3.44 \(\int \frac {(a+b x^2) \sin (c+d x)}{x} \, dx\)

Optimal. Leaf size=41 \[ a \sin (c) \text {Ci}(d x)+a \cos (c) \text {Si}(d x)+\frac {b \sin (c+d x)}{d^2}-\frac {b x \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3339, 3303, 3299, 3302, 3296, 2637} \[ a \sin (c) \text {CosIntegral}(d x)+a \cos (c) \text {Si}(d x)+\frac {b \sin (c+d x)}{d^2}-\frac {b x \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*Sin[c + d*x])/x,x]

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

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

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 3339

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegran
d[Sin[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 54, normalized size = 1.32 \[ a \sin (c) \text {Ci}(d x)+a \cos (c) \text {Si}(d x)-\frac {b \cos (d x) (d x \cos (c)-\sin (c))}{d^2}+\frac {b \sin (d x) (d x \sin (c)+\cos (c))}{d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*Sin[c + d*x])/x,x]

[Out]

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

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fricas [A]  time = 0.76, size = 61, normalized size = 1.49 \[ \frac {2 \, a d^{2} \cos \relax (c) \operatorname {Si}\left (d x\right ) - 2 \, b d x \cos \left (d x + c\right ) + 2 \, b \sin \left (d x + c\right ) + {\left (a d^{2} \operatorname {Ci}\left (d x\right ) + a d^{2} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{2 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*sin(d*x+c)/x,x, algorithm="fricas")

[Out]

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

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giac [C]  time = 0.77, size = 432, normalized size = 10.54 \[ -\frac {a d^{2} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{2} \Re \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{2} \Re \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 2 \, b d x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{2} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{2} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, b d x \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d^{2} \Re \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{2} \Re \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 8 \, b d x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, b d x \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) + a d^{2} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) - 2 \, a d^{2} \operatorname {Si}\left (d x\right ) + 4 \, b \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 4 \, b \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b d x - 4 \, b \tan \left (\frac {1}{2} \, d x\right ) - 4 \, b \tan \left (\frac {1}{2} \, c\right )}{2 \, {\left (d^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} + d^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*sin(d*x+c)/x,x, algorithm="giac")

[Out]

-1/2*(a*d^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*imag_part(cos_integral(-d*x))*tan
(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^2*real_part(cos_integ
ral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^2*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 2*b*d*
x*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a*d^2*imag_part(cos_integr
al(-d*x))*tan(1/2*d*x)^2 - 2*a*d^2*sin_integral(d*x)*tan(1/2*d*x)^2 + a*d^2*imag_part(cos_integral(d*x))*tan(1
/2*c)^2 - a*d^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a*d^2*sin_integral(d*x)*tan(1/2*c)^2 - 2*b*d*x*
tan(1/2*d*x)^2 - 2*a*d^2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^2*real_part(cos_integral(-d*x))*tan(1
/2*c) - 8*b*d*x*tan(1/2*d*x)*tan(1/2*c) - 2*b*d*x*tan(1/2*c)^2 - a*d^2*imag_part(cos_integral(d*x)) + a*d^2*im
ag_part(cos_integral(-d*x)) - 2*a*d^2*sin_integral(d*x) + 4*b*tan(1/2*d*x)^2*tan(1/2*c) + 4*b*tan(1/2*d*x)*tan
(1/2*c)^2 + 2*b*d*x - 4*b*tan(1/2*d*x) - 4*b*tan(1/2*c))/(d^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + d^2*tan(1/2*d*x)^2
+ d^2*tan(1/2*c)^2 + d^2)

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maple [A]  time = 0.03, size = 60, normalized size = 1.46 \[ \frac {\left (1+c \right ) b \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}+\frac {2 c b \cos \left (d x +c \right )}{d^{2}}+a \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*sin(d*x+c)/x,x)

[Out]

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

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maxima [C]  time = 0.64, size = 66, normalized size = 1.61 \[ -\frac {2 \, b d x \cos \left (d x + c\right ) - {\left (a {\left (-i \, {\rm Ei}\left (i \, d x\right ) + i \, {\rm Ei}\left (-i \, d x\right )\right )} \cos \relax (c) + a {\left ({\rm Ei}\left (i \, d x\right ) + {\rm Ei}\left (-i \, d x\right )\right )} \sin \relax (c)\right )} d^{2} - 2 \, b \sin \left (d x + c\right )}{2 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*sin(d*x+c)/x,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ a\,\mathrm {cosint}\left (d\,x\right )\,\sin \relax (c)+a\,\mathrm {sinint}\left (d\,x\right )\,\cos \relax (c)+\frac {b\,\left (\sin \left (c+d\,x\right )-d\,x\,\cos \left (c+d\,x\right )\right )}{d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^2))/x,x)

[Out]

a*cosint(d*x)*sin(c) + a*sinint(d*x)*cos(c) + (b*(sin(c + d*x) - d*x*cos(c + d*x)))/d^2

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sympy [A]  time = 4.90, size = 63, normalized size = 1.54 \[ a \sin {\relax (c )} \operatorname {Ci}{\left (d x \right )} + a \cos {\relax (c )} \operatorname {Si}{\left (d x \right )} + b x \left (\begin {cases} - \cos {\relax (c )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) - b \left (\begin {cases} - x \cos {\relax (c )} & \text {for}\: d = 0 \\- \frac {\begin {cases} \frac {\sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \cos {\relax (c )} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*sin(d*x+c)/x,x)

[Out]

a*sin(c)*Ci(d*x) + a*cos(c)*Si(d*x) + b*x*Piecewise((-cos(c), Eq(d, 0)), (-cos(c + d*x)/d, True)) - b*Piecewis
e((-x*cos(c), Eq(d, 0)), (-Piecewise((sin(c + d*x)/d, Ne(d, 0)), (x*cos(c), True))/d, True))

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