3.1.13 \(\int \text {Si}(b x)^2 \, dx\) [13]

Optimal. Leaf size=32 \[ \frac {2 \cos (b x) \text {Si}(b x)}{b}+x \text {Si}(b x)^2-\frac {\text {Si}(2 b x)}{b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6640, 6646, 12, 4491, 3380} \begin {gather*} x \text {Si}(b x)^2-\frac {\text {Si}(2 b x)}{b}+\frac {2 \text {Si}(b x) \cos (b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[SinIntegral[b*x]^2,x]

[Out]

(2*Cos[b*x]*SinIntegral[b*x])/b + x*SinIntegral[b*x]^2 - SinIntegral[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}(b x)^2 \, dx &=x \text {Si}(b x)^2-2 \int \sin (b x) \text {Si}(b x) \, dx\\ &=\frac {2 \cos (b x) \text {Si}(b x)}{b}+x \text {Si}(b x)^2-2 \int \frac {\cos (b x) \sin (b x)}{b x} \, dx\\ &=\frac {2 \cos (b x) \text {Si}(b x)}{b}+x \text {Si}(b x)^2-\frac {2 \int \frac {\cos (b x) \sin (b x)}{x} \, dx}{b}\\ &=\frac {2 \cos (b x) \text {Si}(b x)}{b}+x \text {Si}(b x)^2-\frac {2 \int \frac {\sin (2 b x)}{2 x} \, dx}{b}\\ &=\frac {2 \cos (b x) \text {Si}(b x)}{b}+x \text {Si}(b x)^2-\frac {\int \frac {\sin (2 b x)}{x} \, dx}{b}\\ &=\frac {2 \cos (b x) \text {Si}(b x)}{b}+x \text {Si}(b x)^2-\frac {\text {Si}(2 b x)}{b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 32, normalized size = 1.00 \begin {gather*} \frac {2 \cos (b x) \text {Si}(b x)}{b}+x \text {Si}(b x)^2-\frac {\text {Si}(2 b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[SinIntegral[b*x]^2,x]

[Out]

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

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Maple [A]
time = 0.36, size = 32, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x \right )^{2} b x +2 \cos \left (b x \right ) \sinIntegral \left (b x \right )-\sinIntegral \left (2 b x \right )}{b}\)	\(32\)
default	\(\frac {\sinIntegral \left (b x \right )^{2} b x +2 \cos \left (b x \right ) \sinIntegral \left (b x \right )-\sinIntegral \left (2 b x \right )}{b}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2,x, algorithm="maxima")

[Out]

integrate(sin_integral(b*x)^2, x)

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Fricas [A]
time = 0.37, size = 31, normalized size = 0.97 \begin {gather*} \frac {b x \operatorname {Si}\left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) - \operatorname {Si}\left (2 \, b x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(Si(b*x)**2, x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.40, size = 49, normalized size = 1.53 \begin {gather*} x \operatorname {Si}\left (b x\right )^{2} + \frac {2 \, \cos \left (b x\right ) \operatorname {Si}\left (b x\right )}{b} - \frac {\Im \left ( \operatorname {Ci}\left (2 \, b x\right ) \right ) - \Im \left ( \operatorname {Ci}\left (-2 \, b x\right ) \right ) + 2 \, \operatorname {Si}\left (2 \, b x\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*sin_integral(b*x)^2 + 2*cos(b*x)*sin_integral(b*x)/b - 1/2*(imag_part(cos_integral(2*b*x)) - imag_part(cos_i
ntegral(-2*b*x)) + 2*sin_integral(2*b*x))/b

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int {\mathrm {sinint}\left (b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinint(b*x)^2,x)

[Out]

int(sinint(b*x)^2, x)

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