3.3 \(\int F^{c (a+b x)} \sin ^2(d+e x) \, dx\)
Optimal. Leaf size=128 \[ \frac{b c \log (F) \sin ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac{2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac{2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )} \]
[Out]
(2*e^2*F^(c*(a + b*x)))/(b*c*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) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^2)/(4*e^2 + b^2*c^2*Log[F]^2)
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Rubi [A] time = 0.0520328, antiderivative size = 128, 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 =
{4434, 2194} \[ \frac{b c \log (F) \sin ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac{2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac{2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[F^(c*(a + b*x))*Sin[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)) - (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) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^2)/(4*e^2 + b^2*c^2*Log[F]^2)
Rule 4434
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*
x))*Sin[d + e*x]^n)/(e^2*n^2 + b^2*c^2*Log[F]^2), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x] - Simp[(e*n*F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x]^(n - 1))/(
e^2*n^2 + b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
n, 1]
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]
Rubi steps
\begin{align*} \int F^{c (a+b x)} \sin ^2(d+e x) \, dx &=-\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{b c F^{c (a+b x)} \log (F) \sin ^2(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{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{b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}\\ \end{align*}
Mathematica [A] time = 0.212046, size = 86, normalized size = 0.67 \[ \frac{F^{c (a+b x)} \left (-b^2 c^2 \log ^2(F) \cos (2 (d+e x))+b^2 c^2 \log ^2(F)-2 b c e \log (F) \sin (2 (d+e x))+4 e^2\right )}{2 b^3 c^3 \log ^3(F)+8 b c e^2 \log (F)} \]
Antiderivative was successfully verified.
[In]
Integrate[F^(c*(a + b*x))*Sin[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)
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Maple [A] time = 0.049, size = 153, normalized size = 1.2 \begin{align*}{\frac{{F}^{c \left ( bx+a \right ) }}{2\,bc\ln \left ( F \right ) }}-{\frac{1}{2+2\, \left ( \tan \left ( ex+d \right ) \right ) ^{2}} \left ({\frac{bc\ln \left ( F \right ){{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}}{4\,{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}}+4\,{\frac{e{{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}\tan \left ( ex+d \right ) }{4\,{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 ) }} \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{4\,{e}^{2}+{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(b*x+a))*sin(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)*ln(F)*b*c*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)*ln(F)*b*c*exp(c*(b*x+a)*ln(F))*tan(e*x+d)^2)
/(1+tan(e*x+d)^2)
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Maxima [B] time = 1.2013, size = 481, normalized size = 3.76 \begin{align*} -\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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*sin(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*l
og(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)
________________________________________________________________________________________
Fricas [A] time = 0.49243, size = 207, normalized size = 1.62 \begin{align*} -\frac{{\left (2 \, b c e \cos \left (e x + d\right ) \log \left (F\right ) \sin \left (e x + d\right ) +{\left (b^{2} c^{2} \cos \left (e x + d\right )^{2} - b^{2} c^{2}\right )} \log \left (F\right )^{2} - 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 )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*sin(e*x+d)^2,x, algorithm="fricas")
[Out]
-(2*b*c*e*cos(e*x + d)*log(F)*sin(e*x + d) + (b^2*c^2*cos(e*x + d)^2 - b^2*c^2)*log(F)^2 - 2*e^2)*F^(b*c*x + a
*c)/(b^3*c^3*log(F)^3 + 4*b*c*e^2*log(F))
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Sympy [A] time = 159.735, size = 627, normalized size = 4.9 \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))*sin(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*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)*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))
________________________________________________________________________________________
Giac [C] time = 1.25514, size = 1260, normalized size = 9.84 \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))*sin(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) -
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))) + (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*l
og(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*lo
g(abs(F)) + a*c*log(abs(F)))