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

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

[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)

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Rubi [A]
time = 0.01, 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 = {4517} \begin {gather*} \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} \end {gather*}

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 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*} \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 \begin {gather*} \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)} \end {gather*}

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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Maple [A]
time = 0.08, size = 74, normalized size = 1.01

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (74) = 148\).
time = 0.28, size = 197, normalized size = 2.70 \begin {gather*} -\frac {{\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} \cos \left (d\right ) e\right )} F^{b c x} \cos \left (x e + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} \cos \left (d\right ) e\right )} F^{b c x} \cos \left (x e\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 (x e + 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 (x e\right )}{2 \, {\left ({\left (b^{2} c^{2} \log \left (F\right )^{2} + e^{2}\right )} \cos \left (d\right )^{2} + {\left (b^{2} c^{2} \log \left (F\right )^{2} + e^{2}\right )} \sin \left (d\right )^{2}\right )}} \end {gather*}

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

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Fricas [A]
time = 2.24, size = 51, normalized size = 0.70 \begin {gather*} \frac {{\left (b c \log \left (F\right ) \sin \left (x e + d\right ) - \cos \left (x e + d\right ) e\right )} F^{b c x + a c}}{b^{2} c^{2} \log \left (F\right )^{2} + e^{2}} \end {gather*}

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

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Sympy [C] Result contains complex when optimal does not.
time = 2.64, size = 462, normalized size = 6.33 \begin {gather*} \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 }} i \sin {\left (d + e x \right )}}{2 e} - \frac {\left (-1\right )^{a c} \left (-1\right )^{\frac {e x}{\pi }} \cos {\left (d + e x \right )}}{e} & \text {for}\: F = -1 \wedge b = \frac {e}{\pi c} \\x \sin {\left (d \right )} & \text {for}\: F = 1 \wedge e = 0 \\\frac {b c \left (e^{- \frac {i e}{b c}}\right )^{a c} \left (e^{- \frac {i e}{b c}}\right )^{b c x} \log {\left (e^{- \frac {i e}{b c}} \right )} \sin {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (e^{- \frac {i e}{b c}} \right )}^{2} + e^{2}} - \frac {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 )}}{b^{2} c^{2} \log {\left (e^{- \frac {i e}{b c}} \right )}^{2} + e^{2}} & \text {for}\: F = e^{- \frac {i e}{b c}} \\\frac {b c \left (e^{\frac {i e}{b c}}\right )^{a c} \left (e^{\frac {i e}{b c}}\right )^{b c x} \log {\left (e^{\frac {i e}{b c}} \right )} \sin {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (e^{\frac {i e}{b c}} \right )}^{2} + e^{2}} - \frac {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 )}}{b^{2} c^{2} \log {\left (e^{\frac {i e}{b c}} \right )}^{2} + e^{2}} & \text {for}\: F = e^{\frac {i e}{b c}} \\\frac {F^{a c} F^{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} F^{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} \end {gather*}

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

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Giac [C] Result contains complex when optimal does not.
time = 0.41, size = 634, normalized size = 8.68 \begin {gather*} {\left (\frac {2 \, b c \log \left ({\left | F \right |}\right ) \sin \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c + e x + d\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c + 2 \, e\right )}^{2}} - \frac {{\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c + 2 \, e\right )} \cos \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c + e x + d\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c + 2 \, e\right )}^{2}}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} - {\left (\frac {2 \, b c \log \left ({\left | F \right |}\right ) \sin \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c - e x - d\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c - 2 \, e\right )}^{2}} - \frac {{\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c - 2 \, e\right )} \cos \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c - e x - d\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c - 2 \, e\right )}^{2}}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} - {\left (-\frac {i \, e^{\left (\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi b c x + \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi a c + i \, e x + i \, d\right )}}{2 i \, \pi b c \mathrm {sgn}\left (F\right ) - 2 i \, \pi b c + 4 \, b c \log \left ({\left | F \right |}\right ) + 4 i \, e} - \frac {i \, e^{\left (-\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi b c x - \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi a c - i \, e x - i \, d\right )}}{-2 i \, \pi b c \mathrm {sgn}\left (F\right ) + 2 i \, \pi b c + 4 \, b c \log \left ({\left | F \right |}\right ) - 4 i \, e}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} - {\left (\frac {i \, e^{\left (\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi b c x + \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi a c - i \, e x - i \, d\right )}}{2 i \, \pi b c \mathrm {sgn}\left (F\right ) - 2 i \, \pi b c + 4 \, b c \log \left ({\left | F \right |}\right ) - 4 i \, e} + \frac {i \, e^{\left (-\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi b c x - \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi a c + i \, e x + i \, d\right )}}{-2 i \, \pi b c \mathrm {sgn}\left (F\right ) + 2 i \, \pi b c + 4 \, b c \log \left ({\left | F \right |}\right ) + 4 i \, e}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} \end {gather*}

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 + 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)))

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Mupad [B]
time = 2.40, size = 50, normalized size = 0.68 \begin {gather*} -\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} \end {gather*}

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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