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

Optimal. Leaf size=265 \[ \frac{a^3 d^2 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{2 b^6}+\frac{3 a^2 d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^5}+\frac{a^3 d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{2 b^6}-\frac{3 a^2 d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^5}+\frac{a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac{3 a^2 \sin (c+d x)}{b^4 (a+b x)}+\frac{a^3 d \cos (c+d x)}{2 b^5 (a+b x)}-\frac{3 a \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^4}-\frac{3 a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^4}-\frac{\cos (c+d x)}{b^3 d} \]

[Out]

-(Cos[c + d*x]/(b^3*d)) + (a^3*d*Cos[c + d*x])/(2*b^5*(a + b*x)) + (3*a^2*d*Cos[c - (a*d)/b]*CosIntegral[(a*d)
/b + d*x])/b^5 - (3*a*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^4 + (a^3*d^2*CosIntegral[(a*d)/b + d*x]*S
in[c - (a*d)/b])/(2*b^6) + (a^3*Sin[c + d*x])/(2*b^4*(a + b*x)^2) - (3*a^2*Sin[c + d*x])/(b^4*(a + b*x)) - (3*
a*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^4 + (a^3*d^2*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/(2*
b^6) - (3*a^2*d*Sin[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^5

________________________________________________________________________________________

Rubi [A]  time = 0.609746, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {6742, 2638, 3297, 3303, 3299, 3302} \[ \frac{a^3 d^2 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{2 b^6}+\frac{3 a^2 d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^5}+\frac{a^3 d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{2 b^6}-\frac{3 a^2 d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^5}+\frac{a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac{3 a^2 \sin (c+d x)}{b^4 (a+b x)}+\frac{a^3 d \cos (c+d x)}{2 b^5 (a+b x)}-\frac{3 a \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^4}-\frac{3 a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^4}-\frac{\cos (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Sin[c + d*x])/(a + b*x)^3,x]

[Out]

-(Cos[c + d*x]/(b^3*d)) + (a^3*d*Cos[c + d*x])/(2*b^5*(a + b*x)) + (3*a^2*d*Cos[c - (a*d)/b]*CosIntegral[(a*d)
/b + d*x])/b^5 - (3*a*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^4 + (a^3*d^2*CosIntegral[(a*d)/b + d*x]*S
in[c - (a*d)/b])/(2*b^6) + (a^3*Sin[c + d*x])/(2*b^4*(a + b*x)^2) - (3*a^2*Sin[c + d*x])/(b^4*(a + b*x)) - (3*
a*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^4 + (a^3*d^2*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/(2*
b^6) - (3*a^2*d*Sin[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^5

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 3297

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

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)^3} \, dx &=\int \left (\frac{\sin (c+d x)}{b^3}-\frac{a^3 \sin (c+d x)}{b^3 (a+b x)^3}+\frac{3 a^2 \sin (c+d x)}{b^3 (a+b x)^2}-\frac{3 a \sin (c+d x)}{b^3 (a+b x)}\right ) \, dx\\ &=\frac{\int \sin (c+d x) \, dx}{b^3}-\frac{(3 a) \int \frac{\sin (c+d x)}{a+b x} \, dx}{b^3}+\frac{\left (3 a^2\right ) \int \frac{\sin (c+d x)}{(a+b x)^2} \, dx}{b^3}-\frac{a^3 \int \frac{\sin (c+d x)}{(a+b x)^3} \, dx}{b^3}\\ &=-\frac{\cos (c+d x)}{b^3 d}+\frac{a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac{3 a^2 \sin (c+d x)}{b^4 (a+b x)}+\frac{\left (3 a^2 d\right ) \int \frac{\cos (c+d x)}{a+b x} \, dx}{b^4}-\frac{\left (a^3 d\right ) \int \frac{\cos (c+d x)}{(a+b x)^2} \, dx}{2 b^4}-\frac{\left (3 a \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 (3 a \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{\cos (c+d x)}{b^3 d}+\frac{a^3 d \cos (c+d x)}{2 b^5 (a+b x)}-\frac{3 a \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^4}+\frac{a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac{3 a^2 \sin (c+d x)}{b^4 (a+b x)}-\frac{3 a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^4}+\frac{\left (a^3 d^2\right ) \int \frac{\sin (c+d x)}{a+b x} \, dx}{2 b^5}+\frac{\left (3 a^2 d \cos \left (c-\frac{a d}{b}\right )\right ) \int \frac{\cos \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b^4}-\frac{\left (3 a^2 d \sin \left (c-\frac{a d}{b}\right )\right ) \int \frac{\sin \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b^4}\\ &=-\frac{\cos (c+d x)}{b^3 d}+\frac{a^3 d \cos (c+d x)}{2 b^5 (a+b x)}+\frac{3 a^2 d \cos \left (c-\frac{a d}{b}\right ) \text{Ci}\left (\frac{a d}{b}+d x\right )}{b^5}-\frac{3 a \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^4}+\frac{a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac{3 a^2 \sin (c+d x)}{b^4 (a+b x)}-\frac{3 a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^4}-\frac{3 a^2 d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^5}+\frac{\left (a^3 d^2 \cos \left (c-\frac{a d}{b}\right )\right ) \int \frac{\sin \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{2 b^5}+\frac{\left (a^3 d^2 \sin \left (c-\frac{a d}{b}\right )\right ) \int \frac{\cos \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{2 b^5}\\ &=-\frac{\cos (c+d x)}{b^3 d}+\frac{a^3 d \cos (c+d x)}{2 b^5 (a+b x)}+\frac{3 a^2 d \cos \left (c-\frac{a d}{b}\right ) \text{Ci}\left (\frac{a d}{b}+d x\right )}{b^5}-\frac{3 a \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^4}+\frac{a^3 d^2 \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{2 b^6}+\frac{a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac{3 a^2 \sin (c+d x)}{b^4 (a+b x)}-\frac{3 a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^4}+\frac{a^3 d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{2 b^6}-\frac{3 a^2 d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 1.05277, size = 235, normalized size = 0.89 \[ -\frac{-a d (a+b x)^2 \left (\text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right ) \left (\left (a^2 d^2-6 b^2\right ) \sin \left (c-\frac{a d}{b}\right )+6 a b d \cos \left (c-\frac{a d}{b}\right )\right )+\text{Si}\left (d \left (\frac{a}{b}+x\right )\right ) \left (\left (a^2 d^2-6 b^2\right ) \cos \left (c-\frac{a d}{b}\right )-6 a b d \sin \left (c-\frac{a d}{b}\right )\right )\right )+b \cos (d x) \left (a^2 b d \sin (c) (5 a+6 b x)-\cos (c) (a+b x) \left (a^3 d^2-2 a b^2-2 b^3 x\right )\right )+b \sin (d x) \left (\sin (c) (a+b x) \left (a^3 d^2-2 a b^2-2 b^3 x\right )+a^2 b d \cos (c) (5 a+6 b x)\right )}{2 b^6 d (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[d*x]*(-((a + b*x)*(-2*a*b^2 + a^3*d^2 - 2*b^3*x)*Cos[c]) + a^2*b*d*(5*a + 6*b*x)*Sin[c]) + b*(a^2*b*d*
(5*a + 6*b*x)*Cos[c] + (a + b*x)*(-2*a*b^2 + a^3*d^2 - 2*b^3*x)*Sin[c])*Sin[d*x] - a*d*(a + b*x)^2*(CosIntegra
l[d*(a/b + x)]*(6*a*b*d*Cos[c - (a*d)/b] + (-6*b^2 + a^2*d^2)*Sin[c - (a*d)/b]) + ((-6*b^2 + a^2*d^2)*Cos[c -
(a*d)/b] - 6*a*b*d*Sin[c - (a*d)/b])*SinIntegral[d*(a/b + x)]))/(2*b^6*d*(a + b*x)^2)

________________________________________________________________________________________

Maple [B]  time = 0.016, size = 1208, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^4*(-d^3/b^3*cos(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^3/b^3*(-1/2*sin(d*x+c)/((d*x+c)*b+d
*a-c*b)^2/b+1/2*(-cos(d*x+c)/((d*x+c)*b+d*a-c*b)/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)/b)/b)+3/b^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)*d^3*(-sin(d*x+c)/((d*x+c)*b+d*a-c*b)/b+(Si(d*
x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b+Ci(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b)/b)-3/b^3*(a*d-b*c)*d^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^3*c*(a*d-b*c)^2/b^2*(-1/2*sin(d
*x+c)/((d*x+c)*b+d*a-c*b)^2/b+1/2*(-cos(d*x+c)/((d*x+c)*b+d*a-c*b)/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)/b)/b)+6*d^3*c*(a*d-b*c)/b^2*(-sin(d*x+c)/((d*x+c)*b+d*a-c*b)/b+(Si(
d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b+Ci(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b)/b)-3*d^3*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/b*(a*d-b*c)*d^3*c^2*(-1/2*sin(d*x+c)/((
d*x+c)*b+d*a-c*b)^2/b+1/2*(-cos(d*x+c)/((d*x+c)*b+d*a-c*b)/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)/b)/b)+3*d^3*c^2/b*(-sin(d*x+c)/((d*x+c)*b+d*a-c*b)/b+(Si(d*x+c+(a*d-b*c)/b)
*sin((a*d-b*c)/b)/b+Ci(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b)/b)-d^3*c^3*(-1/2*sin(d*x+c)/((d*x+c)*b+d*a-c*b)^
2/b+1/2*(-cos(d*x+c)/((d*x+c)*b+d*a-c*b)/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)/b)/b))

________________________________________________________________________________________

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)^3,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.88886, size = 1115, normalized size = 4.21 \begin{align*} \frac{2 \,{\left (a^{4} b d^{2} - 2 \, b^{5} x^{2} - 2 \, a^{2} b^{3} +{\left (a^{3} b^{2} d^{2} - 4 \, a b^{4}\right )} x\right )} \cos \left (d x + c\right ) + 2 \,{\left (3 \,{\left (a^{2} b^{3} d^{2} x^{2} + 2 \, a^{3} b^{2} d^{2} x + a^{4} b d^{2}\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) + 3 \,{\left (a^{2} b^{3} d^{2} x^{2} + 2 \, a^{3} b^{2} d^{2} x + a^{4} b d^{2}\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) +{\left (a^{5} d^{3} - 6 \, a^{3} b^{2} d +{\left (a^{3} b^{2} d^{3} - 6 \, a b^{4} d\right )} x^{2} + 2 \,{\left (a^{4} b d^{3} - 6 \, a^{2} b^{3} d\right )} x\right )} \operatorname{Si}\left (\frac{b d x + a d}{b}\right )\right )} \cos \left (-\frac{b c - a d}{b}\right ) - 2 \,{\left (6 \, a^{2} b^{3} d x + 5 \, a^{3} b^{2} d\right )} \sin \left (d x + c\right ) -{\left ({\left (a^{5} d^{3} - 6 \, a^{3} b^{2} d +{\left (a^{3} b^{2} d^{3} - 6 \, a b^{4} d\right )} x^{2} + 2 \,{\left (a^{4} b d^{3} - 6 \, a^{2} b^{3} d\right )} x\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) +{\left (a^{5} d^{3} - 6 \, a^{3} b^{2} d +{\left (a^{3} b^{2} d^{3} - 6 \, a b^{4} d\right )} x^{2} + 2 \,{\left (a^{4} b d^{3} - 6 \, a^{2} b^{3} d\right )} x\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) - 12 \,{\left (a^{2} b^{3} d^{2} x^{2} + 2 \, a^{3} b^{2} d^{2} x + a^{4} b d^{2}\right )} \operatorname{Si}\left (\frac{b d x + a d}{b}\right )\right )} \sin \left (-\frac{b c - a d}{b}\right )}{4 \,{\left (b^{8} d x^{2} + 2 \, a b^{7} d x + a^{2} b^{6} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a^4*b*d^2 - 2*b^5*x^2 - 2*a^2*b^3 + (a^3*b^2*d^2 - 4*a*b^4)*x)*cos(d*x + c) + 2*(3*(a^2*b^3*d^2*x^2 +
2*a^3*b^2*d^2*x + a^4*b*d^2)*cos_integral((b*d*x + a*d)/b) + 3*(a^2*b^3*d^2*x^2 + 2*a^3*b^2*d^2*x + a^4*b*d^2)
*cos_integral(-(b*d*x + a*d)/b) + (a^5*d^3 - 6*a^3*b^2*d + (a^3*b^2*d^3 - 6*a*b^4*d)*x^2 + 2*(a^4*b*d^3 - 6*a^
2*b^3*d)*x)*sin_integral((b*d*x + a*d)/b))*cos(-(b*c - a*d)/b) - 2*(6*a^2*b^3*d*x + 5*a^3*b^2*d)*sin(d*x + c)
- ((a^5*d^3 - 6*a^3*b^2*d + (a^3*b^2*d^3 - 6*a*b^4*d)*x^2 + 2*(a^4*b*d^3 - 6*a^2*b^3*d)*x)*cos_integral((b*d*x
+ a*d)/b) + (a^5*d^3 - 6*a^3*b^2*d + (a^3*b^2*d^3 - 6*a*b^4*d)*x^2 + 2*(a^4*b*d^3 - 6*a^2*b^3*d)*x)*cos_integ
ral(-(b*d*x + a*d)/b) - 12*(a^2*b^3*d^2*x^2 + 2*a^3*b^2*d^2*x + a^4*b*d^2)*sin_integral((b*d*x + a*d)/b))*sin(
-(b*c - a*d)/b))/(b^8*d*x^2 + 2*a*b^7*d*x + a^2*b^6*d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [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)^3,x, algorithm="giac")

[Out]

Timed out