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

Optimal. Leaf size=188 \[ -\frac{2 b \sin (c) \text{CosIntegral}(d x)}{a^3}+\frac{2 b \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{a^3}+\frac{d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{a^2}-\frac{d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{a^2}-\frac{2 b \cos (c) \text{Si}(d x)}{a^3}+\frac{2 b \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{a^3}-\frac{b \sin (c+d x)}{a^2 (a+b x)}+\frac{d \cos (c) \text{CosIntegral}(d x)}{a^2}-\frac{d \sin (c) \text{Si}(d x)}{a^2}-\frac{\sin (c+d x)}{a^2 x} \]

[Out]

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

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Rubi [A]  time = 0.513687, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 16, 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} \[ -\frac{2 b \sin (c) \text{CosIntegral}(d x)}{a^3}+\frac{2 b \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{a^3}+\frac{d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{a^2}-\frac{d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{a^2}-\frac{2 b \cos (c) \text{Si}(d x)}{a^3}+\frac{2 b \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{a^3}-\frac{b \sin (c+d x)}{a^2 (a+b x)}+\frac{d \cos (c) \text{CosIntegral}(d x)}{a^2}-\frac{d \sin (c) \text{Si}(d x)}{a^2}-\frac{\sin (c+d x)}{a^2 x} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(x^2*(a + b*x)^2),x]

[Out]

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

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

Mathematica [A]  time = 1.95359, size = 184, normalized size = 0.98 \[ -\frac{-2 b \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right )-a d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right )+a d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )-2 b \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+\frac{a \sin (c) (a+2 b x) \cos (d x)}{x (a+b x)}+\frac{a \cos (c) (a+2 b x) \sin (d x)}{x (a+b x)}-a d \cos (c) \text{CosIntegral}(d x)+a d \sin (c) \text{Si}(d x)+2 b \sin (c) \text{CosIntegral}(d x)+2 b \cos (c) \text{Si}(d x)}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 256, normalized size = 1.4 \begin{align*} d \left ({\frac{1}{{a}^{2}} \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 ) }+{\frac{{b}^{2}}{{a}^{2}} \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 ) }+2\,{\frac{{b}^{2}}{d{a}^{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 ) }-2\,{\frac{b \left ({\it Si} \left ( dx \right ) \cos \left ( c \right ) +{\it Ci} \left ( dx \right ) \sin \left ( c \right ) \right ) }{d{a}^{3}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(1/a^2*(-sin(d*x+c)/x/d-Si(d*x)*sin(c)+Ci(d*x)*cos(c))+b^2/a^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)+2/d*b^2/a^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)-2/d/a^3*b*(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 )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.89511, size = 923, normalized size = 4.91 \begin{align*} \frac{{\left ({\left (a b d x^{2} + a^{2} d x\right )} \operatorname{Ci}\left (d x\right ) +{\left (a b d x^{2} + a^{2} d x\right )} \operatorname{Ci}\left (-d x\right ) - 4 \,{\left (b^{2} x^{2} + a b x\right )} \operatorname{Si}\left (d x\right )\right )} \cos \left (c\right ) +{\left ({\left (a b d x^{2} + a^{2} d x\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) +{\left (a b d x^{2} + a^{2} d x\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) + 4 \,{\left (b^{2} x^{2} + a b 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 (2 \, a b x + a^{2}\right )} \sin \left (d x + c\right ) - 2 \,{\left ({\left (b^{2} x^{2} + a b x\right )} \operatorname{Ci}\left (d x\right ) +{\left (b^{2} x^{2} + a b x\right )} \operatorname{Ci}\left (-d x\right ) +{\left (a b d x^{2} + a^{2} d x\right )} \operatorname{Si}\left (d x\right )\right )} \sin \left (c\right ) - 2 \,{\left ({\left (b^{2} x^{2} + a b x\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) +{\left (b^{2} x^{2} + a b x\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) -{\left (a b d x^{2} + a^{2} d x\right )} \operatorname{Si}\left (\frac{b d x + a d}{b}\right )\right )} \sin \left (-\frac{b c - a d}{b}\right )}{2 \,{\left (a^{3} b x^{2} + a^{4} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (c + d x \right )}}{x^{2} \left (a + b x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2/(b*x+a)^2,x, algorithm="giac")

[Out]

Timed out