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]*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)

________________________________________________________________________________________

Rubi [A]  time = 0.04918, antiderivative size = 84, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0525414, size = 50, normalized size = 0.6 \[ \frac{d^x \left (\sin (x) \left (x \log ^3(d)+x \log (d)-\log ^2(d)+1\right )-\cos (x) \left (x \log ^2(d)-2 \log (d)+x\right )\right )}{\left (\log ^2(d)+1\right )^2} \]

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

________________________________________________________________________________________

Maple [A]  time = 0.02, size = 137, normalized size = 1.6 \begin{align*}{ \left ({\frac{x{{\rm e}^{x\ln \left ( d \right ) }}}{1+ \left ( \ln \left ( d \right ) \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}}+2\,{\frac{\ln \left ( d \right ){{\rm e}^{x\ln \left ( d \right ) }}}{ \left ( 1+ \left ( \ln \left ( d \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{x{{\rm e}^{x\ln \left ( d \right ) }}}{1+ \left ( \ln \left ( d \right ) \right ) ^{2}}}-2\,{\frac{\ln \left ( d \right ){{\rm e}^{x\ln \left ( d \right ) }} \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{ \left ( 1+ \left ( \ln \left ( d \right ) \right ) ^{2} \right ) ^{2}}}-2\,{\frac{ \left ( \left ( \ln \left ( d \right ) \right ) ^{2}-1 \right ){{\rm e}^{x\ln \left ( d \right ) }}\tan \left ( x/2 \right ) }{ \left ( 1+ \left ( \ln \left ( d \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{x\ln \left ( d \right ){{\rm e}^{x\ln \left ( d \right ) }}\tan \left ( x/2 \right ) }{1+ \left ( \ln \left ( d \right ) \right ) ^{2}}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.979189, size = 81, normalized size = 0.96 \begin{align*} -\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{align*}

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)

________________________________________________________________________________________

Fricas [A]  time = 1.86185, size = 177, normalized size = 2.11 \begin{align*} -\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{align*}

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)

________________________________________________________________________________________

Sympy [A]  time = 3.50037, size = 308, normalized size = 3.67 \begin{align*} \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{align*}

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))

________________________________________________________________________________________

Giac [C]  time = 1.13661, size = 1574, normalized size = 18.74 \begin{align*} \text{result too large to display} \end{align*}

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))