3.124 \(\int x \text {Ci}(a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=109 \[ -\frac {a \text {Ci}(2 a+2 b x)}{2 b^2}+\frac {\text {Ci}(a+b x) \sin (a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}-\frac {a \log (a+b x)}{2 b^2}+\frac {\sin (a+b x) \cos (a+b x)}{2 b^2}-\frac {x \text {Ci}(a+b x) \cos (a+b x)}{b}+\frac {x}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6520, 6742, 2635, 8, 3312, 3302, 6512, 4406, 12, 3299} \[ -\frac {a \text {CosIntegral}(2 a+2 b x)}{2 b^2}+\frac {\text {CosIntegral}(a+b x) \sin (a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}-\frac {a \log (a+b x)}{2 b^2}+\frac {\sin (a+b x) \cos (a+b x)}{2 b^2}-\frac {x \text {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {x}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

Rule 3302

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 3312

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 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 6512

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

Rule 6520

Int[CosIntegral[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> -Simp[((e
+ f*x)^m*Cos[a + b*x]*CosIntegral[c + d*x])/b, x] + (Dist[d/b, Int[((e + f*x)^m*Cos[a + b*x]*Cos[c + d*x])/(c
+ d*x), x], x] + Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Cos[a + b*x]*CosIntegral[c + d*x], x], x]) /; FreeQ[{a,
b, c, d, e, f}, x] && IGtQ[m, 0]

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 76, normalized size = 0.70 \[ \frac {-2 a \text {Ci}(2 (a+b x))+\text {Ci}(a+b x) (4 \sin (a+b x)-4 b x \cos (a+b x))-2 \text {Si}(2 (a+b x))-2 a \log (a+b x)+\sin (2 (a+b x))+2 b x}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {Ci}\left (b x + a\right ) \sin \left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*cos_integral(b*x + a)*sin(b*x + a), x)

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giac [C]  time = 0.22, size = 538, normalized size = 4.94 \[ -{\left (\frac {x \cos \left (b x + a\right )}{b} - \frac {\sin \left (b x + a\right )}{b^{2}}\right )} \operatorname {Ci}\left (b x + a\right ) + \frac {2 \, b x \tan \left (b x\right )^{2} \tan \relax (a)^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + 2 \, b x \tan \left (b x\right )^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} - a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} - a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + 2 \, b x \tan \relax (a)^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \relax (a)^{2} - a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \relax (a)^{2} - a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \relax (a)^{2} - \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} - 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} - 2 \, \tan \left (b x\right )^{2} \tan \relax (a) - \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \relax (a)^{2} + \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \relax (a)^{2} - 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \relax (a)^{2} - 2 \, \tan \left (b x\right ) \tan \relax (a)^{2} + 2 \, b x - 2 \, a \log \left ({\left | b x + a \right |}\right ) - a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) - \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 ) + 2 \, \tan \left (b x\right ) + 2 \, \tan \relax (a)}{4 \, {\left (b^{2} \tan \left (b x\right )^{2} \tan \relax (a)^{2} + b^{2} \tan \left (b x\right )^{2} + b^{2} \tan \relax (a)^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*cos(b*x + a)/b - sin(b*x + a)/b^2)*cos_integral(b*x + a) + 1/4*(2*b*x*tan(b*x)^2*tan(a)^2 - 2*a*log(abs(b*
x + a))*tan(b*x)^2*tan(a)^2 - a*real_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 - a*real_part(cos_int
egral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 - imag_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 + imag_par
t(cos_integral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 - 2*sin_integral(2*b*x + 2*a)*tan(b*x)^2*tan(a)^2 + 2*b*x*ta
n(b*x)^2 - 2*a*log(abs(b*x + a))*tan(b*x)^2 - a*real_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2 - a*real_part(
cos_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*b*x*tan(a)^2 - 2*a*log(abs(b*x + a))*tan(a)^2 - a*real_part(cos_int
egral(2*b*x + 2*a))*tan(a)^2 - a*real_part(cos_integral(-2*b*x - 2*a))*tan(a)^2 - imag_part(cos_integral(2*b*x
+ 2*a))*tan(b*x)^2 + imag_part(cos_integral(-2*b*x - 2*a))*tan(b*x)^2 - 2*sin_integral(2*b*x + 2*a)*tan(b*x)^
2 - 2*tan(b*x)^2*tan(a) - imag_part(cos_integral(2*b*x + 2*a))*tan(a)^2 + imag_part(cos_integral(-2*b*x - 2*a)
)*tan(a)^2 - 2*sin_integral(2*b*x + 2*a)*tan(a)^2 - 2*tan(b*x)*tan(a)^2 + 2*b*x - 2*a*log(abs(b*x + a)) - a*re
al_part(cos_integral(2*b*x + 2*a)) - a*real_part(cos_integral(-2*b*x - 2*a)) - 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) + 2*tan(b*x) + 2*tan(a))/(b^2*tan(
b*x)^2*tan(a)^2 + b^2*tan(b*x)^2 + b^2*tan(a)^2 + b^2)

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maple [A]  time = 0.03, size = 106, normalized size = 0.97 \[ -\frac {x \Ci \left (b x +a \right ) \cos \left (b x +a \right )}{b}+\frac {\Ci \left (b x +a \right ) \sin \left (b x +a \right )}{b^{2}}-\frac {\Si \left (2 b x +2 a \right )}{2 b^{2}}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2 b^{2}}+\frac {x}{2 b}+\frac {a}{2 b^{2}}-\frac {a \ln \left (b x +a \right )}{2 b^{2}}-\frac {a \Ci \left (2 b x +2 a \right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*Ci(b*x+a)*sin(b*x+a),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x {\rm Ci}\left (b x + a\right ) \sin \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Ci(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {cosint}\left (a+b\,x\right )\,\sin \left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosint(a + b*x)*sin(a + b*x),x)

[Out]

int(x*cosint(a + b*x)*sin(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sin {\left (a + b x \right )} \operatorname {Ci}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(a + b*x)*Ci(a + b*x), x)

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