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

Optimal. Leaf size=149 \[ \frac {a^2 d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {a^2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {a^2 \sin (c+d x)}{b^3 (a+b x)}-\frac {2 a \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{b^3}-\frac {2 a \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^3}-\frac {\cos (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6742, 2638, 3297, 3303, 3299, 3302} \[ \frac {a^2 d \cos \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^4}-\frac {a^2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {a^2 \sin (c+d x)}{b^3 (a+b x)}-\frac {2 a \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^3}-\frac {2 a \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^3}-\frac {\cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sin[c + d*x])/(a + b*x)^2,x]

[Out]

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

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Mathematica [A]  time = 0.84, size = 117, normalized size = 0.79 \[ \frac {b \left (-\frac {a^2 \sin (c+d x)}{a+b x}-\frac {b \cos (c+d x)}{d}\right )+a \text {Ci}\left (d \left (\frac {a}{b}+x\right )\right ) \left (a d \cos \left (c-\frac {a d}{b}\right )-2 b \sin \left (c-\frac {a d}{b}\right )\right )-a \text {Si}\left (d \left (\frac {a}{b}+x\right )\right ) \left (a d \sin \left (c-\frac {a d}{b}\right )+2 b \cos \left (c-\frac {a d}{b}\right )\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 264, normalized size = 1.77 \[ -\frac {2 \, a^{2} b d \sin \left (d x + c\right ) + 2 \, {\left (b^{3} x + a b^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} b d^{2} x + a^{3} d^{2}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left (a^{2} b d^{2} x + a^{3} d^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) - 4 \, {\left (a b^{2} d x + a^{2} b d\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) - 2 \, {\left ({\left (a b^{2} d x + a^{2} b d\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left (a b^{2} d x + a^{2} b d\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) + {\left (a^{2} b d^{2} x + a^{3} d^{2}\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{2 \, {\left (b^{5} d x + a b^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.75, size = 1120, normalized size = 7.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a)*a^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d^2*cos(-(b*c - a*d)/b)*cos_integral(((b*x + a)*(b*c/(b*x +
a) - a*d/(b*x + a) + d) - b*c + a*d)/b) - a^2*b*c*d^2*cos(-(b*c - a*d)/b)*cos_integral(((b*x + a)*(b*c/(b*x +
a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + a^3*d^3*cos(-(b*c - a*d)/b)*cos_integral(((b*x + a)*(b*c/(b*x + a)
- a*d/(b*x + a) + d) - b*c + a*d)/b) + (b*x + a)*a^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d^2*sin(-(b*c - a*d)/
b)*sin_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) - a^2*b*c*d^2*sin(-(b*c - a*d)/
b)*sin_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + a^3*d^3*sin(-(b*c - a*d)/b)*s
in_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + 2*(b*x + a)*a*b*(b*c/(b*x + a) -
a*d/(b*x + a) + d)*d*cos_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*sin(-(b*c - a
*d)/b) - 2*a*b^2*c*d*cos_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*sin(-(b*c - a
*d)/b) + 2*a^2*b*d^2*cos_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*sin(-(b*c - a
*d)/b) - 2*(b*x + a)*a*b*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d*cos(-(b*c - a*d)/b)*sin_integral(((b*x + a)*(b*
c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + 2*a*b^2*c*d*cos(-(b*c - a*d)/b)*sin_integral(((b*x + a)*(b*
c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) - 2*a^2*b*d^2*cos(-(b*c - a*d)/b)*sin_integral(((b*x + a)*(b*
c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + a^2*b*d^2*sin(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d
)/b) - (b*x + a)*b^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)*cos(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b)
+ b^3*c*cos(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) - a*b^2*d*cos(-(b*x + a)*(b*c/(b*x + a) - a*d/(
b*x + a) + d)/b))*b^2/(((b*x + a)*b^6*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b^7*c + a*b^6*d)*d)

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maple [B]  time = 0.03, size = 553, normalized size = 3.71 \[ \frac {-\frac {2 \left (d a -c b \right ) d^{2} \left (\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}\right )}{b^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) d^{2} \left (-\frac {\sin \left (d x +c \right )}{\left (\left (d x +c \right ) b +d a -c b \right ) b}+\frac {\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}+\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}}{b}\right )}{b^{2}}-\frac {d^{2} \cos \left (d x +c \right )}{b^{2}}-\frac {2 d^{2} c \left (\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}\right )}{b}+\frac {2 \left (d a -c b \right ) d^{2} c \left (-\frac {\sin \left (d x +c \right )}{\left (\left (d x +c \right ) b +d a -c b \right ) b}+\frac {\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}+\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}}{b}\right )}{b}+d^{2} c^{2} \left (-\frac {\sin \left (d x +c \right )}{\left (\left (d x +c \right ) b +d a -c b \right ) b}+\frac {\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}+\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}}{b}\right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^3*(-2/b^2*(a*d-b*c)*d^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
)+(a^2*d^2-2*a*b*c*d+b^2*c^2)*d^2/b^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)-d^2/b^2*cos(d*x+c)-2*d^2*c/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)+2/b*(a*d-b*c)*d^2*c*(-sin(d*x+c)/((d*x+c)*b+d*a-c*b)/b+(S
i(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^2*c^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))

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\sin \left (c+d\,x\right )}{{\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*sin(c + d*x))/(a + b*x)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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