3.1.28 \(\int \text {Si}(a+b x)^2 \, dx\) [28]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 = {6640, 6646, 4491, 12, 3380} \begin {gather*} \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} \end {gather*}

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 12

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

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*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.88 \begin {gather*} \frac {2 \cos (a+b x) \text {Si}(a+b x)+(a+b x) \text {Si}(a+b x)^2-\text {Si}(2 (a+b x))}{b} \end {gather*}

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 = 0.35, size = 45, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x +a \right )^{2} \left (b x +a \right )+2 \cos \left (b x +a \right ) \sinIntegral \left (b x +a \right )-\sinIntegral \left (2 b x +2 a \right )}{b}\)	\(45\)
default	\(\frac {\sinIntegral \left (b x +a \right )^{2} \left (b x +a \right )+2 \cos \left (b x +a \right ) \sinIntegral \left (b x +a \right )-\sinIntegral \left (2 b x +2 a \right )}{b}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(sin_integral(b*x + a)^2, x)

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Fricas [A]
time = 0.38, size = 44, normalized size = 0.90 \begin {gather*} \frac {{\left (b x + a\right )} \operatorname {Si}\left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) - \operatorname {Si}\left (2 \, b x + 2 \, a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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 = 0.00, 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(sin_integral(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(sin_integral(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\mathrm {sinint}\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinint(a + b*x)^2,x)

[Out]

int(sinint(a + b*x)^2, x)

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