3.1.27 \(\int x \text {Si}(a+b x)^2 \, dx\) [27]

Optimal. Leaf size=154 \[ \frac {\cos (2 a+2 b x)}{4 b^2}-\frac {\text {CosIntegral}(2 a+2 b x)}{2 b^2}+\frac {\log (a+b x)}{2 b^2}-\frac {a \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}+\frac {a \text {Si}(2 a+2 b x)}{b^2} \]

[Out]

-1/2*Ci(2*b*x+2*a)/b^2+1/4*cos(2*b*x+2*a)/b^2+1/2*ln(b*x+a)/b^2-a*cos(b*x+a)*Si(b*x+a)/b^2+x*cos(b*x+a)*Si(b*x
+a)/b-1/2*a*(b*x+a)*Si(b*x+a)^2/b^2+1/2*x*(b*x+a)*Si(b*x+a)^2/b+a*Si(2*b*x+2*a)/b^2-Si(b*x+a)*sin(b*x+a)/b^2

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Rubi [A]
time = 0.22, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.400, Rules used = {6644, 6648, 4669, 6873, 6874, 2718, 3380, 6652, 3393, 3383, 6640, 6646, 4491, 12} \begin {gather*} -\frac {\text {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {a \text {Si}(2 a+2 b x)}{b^2}-\frac {\text {Si}(a+b x) \sin (a+b x)}{b^2}-\frac {a \text {Si}(a+b x) \cos (a+b x)}{b^2}+\frac {\log (a+b x)}{2 b^2}+\frac {\cos (2 a+2 b x)}{4 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}+\frac {x \text {Si}(a+b x) \cos (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*SinIntegral[a + b*x]^2,x]

[Out]

Cos[2*a + 2*b*x]/(4*b^2) - CosIntegral[2*a + 2*b*x]/(2*b^2) + Log[a + b*x]/(2*b^2) - (a*Cos[a + b*x]*SinIntegr
al[a + b*x])/b^2 + (x*Cos[a + b*x]*SinIntegral[a + b*x])/b - (Sin[a + b*x]*SinIntegral[a + b*x])/b^2 - (a*(a +
b*x)*SinIntegral[a + b*x]^2)/(2*b^2) + (x*(a + b*x)*SinIntegral[a + b*x]^2)/(2*b) + (a*SinIntegral[2*a + 2*b*
x])/b^2

Rule 12

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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])
)

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4669

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

Rule 6640

Int[SinIntegral[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(SinIntegral[a + b*x]^2/b), x] - Dist[2, In
t[Sin[a + b*x]*SinIntegral[a + b*x], x], x] /; FreeQ[{a, b}, x]

Rule 6644

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(c + d*x)^m*(SinI
ntegral[a + b*x]^2/(b*(m + 1))), x] + (-Dist[2/(m + 1), Int[(c + d*x)^m*Sin[a + b*x]*SinIntegral[a + b*x], x],
x] + Dist[(b*c - a*d)*(m/(b*(m + 1))), Int[(c + d*x)^(m - 1)*SinIntegral[a + b*x]^2, x], x]) /; FreeQ[{a, b,
c, d}, x] && IGtQ[m, 0]

Rule 6646

Int[Sin[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(-Cos[a + b*x])*(SinIntegral[c
+ d*x]/b), x] + Dist[d/b, Int[Cos[a + b*x]*(Sin[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]

Rule 6648

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] + (Dist[d/b, Int[(e + f*x)^m*Cos[a + b*x]*(Sin[c + d*x]/(
c + d*x)), x], x] + Dist[f*(m/b), Int[(e + f*x)^(m - 1)*Cos[a + b*x]*SinIntegral[c + d*x], x], x]) /; FreeQ[{a
, b, c, d, e, f}, x] && IGtQ[m, 0]

Rule 6652

Int[Cos[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[a + b*x]*(SinIntegral[c + d
*x]/b), x] - Dist[d/b, Int[Sin[a + b*x]*(Sin[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int x \text {Si}(a+b x)^2 \, dx &=\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}-\frac {a \int \text {Si}(a+b x)^2 \, dx}{2 b}-\int x \sin (a+b x) \text {Si}(a+b x) \, dx\\ &=\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}-\frac {\int \cos (a+b x) \text {Si}(a+b x) \, dx}{b}+\frac {a \int \sin (a+b x) \text {Si}(a+b x) \, dx}{b}-\int \frac {x \cos (a+b x) \sin (a+b x)}{a+b x} \, dx\\ &=-\frac {a \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}-\frac {1}{2} \int \frac {x \sin (2 (a+b x))}{a+b x} \, dx+\frac {\int \frac {\sin ^2(a+b x)}{a+b x} \, dx}{b}+\frac {a \int \frac {\cos (a+b x) \sin (a+b x)}{a+b x} \, dx}{b}\\ &=-\frac {a \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}-\frac {1}{2} \int \frac {x \sin (2 a+2 b x)}{a+b x} \, dx+\frac {\int \left (\frac {1}{2 (a+b x)}-\frac {\cos (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b}+\frac {a \int \frac {\sin (2 a+2 b x)}{2 (a+b x)} \, dx}{b}\\ &=\frac {\log (a+b x)}{2 b^2}-\frac {a \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}-\frac {1}{2} \int \left (\frac {\sin (2 a+2 b x)}{b}+\frac {a \sin (2 a+2 b x)}{b (-a-b x)}\right ) \, dx-\frac {\int \frac {\cos (2 a+2 b x)}{a+b x} \, dx}{2 b}+\frac {a \int \frac {\sin (2 a+2 b x)}{a+b x} \, dx}{2 b}\\ &=-\frac {\text {Ci}(2 a+2 b x)}{2 b^2}+\frac {\log (a+b x)}{2 b^2}-\frac {a \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}+\frac {a \text {Si}(2 a+2 b x)}{2 b^2}-\frac {\int \sin (2 a+2 b x) \, dx}{2 b}-\frac {a \int \frac {\sin (2 a+2 b x)}{-a-b x} \, dx}{2 b}\\ &=\frac {\cos (2 a+2 b x)}{4 b^2}-\frac {\text {Ci}(2 a+2 b x)}{2 b^2}+\frac {\log (a+b x)}{2 b^2}-\frac {a \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}+\frac {a \text {Si}(2 a+2 b x)}{b^2}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 95, normalized size = 0.62 \begin {gather*} \frac {\cos (2 (a+b x))-2 \text {CosIntegral}(2 (a+b x))+2 \log (a+b x)-4 ((a-b x) \cos (a+b x)+\sin (a+b x)) \text {Si}(a+b x)-2 \left (a^2-b^2 x^2\right ) \text {Si}(a+b x)^2+4 a \text {Si}(2 (a+b x))}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*SinIntegral[a + b*x]^2,x]

[Out]

(Cos[2*(a + b*x)] - 2*CosIntegral[2*(a + b*x)] + 2*Log[a + b*x] - 4*((a - b*x)*Cos[a + b*x] + Sin[a + b*x])*Si
nIntegral[a + b*x] - 2*(a^2 - b^2*x^2)*SinIntegral[a + b*x]^2 + 4*a*SinIntegral[2*(a + b*x)])/(4*b^2)

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Maple [A]
time = 0.57, size = 111, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x +a \right )^{2} \left (-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{2}\right )-2 \sinIntegral \left (b x +a \right ) \left (a \cos \left (b x +a \right )+\frac {\sin \left (b x +a \right )}{2}-\frac {\left (b x +a \right ) \cos \left (b x +a \right )}{2}\right )+a \sinIntegral \left (2 b x +2 a \right )+\frac {\ln \left (b x +a \right )}{2}-\frac {\cosineIntegral \left (2 b x +2 a \right )}{2}+\frac {\left (\cos ^{2}\left (b x +a \right )\right )}{2}}{b^{2}}\)	\(111\)
default	\(\frac {\sinIntegral \left (b x +a \right )^{2} \left (-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{2}\right )-2 \sinIntegral \left (b x +a \right ) \left (a \cos \left (b x +a \right )+\frac {\sin \left (b x +a \right )}{2}-\frac {\left (b x +a \right ) \cos \left (b x +a \right )}{2}\right )+a \sinIntegral \left (2 b x +2 a \right )+\frac {\ln \left (b x +a \right )}{2}-\frac {\cosineIntegral \left (2 b x +2 a \right )}{2}+\frac {\left (\cos ^{2}\left (b x +a \right )\right )}{2}}{b^{2}}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*Si(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b^2*(Si(b*x+a)^2*(-a*(b*x+a)+1/2*(b*x+a)^2)-2*Si(b*x+a)*(a*cos(b*x+a)+1/2*sin(b*x+a)-1/2*(b*x+a)*cos(b*x+a))
+a*Si(2*b*x+2*a)+1/2*ln(b*x+a)-1/2*Ci(2*b*x+2*a)+1/2*cos(b*x+a)^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin_integral(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate(x*sin_integral(b*x + a)^2, x)

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Fricas [A]
time = 0.39, size = 116, normalized size = 0.75 \begin {gather*} \frac {4 \, {\left (b x - a\right )} \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) + 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \operatorname {Si}\left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right )^{2} + 4 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - 4 \, \sin \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) - \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) - \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) + 2 \, \log \left (b x + a\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin_integral(b*x+a)^2,x, algorithm="fricas")

[Out]

1/4*(4*(b*x - a)*cos(b*x + a)*sin_integral(b*x + a) + 2*(b^2*x^2 - a^2)*sin_integral(b*x + a)^2 + 2*cos(b*x +
a)^2 + 4*a*sin_integral(2*b*x + 2*a) - 4*sin(b*x + a)*sin_integral(b*x + a) - cos_integral(2*b*x + 2*a) - cos_
integral(-2*b*x - 2*a) + 2*log(b*x + a))/b^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {Si}^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Si(b*x+a)**2,x)

[Out]

Integral(x*Si(a + b*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin_integral(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x*sin_integral(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\mathrm {sinint}\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinint(a + b*x)^2,x)

[Out]

int(x*sinint(a + b*x)^2, x)

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