3.4 \(\int F^{c (a+b x)} \sin (d+e x) \, dx\)

Optimal. Leaf size=73 \[ \frac {b c \log (F) \sin (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4432} \[ \frac {b c \log (F) \sin (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sin[d + e*x],x]

[Out]

-((e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x])/(e^2
+ b^2*c^2*Log[F]^2)

Rule 4432

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[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)} \sin (d+e x) \, dx &=-\frac {e F^{c (a+b x)} \cos (d+e x)}{e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin (d+e x)}{e^2+b^2 c^2 \log ^2(F)}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 48, normalized size = 0.66 \[ \frac {F^{c (a+b x)} (b c \log (F) \sin (d+e x)-e \cos (d+e x))}{b^2 c^2 \log ^2(F)+e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(c*(a + b*x))*(-(e*Cos[d + e*x]) + b*c*Log[F]*Sin[d + e*x]))/(e^2 + b^2*c^2*Log[F]^2)

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fricas [A]  time = 0.92, size = 49, normalized size = 0.67 \[ \frac {{\left (b c \log \relax (F) \sin \left (e x + d\right ) - e \cos \left (e x + d\right )\right )} F^{b c x + a c}}{b^{2} c^{2} \log \relax (F)^{2} + e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.21, size = 652, normalized size = 8.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*c*log(abs(F))*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) - (pi*b*c*sgn(F) - pi*b*c + 2*e)*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)/(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))) - (2*b*c*log(abs(F))*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) - (p
i*b*c*sgn(F) - pi*b*c - 2*e)*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
)/(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))) + 1/2*
(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 + I*x*e + I*d)/(2*I*pi*b*c
*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) + 4*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 - I*x*e - I*d)/(-2*I*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*I*e))*e^(b*c
*x*log(abs(F)) + a*c*log(abs(F))) + 1/2*(-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 - I*x*e - I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*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 + I*x*e + I*d)/(-2*I*pi*b*c*sgn(F) + 2*I*pi
*b*c + 4*b*c*log(abs(F)) + 4*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)))

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maple [A]  time = 0.04, size = 130, normalized size = 1.78 \[ \frac {\frac {e \,{\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )} \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{e^{2}+b^{2} c^{2} \ln \relax (F )^{2}}-\frac {e \,{\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )}}{e^{2}+b^{2} c^{2} \ln \relax (F )^{2}}+\frac {2 b c \ln \relax (F ) {\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \relax (F )^{2}}}{1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.13, size = 194, normalized size = 2.66 \[ -\frac {{\left (F^{a c} b c \log \relax (F) \sin \relax (d) + F^{a c} e \cos \relax (d)\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \relax (F) \sin \relax (d) - F^{a c} e \cos \relax (d)\right )} F^{b c x} \cos \left (e x\right ) - {\left (F^{a c} b c \cos \relax (d) \log \relax (F) - F^{a c} e \sin \relax (d)\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \cos \relax (d) \log \relax (F) + F^{a c} e \sin \relax (d)\right )} F^{b c x} \sin \left (e x\right )}{2 \, {\left (b^{2} c^{2} \cos \relax (d)^{2} \log \relax (F)^{2} + b^{2} c^{2} \log \relax (F)^{2} \sin \relax (d)^{2} + {\left (\cos \relax (d)^{2} + \sin \relax (d)^{2}\right )} e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((F^(a*c)*b*c*log(F)*sin(d) + F^(a*c)*e*cos(d))*F^(b*c*x)*cos(e*x + 2*d) - (F^(a*c)*b*c*log(F)*sin(d) - F
^(a*c)*e*cos(d))*F^(b*c*x)*cos(e*x) - (F^(a*c)*b*c*cos(d)*log(F) - F^(a*c)*e*sin(d))*F^(b*c*x)*sin(e*x + 2*d)
- (F^(a*c)*b*c*cos(d)*log(F) + F^(a*c)*e*sin(d))*F^(b*c*x)*sin(e*x))/(b^2*c^2*cos(d)^2*log(F)^2 + b^2*c^2*log(
F)^2*sin(d)^2 + (cos(d)^2 + sin(d)^2)*e^2)

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mupad [B]  time = 2.40, size = 50, normalized size = 0.68 \[ -\frac {F^{a\,c+b\,c\,x}\,\left (e\,\cos \left (d+e\,x\right )-b\,c\,\sin \left (d+e\,x\right )\,\ln \relax (F)\right )}{b^2\,c^2\,{\ln \relax (F)}^2+e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))*sin(d + e*x),x)

[Out]

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

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sympy [A]  time = 7.84, size = 326, normalized size = 4.47 \[ \begin {cases} \frac {\left (-1\right )^{a c} \left (-1\right )^{\frac {e x}{\pi }} x \sin {\left (d + e x \right )}}{2} + \frac {\left (-1\right )^{a c} \left (-1\right )^{\frac {e x}{\pi }} i x \cos {\left (d + e x \right )}}{2} - \frac {\left (-1\right )^{a c} \left (-1\right )^{\frac {e x}{\pi }} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: F = -1 \wedge b = \frac {e}{\pi c} \\x \sin {\relax (d )} & \text {for}\: F = 1 \wedge e = 0 \\\tilde {\infty } e \left (e^{- \frac {i e}{b c}}\right )^{a c} \left (e^{- \frac {i e}{b c}}\right )^{b c x} \sin {\left (d + e x \right )} + \tilde {\infty } e \left (e^{- \frac {i e}{b c}}\right )^{a c} \left (e^{- \frac {i e}{b c}}\right )^{b c x} \cos {\left (d + e x \right )} & \text {for}\: F = e^{- \frac {i e}{b c}} \\\tilde {\infty } e \left (e^{\frac {i e}{b c}}\right )^{a c} \left (e^{\frac {i e}{b c}}\right )^{b c x} \sin {\left (d + e x \right )} + \tilde {\infty } e \left (e^{\frac {i e}{b c}}\right )^{a c} \left (e^{\frac {i e}{b c}}\right )^{b c x} \cos {\left (d + e x \right )} & \text {for}\: F = e^{\frac {i e}{b c}} \\\frac {F^{a c} F^{b c x} b c \log {\relax (F )} \sin {\left (d + e x \right )}}{b^{2} c^{2} \log {\relax (F )}^{2} + e^{2}} - \frac {F^{a c} F^{b c x} e \cos {\left (d + e x \right )}}{b^{2} c^{2} \log {\relax (F )}^{2} + e^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-1)**(a*c)*(-1)**(e*x/pi)*x*sin(d + e*x)/2 + (-1)**(a*c)*(-1)**(e*x/pi)*I*x*cos(d + e*x)/2 - (-1)*
*(a*c)*(-1)**(e*x/pi)*cos(d + e*x)/(2*e), Eq(F, -1) & Eq(b, e/(pi*c))), (x*sin(d), Eq(F, 1) & Eq(e, 0)), (zoo*
e*exp(-I*e/(b*c))**(a*c)*exp(-I*e/(b*c))**(b*c*x)*sin(d + e*x) + zoo*e*exp(-I*e/(b*c))**(a*c)*exp(-I*e/(b*c))*
*(b*c*x)*cos(d + e*x), Eq(F, exp(-I*e/(b*c)))), (zoo*e*exp(I*e/(b*c))**(a*c)*exp(I*e/(b*c))**(b*c*x)*sin(d + e
*x) + zoo*e*exp(I*e/(b*c))**(a*c)*exp(I*e/(b*c))**(b*c*x)*cos(d + e*x), Eq(F, exp(I*e/(b*c)))), (F**(a*c)*F**(
b*c*x)*b*c*log(F)*sin(d + e*x)/(b**2*c**2*log(F)**2 + e**2) - F**(a*c)*F**(b*c*x)*e*cos(d + e*x)/(b**2*c**2*lo
g(F)**2 + e**2), True))

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