3.38 \(\int \frac {\sin (c+d x)}{x^2 (a+b x)^3} \, dx\)

Optimal. Leaf size=299 \[ -\frac {3 b \sin (c) \text {Ci}(d x)}{a^4}+\frac {3 b \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{a^4}-\frac {3 b \cos (c) \text {Si}(d x)}{a^4}+\frac {3 b \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^4}+\frac {2 d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{a^3}-\frac {2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^3}-\frac {2 b \sin (c+d x)}{a^3 (a+b x)}+\frac {d \cos (c) \text {Ci}(d x)}{a^3}-\frac {d \sin (c) \text {Si}(d x)}{a^3}-\frac {\sin (c+d x)}{a^3 x}-\frac {d^2 \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{2 a^2 b}-\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 a^2 b}-\frac {b \sin (c+d x)}{2 a^2 (a+b x)^2}-\frac {d \cos (c+d x)}{2 a^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.67, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 21, 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} \[ -\frac {d^2 \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{2 a^2 b}-\frac {3 b \sin (c) \text {CosIntegral}(d x)}{a^4}+\frac {3 b \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{a^4}+\frac {2 d \cos \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{a^3}-\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 a^2 b}-\frac {2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^3}-\frac {3 b \cos (c) \text {Si}(d x)}{a^4}+\frac {3 b \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^4}-\frac {2 b \sin (c+d x)}{a^3 (a+b x)}-\frac {b \sin (c+d x)}{2 a^2 (a+b x)^2}-\frac {d \cos (c+d x)}{2 a^2 (a+b x)}+\frac {d \cos (c) \text {CosIntegral}(d x)}{a^3}-\frac {d \sin (c) \text {Si}(d x)}{a^3}-\frac {\sin (c+d x)}{a^3 x} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(x^2*(a + b*x)^3),x]

[Out]

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

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

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Mathematica [A]  time = 2.07, size = 540, normalized size = 1.81 \[ -\frac {a^4 d^2 x \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+2 a^3 b d^2 x^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+2 a^3 b d x \sin (c) \text {Si}(d x)+4 a^3 b d x \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+2 a^3 b \sin (c+d x)+a^3 b d x \cos (c+d x)+x (a+b x)^2 \text {Ci}\left (d \left (\frac {a}{b}+x\right )\right ) \left (\left (a^2 d^2-6 b^2\right ) \sin \left (c-\frac {a d}{b}\right )-4 a b d \cos \left (c-\frac {a d}{b}\right )\right )+a^2 b^2 d^2 x^3 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+4 a^2 b^2 d x^2 \sin (c) \text {Si}(d x)+8 a^2 b^2 d x^2 \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+6 a^2 b^2 x \cos (c) \text {Si}(d x)-6 a^2 b^2 x \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+a^2 b^2 d x^2 \cos (c+d x)+9 a^2 b^2 x \sin (c+d x)-6 b^4 x^3 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+2 a b^3 d x^3 \sin (c) \text {Si}(d x)+4 a b^3 d x^3 \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+12 a b^3 x^2 \cos (c) \text {Si}(d x)-12 a b^3 x^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+6 a b^3 x^2 \sin (c+d x)+2 b x (a+b x)^2 \text {Ci}(d x) (3 b \sin (c)-a d \cos (c))+6 b^4 x^3 \cos (c) \text {Si}(d x)}{2 a^4 b x (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]/(x^2*(a + b*x)^3),x]

[Out]

-1/2*(a^3*b*d*x*Cos[c + d*x] + a^2*b^2*d*x^2*Cos[c + d*x] + 2*b*x*(a + b*x)^2*CosIntegral[d*x]*(-(a*d*Cos[c])
+ 3*b*Sin[c]) + x*(a + b*x)^2*CosIntegral[d*(a/b + x)]*(-4*a*b*d*Cos[c - (a*d)/b] + (-6*b^2 + a^2*d^2)*Sin[c -
(a*d)/b]) + 2*a^3*b*Sin[c + d*x] + 9*a^2*b^2*x*Sin[c + d*x] + 6*a*b^3*x^2*Sin[c + d*x] + 6*a^2*b^2*x*Cos[c]*S
inIntegral[d*x] + 12*a*b^3*x^2*Cos[c]*SinIntegral[d*x] + 6*b^4*x^3*Cos[c]*SinIntegral[d*x] + 2*a^3*b*d*x*Sin[c
]*SinIntegral[d*x] + 4*a^2*b^2*d*x^2*Sin[c]*SinIntegral[d*x] + 2*a*b^3*d*x^3*Sin[c]*SinIntegral[d*x] - 6*a^2*b
^2*x*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + a^4*d^2*x*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] - 12*a*b^
3*x^2*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + 2*a^3*b*d^2*x^2*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] -
6*b^4*x^3*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + a^2*b^2*d^2*x^3*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)
] + 4*a^3*b*d*x*Sin[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + 8*a^2*b^2*d*x^2*Sin[c - (a*d)/b]*SinIntegral[d*(a/
b + x)] + 4*a*b^3*d*x^3*Sin[c - (a*d)/b]*SinIntegral[d*(a/b + x)])/(a^4*b*x*(a + b*x)^2)

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fricas [B]  time = 0.73, size = 689, normalized size = 2.30 \[ -\frac {2 \, {\left (a^{2} b^{2} d x^{2} + a^{3} b d x\right )} \cos \left (d x + c\right ) - 2 \, {\left ({\left (a b^{3} d x^{3} + 2 \, a^{2} b^{2} d x^{2} + a^{3} b d x\right )} \operatorname {Ci}\left (d x\right ) + {\left (a b^{3} d x^{3} + 2 \, a^{2} b^{2} d x^{2} + a^{3} b d x\right )} \operatorname {Ci}\left (-d x\right ) - 6 \, {\left (b^{4} x^{3} + 2 \, a b^{3} x^{2} + a^{2} b^{2} x\right )} \operatorname {Si}\left (d x\right )\right )} \cos \relax (c) - 2 \, {\left (2 \, {\left (a b^{3} d x^{3} + 2 \, a^{2} b^{2} d x^{2} + a^{3} b d x\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + 2 \, {\left (a b^{3} d x^{3} + 2 \, a^{2} b^{2} d x^{2} + a^{3} b d x\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) - {\left ({\left (a^{2} b^{2} d^{2} - 6 \, b^{4}\right )} x^{3} + 2 \, {\left (a^{3} b d^{2} - 6 \, a b^{3}\right )} x^{2} + {\left (a^{4} d^{2} - 6 \, a^{2} b^{2}\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 b^{3} x^{2} + 9 \, a^{2} b^{2} x + 2 \, a^{3} b\right )} \sin \left (d x + c\right ) + 2 \, {\left (3 \, {\left (b^{4} x^{3} + 2 \, a b^{3} x^{2} + a^{2} b^{2} x\right )} \operatorname {Ci}\left (d x\right ) + 3 \, {\left (b^{4} x^{3} + 2 \, a b^{3} x^{2} + a^{2} b^{2} x\right )} \operatorname {Ci}\left (-d x\right ) + 2 \, {\left (a b^{3} d x^{3} + 2 \, a^{2} b^{2} d x^{2} + a^{3} b d x\right )} \operatorname {Si}\left (d x\right )\right )} \sin \relax (c) - {\left ({\left ({\left (a^{2} b^{2} d^{2} - 6 \, b^{4}\right )} x^{3} + 2 \, {\left (a^{3} b d^{2} - 6 \, a b^{3}\right )} x^{2} + {\left (a^{4} d^{2} - 6 \, a^{2} b^{2}\right )} x\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left ({\left (a^{2} b^{2} d^{2} - 6 \, b^{4}\right )} x^{3} + 2 \, {\left (a^{3} b d^{2} - 6 \, a b^{3}\right )} x^{2} + {\left (a^{4} d^{2} - 6 \, a^{2} b^{2}\right )} x\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) + 8 \, {\left (a b^{3} d x^{3} + 2 \, a^{2} b^{2} d x^{2} + a^{3} b d x\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^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/x^2/(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/4*(2*(a^2*b^2*d*x^2 + a^3*b*d*x)*cos(d*x + c) - 2*((a*b^3*d*x^3 + 2*a^2*b^2*d*x^2 + a^3*b*d*x)*cos_integral
(d*x) + (a*b^3*d*x^3 + 2*a^2*b^2*d*x^2 + a^3*b*d*x)*cos_integral(-d*x) - 6*(b^4*x^3 + 2*a*b^3*x^2 + a^2*b^2*x)
*sin_integral(d*x))*cos(c) - 2*(2*(a*b^3*d*x^3 + 2*a^2*b^2*d*x^2 + a^3*b*d*x)*cos_integral((b*d*x + a*d)/b) +
2*(a*b^3*d*x^3 + 2*a^2*b^2*d*x^2 + a^3*b*d*x)*cos_integral(-(b*d*x + a*d)/b) - ((a^2*b^2*d^2 - 6*b^4)*x^3 + 2*
(a^3*b*d^2 - 6*a*b^3)*x^2 + (a^4*d^2 - 6*a^2*b^2)*x)*sin_integral((b*d*x + a*d)/b))*cos(-(b*c - a*d)/b) + 2*(6
*a*b^3*x^2 + 9*a^2*b^2*x + 2*a^3*b)*sin(d*x + c) + 2*(3*(b^4*x^3 + 2*a*b^3*x^2 + a^2*b^2*x)*cos_integral(d*x)
+ 3*(b^4*x^3 + 2*a*b^3*x^2 + a^2*b^2*x)*cos_integral(-d*x) + 2*(a*b^3*d*x^3 + 2*a^2*b^2*d*x^2 + a^3*b*d*x)*sin
_integral(d*x))*sin(c) - (((a^2*b^2*d^2 - 6*b^4)*x^3 + 2*(a^3*b*d^2 - 6*a*b^3)*x^2 + (a^4*d^2 - 6*a^2*b^2)*x)*
cos_integral((b*d*x + a*d)/b) + ((a^2*b^2*d^2 - 6*b^4)*x^3 + 2*(a^3*b*d^2 - 6*a*b^3)*x^2 + (a^4*d^2 - 6*a^2*b^
2)*x)*cos_integral(-(b*d*x + a*d)/b) + 8*(a*b^3*d*x^3 + 2*a^2*b^2*d*x^2 + a^3*b*d*x)*sin_integral((b*d*x + a*d
)/b))*sin(-(b*c - a*d)/b))/(a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x)

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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^2/(b*x+a)^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.03, size = 405, normalized size = 1.35 \[ d \left (\frac {-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )}{a^{3}}+\frac {d \,b^{2} \left (-\frac {\sin \left (d x +c \right )}{2 \left (\left (d x +c \right ) b +d a -c b \right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (\left (d x +c \right ) b +d a -c b \right ) b}-\frac {\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}}{b}}{2 b}\right )}{a^{2}}+\frac {3 b^{2} \left (\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}\right )}{d \,a^{4}}-\frac {3 b \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right )}{d \,a^{4}}+\frac {2 b^{2} \left (-\frac {\sin \left (d x +c \right )}{\left (\left (d x +c \right ) b +d a -c b \right ) b}+\frac {\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}+\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}}{b}\right )}{a^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)/x^2/(b*x+a)^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )}{{\left (b x + a\right )}^{3} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/x^2/(b*x+a)^3,x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)/((b*x + a)^3*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (c+d\,x\right )}{x^2\,{\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)/(x^2*(a + b*x)^3),x)

[Out]

int(sin(c + d*x)/(x^2*(a + b*x)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (c + d x \right )}}{x^{2} \left (a + b x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/x**2/(b*x+a)**3,x)

[Out]

Integral(sin(c + d*x)/(x**2*(a + b*x)**3), x)

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