3.19 \(\int \frac{x^3 \sin (c+d x)}{a+b x} \, dx\)

Optimal. Leaf size=152 \[ -\frac{a^3 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^4}-\frac{a^3 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^4}-\frac{a^2 \cos (c+d x)}{b^3 d}-\frac{a \sin (c+d x)}{b^2 d^2}+\frac{a x \cos (c+d x)}{b^2 d}+\frac{2 x \sin (c+d x)}{b d^2}+\frac{2 \cos (c+d x)}{b d^3}-\frac{x^2 \cos (c+d x)}{b d} \]

[Out]

(2*Cos[c + d*x])/(b*d^3) - (a^2*Cos[c + d*x])/(b^3*d) + (a*x*Cos[c + d*x])/(b^2*d) - (x^2*Cos[c + d*x])/(b*d)
- (a^3*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^4 - (a*Sin[c + d*x])/(b^2*d^2) + (2*x*Sin[c + d*x])/(b*d
^2) - (a^3*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^4

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Rubi [A]  time = 0.30645, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {6742, 2638, 3296, 2637, 3303, 3299, 3302} \[ -\frac{a^3 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^4}-\frac{a^3 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^4}-\frac{a^2 \cos (c+d x)}{b^3 d}-\frac{a \sin (c+d x)}{b^2 d^2}+\frac{a x \cos (c+d x)}{b^2 d}+\frac{2 x \sin (c+d x)}{b d^2}+\frac{2 \cos (c+d x)}{b d^3}-\frac{x^2 \cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Sin[c + d*x])/(a + b*x),x]

[Out]

(2*Cos[c + d*x])/(b*d^3) - (a^2*Cos[c + d*x])/(b^3*d) + (a*x*Cos[c + d*x])/(b^2*d) - (x^2*Cos[c + d*x])/(b*d)
- (a^3*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^4 - (a*Sin[c + d*x])/(b^2*d^2) + (2*x*Sin[c + d*x])/(b*d
^2) - (a^3*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^4

Rule 6742

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 2637

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

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 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]

Rubi steps

\begin{align*} \int \frac{x^3 \sin (c+d x)}{a+b x} \, dx &=\int \left (\frac{a^2 \sin (c+d x)}{b^3}-\frac{a x \sin (c+d x)}{b^2}+\frac{x^2 \sin (c+d x)}{b}-\frac{a^3 \sin (c+d x)}{b^3 (a+b x)}\right ) \, dx\\ &=\frac{a^2 \int \sin (c+d x) \, dx}{b^3}-\frac{a^3 \int \frac{\sin (c+d x)}{a+b x} \, dx}{b^3}-\frac{a \int x \sin (c+d x) \, dx}{b^2}+\frac{\int x^2 \sin (c+d x) \, dx}{b}\\ &=-\frac{a^2 \cos (c+d x)}{b^3 d}+\frac{a x \cos (c+d x)}{b^2 d}-\frac{x^2 \cos (c+d x)}{b d}-\frac{a \int \cos (c+d x) \, dx}{b^2 d}+\frac{2 \int x \cos (c+d x) \, dx}{b d}-\frac{\left (a^3 \cos \left (c-\frac{a d}{b}\right )\right ) \int \frac{\sin \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b^3}-\frac{\left (a^3 \sin \left (c-\frac{a d}{b}\right )\right ) \int \frac{\cos \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b^3}\\ &=-\frac{a^2 \cos (c+d x)}{b^3 d}+\frac{a x \cos (c+d x)}{b^2 d}-\frac{x^2 \cos (c+d x)}{b d}-\frac{a^3 \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^4}-\frac{a \sin (c+d x)}{b^2 d^2}+\frac{2 x \sin (c+d x)}{b d^2}-\frac{a^3 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^4}-\frac{2 \int \sin (c+d x) \, dx}{b d^2}\\ &=\frac{2 \cos (c+d x)}{b d^3}-\frac{a^2 \cos (c+d x)}{b^3 d}+\frac{a x \cos (c+d x)}{b^2 d}-\frac{x^2 \cos (c+d x)}{b d}-\frac{a^3 \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^4}-\frac{a \sin (c+d x)}{b^2 d^2}+\frac{2 x \sin (c+d x)}{b d^2}-\frac{a^3 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.596237, size = 117, normalized size = 0.77 \[ -\frac{b \left (\left (a^2 d^2-a b d^2 x+b^2 \left (d^2 x^2-2\right )\right ) \cos (c+d x)+b d (a-2 b x) \sin (c+d x)\right )+a^3 d^3 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right )+a^3 d^3 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )}{b^4 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Sin[c + d*x])/(a + b*x),x]

[Out]

-((a^3*d^3*CosIntegral[d*(a/b + x)]*Sin[c - (a*d)/b] + b*((a^2*d^2 - a*b*d^2*x + b^2*(-2 + d^2*x^2))*Cos[c + d
*x] + b*d*(a - 2*b*x)*Sin[c + d*x]) + a^3*d^3*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)])/(b^4*d^3))

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Maple [B]  time = 0.012, size = 514, normalized size = 3.4 \begin{align*}{\frac{1}{{d}^{4}} \left ({\frac{ \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2}-bad+{b}^{2}c+{b}^{2} \right ) d \left ( - \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) +2\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) }{{b}^{3}}}-{\frac{ \left ({a}^{3}{d}^{3}-3\,{a}^{2}bc{d}^{2}+3\,a{b}^{2}{c}^{2}d-{b}^{3}{c}^{3} \right ) d}{{b}^{3}} \left ({\frac{1}{b}{\it Si} \left ( dx+c+{\frac{da-cb}{b}} \right ) \cos \left ({\frac{da-cb}{b}} \right ) }-{\frac{1}{b}{\it Ci} \left ( dx+c+{\frac{da-cb}{b}} \right ) \sin \left ({\frac{da-cb}{b}} \right ) } \right ) }-3\,{\frac{dc \left ( -da+cb+b \right ) \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{b}^{2}}}-3\,{\frac{ \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) dc}{{b}^{2}} \left ({\frac{1}{b}{\it Si} \left ( dx+c+{\frac{da-cb}{b}} \right ) \cos \left ({\frac{da-cb}{b}} \right ) }-{\frac{1}{b}{\it Ci} \left ( dx+c+{\frac{da-cb}{b}} \right ) \sin \left ({\frac{da-cb}{b}} \right ) } \right ) }-3\,{\frac{d{c}^{2}\cos \left ( dx+c \right ) }{b}}-3\,{\frac{ \left ( da-cb \right ) d{c}^{2}}{b} \left ({\frac{1}{b}{\it Si} \left ( dx+c+{\frac{da-cb}{b}} \right ) \cos \left ({\frac{da-cb}{b}} \right ) }-{\frac{1}{b}{\it Ci} \left ( dx+c+{\frac{da-cb}{b}} \right ) \sin \left ({\frac{da-cb}{b}} \right ) } \right ) }-d{c}^{3} \left ({\frac{1}{b}{\it Si} \left ( dx+c+{\frac{da-cb}{b}} \right ) \cos \left ({\frac{da-cb}{b}} \right ) }-{\frac{1}{b}{\it Ci} \left ( dx+c+{\frac{da-cb}{b}} \right ) \sin \left ({\frac{da-cb}{b}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sin(d*x+c)/(b*x+a),x)

[Out]

1/d^4*((a^2*d^2-2*a*b*c*d+b^2*c^2-a*b*d+b^2*c+b^2)*d/b^3*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x
+c))-(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)*d/b^3*(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b-Ci(d*x+c+(a
*d-b*c)/b)*sin((a*d-b*c)/b)/b)-3*d*c*(-a*d+b*c+b)/b^2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-3*(a^2*d^2-2*a*b*c*d+b^2
*c^2)*d*c/b^2*(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b-Ci(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b)-3*d*c^2/b*co
s(d*x+c)-3*(a*d-b*c)*d*c^2/b*(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b-Ci(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/
b)-d*c^3*(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b-Ci(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin(d*x+c)/(b*x+a),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.81177, size = 385, normalized size = 2.53 \begin{align*} -\frac{2 \, a^{3} d^{3} \cos \left (-\frac{b c - a d}{b}\right ) \operatorname{Si}\left (\frac{b d x + a d}{b}\right ) + 2 \,{\left (b^{3} d^{2} x^{2} - a b^{2} d^{2} x + a^{2} b d^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right ) - 2 \,{\left (2 \, b^{3} d x - a b^{2} d\right )} \sin \left (d x + c\right ) -{\left (a^{3} d^{3} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) + a^{3} d^{3} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right )\right )} \sin \left (-\frac{b c - a d}{b}\right )}{2 \, b^{4} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin(d*x+c)/(b*x+a),x, algorithm="fricas")

[Out]

-1/2*(2*a^3*d^3*cos(-(b*c - a*d)/b)*sin_integral((b*d*x + a*d)/b) + 2*(b^3*d^2*x^2 - a*b^2*d^2*x + a^2*b*d^2 -
2*b^3)*cos(d*x + c) - 2*(2*b^3*d*x - a*b^2*d)*sin(d*x + c) - (a^3*d^3*cos_integral((b*d*x + a*d)/b) + a^3*d^3
*cos_integral(-(b*d*x + a*d)/b))*sin(-(b*c - a*d)/b))/(b^4*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sin{\left (c + d x \right )}}{a + b x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*sin(d*x+c)/(b*x+a),x)

[Out]

Integral(x**3*sin(c + d*x)/(a + b*x), x)

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Giac [C]  time = 1.16905, size = 911, normalized size = 5.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin(d*x+c)/(b*x+a),x, algorithm="giac")

[Out]

-1/2*(a^3*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - a^3*imag_part(cos_integral(-d*x
- a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a^3*sin_integral((b*d*x + a*d)/b)*tan(1/2*c)^2*tan(1/2*a*d/b)^2 +
2*a^3*real_part(cos_integral(d*x + a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b) + 2*a^3*real_part(cos_integral(-d*x -
a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b) - 2*a^3*real_part(cos_integral(d*x + a*d/b))*tan(1/2*c)*tan(1/2*a*d/b)^2 -
2*a^3*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*c)*tan(1/2*a*d/b)^2 - a^3*imag_part(cos_integral(d*x + a*
d/b))*tan(1/2*c)^2 + a^3*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*c)^2 - 2*a^3*sin_integral((b*d*x + a*d)
/b)*tan(1/2*c)^2 + 4*a^3*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*c)*tan(1/2*a*d/b) - 4*a^3*imag_part(cos_
integral(-d*x - a*d/b))*tan(1/2*c)*tan(1/2*a*d/b) + 8*a^3*sin_integral((b*d*x + a*d)/b)*tan(1/2*c)*tan(1/2*a*d
/b) - a^3*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*a*d/b)^2 + a^3*imag_part(cos_integral(-d*x - a*d/b))*ta
n(1/2*a*d/b)^2 - 2*a^3*sin_integral((b*d*x + a*d)/b)*tan(1/2*a*d/b)^2 + 2*a^3*real_part(cos_integral(d*x + a*d
/b))*tan(1/2*c) + 2*a^3*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*c) - 2*a^3*real_part(cos_integral(d*x +
a*d/b))*tan(1/2*a*d/b) - 2*a^3*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*a*d/b) + a^3*imag_part(cos_integr
al(d*x + a*d/b)) - a^3*imag_part(cos_integral(-d*x - a*d/b)) + 2*a^3*sin_integral((b*d*x + a*d)/b))/(b^4*tan(1
/2*c)^2*tan(1/2*a*d/b)^2 + b^4*tan(1/2*c)^2 + b^4*tan(1/2*a*d/b)^2 + b^4)