3.2.36 \(\int d^x x \sin (x) \, dx\) [136]

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

[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 = 0.03, 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 = {4517, 4553, 4518} \begin {gather*} \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} \end {gather*}

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 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]

Rule 4518

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 4553

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 = 0.04, size = 50, normalized size = 0.60 \begin {gather*} \frac {d^x \left (-\cos (x) \left (x-2 \log (d)+x \log ^2(d)\right )+\left (1+x \log (d)-\log ^2(d)+x \log ^3(d)\right ) \sin (x)\right )}{\left (1+\log ^2(d)\right )^2} \end {gather*}

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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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 3.64, size = 249, normalized size = 2.96 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {\left (-x \text {Cos}\left [x\right ]+I x \text {Sin}\left [x\right ]-I x^2 \text {Cos}\left [x\right ]+x^2 \text {Sin}\left [x\right ]+I \text {Cos}\left [x\right ]\right ) E^{-I x}}{4},d\text {==}E^{-I}\right \},\left \{\frac {\left (-x \text {Cos}\left [x\right ]-I x \text {Sin}\left [x\right ]+I x^2 \text {Cos}\left [x\right ]+x^2 \text {Sin}\left [x\right ]-I \text {Cos}\left [x\right ]\right ) E^{I x}}{4},d\text {==}E^I\right \}\right \},-\frac {x \text {Cos}\left [x\right ] d^x \text {Log}\left [d\right ]^2}{1+\text {Log}\left [d\right ]^4+2 \text {Log}\left [d\right ]^2}-\frac {x \text {Cos}\left [x\right ] d^x}{1+\text {Log}\left [d\right ]^4+2 \text {Log}\left [d\right ]^2}+\frac {x \text {Log}\left [d\right ] d^x \text {Sin}\left [x\right ]}{1+\text {Log}\left [d\right ]^4+2 \text {Log}\left [d\right ]^2}+\frac {x d^x \text {Log}\left [d\right ]^3 \text {Sin}\left [x\right ]}{1+\text {Log}\left [d\right ]^4+2 \text {Log}\left [d\right ]^2}-\frac {d^x \text {Log}\left [d\right ]^2 \text {Sin}\left [x\right ]}{1+\text {Log}\left [d\right ]^4+2 \text {Log}\left [d\right ]^2}+\frac {2 \text {Cos}\left [x\right ] \text {Log}\left [d\right ] d^x}{1+\text {Log}\left [d\right ]^4+2 \text {Log}\left [d\right ]^2}+\frac {d^x \text {Sin}\left [x\right ]}{1+\text {Log}\left [d\right ]^4+2 \text {Log}\left [d\right ]^2}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x*d^x*Sin[x],x]')

[Out]

Piecewise[{{(-x Cos[x] + I x Sin[x] - I x ^ 2 Cos[x] + x ^ 2 Sin[x] + I Cos[x]) E ^ (-I x) / 4, d == E ^ (-I)}
, {(-x Cos[x] - I x Sin[x] + I x ^ 2 Cos[x] + x ^ 2 Sin[x] - I Cos[x]) E ^ (I x) / 4, d == E ^ I}}, -x Cos[x]
d ^ x Log[d] ^ 2 / (1 + Log[d] ^ 4 + 2 Log[d] ^ 2) - x Cos[x] d ^ x / (1 + Log[d] ^ 4 + 2 Log[d] ^ 2) + x Log[
d] d ^ x Sin[x] / (1 + Log[d] ^ 4 + 2 Log[d] ^ 2) + x d ^ x Log[d] ^ 3 Sin[x] / (1 + Log[d] ^ 4 + 2 Log[d] ^ 2
) - d ^ x Log[d] ^ 2 Sin[x] / (1 + Log[d] ^ 4 + 2 Log[d] ^ 2) + 2 Cos[x] Log[d] d ^ x / (1 + Log[d] ^ 4 + 2 Lo
g[d] ^ 2) + d ^ x Sin[x] / (1 + Log[d] ^ 4 + 2 Log[d] ^ 2)]

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Maple [C] Result contains complex when optimal does not.
time = 0.04, size = 58, normalized size = 0.69

method	result	size
risch	\(-\frac {i \left (-1+x \ln \left (d \right )+i x \right ) d^{x} {\mathrm e}^{i x}}{2 \left (\ln \left (d \right )+i\right )^{2}}+\frac {i \left (-1+x \ln \left (d \right )-i x \right ) d^{x} {\mathrm e}^{-i x}}{2 \left (\ln \left (d \right )-i\right )^{2}}\)	\(58\)
norman	\(\frac {\frac {x \,{\mathrm e}^{x \ln \left (d \right )} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{1+\ln \left (d \right )^{2}}+\frac {2 \ln \left (d \right ) {\mathrm e}^{x \ln \left (d \right )}}{\left (1+\ln \left (d \right )^{2}\right )^{2}}-\frac {x \,{\mathrm e}^{x \ln \left (d \right )}}{1+\ln \left (d \right )^{2}}-\frac {2 \ln \left (d \right ) {\mathrm e}^{x \ln \left (d \right )} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{\left (1+\ln \left (d \right )^{2}\right )^{2}}-\frac {2 \left (\ln \left (d \right )^{2}-1\right ) {\mathrm e}^{x \ln \left (d \right )} \tan \left (\frac {x}{2}\right )}{\left (1+\ln \left (d \right )^{2}\right )^{2}}+\frac {2 \ln \left (d \right ) x \,{\mathrm e}^{x \ln \left (d \right )} \tan \left (\frac {x}{2}\right )}{1+\ln \left (d \right )^{2}}}{1+\tan ^{2}\left (\frac {x}{2}\right )}\)	\(137\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(d^x*x*sin(x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 60, normalized size = 0.71 \begin {gather*} -\frac {{\left ({\left (\log \left (d\right )^{2} + 1\right )} x - 2 \, \log \left (d\right )\right )} d^{x} \cos \left (x\right ) - {\left ({\left (\log \left (d\right )^{3} + \log \left (d\right )\right )} x - \log \left (d\right )^{2} + 1\right )} d^{x} \sin \left (x\right )}{\log \left (d\right )^{4} + 2 \, \log \left (d\right )^{2} + 1} \end {gather*}

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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Fricas [A]
time = 0.32, size = 60, normalized size = 0.71 \begin {gather*} -\frac {{\left (x \cos \left (x\right ) \log \left (d\right )^{2} + x \cos \left (x\right ) - 2 \, \cos \left (x\right ) \log \left (d\right ) - {\left (x \log \left (d\right )^{3} + x \log \left (d\right ) - \log \left (d\right )^{2} + 1\right )} \sin \left (x\right )\right )} d^{x}}{\log \left (d\right )^{4} + 2 \, \log \left (d\right )^{2} + 1} \end {gather*}

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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Sympy [A]
time = 0.51, size = 308, normalized size = 3.67 \begin {gather*} \begin {cases} \frac {x^{2} e^{- i x} \sin {\left (x \right )}}{4} - \frac {i x^{2} e^{- i x} \cos {\left (x \right )}}{4} + \frac {i x e^{- i x} \sin {\left (x \right )}}{4} - \frac {x e^{- i x} \cos {\left (x \right )}}{4} + \frac {i e^{- i x} \cos {\left (x \right )}}{4} & \text {for}\: d = e^{- i} \\\frac {x^{2} e^{i x} \sin {\left (x \right )}}{4} + \frac {i x^{2} e^{i x} \cos {\left (x \right )}}{4} - \frac {i x e^{i x} \sin {\left (x \right )}}{4} - \frac {x e^{i x} \cos {\left (x \right )}}{4} - \frac {i e^{i x} \cos {\left (x \right )}}{4} & \text {for}\: d = e^{i} \\\frac {d^{x} x \log {\left (d \right )}^{3} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} - \frac {d^{x} x \log {\left (d \right )}^{2} \cos {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} + \frac {d^{x} x \log {\left (d \right )} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} - \frac {d^{x} x \cos {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} - \frac {d^{x} \log {\left (d \right )}^{2} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} + \frac {2 d^{x} \log {\left (d \right )} \cos {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} + \frac {d^{x} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} & \text {otherwise} \end {cases} \end {gather*}

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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Giac [C] Result contains complex when optimal does not.
time = 0.00, size = 1369, normalized size = 16.30

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 - 1/4*((pi*x*sgn(d) - pi*x - 2*I*x*log(abs(d)) + 2*x + 2
*I)*e^(1/2*I*pi*x*sgn(d) - 1/2*I*pi*x + I*x)/(2*pi + pi^2*sgn(d) + 2*I*pi*log(abs(d))*sgn(d) - pi^2 - 2*I*pi*l
og(abs(d)) + 2*log(abs(d))^2 - 2*pi*sgn(d) + 4*I*log(abs(d)) - 2) - (pi*x*sgn(d) - pi*x + 2*I*x*log(abs(d)) +
2*x - 2*I)*e^(-1/2*I*pi*x*sgn(d) + 1/2*I*pi*x - I*x)/(2*pi + pi^2*sgn(d) - 2*I*pi*log(abs(d))*sgn(d) - pi^2 +
2*I*pi*log(abs(d)) + 2*log(abs(d))^2 - 2*pi*sgn(d) - 4*I*log(abs(d)) - 2))*abs(d)^x - 1/4*((pi*x*sgn(d) - pi*x
- 2*I*x*log(abs(d)) - 2*x + 2*I)*e^(1/2*I*pi*x*sgn(d) - 1/2*I*pi*x - I*x)/(2*pi - pi^2*sgn(d) - 2*I*pi*log(ab
s(d))*sgn(d) + pi^2 + 2*I*pi*log(abs(d)) - 2*log(abs(d))^2 - 2*pi*sgn(d) + 4*I*log(abs(d)) + 2) - (pi*x*sgn(d)
- pi*x + 2*I*x*log(abs(d)) - 2*x - 2*I)*e^(-1/2*I*pi*x*sgn(d) + 1/2*I*pi*x + I*x)/(2*pi - pi^2*sgn(d) + 2*I*p
i*log(abs(d))*sgn(d) + pi^2 - 2*I*pi*log(abs(d)) - 2*log(abs(d))^2 - 2*pi*sgn(d) - 4*I*log(abs(d)) + 2))*abs(d
)^x

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

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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