
Antiderivative was successfully verified.
[In]
Int[F^(c*(a + b*x))*Cos[d + e*x]^3,x]
[Out]
(b*c*F^(c*(a + b*x))*Cos[d + e*x]^3*Log[F])/(9*e^2 + b^2*c^2*Log[F]^2) + (6*b*c*e^2*F^(c*(a + b*x))*Cos[d + e*
x]*Log[F])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) + (3*e*F^(c*(a + b*x))*Cos[d + e*x]^2*Sin[d +
e*x])/(9*e^2 + b^2*c^2*Log[F]^2) + (6*e^3*F^(c*(a + b*x))*Sin[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4
*c^4*Log[F]^4)
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]
Rule 4433
Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]
Rubi steps
\begin{align*} \int F^{c (a+b x)} \cos ^3(d+e x) \, dx &=\frac{b c F^{c (a+b x)} \cos ^3(d+e x) \log (F)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac{3 e F^{c (a+b x)} \cos ^2(d+e x) \sin (d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac{\left (6 e^2\right ) \int F^{c (a+b x)} \cos (d+e x) \, dx}{9 e^2+b^2 c^2 \log ^2(F)}\\ &=\frac{b c F^{c (a+b x)} \cos ^3(d+e x) \log (F)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac{6 b c e^2 F^{c (a+b x)} \cos (d+e x) \log (F)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}+\frac{3 e F^{c (a+b x)} \cos ^2(d+e x) \sin (d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac{6 e^3 F^{c (a+b x)} \sin (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}\\ \end{align*}
Mathematica [A] time = 0.653698, size = 155, normalized size = 0.78 \[ \frac{F^{c (a+b x)} \left (b c \log (F) \cos (3 (d+e x)) \left (b^2 c^2 \log ^2(F)+e^2\right )+3 b c \log (F) \cos (d+e x) \left (b^2 c^2 \log ^2(F)+9 e^2\right )+6 e \sin (d+e x) \left (\cos (2 (d+e x)) \left (b^2 c^2 \log ^2(F)+e^2\right )+b^2 c^2 \log ^2(F)+5 e^2\right )\right )}{4 \left (10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[F^(c*(a + b*x))*Cos[d + e*x]^3,x]
[Out]
(F^(c*(a + b*x))*(b*c*Cos[3*(d + e*x)]*Log[F]*(e^2 + b^2*c^2*Log[F]^2) + 3*b*c*Cos[d + e*x]*Log[F]*(9*e^2 + b^
2*c^2*Log[F]^2) + 6*e*(5*e^2 + b^2*c^2*Log[F]^2 + Cos[2*(d + e*x)]*(e^2 + b^2*c^2*Log[F]^2))*Sin[d + e*x]))/(4
*(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4))
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Maple [A] time = 0.11, size = 274, normalized size = 1.4 \begin{align*}{\frac{1}{4} \left ({\frac{bc\ln \left ( F \right ){{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}}{9\,{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}}+6\,{\frac{e{{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}\tan \left ( 3/2\,ex+3/2\,d \right ) }{9\,{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}}-{\frac{bc\ln \left ( F \right ){{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}}{9\,{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( \tan \left ({\frac{3\,ex}{2}}+{\frac{3\,d}{2}} \right ) \right ) ^{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{3\,ex}{2}}+{\frac{3\,d}{2}} \right ) \right ) ^{2} \right ) ^{-1}}+{\frac{3}{4} \left ({\frac{bc\ln \left ( F \right ){{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}}{{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}}+2\,{\frac{e{{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}\tan \left ( d/2+1/2\,ex \right ) }{{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}}-{\frac{bc\ln \left ( F \right ){{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}}{{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(b*x+a))*cos(e*x+d)^3,x)
[Out]
1/4*(ln(F)*b*c/(9*e^2+b^2*c^2*ln(F)^2)*exp(c*(b*x+a)*ln(F))+6/(9*e^2+b^2*c^2*ln(F)^2)*e*exp(c*(b*x+a)*ln(F))*t
an(3/2*e*x+3/2*d)-ln(F)*b*c/(9*e^2+b^2*c^2*ln(F)^2)*exp(c*(b*x+a)*ln(F))*tan(3/2*e*x+3/2*d)^2)/(1+tan(3/2*e*x+
3/2*d)^2)+3/4*(ln(F)*b*c/(e^2+b^2*c^2*ln(F)^2)*exp(c*(b*x+a)*ln(F))+2/(e^2+b^2*c^2*ln(F)^2)*e*exp(c*(b*x+a)*ln
(F))*tan(1/2*d+1/2*e*x)-ln(F)*b*c/(e^2+b^2*c^2*ln(F)^2)*exp(c*(b*x+a)*ln(F))*tan(1/2*d+1/2*e*x)^2)/(1+tan(1/2*
d+1/2*e*x)^2)
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Maxima [B] time = 1.23256, size = 1098, normalized size = 5.52 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*cos(e*x+d)^3,x, algorithm="maxima")
[Out]
1/8*((F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 + 3*F^(a*c)*b^2*c^2*e*log(F)^2*sin(3*d) + F^(a*c)*b*c*e^2*cos(3*d)*log
(F) + 3*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*cos(3*e*x) + (F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 - 3*F^(a*c)*b^2*c^2*e*
log(F)^2*sin(3*d) + F^(a*c)*b*c*e^2*cos(3*d)*log(F) - 3*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*cos(3*e*x + 6*d) + 3*(
F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 - F^(a*c)*b^2*c^2*e*log(F)^2*sin(3*d) + 9*F^(a*c)*b*c*e^2*cos(3*d)*log(F) -
9*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*cos(e*x + 4*d) + 3*(F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 + F^(a*c)*b^2*c^2*e*lo
g(F)^2*sin(3*d) + 9*F^(a*c)*b*c*e^2*cos(3*d)*log(F) + 9*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*cos(e*x - 2*d) - (F^(a
*c)*b^3*c^3*log(F)^3*sin(3*d) - 3*F^(a*c)*b^2*c^2*e*cos(3*d)*log(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(3*d) - 3*F^
(a*c)*e^3*cos(3*d))*F^(b*c*x)*sin(3*e*x) + (F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) + 3*F^(a*c)*b^2*c^2*e*cos(3*d)*l
og(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(3*d) + 3*F^(a*c)*e^3*cos(3*d))*F^(b*c*x)*sin(3*e*x + 6*d) + 3*(F^(a*c)*b^
3*c^3*log(F)^3*sin(3*d) + F^(a*c)*b^2*c^2*e*cos(3*d)*log(F)^2 + 9*F^(a*c)*b*c*e^2*log(F)*sin(3*d) + 9*F^(a*c)*
e^3*cos(3*d))*F^(b*c*x)*sin(e*x + 4*d) - 3*(F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) - F^(a*c)*b^2*c^2*e*cos(3*d)*log
(F)^2 + 9*F^(a*c)*b*c*e^2*log(F)*sin(3*d) - 9*F^(a*c)*e^3*cos(3*d))*F^(b*c*x)*sin(e*x - 2*d))/(b^4*c^4*cos(3*d
)^2*log(F)^4 + b^4*c^4*log(F)^4*sin(3*d)^2 + 9*(cos(3*d)^2 + sin(3*d)^2)*e^4 + 10*(b^2*c^2*cos(3*d)^2*log(F)^2
+ b^2*c^2*log(F)^2*sin(3*d)^2)*e^2)
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Fricas [A] time = 0.496551, size = 331, normalized size = 1.66 \begin{align*} \frac{{\left (b^{3} c^{3} \cos \left (e x + d\right )^{3} \log \left (F\right )^{3} +{\left (b c e^{2} \cos \left (e x + d\right )^{3} + 6 \, b c e^{2} \cos \left (e x + d\right )\right )} \log \left (F\right ) + 3 \,{\left (b^{2} c^{2} e \cos \left (e x + d\right )^{2} \log \left (F\right )^{2} + e^{3} \cos \left (e x + d\right )^{2} + 2 \, e^{3}\right )} \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{4} c^{4} \log \left (F\right )^{4} + 10 \, b^{2} c^{2} e^{2} \log \left (F\right )^{2} + 9 \, e^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*cos(e*x+d)^3,x, algorithm="fricas")
[Out]
(b^3*c^3*cos(e*x + d)^3*log(F)^3 + (b*c*e^2*cos(e*x + d)^3 + 6*b*c*e^2*cos(e*x + d))*log(F) + 3*(b^2*c^2*e*cos
(e*x + d)^2*log(F)^2 + e^3*cos(e*x + d)^2 + 2*e^3)*sin(e*x + d))*F^(b*c*x + a*c)/(b^4*c^4*log(F)^4 + 10*b^2*c^
2*e^2*log(F)^2 + 9*e^4)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F**(c*(b*x+a))*cos(e*x+d)**3,x)
[Out]
Timed out
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Giac [C] time = 1.31953, size = 1764, normalized size = 8.86 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*cos(e*x+d)^3,x, algorithm="giac")
[Out]
1/4*(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 + 3*x*e + 3*d)*log(abs(F))/
(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 6*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 6*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 + 3*x*e + 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(
F) - pi*b*c + 6*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 3/4*(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 + x*e + d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*
c + 2*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 2*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 + x*e + d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log
(abs(F))) + 3/4*(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 - x*e - d)*log(
abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 2*e)^2) + (pi*b*c*sgn(F) - pi*b*c - 2*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 - x*e - d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*
sgn(F) - pi*b*c - 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 1/4*(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 - 3*x*e - 3*d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F)
- pi*b*c - 6*e)^2) + (pi*b*c*sgn(F) - pi*b*c - 6*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 - 3*x*e - 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 6*e)^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 + 3*I*x*e + 3*I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) + 48*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 - 3*I*x*e - 3*I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi
*b*c + 16*b*c*log(abs(F)) - 48*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - 1/2*I*(-6*I*e^(1/2*I*pi*b*c*x*s
gn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c + I*x*e + I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 1
6*b*c*log(abs(F)) + 16*I*e) + 6*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 - I*x*e - I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) - 16*I*e))*e^(b*c*x*log(abs(F)) + a*c
*log(abs(F))) - 1/2*I*(-6*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 - I
*x*e - I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) - 16*I*e) + 6*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 + I*x*e + I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*lo
g(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 - 3*I*x*e - 3*I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(ab
s(F)) - 48*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 + 3*I*x*
e + 3*I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) + 48*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs
(F)))