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

Optimal. Leaf size=179 \[ \frac{a d^2 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{2 b^4}+\frac{d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^3}+\frac{a d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{2 b^4}-\frac{d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^3}-\frac{\sin (c+d x)}{b^2 (a+b x)}+\frac{a \sin (c+d x)}{2 b^2 (a+b x)^2}+\frac{a d \cos (c+d x)}{2 b^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.349956, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6742, 3297, 3303, 3299, 3302} \[ \frac{a d^2 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{2 b^4}+\frac{d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^3}+\frac{a d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{2 b^4}-\frac{d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^3}-\frac{\sin (c+d x)}{b^2 (a+b x)}+\frac{a \sin (c+d x)}{2 b^2 (a+b x)^2}+\frac{a d \cos (c+d x)}{2 b^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sin[c + d*x])/(a + b*x)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.580294, size = 157, normalized size = 0.88 \[ \frac{d (a+b x)^2 \left (\text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right ) \left (a d \sin \left (c-\frac{a d}{b}\right )+2 b \cos \left (c-\frac{a d}{b}\right )\right )+\text{Si}\left (d \left (\frac{a}{b}+x\right )\right ) \left (a d \cos \left (c-\frac{a d}{b}\right )-2 b \sin \left (c-\frac{a d}{b}\right )\right )\right )+b \cos (d x) (a d \cos (c) (a+b x)-b \sin (c) (a+2 b x))-b \sin (d x) (a d \sin (c) (a+b x)+b \cos (c) (a+2 b x))}{2 b^4 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 419, normalized size = 2.3 \begin{align*}{\frac{1}{{d}^{2}} \left ( -{\frac{{d}^{3} \left ( da-cb \right ) }{b} \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 ) }+{\frac{{d}^{3}}{b} \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 ) }-{d}^{3}c \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2*(-d^3*(a*d-b*c)/b*(-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)+d^3/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*(-
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))

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

[Out]

Timed out

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Fricas [A]  time = 1.33687, size = 792, normalized size = 4.42 \begin{align*} \frac{2 \,{\left (a b^{2} d x + a^{2} b d\right )} \cos \left (d x + c\right ) + 2 \,{\left ({\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) +{\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) +{\left (a b^{2} d^{2} x^{2} + 2 \, a^{2} b d^{2} x + a^{3} d^{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 (2 \, b^{3} x + a b^{2}\right )} \sin \left (d x + c\right ) -{\left ({\left (a b^{2} d^{2} x^{2} + 2 \, a^{2} b d^{2} x + a^{3} d^{2}\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) +{\left (a b^{2} d^{2} x^{2} + 2 \, a^{2} b d^{2} x + a^{3} d^{2}\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) - 4 \,{\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\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^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out