\(\int F^{c (a+b x)} \sin (d+e x) \, dx\) [4]

Optimal result

Integrand size = 16, antiderivative size = 73 \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=-\frac {e F^{c (a+b x)} \cos (d+e x)}{e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin (d+e x)}{e^2+b^2 c^2 \log ^2(F)} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4517} \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=\frac {b c \log (F) \sin (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2} \]

[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 4517

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=\frac {F^{c (a+b x)} (-e \cos (d+e x)+b c \log (F) \sin (d+e x))}{e^2+b^2 c^2 \log ^2(F)} \]

[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)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {F^{c \left (x b +a \right )} \left (b c \ln \left (F \right ) \sin \left (e x +d \right )-e \cos \left (e x +d \right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(49\)
risch	\(-\frac {e \,F^{c \left (x b +a \right )} \cos \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {b c \,F^{c \left (x b +a \right )} \ln \left (F \right ) \sin \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(74\)
norman	\(\frac {\frac {e \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 b c \ln \left (F \right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}\)	\(130\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=\frac {{\left (b c \log \left (F\right ) \sin \left (e x + d\right ) - e \cos \left (e x + d\right )\right )} F^{b c x + a c}}{b^{2} c^{2} \log \left (F\right )^{2} + e^{2}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 347, normalized size of antiderivative = 4.75 \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=\begin {cases} x \sin {\left (d \right )} & \text {for}\: F = 1 \wedge e = 0 \\F^{a c} x \sin {\left (d \right )} & \text {for}\: b = 0 \wedge e = 0 \\x \sin {\left (d \right )} & \text {for}\: c = 0 \wedge e = 0 \\- \frac {F^{a c + b c x} x \sin {\left (i b c x \log {\left (F \right )} - d \right )}}{2} + \frac {i F^{a c + b c x} x \cos {\left (i b c x \log {\left (F \right )} - d \right )}}{2} + \frac {F^{a c + b c x} \sin {\left (i b c x \log {\left (F \right )} - d \right )}}{2 b c \log {\left (F \right )}} - \frac {i F^{a c + b c x} \cos {\left (i b c x \log {\left (F \right )} - d \right )}}{b c \log {\left (F \right )}} & \text {for}\: e = - i b c \log {\left (F \right )} \\\frac {F^{a c + b c x} x \sin {\left (i b c x \log {\left (F \right )} + d \right )}}{2} - \frac {i F^{a c + b c x} x \cos {\left (i b c x \log {\left (F \right )} + d \right )}}{2} - \frac {F^{a c + b c x} \sin {\left (i b c x \log {\left (F \right )} + d \right )}}{2 b c \log {\left (F \right )}} + \frac {i F^{a c + b c x} \cos {\left (i b c x \log {\left (F \right )} + d \right )}}{b c \log {\left (F \right )}} & \text {for}\: e = i b c \log {\left (F \right )} \\\frac {F^{a c + b c x} b c \log {\left (F \right )} \sin {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} + e^{2}} - \frac {F^{a c + b c x} e \cos {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} + e^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (73) = 146\).

Time = 0.22 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.66 \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=-\frac {{\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} e \cos \left (d\right )\right )} F^{b c x} \cos \left (e x\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) - F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) + F^{a c} e \sin \left (d\right )\right )} F^{b c x} \sin \left (e x\right )}{2 \, {\left (b^{2} c^{2} \cos \left (d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (d\right )^{2} + {\left (\cos \left (d\right )^{2} + \sin \left (d\right )^{2}\right )} e^{2}\right )}} \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 634, normalized size of antiderivative = 8.68 \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=\text {Too large to display} \]

[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 + e*x + 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 + e*x + 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 - e*x - 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 - e*x - 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))) - (-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*e*x + I*d)/(2*I*pi*b*c*sgn(
F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) + 4*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 - I*e*x - 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))) - (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*e*x - I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*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 + I*e*x + 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)))

Mupad [B] (verification not implemented)

Time = 28.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int F^{c (a+b x)} \sin (d+e x) \, dx=-\frac {F^{a\,c+b\,c\,x}\,\left (e\,\cos \left (d+e\,x\right )-b\,c\,\sin \left (d+e\,x\right )\,\ln \left (F\right )\right )}{b^2\,c^2\,{\ln \left (F\right )}^2+e^2} \]

[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)