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

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

[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

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Rubi [A]  time = 0.804205, antiderivative size = 377, normalized size of antiderivative = 1., 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 6742

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

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{\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*}

Mathematica [A]  time = 2.24952, size = 630, normalized size = 1.67 \[ \frac{-x^2 (a+b x)^2 \text{CosIntegral}(d x) \left (\sin (c) \left (a^2 d^2-12 b^2\right )+6 a b d \cos (c)\right )+x^2 (a+b x)^2 \text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right ) \left (\left (a^2 d^2-12 b^2\right ) \sin \left (c-\frac{a d}{b}\right )-6 a b d \cos \left (c-\frac{a d}{b}\right )\right )-a^2 b^2 d^2 x^4 \cos (c) \text{Si}(d x)+a^2 b^2 d^2 x^4 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+12 a^2 b^2 d x^3 \sin (c) \text{Si}(d x)+12 a^2 b^2 d x^3 \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+12 a^2 b^2 x^2 \cos (c) \text{Si}(d x)-12 a^2 b^2 x^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+18 a^2 b^2 x^2 \sin (c+d x)+a^4 d^2 x^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )-2 a^3 b d^2 x^3 \cos (c) \text{Si}(d x)+2 a^3 b d^2 x^3 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+6 a^3 b d x^2 \sin (c) \text{Si}(d x)+6 a^3 b d x^2 \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )-a^3 b d x^2 \cos (c+d x)+4 a^3 b x \sin (c+d x)-a^4 d^2 x^2 \cos (c) \text{Si}(d x)-a^4 \sin (c+d x)+a^4 (-d) x \cos (c+d x)+6 a b^3 d x^4 \sin (c) \text{Si}(d x)+6 a b^3 d x^4 \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+24 a b^3 x^3 \cos (c) \text{Si}(d x)-24 a b^3 x^3 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )-12 b^4 x^4 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+12 a b^3 x^3 \sin (c+d x)+12 b^4 x^4 \cos (c) \text{Si}(d x)}{2 a^5 x^2 (a+b x)^2} \]

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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Maple [A]  time = 0.013, size = 466, normalized size = 1.2 \begin{align*}{d}^{2} \left ( -{\frac{{b}^{3}}{{a}^{3}} \left ( -{\frac{\sin \left ( dx+c \right ) }{2\, \left ( \left ( dx+c \right ) b+da-cb \right ) ^{2}b}}+{\frac{1}{2\,b} \left ( -{\frac{\cos \left ( dx+c \right ) }{ \left ( \left ( dx+c \right ) b+da-cb \right ) b}}-{\frac{1}{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 ) } \right ) } \right ) }-3\,{\frac{b}{d{a}^{4}} \left ( -{\frac{\sin \left ( dx+c \right ) }{dx}}-{\it Si} \left ( dx \right ) \sin \left ( c \right ) +{\it Ci} \left ( dx \right ) \cos \left ( c \right ) \right ) }-3\,{\frac{{b}^{3}}{d{a}^{4}} \left ( -{\frac{\sin \left ( dx+c \right ) }{ \left ( \left ( dx+c \right ) b+da-cb \right ) b}}+{\frac{1}{b} \left ({\frac{1}{b}{\it Si} \left ( dx+c+{\frac{da-cb}{b}} \right ) \sin \left ({\frac{da-cb}{b}} \right ) }+{\frac{1}{b}{\it Ci} \left ( dx+c+{\frac{da-cb}{b}} \right ) \cos \left ({\frac{da-cb}{b}} \right ) } \right ) } \right ) }+{\frac{1}{{a}^{3}} \left ( -{\frac{\sin \left ( dx+c \right ) }{2\,{d}^{2}{x}^{2}}}-{\frac{\cos \left ( dx+c \right ) }{2\,dx}}-{\frac{{\it Si} \left ( dx \right ) \cos \left ( c \right ) }{2}}-{\frac{{\it Ci} \left ( dx \right ) \sin \left ( c \right ) }{2}} \right ) }-6\,{\frac{{b}^{3}}{{d}^{2}{a}^{5}} \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 ) }+6\,{\frac{{b}^{2} \left ({\it Si} \left ( dx \right ) \cos \left ( c \right ) +{\it Ci} \left ( dx \right ) \sin \left ( c \right ) \right ) }{{d}^{2}{a}^{5}}} \right ) \end{align*}

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

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

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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Fricas [B]  time = 1.82921, size = 1833, normalized size = 4.86 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

Timed out

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

[Out]

Timed out