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

Optimal. Leaf size=84 \[ \frac {x d^x \log (d) \sin (x)}{\log ^2(d)+1}-\frac {d^x \log ^2(d) \sin (x)}{\left (\log ^2(d)+1\right )^2}+\frac {d^x \sin (x)}{\left (\log ^2(d)+1\right )^2}-\frac {x d^x \cos (x)}{\log ^2(d)+1}+\frac {2 d^x \log (d) \cos (x)}{\left (\log ^2(d)+1\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4432, 4465, 4433} \[ \frac {x d^x \log (d) \sin (x)}{\log ^2(d)+1}-\frac {d^x \log ^2(d) \sin (x)}{\left (\log ^2(d)+1\right )^2}+\frac {d^x \sin (x)}{\left (\log ^2(d)+1\right )^2}-\frac {x d^x \cos (x)}{\log ^2(d)+1}+\frac {2 d^x \log (d) \cos (x)}{\left (\log ^2(d)+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4433

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[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]

Rule 4465

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_.)*Sin[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Module[{u
= IntHide[F^(c*(a + b*x))*Sin[d + e*x]^n, x]}, Dist[(f*x)^m, u, x] - Dist[f*m, Int[(f*x)^(m - 1)*u, x], x]] /
; FreeQ[{F, a, b, c, d, e, f}, x] && IGtQ[n, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int d^x x \sin (x) \, dx &=-\frac {d^x x \cos (x)}{1+\log ^2(d)}+\frac {d^x x \log (d) \sin (x)}{1+\log ^2(d)}-\int \left (-\frac {d^x \cos (x)}{1+\log ^2(d)}+\frac {d^x \log (d) \sin (x)}{1+\log ^2(d)}\right ) \, dx\\ &=-\frac {d^x x \cos (x)}{1+\log ^2(d)}+\frac {d^x x \log (d) \sin (x)}{1+\log ^2(d)}+\frac {\int d^x \cos (x) \, dx}{1+\log ^2(d)}-\frac {\log (d) \int d^x \sin (x) \, dx}{1+\log ^2(d)}\\ &=\frac {2 d^x \cos (x) \log (d)}{\left (1+\log ^2(d)\right )^2}-\frac {d^x x \cos (x)}{1+\log ^2(d)}+\frac {d^x \sin (x)}{\left (1+\log ^2(d)\right )^2}-\frac {d^x \log ^2(d) \sin (x)}{\left (1+\log ^2(d)\right )^2}+\frac {d^x x \log (d) \sin (x)}{1+\log ^2(d)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 50, normalized size = 0.60 \[ \frac {d^x \left (\sin (x) \left (x \log ^3(d)+x \log (d)-\log ^2(d)+1\right )-\cos (x) \left (x \log ^2(d)-2 \log (d)+x\right )\right )}{\left (\log ^2(d)+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^x*(-(Cos[x]*(x - 2*Log[d] + x*Log[d]^2)) + (1 + x*Log[d] - Log[d]^2 + x*Log[d]^3)*Sin[x]))/(1 + Log[d]^2)^2

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fricas [A]  time = 0.43, size = 60, normalized size = 0.71 \[ -\frac {{\left (x \cos \relax (x) \log \relax (d)^{2} + x \cos \relax (x) - 2 \, \cos \relax (x) \log \relax (d) - {\left (x \log \relax (d)^{3} + x \log \relax (d) - \log \relax (d)^{2} + 1\right )} \sin \relax (x)\right )} d^{x}}{\log \relax (d)^{4} + 2 \, \log \relax (d)^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*cos(x)*log(d)^2 + x*cos(x) - 2*cos(x)*log(d) - (x*log(d)^3 + x*log(d) - log(d)^2 + 1)*sin(x))*d^x/(log(d)^
4 + 2*log(d)^2 + 1)

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giac [C]  time = 1.28, size = 1166, normalized size = 13.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(((2*pi + pi^2*sgn(d) - pi^2 + 2*log(abs(d))^2 - 2*pi*sgn(d) - 2)*(pi*x*sgn(d) - pi*x + 2*x)/((2*pi + pi^2
*sgn(d) - pi^2 + 2*log(abs(d))^2 - 2*pi*sgn(d) - 2)^2 + 4*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) + 2*log(abs(
d)))^2) - 4*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) + 2*log(abs(d)))*(x*log(abs(d)) - 1)/((2*pi + pi^2*sgn(d)
- pi^2 + 2*log(abs(d))^2 - 2*pi*sgn(d) - 2)^2 + 4*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) + 2*log(abs(d)))^2))
*cos(1/2*pi*x*sgn(d) - 1/2*pi*x + x) + 2*((pi*x*sgn(d) - pi*x + 2*x)*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) +
2*log(abs(d)))/((2*pi + pi^2*sgn(d) - pi^2 + 2*log(abs(d))^2 - 2*pi*sgn(d) - 2)^2 + 4*(pi*log(abs(d))*sgn(d)
- pi*log(abs(d)) + 2*log(abs(d)))^2) + (2*pi + pi^2*sgn(d) - pi^2 + 2*log(abs(d))^2 - 2*pi*sgn(d) - 2)*(x*log(
abs(d)) - 1)/((2*pi + pi^2*sgn(d) - pi^2 + 2*log(abs(d))^2 - 2*pi*sgn(d) - 2)^2 + 4*(pi*log(abs(d))*sgn(d) - p
i*log(abs(d)) + 2*log(abs(d)))^2))*sin(1/2*pi*x*sgn(d) - 1/2*pi*x + x))*abs(d)^x + 1/2*(((2*pi - pi^2*sgn(d) +
pi^2 - 2*log(abs(d))^2 - 2*pi*sgn(d) + 2)*(pi*x*sgn(d) - pi*x - 2*x)/((2*pi - pi^2*sgn(d) + pi^2 - 2*log(abs(
d))^2 - 2*pi*sgn(d) + 2)^2 + 4*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) - 2*log(abs(d)))^2) + 4*(pi*log(abs(d))
*sgn(d) - pi*log(abs(d)) - 2*log(abs(d)))*(x*log(abs(d)) - 1)/((2*pi - pi^2*sgn(d) + pi^2 - 2*log(abs(d))^2 -
2*pi*sgn(d) + 2)^2 + 4*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) - 2*log(abs(d)))^2))*cos(1/2*pi*x*sgn(d) - 1/2*
pi*x - x) - 2*((pi*x*sgn(d) - pi*x - 2*x)*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) - 2*log(abs(d)))/((2*pi - pi
^2*sgn(d) + pi^2 - 2*log(abs(d))^2 - 2*pi*sgn(d) + 2)^2 + 4*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) - 2*log(ab
s(d)))^2) - (2*pi - pi^2*sgn(d) + pi^2 - 2*log(abs(d))^2 - 2*pi*sgn(d) + 2)*(x*log(abs(d)) - 1)/((2*pi - pi^2*
sgn(d) + pi^2 - 2*log(abs(d))^2 - 2*pi*sgn(d) + 2)^2 + 4*(pi*log(abs(d))*sgn(d) - pi*log(abs(d)) - 2*log(abs(d
)))^2))*sin(1/2*pi*x*sgn(d) - 1/2*pi*x - x))*abs(d)^x - 1/2*abs(d)^x*((2*pi*x*sgn(d) - 2*pi*x - 4*I*x*log(abs(
d)) + 4*x + 4*I)*e^(1/2*I*pi*x*sgn(d) - 1/2*I*pi*x + I*x)/(8*pi + 4*pi^2*sgn(d) + 8*I*pi*log(abs(d))*sgn(d) -
4*pi^2 - 8*I*pi*log(abs(d)) + 8*log(abs(d))^2 - 8*pi*sgn(d) + 16*I*log(abs(d)) - 8) - (2*pi*x*sgn(d) - 2*pi*x
+ 4*I*x*log(abs(d)) + 4*x - 4*I)*e^(-1/2*I*pi*x*sgn(d) + 1/2*I*pi*x - I*x)/(8*pi + 4*pi^2*sgn(d) - 8*I*pi*log(
abs(d))*sgn(d) - 4*pi^2 + 8*I*pi*log(abs(d)) + 8*log(abs(d))^2 - 8*pi*sgn(d) - 16*I*log(abs(d)) - 8)) - 1/2*ab
s(d)^x*((2*pi*x*sgn(d) - 2*pi*x - 4*I*x*log(abs(d)) - 4*x + 4*I)*e^(1/2*I*pi*x*sgn(d) - 1/2*I*pi*x - I*x)/(8*p
i - 4*pi^2*sgn(d) - 8*I*pi*log(abs(d))*sgn(d) + 4*pi^2 + 8*I*pi*log(abs(d)) - 8*log(abs(d))^2 - 8*pi*sgn(d) +
16*I*log(abs(d)) + 8) - (2*pi*x*sgn(d) - 2*pi*x + 4*I*x*log(abs(d)) - 4*x - 4*I)*e^(-1/2*I*pi*x*sgn(d) + 1/2*I
*pi*x + I*x)/(8*pi - 4*pi^2*sgn(d) + 8*I*pi*log(abs(d))*sgn(d) + 4*pi^2 - 8*I*pi*log(abs(d)) - 8*log(abs(d))^2
- 8*pi*sgn(d) - 16*I*log(abs(d)) + 8))

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maple [A]  time = 0.04, size = 137, normalized size = 1.63 \[ \frac {\frac {2 x \,{\mathrm e}^{x \ln \relax (d )} \ln \relax (d ) \tan \left (\frac {x}{2}\right )}{\ln \relax (d )^{2}+1}+\frac {x \,{\mathrm e}^{x \ln \relax (d )} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{\ln \relax (d )^{2}+1}-\frac {2 \,{\mathrm e}^{x \ln \relax (d )} \ln \relax (d ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{\left (\ln \relax (d )^{2}+1\right )^{2}}-\frac {x \,{\mathrm e}^{x \ln \relax (d )}}{\ln \relax (d )^{2}+1}+\frac {2 \,{\mathrm e}^{x \ln \relax (d )} \ln \relax (d )}{\left (\ln \relax (d )^{2}+1\right )^{2}}-\frac {2 \left (\ln \relax (d )^{2}-1\right ) {\mathrm e}^{x \ln \relax (d )} \tan \left (\frac {x}{2}\right )}{\left (\ln \relax (d )^{2}+1\right )^{2}}}{\tan ^{2}\left (\frac {x}{2}\right )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 60, normalized size = 0.71 \[ -\frac {{\left ({\left (\log \relax (d)^{2} + 1\right )} x - 2 \, \log \relax (d)\right )} d^{x} \cos \relax (x) - {\left ({\left (\log \relax (d)^{3} + \log \relax (d)\right )} x - \log \relax (d)^{2} + 1\right )} d^{x} \sin \relax (x)}{\log \relax (d)^{4} + 2 \, \log \relax (d)^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((log(d)^2 + 1)*x - 2*log(d))*d^x*cos(x) - ((log(d)^3 + log(d))*x - log(d)^2 + 1)*d^x*sin(x))/(log(d)^4 + 2*
log(d)^2 + 1)

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mupad [B]  time = 0.28, size = 57, normalized size = 0.68 \[ \frac {d^x\,\left (\sin \relax (x)+2\,\ln \relax (d)\,\cos \relax (x)-{\ln \relax (d)}^2\,\sin \relax (x)-x\,\cos \relax (x)+x\,\ln \relax (d)\,\sin \relax (x)-x\,{\ln \relax (d)}^2\,\cos \relax (x)+x\,{\ln \relax (d)}^3\,\sin \relax (x)\right )}{{\left ({\ln \relax (d)}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^x*(sin(x) + 2*log(d)*cos(x) - log(d)^2*sin(x) - x*cos(x) + x*log(d)*sin(x) - x*log(d)^2*cos(x) + x*log(d)^3
*sin(x)))/(log(d)^2 + 1)^2

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sympy [A]  time = 3.28, size = 308, normalized size = 3.67 \[ \begin {cases} \frac {x^{2} e^{- i x} \sin {\relax (x )}}{4} - \frac {i x^{2} e^{- i x} \cos {\relax (x )}}{4} + \frac {i x e^{- i x} \sin {\relax (x )}}{4} - \frac {x e^{- i x} \cos {\relax (x )}}{4} + \frac {i e^{- i x} \cos {\relax (x )}}{4} & \text {for}\: d = e^{- i} \\\frac {x^{2} e^{i x} \sin {\relax (x )}}{4} + \frac {i x^{2} e^{i x} \cos {\relax (x )}}{4} - \frac {i x e^{i x} \sin {\relax (x )}}{4} - \frac {x e^{i x} \cos {\relax (x )}}{4} - \frac {i e^{i x} \cos {\relax (x )}}{4} & \text {for}\: d = e^{i} \\\frac {d^{x} x \log {\relax (d )}^{3} \sin {\relax (x )}}{\log {\relax (d )}^{4} + 2 \log {\relax (d )}^{2} + 1} - \frac {d^{x} x \log {\relax (d )}^{2} \cos {\relax (x )}}{\log {\relax (d )}^{4} + 2 \log {\relax (d )}^{2} + 1} + \frac {d^{x} x \log {\relax (d )} \sin {\relax (x )}}{\log {\relax (d )}^{4} + 2 \log {\relax (d )}^{2} + 1} - \frac {d^{x} x \cos {\relax (x )}}{\log {\relax (d )}^{4} + 2 \log {\relax (d )}^{2} + 1} - \frac {d^{x} \log {\relax (d )}^{2} \sin {\relax (x )}}{\log {\relax (d )}^{4} + 2 \log {\relax (d )}^{2} + 1} + \frac {2 d^{x} \log {\relax (d )} \cos {\relax (x )}}{\log {\relax (d )}^{4} + 2 \log {\relax (d )}^{2} + 1} + \frac {d^{x} \sin {\relax (x )}}{\log {\relax (d )}^{4} + 2 \log {\relax (d )}^{2} + 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*exp(-I*x)*sin(x)/4 - I*x**2*exp(-I*x)*cos(x)/4 + I*x*exp(-I*x)*sin(x)/4 - x*exp(-I*x)*cos(x)/4
+ I*exp(-I*x)*cos(x)/4, Eq(d, exp(-I))), (x**2*exp(I*x)*sin(x)/4 + I*x**2*exp(I*x)*cos(x)/4 - I*x*exp(I*x)*si
n(x)/4 - x*exp(I*x)*cos(x)/4 - I*exp(I*x)*cos(x)/4, Eq(d, exp(I))), (d**x*x*log(d)**3*sin(x)/(log(d)**4 + 2*lo
g(d)**2 + 1) - d**x*x*log(d)**2*cos(x)/(log(d)**4 + 2*log(d)**2 + 1) + d**x*x*log(d)*sin(x)/(log(d)**4 + 2*log
(d)**2 + 1) - d**x*x*cos(x)/(log(d)**4 + 2*log(d)**2 + 1) - d**x*log(d)**2*sin(x)/(log(d)**4 + 2*log(d)**2 + 1
) + 2*d**x*log(d)*cos(x)/(log(d)**4 + 2*log(d)**2 + 1) + d**x*sin(x)/(log(d)**4 + 2*log(d)**2 + 1), True))

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