3.20 \(\int \frac {x^2 \sin (c+d x)}{a+b x} \, dx\)
Optimal. Leaf size=99 \[ \frac {a^2 \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {a \cos (c+d x)}{b^2 d}+\frac {\sin (c+d x)}{b d^2}-\frac {x \cos (c+d x)}{b d} \]
[Out]
a*cos(d*x+c)/b^2/d-x*cos(d*x+c)/b/d+a^2*cos(-c+a*d/b)*Si(a*d/b+d*x)/b^3-a^2*Ci(a*d/b+d*x)*sin(-c+a*d/b)/b^3+si
n(d*x+c)/b/d^2
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Rubi [A] time = 0.26, antiderivative size = 99, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used =
{6742, 2638, 3296, 2637, 3303, 3299, 3302} \[ \frac {a^2 \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^3}+\frac {a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {a \cos (c+d x)}{b^2 d}+\frac {\sin (c+d x)}{b d^2}-\frac {x \cos (c+d x)}{b d} \]
Antiderivative was successfully verified.
[In]
Int[(x^2*Sin[c + d*x])/(a + b*x),x]
[Out]
(a*Cos[c + d*x])/(b^2*d) - (x*Cos[c + d*x])/(b*d) + (a^2*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^3 + Si
n[c + d*x]/(b*d^2) + (a^2*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^3
Rule 2637
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Rule 2638
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Rule 3296
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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]
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} \, dx &=\int \left (-\frac {a \sin (c+d x)}{b^2}+\frac {x \sin (c+d x)}{b}+\frac {a^2 \sin (c+d x)}{b^2 (a+b x)}\right ) \, dx\\ &=-\frac {a \int \sin (c+d x) \, dx}{b^2}+\frac {a^2 \int \frac {\sin (c+d x)}{a+b x} \, dx}{b^2}+\frac {\int x \sin (c+d x) \, dx}{b}\\ &=\frac {a \cos (c+d x)}{b^2 d}-\frac {x \cos (c+d x)}{b d}+\frac {\int \cos (c+d x) \, dx}{b d}+\frac {\left (a^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}{b^2}+\frac {\left (a^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}{b^2}\\ &=\frac {a \cos (c+d x)}{b^2 d}-\frac {x \cos (c+d x)}{b d}+\frac {a^2 \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^3}+\frac {\sin (c+d x)}{b d^2}+\frac {a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^3}\\ \end {align*}
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Mathematica [A] time = 0.32, size = 87, normalized size = 0.88 \[ \frac {a^2 d^2 \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (d \left (\frac {a}{b}+x\right )\right )+a^2 d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+b (d (a-b x) \cos (c+d x)+b \sin (c+d x))}{b^3 d^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(x^2*Sin[c + d*x])/(a + b*x),x]
[Out]
(a^2*d^2*CosIntegral[d*(a/b + x)]*Sin[c - (a*d)/b] + b*(d*(a - b*x)*Cos[c + d*x] + b*Sin[c + d*x]) + a^2*d^2*C
os[c - (a*d)/b]*SinIntegral[d*(a/b + x)])/(b^3*d^2)
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fricas [A] time = 0.65, size = 133, normalized size = 1.34 \[ \frac {2 \, a^{2} d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {b d x + a d}{b}\right ) + 2 \, b^{2} \sin \left (d x + c\right ) - 2 \, {\left (b^{2} d x - a b d\right )} \cos \left (d x + c\right ) - {\left (a^{2} d^{2} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + a^{2} d^{2} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{2 \, b^{3} d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*sin(d*x+c)/(b*x+a),x, algorithm="fricas")
[Out]
1/2*(2*a^2*d^2*cos(-(b*c - a*d)/b)*sin_integral((b*d*x + a*d)/b) + 2*b^2*sin(d*x + c) - 2*(b^2*d*x - a*b*d)*co
s(d*x + c) - (a^2*d^2*cos_integral((b*d*x + a*d)/b) + a^2*d^2*cos_integral(-(b*d*x + a*d)/b))*sin(-(b*c - a*d)
/b))/(b^3*d^2)
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giac [C] time = 0.93, size = 2205, normalized size = 22.27 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*sin(d*x+c)/(b*x+a),x, algorithm="giac")
[Out]
1/2*(a^2*d^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - a^2*d
^2*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a^2*d^2*sin_
integral((b*d*x + a*d)/b)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a^2*d^2*real_part(cos_integ
ral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b) + 2*a^2*d^2*real_part(cos_integral(-d*x -
a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b) - 2*a^2*d^2*real_part(cos_integral(d*x + a*d/b))*t
an(1/2*d*x + 1/2*c)^2*tan(1/2*c)*tan(1/2*a*d/b)^2 - 2*a^2*d^2*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*
x + 1/2*c)^2*tan(1/2*c)*tan(1/2*a*d/b)^2 + 2*b^2*d*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - a^
2*d^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + a^2*d^2*imag_part(cos_integra
l(-d*x - a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*d*x + 1
/2*c)^2*tan(1/2*c)^2 + 4*a^2*d^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)*tan(1/
2*a*d/b) - 4*a^2*d^2*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)*tan(1/2*a*d/b) +
8*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)*tan(1/2*a*d/b) - a^2*d^2*imag_part(c
os_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/b)^2 + a^2*d^2*imag_part(cos_integral(-d*x - a*d/
b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/b)^2 - 2*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*d*x + 1/2*c)^2*t
an(1/2*a*d/b)^2 + a^2*d^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - a^2*d^2*imag_pa
rt(cos_integral(-d*x - a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2
*c)^2*tan(1/2*a*d/b)^2 - 2*a*b*d*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a^2*d^2*real_part(co
s_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 2*a^2*d^2*real_part(cos_integral(-d*x - a*d/b))*t
an(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 2*b^2*d*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*a^2*d^2*real_part(cos_int
egral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/b) - 2*a^2*d^2*real_part(cos_integral(-d*x - a*d/b))*ta
n(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/b) + 2*a^2*d^2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/
b) + 2*a^2*d^2*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b) + 2*b^2*d*x*tan(1/2*d*x + 1/2
*c)^2*tan(1/2*a*d/b)^2 - 2*a^2*d^2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*c)*tan(1/2*a*d/b)^2 - 2*a^2*d^
2*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*c)*tan(1/2*a*d/b)^2 - 2*b^2*d*x*tan(1/2*c)^2*tan(1/2*a*d/b)^2
+ a^2*d^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2 - a^2*d^2*imag_part(cos_integral(-d*x -
a*d/b))*tan(1/2*d*x + 1/2*c)^2 + 2*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*d*x + 1/2*c)^2 - a^2*d^2*imag
_part(cos_integral(d*x + a*d/b))*tan(1/2*c)^2 + a^2*d^2*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*c)^2 - 2
*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*c)^2 - 2*a*b*d*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 4*a^2*d^2*
imag_part(cos_integral(d*x + a*d/b))*tan(1/2*c)*tan(1/2*a*d/b) - 4*a^2*d^2*imag_part(cos_integral(-d*x - a*d/b
))*tan(1/2*c)*tan(1/2*a*d/b) + 8*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*c)*tan(1/2*a*d/b) - a^2*d^2*ima
g_part(cos_integral(d*x + a*d/b))*tan(1/2*a*d/b)^2 + a^2*d^2*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*a*d
/b)^2 - 2*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*a*d/b)^2 - 2*a*b*d*tan(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/
b)^2 + 2*a*b*d*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 4*b^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*b^
2*d*x*tan(1/2*d*x + 1/2*c)^2 + 2*a^2*d^2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*c) + 2*a^2*d^2*real_part
(cos_integral(-d*x - a*d/b))*tan(1/2*c) - 2*b^2*d*x*tan(1/2*c)^2 - 2*a^2*d^2*real_part(cos_integral(d*x + a*d/
b))*tan(1/2*a*d/b) - 2*a^2*d^2*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*a*d/b) - 2*b^2*d*x*tan(1/2*a*d/b)
^2 + a^2*d^2*imag_part(cos_integral(d*x + a*d/b)) - a^2*d^2*imag_part(cos_integral(-d*x - a*d/b)) + 2*a^2*d^2*
sin_integral((b*d*x + a*d)/b) - 2*a*b*d*tan(1/2*d*x + 1/2*c)^2 + 2*a*b*d*tan(1/2*c)^2 + 4*b^2*tan(1/2*d*x + 1/
2*c)*tan(1/2*c)^2 + 2*a*b*d*tan(1/2*a*d/b)^2 + 4*b^2*tan(1/2*d*x + 1/2*c)*tan(1/2*a*d/b)^2 - 2*b^2*d*x + 2*a*b
*d + 4*b^2*tan(1/2*d*x + 1/2*c))/(b^3*d^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + b^3*d^2*tan(1
/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + b^3*d^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/b)^2 + b^3*d^2*tan(1/2*c)^2*tan(1/
2*a*d/b)^2 + b^3*d^2*tan(1/2*d*x + 1/2*c)^2 + b^3*d^2*tan(1/2*c)^2 + b^3*d^2*tan(1/2*a*d/b)^2 + b^3*d^2)
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maple [B] time = 0.03, size = 315, normalized size = 3.18 \[ \frac {\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) d \left (\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}\right )}{b^{2}}+\frac {\left (-d a +c b +b \right ) d \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{b^{2}}+\frac {2 \left (d a -c b \right ) d c \left (\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}\right )}{b}+\frac {2 d c \cos \left (d x +c \right )}{b}+d \,c^{2} \left (\frac {\Si \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\Ci \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}\right )}{d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*sin(d*x+c)/(b*x+a),x)
[Out]
1/d^3*((a^2*d^2-2*a*b*c*d+b^2*c^2)*d/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)+(-a*d+b*c+b)*d/b^2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+2*(a*d-b*c)*d*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*d*c/b*cos(d*x+c)+d*c^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))
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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),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\sin \left (c+d\,x\right )}{a+b\,x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^2*sin(c + d*x))/(a + b*x),x)
[Out]
int((x^2*sin(c + d*x))/(a + b*x), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sin {\left (c + d x \right )}}{a + b x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*sin(d*x+c)/(b*x+a),x)
[Out]
Integral(x**2*sin(c + d*x)/(a + b*x), x)
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