3.134 \(\int x \cos (a+b x) \text {Ci}(c+d x) \, dx\)

Optimal. Leaf size=370 \[ -\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b^2}+\frac {\cos (a+b x) \text {Ci}(c+d x)}{b^2}-\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b^2}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b^2}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b^2}+\frac {c \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b d}+\frac {c \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b d}+\frac {x \sin (a+b x) \text {Ci}(c+d x)}{b}+\frac {c \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b d}+\frac {c \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b d}+\frac {\cos (a+x (b-d)-c)}{2 b (b-d)}+\frac {\cos (a+x (b+d)+c)}{2 b (b+d)} \]

[Out]

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

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Rubi [A]  time = 0.80, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6514, 6742, 4574, 2638, 4430, 3303, 3299, 3302, 6518, 4429} \[ -\frac {\cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {c (b-d)}{d}+x (b-d)\right )}{2 b^2}+\frac {\cos (a+b x) \text {CosIntegral}(c+d x)}{b^2}-\frac {\cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {c (b+d)}{d}+x (b+d)\right )}{2 b^2}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b^2}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b^2}+\frac {c \sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {c (b-d)}{d}+x (b-d)\right )}{2 b d}+\frac {c \sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {c (b+d)}{d}+x (b+d)\right )}{2 b d}+\frac {x \sin (a+b x) \text {CosIntegral}(c+d x)}{b}+\frac {c \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b d}+\frac {c \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b d}+\frac {\cos (a+x (b-d)-c)}{2 b (b-d)}+\frac {\cos (a+x (b+d)+c)}{2 b (b+d)} \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x]*CosIntegral[c + d*x],x]

[Out]

Cos[a - c + (b - d)*x]/(2*b*(b - d)) + Cos[a + c + (b + d)*x]/(2*b*(b + d)) - (Cos[a - (b*c)/d]*CosIntegral[(c
*(b - d))/d + (b - d)*x])/(2*b^2) + (Cos[a + b*x]*CosIntegral[c + d*x])/b^2 - (Cos[a - (b*c)/d]*CosIntegral[(c
*(b + d))/d + (b + d)*x])/(2*b^2) + (c*CosIntegral[(c*(b - d))/d + (b - d)*x]*Sin[a - (b*c)/d])/(2*b*d) + (c*C
osIntegral[(c*(b + d))/d + (b + d)*x]*Sin[a - (b*c)/d])/(2*b*d) + (x*CosIntegral[c + d*x]*Sin[a + b*x])/b + (c
*Cos[a - (b*c)/d]*SinIntegral[(c*(b - d))/d + (b - d)*x])/(2*b*d) + (Sin[a - (b*c)/d]*SinIntegral[(c*(b - d))/
d + (b - d)*x])/(2*b^2) + (c*Cos[a - (b*c)/d]*SinIntegral[(c*(b + d))/d + (b + d)*x])/(2*b*d) + (Sin[a - (b*c)
/d]*SinIntegral[(c*(b + d))/d + (b + d)*x])/(2*b^2)

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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]

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 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 4429

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*Cos[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Int[E
xpandTrigReduce[(e + f*x)^m, Cos[a + b*x]^p*Cos[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p,
0] && IGtQ[q, 0] && IntegerQ[m]

Rule 4430

Int[Cos[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int[E
xpandTrigReduce[(e + f*x)^m, Sin[a + b*x]^p*Cos[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IGtQ[q, 0]

Rule 4574

Int[Cos[w_]^(q_.)*Sin[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p*Cos[w]^q, x], x] /; IGtQ[p, 0] &&
IGtQ[q, 0] && ((PolynomialQ[v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w],
x]))

Rule 6514

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

Rule 6518

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

Rule 6742

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

Rubi steps

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

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Mathematica [C]  time = 3.56, size = 394, normalized size = 1.06 \[ \frac {\text {Ci}(c+d x) (b x \sin (a+b x)+\cos (a+b x))}{b^2}+\frac {i \exp \left (-\frac {i (d (a+c+d x)+b (c+d x))}{d}\right ) \left (e^{\frac {i b c}{d}} \left (\left (b^2-d^2\right ) (b c+i d) e^{\frac {i (b+d) (c+d x)}{d}} \text {Ei}\left (-\frac {i (b+d) (c+d x)}{d}\right )-i b d \left (d \left (-1+e^{2 i (a+b x)}\right )+b \left (1+e^{2 i (a+b x)}\right )\right )\right )-\left (b^2-d^2\right ) (b c-i d) e^{i (2 a+x (b+d)+c)} \text {Ei}\left (\frac {i (b-d) (c+d x)}{d}\right )\right )}{4 b^2 d (b-d) (b+d)}+\frac {i e^{-i (a-c)} \left ((-b c+i d) e^{2 i a-\frac {i c (b+d)}{d}} \text {Ei}\left (\frac {i (b+d) (c+d x)}{d}\right )-i b d \left (\frac {e^{i (2 a+x (b+d))}}{b+d}+\frac {e^{-i x (b-d)}}{b-d}\right )+(b c+i d) e^{\frac {i c (b-d)}{d}} \text {Ei}\left (-\frac {i (b-d) (c+d x)}{d}\right )\right )}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x]*CosIntegral[c + d*x],x]

[Out]

((I/4)*(-((b*c - I*d)*(b^2 - d^2)*E^(I*(2*a + c + (b + d)*x))*ExpIntegralEi[(I*(b - d)*(c + d*x))/d]) + E^((I*
b*c)/d)*((-I)*b*d*(d*(-1 + E^((2*I)*(a + b*x))) + b*(1 + E^((2*I)*(a + b*x)))) + (b*c + I*d)*(b^2 - d^2)*E^((I
*(b + d)*(c + d*x))/d)*ExpIntegralEi[((-I)*(b + d)*(c + d*x))/d])))/(b^2*(b - d)*d*(b + d)*E^((I*(b*(c + d*x)
+ d*(a + c + d*x)))/d)) + ((I/4)*((-I)*b*d*(1/((b - d)*E^(I*(b - d)*x)) + E^(I*(2*a + (b + d)*x))/(b + d)) + (
b*c + I*d)*E^((I*c*(b - d))/d)*ExpIntegralEi[((-I)*(b - d)*(c + d*x))/d] + (-(b*c) + I*d)*E^((2*I)*a - (I*c*(b
 + d))/d)*ExpIntegralEi[(I*(b + d)*(c + d*x))/d]))/(b^2*d*E^(I*(a - c))) + (CosIntegral[c + d*x]*(Cos[a + b*x]
 + b*x*Sin[a + b*x]))/b^2

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \cos \left (b x + a\right ) \operatorname {Ci}\left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Ci(d*x+c)*cos(b*x+a),x, algorithm="fricas")

[Out]

integral(x*cos(b*x + a)*cos_integral(d*x + c), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Ci(d*x+c)*cos(b*x+a),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.06, size = 1212, normalized size = 3.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*Ci(d*x+c)*cos(b*x+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Ci(d*x+c)*cos(b*x+a),x, algorithm="maxima")

[Out]

integrate(x*Ci(d*x + c)*cos(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosint(c + d*x)*cos(a + b*x),x)

[Out]

int(x*cosint(c + d*x)*cos(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Ci(d*x+c)*cos(b*x+a),x)

[Out]

Integral(x*cos(a + b*x)*Ci(c + d*x), x)

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