3.134 \(\int d^x \sin (x) \, dx\)

Optimal. Leaf size=32 \[ \frac {d^x \log (d) \sin (x)}{\log ^2(d)+1}-\frac {d^x \cos (x)}{\log ^2(d)+1} \]

[Out]

-d^x*cos(x)/(1+ln(d)^2)+d^x*ln(d)*sin(x)/(1+ln(d)^2)

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Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4432} \[ \frac {d^x \log (d) \sin (x)}{\log ^2(d)+1}-\frac {d^x \cos (x)}{\log ^2(d)+1} \]

Antiderivative was successfully verified.

[In]

Int[d^x*Sin[x],x]

[Out]

-((d^x*Cos[x])/(1 + Log[d]^2)) + (d^x*Log[d]*Sin[x])/(1 + Log[d]^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 d^x \sin (x) \, dx &=-\frac {d^x \cos (x)}{1+\log ^2(d)}+\frac {d^x \log (d) \sin (x)}{1+\log ^2(d)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.69 \[ \frac {d^x (\log (d) \sin (x)-\cos (x))}{\log ^2(d)+1} \]

Antiderivative was successfully verified.

[In]

Integrate[d^x*Sin[x],x]

[Out]

(d^x*(-Cos[x] + Log[d]*Sin[x]))/(1 + Log[d]^2)

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fricas [A]  time = 0.44, size = 22, normalized size = 0.69 \[ \frac {{\left (\log \relax (d) \sin \relax (x) - \cos \relax (x)\right )} d^{x}}{\log \relax (d)^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(d^x*sin(x),x, algorithm="fricas")

[Out]

(log(d)*sin(x) - cos(x))*d^x/(log(d)^2 + 1)

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giac [C]  time = 1.33, size = 328, normalized size = 10.25 \[ {\left | d \right |}^{x} {\left (\frac {{\left (\pi - \pi \mathrm {sgn}\relax (d) - 2\right )} \cos \left (\frac {1}{2} \, \pi x \mathrm {sgn}\relax (d) - \frac {1}{2} \, \pi x + x\right )}{{\left (\pi - \pi \mathrm {sgn}\relax (d) - 2\right )}^{2} + 4 \, \log \left ({\left | d \right |}\right )^{2}} + \frac {2 \, \log \left ({\left | d \right |}\right ) \sin \left (\frac {1}{2} \, \pi x \mathrm {sgn}\relax (d) - \frac {1}{2} \, \pi x + x\right )}{{\left (\pi - \pi \mathrm {sgn}\relax (d) - 2\right )}^{2} + 4 \, \log \left ({\left | d \right |}\right )^{2}}\right )} - {\left | d \right |}^{x} {\left (\frac {{\left (\pi - \pi \mathrm {sgn}\relax (d) + 2\right )} \cos \left (\frac {1}{2} \, \pi x \mathrm {sgn}\relax (d) - \frac {1}{2} \, \pi x - x\right )}{{\left (\pi - \pi \mathrm {sgn}\relax (d) + 2\right )}^{2} + 4 \, \log \left ({\left | d \right |}\right )^{2}} + \frac {2 \, \log \left ({\left | d \right |}\right ) \sin \left (\frac {1}{2} \, \pi x \mathrm {sgn}\relax (d) - \frac {1}{2} \, \pi x - x\right )}{{\left (\pi - \pi \mathrm {sgn}\relax (d) + 2\right )}^{2} + 4 \, \log \left ({\left | d \right |}\right )^{2}}\right )} + \frac {1}{2} \, {\left | d \right |}^{x} {\left (\frac {2 i \, e^{\left (\frac {1}{2} i \, \pi x \mathrm {sgn}\relax (d) - \frac {1}{2} i \, \pi x + i \, x\right )}}{-2 i \, \pi + 2 i \, \pi \mathrm {sgn}\relax (d) + 4 \, \log \left ({\left | d \right |}\right ) + 4 i} + \frac {2 i \, e^{\left (-\frac {1}{2} i \, \pi x \mathrm {sgn}\relax (d) + \frac {1}{2} i \, \pi x - i \, x\right )}}{2 i \, \pi - 2 i \, \pi \mathrm {sgn}\relax (d) + 4 \, \log \left ({\left | d \right |}\right ) - 4 i}\right )} + \frac {1}{2} \, {\left | d \right |}^{x} {\left (-\frac {2 i \, e^{\left (\frac {1}{2} i \, \pi x \mathrm {sgn}\relax (d) - \frac {1}{2} i \, \pi x - i \, x\right )}}{-2 i \, \pi + 2 i \, \pi \mathrm {sgn}\relax (d) + 4 \, \log \left ({\left | d \right |}\right ) - 4 i} - \frac {2 i \, e^{\left (-\frac {1}{2} i \, \pi x \mathrm {sgn}\relax (d) + \frac {1}{2} i \, \pi x + i \, x\right )}}{2 i \, \pi - 2 i \, \pi \mathrm {sgn}\relax (d) + 4 \, \log \left ({\left | d \right |}\right ) + 4 i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(d^x*sin(x),x, algorithm="giac")

[Out]

abs(d)^x*((pi - pi*sgn(d) - 2)*cos(1/2*pi*x*sgn(d) - 1/2*pi*x + x)/((pi - pi*sgn(d) - 2)^2 + 4*log(abs(d))^2)
+ 2*log(abs(d))*sin(1/2*pi*x*sgn(d) - 1/2*pi*x + x)/((pi - pi*sgn(d) - 2)^2 + 4*log(abs(d))^2)) - abs(d)^x*((p
i - pi*sgn(d) + 2)*cos(1/2*pi*x*sgn(d) - 1/2*pi*x - x)/((pi - pi*sgn(d) + 2)^2 + 4*log(abs(d))^2) + 2*log(abs(
d))*sin(1/2*pi*x*sgn(d) - 1/2*pi*x - x)/((pi - pi*sgn(d) + 2)^2 + 4*log(abs(d))^2)) + 1/2*abs(d)^x*(2*I*e^(1/2
*I*pi*x*sgn(d) - 1/2*I*pi*x + I*x)/(-2*I*pi + 2*I*pi*sgn(d) + 4*log(abs(d)) + 4*I) + 2*I*e^(-1/2*I*pi*x*sgn(d)
+ 1/2*I*pi*x - I*x)/(2*I*pi - 2*I*pi*sgn(d) + 4*log(abs(d)) - 4*I)) + 1/2*abs(d)^x*(-2*I*e^(1/2*I*pi*x*sgn(d)
- 1/2*I*pi*x - I*x)/(-2*I*pi + 2*I*pi*sgn(d) + 4*log(abs(d)) - 4*I) - 2*I*e^(-1/2*I*pi*x*sgn(d) + 1/2*I*pi*x
+ I*x)/(2*I*pi - 2*I*pi*sgn(d) + 4*log(abs(d)) + 4*I))

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maple [B]  time = 0.04, size = 69, normalized size = 2.16 \[ \frac {\frac {2 \,{\mathrm e}^{x \ln \relax (d )} \ln \relax (d ) \tan \left (\frac {x}{2}\right )}{\ln \relax (d )^{2}+1}+\frac {{\mathrm e}^{x \ln \relax (d )} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{\ln \relax (d )^{2}+1}-\frac {{\mathrm e}^{x \ln \relax (d )}}{\ln \relax (d )^{2}+1}}{\tan ^{2}\left (\frac {x}{2}\right )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(d^x*sin(x),x)

[Out]

(1/(1+ln(d)^2)*exp(x*ln(d))*tan(1/2*x)^2-1/(1+ln(d)^2)*exp(x*ln(d))+2*ln(d)/(1+ln(d)^2)*exp(x*ln(d))*tan(1/2*x
))/(tan(1/2*x)^2+1)

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maxima [A]  time = 0.43, size = 25, normalized size = 0.78 \[ \frac {d^{x} \log \relax (d) \sin \relax (x) - d^{x} \cos \relax (x)}{\log \relax (d)^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(d^x*sin(x),x, algorithm="maxima")

[Out]

(d^x*log(d)*sin(x) - d^x*cos(x))/(log(d)^2 + 1)

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mupad [B]  time = 0.02, size = 22, normalized size = 0.69 \[ -\frac {d^x\,\left (\cos \relax (x)-\ln \relax (d)\,\sin \relax (x)\right )}{{\ln \relax (d)}^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(d^x*sin(x),x)

[Out]

-(d^x*(cos(x) - log(d)*sin(x)))/(log(d)^2 + 1)

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sympy [A]  time = 1.04, size = 104, normalized size = 3.25 \[ \begin {cases} \frac {x e^{- i x} \sin {\relax (x )}}{2} - \frac {i x e^{- i x} \cos {\relax (x )}}{2} - \frac {e^{- i x} \cos {\relax (x )}}{2} & \text {for}\: d = e^{- i} \\\frac {x e^{i x} \sin {\relax (x )}}{2} + \frac {i x e^{i x} \cos {\relax (x )}}{2} - \frac {e^{i x} \cos {\relax (x )}}{2} & \text {for}\: d = e^{i} \\\frac {d^{x} \log {\relax (d )} \sin {\relax (x )}}{\log {\relax (d )}^{2} + 1} - \frac {d^{x} \cos {\relax (x )}}{\log {\relax (d )}^{2} + 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(d**x*sin(x),x)

[Out]

Piecewise((x*exp(-I*x)*sin(x)/2 - I*x*exp(-I*x)*cos(x)/2 - exp(-I*x)*cos(x)/2, Eq(d, exp(-I))), (x*exp(I*x)*si
n(x)/2 + I*x*exp(I*x)*cos(x)/2 - exp(I*x)*cos(x)/2, Eq(d, exp(I))), (d**x*log(d)*sin(x)/(log(d)**2 + 1) - d**x
*cos(x)/(log(d)**2 + 1), True))

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