3.28 \(\int \text{Si}(a+b x)^2 \, dx\)

Optimal. Leaf size=49 \[ \frac{(a+b x) \text{Si}(a+b x)^2}{b}-\frac{\text{Si}(2 a+2 b x)}{b}+\frac{2 \text{Si}(a+b x) \cos (a+b x)}{b} \]

[Out]

(2*Cos[a + b*x]*SinIntegral[a + b*x])/b + ((a + b*x)*SinIntegral[a + b*x]^2)/b - SinIntegral[2*a + 2*b*x]/b

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Rubi [A]  time = 0.0646836, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {6505, 6511, 4406, 12, 3299} \[ \frac{(a+b x) \text{Si}(a+b x)^2}{b}-\frac{\text{Si}(2 a+2 b x)}{b}+\frac{2 \text{Si}(a+b x) \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[SinIntegral[a + b*x]^2,x]

[Out]

(2*Cos[a + b*x]*SinIntegral[a + b*x])/b + ((a + b*x)*SinIntegral[a + b*x]^2)/b - SinIntegral[2*a + 2*b*x]/b

Rule 6505

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 6511

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 4406

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 12

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

Rule 3299

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]

Rubi steps

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

Mathematica [A]  time = 0.0089927, size = 43, normalized size = 0.88 \[ \frac{(a+b x) \text{Si}(a+b x)^2-\text{Si}(2 (a+b x))+2 \text{Si}(a+b x) \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[SinIntegral[a + b*x]^2,x]

[Out]

(2*Cos[a + b*x]*SinIntegral[a + b*x] + (a + b*x)*SinIntegral[a + b*x]^2 - SinIntegral[2*(a + b*x)])/b

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Maple [A]  time = 0.049, size = 45, normalized size = 0.9 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({\it Si} \left ( bx+a \right ) \right ) ^{2}+2\,\cos \left ( bx+a \right ){\it Si} \left ( bx+a \right ) -{\it Si} \left ( 2\,bx+2\,a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\rm Si}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Si(b*x + a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\operatorname{Si}\left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin_integral(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(Si(a + b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\rm Si}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Si(b*x + a)^2, x)