3.1.57 \(\int \sin (a+b x) \text {Si}(a+b x) \, dx\) [57]

Optimal. Leaf size=34 \[ -\frac {\cos (a+b x) \text {Si}(a+b x)}{b}+\frac {\text {Si}(2 a+2 b x)}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6646, 4491, 12, 3380} \begin {gather*} \frac {\text {Si}(2 a+2 b x)}{2 b}-\frac {\text {Si}(a+b x) \cos (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*SinIntegral[a + b*x],x]

[Out]

-((Cos[a + b*x]*SinIntegral[a + b*x])/b) + SinIntegral[2*a + 2*b*x]/(2*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 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 \sin (a+b x) \text {Si}(a+b x) \, dx &=-\frac {\cos (a+b x) \text {Si}(a+b x)}{b}+\int \frac {\cos (a+b x) \sin (a+b x)}{a+b x} \, dx\\ &=-\frac {\cos (a+b x) \text {Si}(a+b x)}{b}+\int \frac {\sin (2 a+2 b x)}{2 (a+b x)} \, dx\\ &=-\frac {\cos (a+b x) \text {Si}(a+b x)}{b}+\frac {1}{2} \int \frac {\sin (2 a+2 b x)}{a+b x} \, dx\\ &=-\frac {\cos (a+b x) \text {Si}(a+b x)}{b}+\frac {\text {Si}(2 a+2 b x)}{2 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 33, normalized size = 0.97 \begin {gather*} -\frac {\cos (a+b x) \text {Si}(a+b x)}{b}+\frac {\text {Si}(2 (a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]*SinIntegral[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.19, size = 31, normalized size = 0.91

method	result	size
derivativedivides	\(\frac {-\cos \left (b x +a \right ) \sinIntegral \left (b x +a \right )+\frac {\sinIntegral \left (2 b x +2 a \right )}{2}}{b}\)	\(31\)
default	\(\frac {-\cos \left (b x +a \right ) \sinIntegral \left (b x +a \right )+\frac {\sinIntegral \left (2 b x +2 a \right )}{2}}{b}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-cos(b*x+a)*Si(b*x+a)+1/2*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)*sin(b*x+a),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)*sin_integral(b*x + a), x)

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Fricas [A]
time = 0.33, size = 31, normalized size = 0.91 \begin {gather*} -\frac {2 \, \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) - \operatorname {Si}\left (2 \, b x + 2 \, a\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(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 \sin {\left (a + b x \right )} \operatorname {Si}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)*Si(a + b*x), x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.43, size = 57, normalized size = 1.68 \begin {gather*} -\frac {\cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right )}{b} + \frac {\Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(b*x + a)*sin_integral(b*x + a)/b + 1/4*(imag_part(cos_integral(2*b*x + 2*a)) - imag_part(cos_integral(-2*
b*x - 2*a)) + 2*sin_integral(2*b*x + 2*a))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinint(a + b*x)*sin(a + b*x),x)

[Out]

int(sinint(a + b*x)*sin(a + b*x), x)

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