\(\int x^m \text {Si}(b x) \, dx\) [1]

Optimal result

Integrand size = 8, antiderivative size = 86 \[ \int x^m \text {Si}(b x) \, dx=\frac {x^m (-i b x)^{-m} \Gamma (1+m,-i b x)}{2 b (1+m)}+\frac {x^m (i b x)^{-m} \Gamma (1+m,i b x)}{2 b (1+m)}+\frac {x^{1+m} \text {Si}(b x)}{1+m} \]

[Out]

1/2*x^m*GAMMA(1+m,-I*b*x)/b/(1+m)/((-I*b*x)^m)+1/2*x^m*GAMMA(1+m,I*b*x)/b/(1+m)/((I*b*x)^m)+x^(1+m)*Si(b*x)/(1
+m)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6638, 12, 3389, 2212} \[ \int x^m \text {Si}(b x) \, dx=\frac {x^{m+1} \text {Si}(b x)}{m+1}+\frac {x^m (-i b x)^{-m} \Gamma (m+1,-i b x)}{2 b (m+1)}+\frac {x^m (i b x)^{-m} \Gamma (m+1,i b x)}{2 b (m+1)} \]

[In]

Int[x^m*SinIntegral[b*x],x]

[Out]

(x^m*Gamma[1 + m, (-I)*b*x])/(2*b*(1 + m)*((-I)*b*x)^m) + (x^m*Gamma[1 + m, I*b*x])/(2*b*(1 + m)*(I*b*x)^m) +
(x^(1 + m)*SinIntegral[b*x])/(1 + m)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 6638

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinIntegr
al[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Sin[a + b*x]/(a + b*x)), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \text {Si}(b x)}{1+m}-\frac {b \int \frac {x^m \sin (b x)}{b} \, dx}{1+m} \\ & = \frac {x^{1+m} \text {Si}(b x)}{1+m}-\frac {\int x^m \sin (b x) \, dx}{1+m} \\ & = \frac {x^{1+m} \text {Si}(b x)}{1+m}-\frac {i \int e^{-i b x} x^m \, dx}{2 (1+m)}+\frac {i \int e^{i b x} x^m \, dx}{2 (1+m)} \\ & = \frac {x^m (-i b x)^{-m} \Gamma (1+m,-i b x)}{2 b (1+m)}+\frac {x^m (i b x)^{-m} \Gamma (1+m,i b x)}{2 b (1+m)}+\frac {x^{1+m} \text {Si}(b x)}{1+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int x^m \text {Si}(b x) \, dx=\frac {x^m \left (b^2 x^2\right )^{-m} \left ((i b x)^m \Gamma (1+m,-i b x)+(-i b x)^m \Gamma (1+m,i b x)+2 b x \left (b^2 x^2\right )^m \text {Si}(b x)\right )}{2 b (1+m)} \]

[In]

Integrate[x^m*SinIntegral[b*x],x]

[Out]

(x^m*((I*b*x)^m*Gamma[1 + m, (-I)*b*x] + ((-I)*b*x)^m*Gamma[1 + m, I*b*x] + 2*b*x*(b^2*x^2)^m*SinIntegral[b*x]
))/(2*b*(1 + m)*(b^2*x^2)^m)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.61 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.43

method	result	size
meijerg	\(\frac {b \,x^{2+m} \operatorname {hypergeom}\left (\left [\frac {1}{2}, 1+\frac {m}{2}\right ], \left [\frac {3}{2}, \frac {3}{2}, 2+\frac {m}{2}\right ], -\frac {b^{2} x^{2}}{4}\right )}{2+m}\)	\(37\)

[In]

int(x^m*Si(b*x),x,method=_RETURNVERBOSE)

[Out]

b/(2+m)*x^(2+m)*hypergeom([1/2,1+1/2*m],[3/2,3/2,2+1/2*m],-1/4*b^2*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int x^m \text {Si}(b x) \, dx=\frac {2 \, b x x^{m} \operatorname {Si}\left (b x\right ) + \frac {\Gamma \left (m + 1, i \, b x\right )}{\left (i \, b\right )^{m}} + \frac {\Gamma \left (m + 1, -i \, b x\right )}{\left (-i \, b\right )^{m}}}{2 \, {\left (b m + b\right )}} \]

[In]

integrate(x^m*sin_integral(b*x),x, algorithm="fricas")

[Out]

1/2*(2*b*x*x^m*sin_integral(b*x) + gamma(m + 1, I*b*x)/(I*b)^m + gamma(m + 1, -I*b*x)/(-I*b)^m)/(b*m + b)

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51 \[ \int x^m \text {Si}(b x) \, dx=\frac {b x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {3}{2}, \frac {3}{2}, \frac {m}{2} + 2 \end {matrix}\middle | {- \frac {b^{2} x^{2}}{4}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} \]

[In]

integrate(x**m*Si(b*x),x)

[Out]

b*x**(m + 2)*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (3/2, 3/2, m/2 + 2), -b**2*x**2/4)/(2*gamma(m/2 + 2))

Maxima [F]

\[ \int x^m \text {Si}(b x) \, dx=\int { x^{m} \operatorname {Si}\left (b x\right ) \,d x } \]

[In]

integrate(x^m*sin_integral(b*x),x, algorithm="maxima")

[Out]

integrate(x^m*sin_integral(b*x), x)

Giac [F]

\[ \int x^m \text {Si}(b x) \, dx=\int { x^{m} \operatorname {Si}\left (b x\right ) \,d x } \]

[In]

integrate(x^m*sin_integral(b*x),x, algorithm="giac")

[Out]

integrate(x^m*sin_integral(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^m \text {Si}(b x) \, dx=\int x^m\,\mathrm {sinint}\left (b\,x\right ) \,d x \]

[In]

int(x^m*sinint(b*x),x)

[Out]

int(x^m*sinint(b*x), x)