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

Optimal. Leaf size=377 \[ \frac {6 b^2 \sin (c) \text {Ci}(d x)}{a^5}-\frac {6 b^2 \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{a^5}+\frac {6 b^2 \cos (c) \text {Si}(d x)}{a^5}-\frac {6 b^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^5}+\frac {3 b^2 \sin (c+d x)}{a^4 (a+b x)}-\frac {3 b d \cos (c) \text {Ci}(d x)}{a^4}-\frac {3 b d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{a^4}+\frac {3 b d \sin (c) \text {Si}(d x)}{a^4}+\frac {3 b d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^4}+\frac {3 b \sin (c+d x)}{a^4 x}+\frac {b^2 \sin (c+d x)}{2 a^3 (a+b x)^2}+\frac {d^2 \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{2 a^3}+\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 a^3}+\frac {b d \cos (c+d x)}{2 a^3 (a+b x)}-\frac {d^2 \sin (c) \text {Ci}(d x)}{2 a^3}-\frac {d^2 \cos (c) \text {Si}(d x)}{2 a^3}-\frac {\sin (c+d x)}{2 a^3 x^2}-\frac {d \cos (c+d x)}{2 a^3 x} \]

[Out]

-3*b*d*Ci(d*x)*cos(c)/a^4-3*b*d*Ci(a*d/b+d*x)*cos(-c+a*d/b)/a^4-1/2*d*cos(d*x+c)/a^3/x+1/2*b*d*cos(d*x+c)/a^3/
(b*x+a)+6*b^2*cos(c)*Si(d*x)/a^5-1/2*d^2*cos(c)*Si(d*x)/a^3-6*b^2*cos(-c+a*d/b)*Si(a*d/b+d*x)/a^5+1/2*d^2*cos(
-c+a*d/b)*Si(a*d/b+d*x)/a^3+6*b^2*Ci(d*x)*sin(c)/a^5-1/2*d^2*Ci(d*x)*sin(c)/a^3+3*b*d*Si(d*x)*sin(c)/a^4+6*b^2
*Ci(a*d/b+d*x)*sin(-c+a*d/b)/a^5-1/2*d^2*Ci(a*d/b+d*x)*sin(-c+a*d/b)/a^3-3*b*d*Si(a*d/b+d*x)*sin(-c+a*d/b)/a^4
-1/2*sin(d*x+c)/a^3/x^2+3*b*sin(d*x+c)/a^4/x+1/2*b^2*sin(d*x+c)/a^3/(b*x+a)^2+3*b^2*sin(d*x+c)/a^4/(b*x+a)

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Rubi [A]  time = 0.80, antiderivative size = 377, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6742, 3297, 3303, 3299, 3302} \[ \frac {6 b^2 \sin (c) \text {CosIntegral}(d x)}{a^5}-\frac {6 b^2 \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{a^5}+\frac {6 b^2 \cos (c) \text {Si}(d x)}{a^5}-\frac {6 b^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^5}+\frac {3 b^2 \sin (c+d x)}{a^4 (a+b x)}+\frac {b^2 \sin (c+d x)}{2 a^3 (a+b x)^2}+\frac {d^2 \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{2 a^3}-\frac {3 b d \cos (c) \text {CosIntegral}(d x)}{a^4}-\frac {3 b d \cos \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{a^4}+\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 a^3}+\frac {3 b d \sin (c) \text {Si}(d x)}{a^4}+\frac {3 b d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^4}+\frac {3 b \sin (c+d x)}{a^4 x}+\frac {b d \cos (c+d x)}{2 a^3 (a+b x)}-\frac {d^2 \sin (c) \text {CosIntegral}(d x)}{2 a^3}-\frac {d^2 \cos (c) \text {Si}(d x)}{2 a^3}-\frac {\sin (c+d x)}{2 a^3 x^2}-\frac {d \cos (c+d x)}{2 a^3 x} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(x^3*(a + b*x)^3),x]

[Out]

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

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 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 6742

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

Rubi steps

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

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Mathematica [A]  time = 2.07, size = 630, normalized size = 1.67 \[ \frac {a^4 d^2 x^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )-a^4 d^2 x^2 \cos (c) \text {Si}(d x)-a^4 \sin (c+d x)+a^4 (-d) x \cos (c+d x)-2 a^3 b d^2 x^3 \cos (c) \text {Si}(d x)+2 a^3 b d^2 x^3 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+6 a^3 b d x^2 \sin (c) \text {Si}(d x)+6 a^3 b d x^2 \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )-a^3 b d x^2 \cos (c+d x)+4 a^3 b x \sin (c+d x)-x^2 (a+b x)^2 \text {Ci}(d x) \left (\sin (c) \left (a^2 d^2-12 b^2\right )+6 a b d \cos (c)\right )+x^2 (a+b x)^2 \text {Ci}\left (d \left (\frac {a}{b}+x\right )\right ) \left (\left (a^2 d^2-12 b^2\right ) \sin \left (c-\frac {a d}{b}\right )-6 a b d \cos \left (c-\frac {a d}{b}\right )\right )-a^2 b^2 d^2 x^4 \cos (c) \text {Si}(d x)+a^2 b^2 d^2 x^4 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+12 a^2 b^2 d x^3 \sin (c) \text {Si}(d x)+12 a^2 b^2 d x^3 \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+12 a^2 b^2 x^2 \cos (c) \text {Si}(d x)-12 a^2 b^2 x^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+18 a^2 b^2 x^2 \sin (c+d x)-12 b^4 x^4 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+6 a b^3 d x^4 \sin (c) \text {Si}(d x)+6 a b^3 d x^4 \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+24 a b^3 x^3 \cos (c) \text {Si}(d x)-24 a b^3 x^3 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+12 a b^3 x^3 \sin (c+d x)+12 b^4 x^4 \cos (c) \text {Si}(d x)}{2 a^5 x^2 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^4*d*x*Cos[c + d*x]) - a^3*b*d*x^2*Cos[c + d*x] - x^2*(a + b*x)^2*CosIntegral[d*x]*(6*a*b*d*Cos[c] + (-12*
b^2 + a^2*d^2)*Sin[c]) + x^2*(a + b*x)^2*CosIntegral[d*(a/b + x)]*(-6*a*b*d*Cos[c - (a*d)/b] + (-12*b^2 + a^2*
d^2)*Sin[c - (a*d)/b]) - a^4*Sin[c + d*x] + 4*a^3*b*x*Sin[c + d*x] + 18*a^2*b^2*x^2*Sin[c + d*x] + 12*a*b^3*x^
3*Sin[c + d*x] + 12*a^2*b^2*x^2*Cos[c]*SinIntegral[d*x] - a^4*d^2*x^2*Cos[c]*SinIntegral[d*x] + 24*a*b^3*x^3*C
os[c]*SinIntegral[d*x] - 2*a^3*b*d^2*x^3*Cos[c]*SinIntegral[d*x] + 12*b^4*x^4*Cos[c]*SinIntegral[d*x] - a^2*b^
2*d^2*x^4*Cos[c]*SinIntegral[d*x] + 6*a^3*b*d*x^2*Sin[c]*SinIntegral[d*x] + 12*a^2*b^2*d*x^3*Sin[c]*SinIntegra
l[d*x] + 6*a*b^3*d*x^4*Sin[c]*SinIntegral[d*x] - 12*a^2*b^2*x^2*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + a^
4*d^2*x^2*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] - 24*a*b^3*x^3*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] +
2*a^3*b*d^2*x^3*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] - 12*b^4*x^4*Cos[c - (a*d)/b]*SinIntegral[d*(a/b +
x)] + a^2*b^2*d^2*x^4*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + 6*a^3*b*d*x^2*Sin[c - (a*d)/b]*SinIntegral[d
*(a/b + x)] + 12*a^2*b^2*d*x^3*Sin[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + 6*a*b^3*d*x^4*Sin[c - (a*d)/b]*SinI
ntegral[d*(a/b + x)])/(2*a^5*x^2*(a + b*x)^2)

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fricas [B]  time = 0.86, size = 816, normalized size = 2.16 \[ -\frac {2 \, {\left (a^{3} b d x^{2} + a^{4} d x\right )} \cos \left (d x + c\right ) + 2 \, {\left (3 \, {\left (a b^{3} d x^{4} + 2 \, a^{2} b^{2} d x^{3} + a^{3} b d x^{2}\right )} \operatorname {Ci}\left (d x\right ) + 3 \, {\left (a b^{3} d x^{4} + 2 \, a^{2} b^{2} d x^{3} + a^{3} b d x^{2}\right )} \operatorname {Ci}\left (-d x\right ) + {\left ({\left (a^{2} b^{2} d^{2} - 12 \, b^{4}\right )} x^{4} + 2 \, {\left (a^{3} b d^{2} - 12 \, a b^{3}\right )} x^{3} + {\left (a^{4} d^{2} - 12 \, a^{2} b^{2}\right )} x^{2}\right )} \operatorname {Si}\left (d x\right )\right )} \cos \relax (c) + 2 \, {\left (3 \, {\left (a b^{3} d x^{4} + 2 \, a^{2} b^{2} d x^{3} + a^{3} b d x^{2}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + 3 \, {\left (a b^{3} d x^{4} + 2 \, a^{2} b^{2} d x^{3} + a^{3} b d x^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) - {\left ({\left (a^{2} b^{2} d^{2} - 12 \, b^{4}\right )} x^{4} + 2 \, {\left (a^{3} b d^{2} - 12 \, a b^{3}\right )} x^{3} + {\left (a^{4} d^{2} - 12 \, a^{2} b^{2}\right )} x^{2}\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) - 2 \, {\left (12 \, a b^{3} x^{3} + 18 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x - a^{4}\right )} \sin \left (d x + c\right ) + {\left ({\left ({\left (a^{2} b^{2} d^{2} - 12 \, b^{4}\right )} x^{4} + 2 \, {\left (a^{3} b d^{2} - 12 \, a b^{3}\right )} x^{3} + {\left (a^{4} d^{2} - 12 \, a^{2} b^{2}\right )} x^{2}\right )} \operatorname {Ci}\left (d x\right ) + {\left ({\left (a^{2} b^{2} d^{2} - 12 \, b^{4}\right )} x^{4} + 2 \, {\left (a^{3} b d^{2} - 12 \, a b^{3}\right )} x^{3} + {\left (a^{4} d^{2} - 12 \, a^{2} b^{2}\right )} x^{2}\right )} \operatorname {Ci}\left (-d x\right ) - 12 \, {\left (a b^{3} d x^{4} + 2 \, a^{2} b^{2} d x^{3} + a^{3} b d x^{2}\right )} \operatorname {Si}\left (d x\right )\right )} \sin \relax (c) + {\left ({\left ({\left (a^{2} b^{2} d^{2} - 12 \, b^{4}\right )} x^{4} + 2 \, {\left (a^{3} b d^{2} - 12 \, a b^{3}\right )} x^{3} + {\left (a^{4} d^{2} - 12 \, a^{2} b^{2}\right )} x^{2}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left ({\left (a^{2} b^{2} d^{2} - 12 \, b^{4}\right )} x^{4} + 2 \, {\left (a^{3} b d^{2} - 12 \, a b^{3}\right )} x^{3} + {\left (a^{4} d^{2} - 12 \, a^{2} b^{2}\right )} x^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) + 12 \, {\left (a b^{3} d x^{4} + 2 \, a^{2} b^{2} d x^{3} + a^{3} b d x^{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 (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(a^3*b*d*x^2 + a^4*d*x)*cos(d*x + c) + 2*(3*(a*b^3*d*x^4 + 2*a^2*b^2*d*x^3 + a^3*b*d*x^2)*cos_integral
(d*x) + 3*(a*b^3*d*x^4 + 2*a^2*b^2*d*x^3 + a^3*b*d*x^2)*cos_integral(-d*x) + ((a^2*b^2*d^2 - 12*b^4)*x^4 + 2*(
a^3*b*d^2 - 12*a*b^3)*x^3 + (a^4*d^2 - 12*a^2*b^2)*x^2)*sin_integral(d*x))*cos(c) + 2*(3*(a*b^3*d*x^4 + 2*a^2*
b^2*d*x^3 + a^3*b*d*x^2)*cos_integral((b*d*x + a*d)/b) + 3*(a*b^3*d*x^4 + 2*a^2*b^2*d*x^3 + a^3*b*d*x^2)*cos_i
ntegral(-(b*d*x + a*d)/b) - ((a^2*b^2*d^2 - 12*b^4)*x^4 + 2*(a^3*b*d^2 - 12*a*b^3)*x^3 + (a^4*d^2 - 12*a^2*b^2
)*x^2)*sin_integral((b*d*x + a*d)/b))*cos(-(b*c - a*d)/b) - 2*(12*a*b^3*x^3 + 18*a^2*b^2*x^2 + 4*a^3*b*x - a^4
)*sin(d*x + c) + (((a^2*b^2*d^2 - 12*b^4)*x^4 + 2*(a^3*b*d^2 - 12*a*b^3)*x^3 + (a^4*d^2 - 12*a^2*b^2)*x^2)*cos
_integral(d*x) + ((a^2*b^2*d^2 - 12*b^4)*x^4 + 2*(a^3*b*d^2 - 12*a*b^3)*x^3 + (a^4*d^2 - 12*a^2*b^2)*x^2)*cos_
integral(-d*x) - 12*(a*b^3*d*x^4 + 2*a^2*b^2*d*x^3 + a^3*b*d*x^2)*sin_integral(d*x))*sin(c) + (((a^2*b^2*d^2 -
12*b^4)*x^4 + 2*(a^3*b*d^2 - 12*a*b^3)*x^3 + (a^4*d^2 - 12*a^2*b^2)*x^2)*cos_integral((b*d*x + a*d)/b) + ((a^
2*b^2*d^2 - 12*b^4)*x^4 + 2*(a^3*b*d^2 - 12*a*b^3)*x^3 + (a^4*d^2 - 12*a^2*b^2)*x^2)*cos_integral(-(b*d*x + a*
d)/b) + 12*(a*b^3*d*x^4 + 2*a^2*b^2*d*x^3 + a^3*b*d*x^2)*sin_integral((b*d*x + a*d)/b))*sin(-(b*c - a*d)/b))/(
a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)

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

[Out]

Timed out

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maple [A]  time = 0.03, size = 466, normalized size = 1.24 \[ d^{2} \left (-\frac {3 b \left (-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )\right )}{d \,a^{4}}-\frac {b^{3} \left (-\frac {\sin \left (d x +c \right )}{2 \left (\left (d x +c \right ) b +d a -c b \right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (\left (d x +c \right ) b +d a -c b \right ) b}-\frac {\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}}{b}}{2 b}\right )}{a^{3}}-\frac {6 b^{3} \left (\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}\right )}{d^{2} a^{5}}+\frac {-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}}{a^{3}}+\frac {6 b^{2} \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right )}{d^{2} a^{5}}-\frac {3 b^{3} \left (-\frac {\sin \left (d x +c \right )}{\left (\left (d x +c \right ) b +d a -c b \right ) b}+\frac {\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}+\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}}{b}\right )}{d \,a^{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(-3/d/a^4*b*(-sin(d*x+c)/x/d-Si(d*x)*sin(c)+Ci(d*x)*cos(c))-b^3/a^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)-6/d^2*b^3/a^5*(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)+1/a^3*(-1/2*sin(d*x+c)/x^2/d^2-1/2*cos(d*x+c)/x/d-1/2*Si(d*x)*cos(c)-1/2*Ci(d*x)*sin(c))+6/d^2/a^5*b
^2*(Si(d*x)*cos(c)+Ci(d*x)*sin(c))-3/d*b^3/a^4*(-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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )}{{\left (b x + a\right )}^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (c+d\,x\right )}{x^3\,{\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)/(x^3*(a + b*x)^3),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (c + d x \right )}}{x^{3} \left (a + b x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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