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

Optimal. Leaf size=241 \[ -\frac{a^2 d^2 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{2 b^5}-\frac{a^2 d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{2 b^5}-\frac{a^2 \sin (c+d x)}{2 b^3 (a+b x)^2}-\frac{a^2 d \cos (c+d x)}{2 b^4 (a+b x)}+\frac{\sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^3}-\frac{2 a d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^4}+\frac{2 a d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^4}+\frac{\cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{2 a \sin (c+d x)}{b^3 (a+b x)} \]

[Out]

-(a^2*d*Cos[c + d*x])/(2*b^4*(a + b*x)) - (2*a*d*Cos[c - (a*d)/b]*CosIntegral[(a*d)/b + d*x])/b^4 + (CosIntegr
al[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^3 - (a^2*d^2*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/(2*b^5) - (a^2
*Sin[c + d*x])/(2*b^3*(a + b*x)^2) + (2*a*Sin[c + d*x])/(b^3*(a + b*x)) + (Cos[c - (a*d)/b]*SinIntegral[(a*d)/
b + d*x])/b^3 - (a^2*d^2*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/(2*b^5) + (2*a*d*Sin[c - (a*d)/b]*SinInt
egral[(a*d)/b + d*x])/b^4

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Rubi [A]  time = 0.535422, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ -\frac{a^2 d^2 \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{2 b^5}-\frac{a^2 d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{2 b^5}-\frac{a^2 \sin (c+d x)}{2 b^3 (a+b x)^2}-\frac{a^2 d \cos (c+d x)}{2 b^4 (a+b x)}+\frac{\sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^3}-\frac{2 a d \cos \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^4}+\frac{2 a d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^4}+\frac{\cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{2 a \sin (c+d x)}{b^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sin[c + d*x])/(a + b*x)^3,x]

[Out]

-(a^2*d*Cos[c + d*x])/(2*b^4*(a + b*x)) - (2*a*d*Cos[c - (a*d)/b]*CosIntegral[(a*d)/b + d*x])/b^4 + (CosIntegr
al[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^3 - (a^2*d^2*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/(2*b^5) - (a^2
*Sin[c + d*x])/(2*b^3*(a + b*x)^2) + (2*a*Sin[c + d*x])/(b^3*(a + b*x)) + (Cos[c - (a*d)/b]*SinIntegral[(a*d)/
b + d*x])/b^3 - (a^2*d^2*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/(2*b^5) + (2*a*d*Sin[c - (a*d)/b]*SinInt
egral[(a*d)/b + d*x])/b^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{x^2 \sin (c+d x)}{(a+b x)^3} \, dx &=\int \left (\frac{a^2 \sin (c+d x)}{b^2 (a+b x)^3}-\frac{2 a \sin (c+d x)}{b^2 (a+b x)^2}+\frac{\sin (c+d x)}{b^2 (a+b x)}\right ) \, dx\\ &=\frac{\int \frac{\sin (c+d x)}{a+b x} \, dx}{b^2}-\frac{(2 a) \int \frac{\sin (c+d x)}{(a+b x)^2} \, dx}{b^2}+\frac{a^2 \int \frac{\sin (c+d x)}{(a+b x)^3} \, dx}{b^2}\\ &=-\frac{a^2 \sin (c+d x)}{2 b^3 (a+b x)^2}+\frac{2 a \sin (c+d x)}{b^3 (a+b x)}-\frac{(2 a d) \int \frac{\cos (c+d x)}{a+b x} \, dx}{b^3}+\frac{\left (a^2 d\right ) \int \frac{\cos (c+d x)}{(a+b x)^2} \, dx}{2 b^3}+\frac{\cos \left (c-\frac{a d}{b}\right ) \int \frac{\sin \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b^2}+\frac{\sin \left (c-\frac{a d}{b}\right ) \int \frac{\cos \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b^2}\\ &=-\frac{a^2 d \cos (c+d x)}{2 b^4 (a+b x)}+\frac{\text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^3}-\frac{a^2 \sin (c+d x)}{2 b^3 (a+b x)^2}+\frac{2 a \sin (c+d x)}{b^3 (a+b x)}+\frac{\cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^3}-\frac{\left (a^2 d^2\right ) \int \frac{\sin (c+d x)}{a+b x} \, dx}{2 b^4}-\frac{\left (2 a 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^3}+\frac{\left (2 a 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^3}\\ &=-\frac{a^2 d \cos (c+d x)}{2 b^4 (a+b x)}-\frac{2 a d \cos \left (c-\frac{a d}{b}\right ) \text{Ci}\left (\frac{a d}{b}+d x\right )}{b^4}+\frac{\text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^3}-\frac{a^2 \sin (c+d x)}{2 b^3 (a+b x)^2}+\frac{2 a \sin (c+d x)}{b^3 (a+b x)}+\frac{\cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^3}+\frac{2 a d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^4}-\frac{\left (a^2 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^4}-\frac{\left (a^2 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^4}\\ &=-\frac{a^2 d \cos (c+d x)}{2 b^4 (a+b x)}-\frac{2 a d \cos \left (c-\frac{a d}{b}\right ) \text{Ci}\left (\frac{a d}{b}+d x\right )}{b^4}+\frac{\text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^3}-\frac{a^2 d^2 \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{2 b^5}-\frac{a^2 \sin (c+d x)}{2 b^3 (a+b x)^2}+\frac{2 a \sin (c+d x)}{b^3 (a+b x)}+\frac{\cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^3}-\frac{a^2 d^2 \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{2 b^5}+\frac{2 a d \sin \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^4}\\ \end{align*}

Mathematica [A]  time = 1.18735, size = 154, normalized size = 0.64 \[ -\frac{-\text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right ) \left (\left (2 b^2-a^2 d^2\right ) \sin \left (c-\frac{a d}{b}\right )-4 a b d \cos \left (c-\frac{a d}{b}\right )\right )+\text{Si}\left (d \left (\frac{a}{b}+x\right )\right ) \left (\left (a^2 d^2-2 b^2\right ) \cos \left (c-\frac{a d}{b}\right )-4 a b d \sin \left (c-\frac{a d}{b}\right )\right )+\frac{a b (a d (a+b x) \cos (c+d x)-b (3 a+4 b x) \sin (c+d x))}{(a+b x)^2}}{2 b^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 779, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Fricas [A]  time = 1.49242, size = 972, normalized size = 4.03 \begin{align*} -\frac{2 \,{\left (a^{2} b^{2} d x + a^{3} b d\right )} \cos \left (d x + c\right ) + 2 \,{\left (2 \,{\left (a b^{3} d x^{2} + 2 \, a^{2} b^{2} d x + a^{3} b d\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) + 2 \,{\left (a b^{3} d x^{2} + 2 \, a^{2} b^{2} d x + a^{3} b d\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) +{\left (a^{4} d^{2} - 2 \, a^{2} b^{2} +{\left (a^{2} b^{2} d^{2} - 2 \, b^{4}\right )} x^{2} + 2 \,{\left (a^{3} b d^{2} - 2 \, a b^{3}\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 (4 \, a b^{3} x + 3 \, a^{2} b^{2}\right )} \sin \left (d x + c\right ) -{\left ({\left (a^{4} d^{2} - 2 \, a^{2} b^{2} +{\left (a^{2} b^{2} d^{2} - 2 \, b^{4}\right )} x^{2} + 2 \,{\left (a^{3} b d^{2} - 2 \, a b^{3}\right )} x\right )} \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) +{\left (a^{4} d^{2} - 2 \, a^{2} b^{2} +{\left (a^{2} b^{2} d^{2} - 2 \, b^{4}\right )} x^{2} + 2 \,{\left (a^{3} b d^{2} - 2 \, a b^{3}\right )} x\right )} \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right ) - 8 \,{\left (a b^{3} d x^{2} + 2 \, a^{2} b^{2} d x + a^{3} 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^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(a^2*b^2*d*x + a^3*b*d)*cos(d*x + c) + 2*(2*(a*b^3*d*x^2 + 2*a^2*b^2*d*x + a^3*b*d)*cos_integral((b*d*
x + a*d)/b) + 2*(a*b^3*d*x^2 + 2*a^2*b^2*d*x + a^3*b*d)*cos_integral(-(b*d*x + a*d)/b) + (a^4*d^2 - 2*a^2*b^2
+ (a^2*b^2*d^2 - 2*b^4)*x^2 + 2*(a^3*b*d^2 - 2*a*b^3)*x)*sin_integral((b*d*x + a*d)/b))*cos(-(b*c - a*d)/b) -
2*(4*a*b^3*x + 3*a^2*b^2)*sin(d*x + c) - ((a^4*d^2 - 2*a^2*b^2 + (a^2*b^2*d^2 - 2*b^4)*x^2 + 2*(a^3*b*d^2 - 2*
a*b^3)*x)*cos_integral((b*d*x + a*d)/b) + (a^4*d^2 - 2*a^2*b^2 + (a^2*b^2*d^2 - 2*b^4)*x^2 + 2*(a^3*b*d^2 - 2*
a*b^3)*x)*cos_integral(-(b*d*x + a*d)/b) - 8*(a*b^3*d*x^2 + 2*a^2*b^2*d*x + a^3*b*d)*sin_integral((b*d*x + a*d
)/b))*sin(-(b*c - a*d)/b))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \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**2*sin(d*x+c)/(b*x+a)**3,x)

[Out]

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

[Out]

Timed out