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

Optimal result

Integrand size = 7, antiderivative size = 84 \[ \int d^x x \sin (x) \, dx=\frac {2 d^x \cos (x) \log (d)}{\left (1+\log ^2(d)\right )^2}-\frac {d^x x \cos (x)}{1+\log ^2(d)}+\frac {d^x \sin (x)}{\left (1+\log ^2(d)\right )^2}-\frac {d^x \log ^2(d) \sin (x)}{\left (1+\log ^2(d)\right )^2}+\frac {d^x x \log (d) \sin (x)}{1+\log ^2(d)} \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4517, 4553, 4518} \[ \int d^x x \sin (x) \, dx=\frac {x d^x \log (d) \sin (x)}{\log ^2(d)+1}-\frac {d^x \log ^2(d) \sin (x)}{\left (\log ^2(d)+1\right )^2}+\frac {d^x \sin (x)}{\left (\log ^2(d)+1\right )^2}-\frac {x d^x \cos (x)}{\log ^2(d)+1}+\frac {2 d^x \log (d) \cos (x)}{\left (\log ^2(d)+1\right )^2} \]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.60 \[ \int d^x x \sin (x) \, dx=\frac {d^x \left (-\cos (x) \left (x-2 \log (d)+x \log ^2(d)\right )+\left (1+x \log (d)-\log ^2(d)+x \log ^3(d)\right ) \sin (x)\right )}{\left (1+\log ^2(d)\right )^2} \]

[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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {d^{x} \left (\ln \left (d \right )^{3} x \sin \left (x \right )+\left (-x \cos \left (x \right )-\sin \left (x \right )\right ) \ln \left (d \right )^{2}+\left (x \sin \left (x \right )+2 \cos \left (x \right )\right ) \ln \left (d \right )-x \cos \left (x \right )+\sin \left (x \right )\right )}{\left (1+\ln \left (d \right )^{2}\right )^{2}}\)	\(56\)
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	\(\frac {\frac {x \,{\mathrm e}^{x \ln \left (d \right )} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{1+\ln \left (d \right )^{2}}+\frac {2 \ln \left (d \right ) {\mathrm e}^{x \ln \left (d \right )}}{\left (1+\ln \left (d \right )^{2}\right )^{2}}-\frac {x \,{\mathrm e}^{x \ln \left (d \right )}}{1+\ln \left (d \right )^{2}}-\frac {2 \ln \left (d \right ) {\mathrm e}^{x \ln \left (d \right )} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{\left (1+\ln \left (d \right )^{2}\right )^{2}}-\frac {2 \left (\ln \left (d \right )^{2}-1\right ) {\mathrm e}^{x \ln \left (d \right )} \tan \left (\frac {x}{2}\right )}{\left (1+\ln \left (d \right )^{2}\right )^{2}}+\frac {2 \ln \left (d \right ) x \,{\mathrm e}^{x \ln \left (d \right )} \tan \left (\frac {x}{2}\right )}{1+\ln \left (d \right )^{2}}}{1+\tan ^{2}\left (\frac {x}{2}\right )}\)	\(137\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int d^x x \sin (x) \, dx=-\frac {{\left (x \cos \left (x\right ) \log \left (d\right )^{2} + x \cos \left (x\right ) - 2 \, \cos \left (x\right ) \log \left (d\right ) - {\left (x \log \left (d\right )^{3} + x \log \left (d\right ) - \log \left (d\right )^{2} + 1\right )} \sin \left (x\right )\right )} d^{x}}{\log \left (d\right )^{4} + 2 \, \log \left (d\right )^{2} + 1} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.67 \[ \int d^x x \sin (x) \, dx=\begin {cases} \frac {x^{2} e^{- i x} \sin {\left (x \right )}}{4} - \frac {i x^{2} e^{- i x} \cos {\left (x \right )}}{4} + \frac {i x e^{- i x} \sin {\left (x \right )}}{4} - \frac {x e^{- i x} \cos {\left (x \right )}}{4} + \frac {i e^{- i x} \cos {\left (x \right )}}{4} & \text {for}\: d = e^{- i} \\\frac {x^{2} e^{i x} \sin {\left (x \right )}}{4} + \frac {i x^{2} e^{i x} \cos {\left (x \right )}}{4} - \frac {i x e^{i x} \sin {\left (x \right )}}{4} - \frac {x e^{i x} \cos {\left (x \right )}}{4} - \frac {i e^{i x} \cos {\left (x \right )}}{4} & \text {for}\: d = e^{i} \\\frac {d^{x} x \log {\left (d \right )}^{3} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} - \frac {d^{x} x \log {\left (d \right )}^{2} \cos {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} + \frac {d^{x} x \log {\left (d \right )} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} - \frac {d^{x} x \cos {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} - \frac {d^{x} \log {\left (d \right )}^{2} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} + \frac {2 d^{x} \log {\left (d \right )} \cos {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} + \frac {d^{x} \sin {\left (x \right )}}{\log {\left (d \right )}^{4} + 2 \log {\left (d \right )}^{2} + 1} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int d^x x \sin (x) \, dx=-\frac {{\left ({\left (\log \left (d\right )^{2} + 1\right )} x - 2 \, \log \left (d\right )\right )} d^{x} \cos \left (x\right ) - {\left ({\left (\log \left (d\right )^{3} + \log \left (d\right )\right )} x - \log \left (d\right )^{2} + 1\right )} d^{x} \sin \left (x\right )}{\log \left (d\right )^{4} + 2 \, \log \left (d\right )^{2} + 1} \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 1156, normalized size of antiderivative = 13.76 \[ \int d^x x \sin (x) \, dx=\text {Too large to display} \]

[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 - 1/4*abs(d)^x*((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) - (pi*x*sgn(d) - pi*x + 2*I*x*log(a
bs(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)) - 1/4*abs(d)^x*((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) - (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))

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.68 \[ \int d^x x \sin (x) \, dx=\frac {d^x\,\left (\sin \left (x\right )+2\,\ln \left (d\right )\,\cos \left (x\right )-{\ln \left (d\right )}^2\,\sin \left (x\right )-x\,\cos \left (x\right )+x\,\ln \left (d\right )\,\sin \left (x\right )-x\,{\ln \left (d\right )}^2\,\cos \left (x\right )+x\,{\ln \left (d\right )}^3\,\sin \left (x\right )\right )}{{\left ({\ln \left (d\right )}^2+1\right )}^2} \]

[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