Integrand size = 10, antiderivative size = 61 \[ \int x \sin (b x) \text {Si}(b x) \, dx=\frac {\operatorname {CosIntegral}(2 b x)}{2 b^2}-\frac {\log (x)}{2 b^2}+\frac {\sin ^2(b x)}{2 b^2}-\frac {x \cos (b x) \text {Si}(b x)}{b}+\frac {\sin (b x) \text {Si}(b x)}{b^2} \] Output:
1/2*Ci(2*b*x)/b^2-1/2*ln(x)/b^2+1/2*sin(b*x)^2/b^2-x*cos(b*x)*Si(b*x)/b+si n(b*x)*Si(b*x)/b^2
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int x \sin (b x) \text {Si}(b x) \, dx=-\frac {\cos (2 b x)-2 \operatorname {CosIntegral}(2 b x)+2 \log (x)+4 (b x \cos (b x)-\sin (b x)) \text {Si}(b x)}{4 b^2} \] Input:
Integrate[x*Sin[b*x]*SinIntegral[b*x],x]
Output:
-1/4*(Cos[2*b*x] - 2*CosIntegral[2*b*x] + 2*Log[x] + 4*(b*x*Cos[b*x] - Sin [b*x])*SinIntegral[b*x])/b^2
Time = 0.48 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {7067, 27, 3042, 3044, 15, 7071, 27, 3042, 3793, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \text {Si}(b x) \sin (b x) \, dx\) |
\(\Big \downarrow \) 7067 |
\(\displaystyle \frac {\int \cos (b x) \text {Si}(b x)dx}{b}+\int \frac {\cos (b x) \sin (b x)}{b}dx-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \cos (b x) \text {Si}(b x)dx}{b}+\frac {\int \cos (b x) \sin (b x)dx}{b}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (b x) \text {Si}(b x)dx}{b}+\frac {\int \cos (b x) \sin (b x)dx}{b}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \sin (b x)d\sin (b x)}{b^2}+\frac {\int \cos (b x) \text {Si}(b x)dx}{b}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\int \cos (b x) \text {Si}(b x)dx}{b}+\frac {\sin ^2(b x)}{2 b^2}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 7071 |
\(\displaystyle \frac {\frac {\text {Si}(b x) \sin (b x)}{b}-\int \frac {\sin ^2(b x)}{b x}dx}{b}+\frac {\sin ^2(b x)}{2 b^2}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\text {Si}(b x) \sin (b x)}{b}-\frac {\int \frac {\sin ^2(b x)}{x}dx}{b}}{b}+\frac {\sin ^2(b x)}{2 b^2}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\text {Si}(b x) \sin (b x)}{b}-\frac {\int \frac {\sin (b x)^2}{x}dx}{b}}{b}+\frac {\sin ^2(b x)}{2 b^2}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {\frac {\text {Si}(b x) \sin (b x)}{b}-\frac {\int \left (\frac {1}{2 x}-\frac {\cos (2 b x)}{2 x}\right )dx}{b}}{b}+\frac {\sin ^2(b x)}{2 b^2}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin ^2(b x)}{2 b^2}+\frac {\frac {\text {Si}(b x) \sin (b x)}{b}-\frac {\frac {\log (x)}{2}-\frac {\operatorname {CosIntegral}(2 b x)}{2}}{b}}{b}-\frac {x \text {Si}(b x) \cos (b x)}{b}\) |
Input:
Int[x*Sin[b*x]*SinIntegral[b*x],x]
Output:
Sin[b*x]^2/(2*b^2) - (x*Cos[b*x]*SinIntegral[b*x])/b + (-((-1/2*CosIntegra l[2*b*x] + Log[x]/2)/b) + (Sin[b*x]*SinIntegral[b*x])/b)/b
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(-(e + f*x)^m)*Cos[a + b*x]*(SinIntegral[c + d*x]/b), x] + (Simp[d/b Int[(e + f*x)^m*Cos[a + b*x]*(Sin[c + d*x]/(c + d*x)), x], x] + Simp[f*(m/b) Int[(e + f*x)^(m - 1)*Cos[a + b*x]*SinIntegr al[c + d*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0]
Int[Cos[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> S imp[Sin[a + b*x]*(SinIntegral[c + d*x]/b), x] - Simp[d/b Int[Sin[a + b*x] *(Sin[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]
Time = 4.51 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(\frac {\operatorname {Si}\left (b x \right ) \left (\sin \left (b x \right )-b x \cos \left (b x \right )\right )-\frac {\cos \left (b x \right )^{2}}{2}-\frac {\ln \left (b x \right )}{2}+\frac {\operatorname {Ci}\left (2 b x \right )}{2}}{b^{2}}\) | \(45\) |
default | \(\frac {\operatorname {Si}\left (b x \right ) \left (\sin \left (b x \right )-b x \cos \left (b x \right )\right )-\frac {\cos \left (b x \right )^{2}}{2}-\frac {\ln \left (b x \right )}{2}+\frac {\operatorname {Ci}\left (2 b x \right )}{2}}{b^{2}}\) | \(45\) |
Input:
int(x*sin(b*x)*Si(b*x),x,method=_RETURNVERBOSE)
Output:
1/b^2*(Si(b*x)*(sin(b*x)-b*x*cos(b*x))-1/2*cos(b*x)^2-1/2*ln(b*x)+1/2*Ci(2 *b*x))
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70 \[ \int x \sin (b x) \text {Si}(b x) \, dx=-\frac {2 \, b x \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) + \cos \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \operatorname {Si}\left (b x\right ) - \operatorname {Ci}\left (2 \, b x\right ) + \log \left (x\right )}{2 \, b^{2}} \] Input:
integrate(x*sin(b*x)*sin_integral(b*x),x, algorithm="fricas")
Output:
-1/2*(2*b*x*cos(b*x)*sin_integral(b*x) + cos(b*x)^2 - 2*sin(b*x)*sin_integ ral(b*x) - cos_integral(2*b*x) + log(x))/b^2
\[ \int x \sin (b x) \text {Si}(b x) \, dx=\int x \sin {\left (b x \right )} \operatorname {Si}{\left (b x \right )}\, dx \] Input:
integrate(x*sin(b*x)*Si(b*x),x)
Output:
Integral(x*sin(b*x)*Si(b*x), x)
\[ \int x \sin (b x) \text {Si}(b x) \, dx=\int { x \sin \left (b x\right ) \operatorname {Si}\left (b x\right ) \,d x } \] Input:
integrate(x*sin(b*x)*sin_integral(b*x),x, algorithm="maxima")
Output:
integrate(x*sin(b*x)*sin_integral(b*x), x)
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int x \sin (b x) \text {Si}(b x) \, dx=-{\left (\frac {x \cos \left (b x\right )}{b} - \frac {\sin \left (b x\right )}{b^{2}}\right )} \operatorname {Si}\left (b x\right ) - \frac {\cos \left (2 \, b x\right ) - \operatorname {Ci}\left (2 \, b x\right ) - \operatorname {Ci}\left (-2 \, b x\right ) + 2 \, \log \left (x\right )}{4 \, b^{2}} \] Input:
integrate(x*sin(b*x)*sin_integral(b*x),x, algorithm="giac")
Output:
-(x*cos(b*x)/b - sin(b*x)/b^2)*sin_integral(b*x) - 1/4*(cos(2*b*x) - cos_i ntegral(2*b*x) - cos_integral(-2*b*x) + 2*log(x))/b^2
Timed out. \[ \int x \sin (b x) \text {Si}(b x) \, dx=\int x\,\mathrm {sinint}\left (b\,x\right )\,\sin \left (b\,x\right ) \,d x \] Input:
int(x*sinint(b*x)*sin(b*x),x)
Output:
int(x*sinint(b*x)*sin(b*x), x)
\[ \int x \sin (b x) \text {Si}(b x) \, dx=\int \mathit {si} \left (b x \right ) \sin \left (b x \right ) x d x \] Input:
int(x*sin(b*x)*Si(b*x),x)
Output:
int(si(b*x)*sin(b*x)*x,x)