3.1 \(\int x^2 \sin (a+b x+c x^2) \, dx\)

Optimal. Leaf size=249 \[ \frac{\sqrt{\frac{\pi }{2}} b^2 \sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{4 c^{5/2}}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{2 c^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sin \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{2 c^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} b^2 \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{4 c^{5/2}}+\frac{b \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac{x \cos \left (a+b x+c x^2\right )}{2 c} \]

[Out]

(b*Cos[a + b*x + c*x^2])/(4*c^2) - (x*Cos[a + b*x + c*x^2])/(2*c) + (Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelC[(b
+ 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/(2*c^(3/2)) + (b^2*Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c
]*Sqrt[2*Pi])])/(4*c^(5/2)) + (b^2*Sqrt[Pi/2]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(
4*c^(5/2)) - (Sqrt[Pi/2]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(2*c^(3/2))

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Rubi [A]  time = 0.242831, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3463, 3448, 3352, 3351, 3461, 3447} \[ \frac{\sqrt{\frac{\pi }{2}} b^2 \sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{4 c^{5/2}}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{2 c^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sin \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{2 c^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} b^2 \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{4 c^{5/2}}+\frac{b \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac{x \cos \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[a + b*x + c*x^2],x]

[Out]

(b*Cos[a + b*x + c*x^2])/(4*c^2) - (x*Cos[a + b*x + c*x^2])/(2*c) + (Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelC[(b
+ 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/(2*c^(3/2)) + (b^2*Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c
]*Sqrt[2*Pi])])/(4*c^(5/2)) + (b^2*Sqrt[Pi/2]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(
4*c^(5/2)) - (Sqrt[Pi/2]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(2*c^(3/2))

Rule 3463

Int[((d_.) + (e_.)*(x_))^(m_)*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> -Simp[(e*(d + e*x)^(m - 1)*
Cos[a + b*x + c*x^2])/(2*c), x] + (Dist[(e^2*(m - 1))/(2*c), Int[(d + e*x)^(m - 2)*Cos[a + b*x + c*x^2], x], x
] - Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Sin[a + b*x + c*x^2], x], x]) /; FreeQ[{a, b, c, d, e}, x]
&& NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rule 3448

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Cos[(b^2 - 4*a*c)/(4*c)], Int[Cos[(b + 2*c*x)^2/
(4*c)], x], x] + Dist[Sin[(b^2 - 4*a*c)/(4*c)], Int[Sin[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] &&
NeQ[b^2 - 4*a*c, 0]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3461

Int[((d_.) + (e_.)*(x_))*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> -Simp[(e*Cos[a + b*x + c*x^2])/(
2*c), x] + Dist[(2*c*d - b*e)/(2*c), Int[Sin[a + b*x + c*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*
d - b*e, 0]

Rule 3447

Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Cos[(b^2 - 4*a*c)/(4*c)], Int[Sin[(b + 2*c*x)^2/
(4*c)], x], x] - Dist[Sin[(b^2 - 4*a*c)/(4*c)], Int[Cos[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] &&
NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int x^2 \sin \left (a+b x+c x^2\right ) \, dx &=-\frac{x \cos \left (a+b x+c x^2\right )}{2 c}+\frac{\int \cos \left (a+b x+c x^2\right ) \, dx}{2 c}-\frac{b \int x \sin \left (a+b x+c x^2\right ) \, dx}{2 c}\\ &=\frac{b \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac{x \cos \left (a+b x+c x^2\right )}{2 c}+\frac{b^2 \int \sin \left (a+b x+c x^2\right ) \, dx}{4 c^2}+\frac{\cos \left (a-\frac{b^2}{4 c}\right ) \int \cos \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx}{2 c}-\frac{\sin \left (a-\frac{b^2}{4 c}\right ) \int \sin \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx}{2 c}\\ &=\frac{b \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac{x \cos \left (a+b x+c x^2\right )}{2 c}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) C\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{2 c^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right ) \sin \left (a-\frac{b^2}{4 c}\right )}{2 c^{3/2}}+\frac{\left (b^2 \cos \left (a-\frac{b^2}{4 c}\right )\right ) \int \sin \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx}{4 c^2}+\frac{\left (b^2 \sin \left (a-\frac{b^2}{4 c}\right )\right ) \int \cos \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx}{4 c^2}\\ &=\frac{b \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac{x \cos \left (a+b x+c x^2\right )}{2 c}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) C\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{2 c^{3/2}}+\frac{b^2 \sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{4 c^{5/2}}+\frac{b^2 \sqrt{\frac{\pi }{2}} C\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right ) \sin \left (a-\frac{b^2}{4 c}\right )}{4 c^{5/2}}-\frac{\sqrt{\frac{\pi }{2}} S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right ) \sin \left (a-\frac{b^2}{4 c}\right )}{2 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.707047, size = 157, normalized size = 0.63 \[ \frac{\sqrt{2 \pi } \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right ) \left (b^2 \sin \left (a-\frac{b^2}{4 c}\right )+2 c \cos \left (a-\frac{b^2}{4 c}\right )\right )+\sqrt{2 \pi } S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right ) \left (b^2 \cos \left (a-\frac{b^2}{4 c}\right )-2 c \sin \left (a-\frac{b^2}{4 c}\right )\right )+2 \sqrt{c} (b-2 c x) \cos (a+x (b+c x))}{8 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[a + b*x + c*x^2],x]

[Out]

(2*Sqrt[c]*(b - 2*c*x)*Cos[a + x*(b + c*x)] + Sqrt[2*Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*(2*c*Cos[a
- b^2/(4*c)] + b^2*Sin[a - b^2/(4*c)]) + Sqrt[2*Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*(b^2*Cos[a - b
^2/(4*c)] - 2*c*Sin[a - b^2/(4*c)]))/(8*c^(5/2))

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Maple [A]  time = 0.009, size = 204, normalized size = 0.8 \begin{align*} -{\frac{x\cos \left ( c{x}^{2}+bx+a \right ) }{2\,c}}-{\frac{b}{2\,c} \left ( -{\frac{\cos \left ( c{x}^{2}+bx+a \right ) }{2\,c}}-{\frac{b\sqrt{2}\sqrt{\pi }}{4} \left ( \cos \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}-ca \right ) } \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx+{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) -\sin \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}-ca \right ) } \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx+{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) \right ){c}^{-{\frac{3}{2}}}} \right ) }+{\frac{\sqrt{2}\sqrt{\pi }}{4} \left ( \cos \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}-ca \right ) } \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx+{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) +\sin \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}-ca \right ) } \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx+{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x*cos(c*x^2+b*x+a)/c-1/2*b/c*(-1/2*cos(c*x^2+b*x+a)/c-1/4*b/c^(3/2)*2^(1/2)*Pi^(1/2)*(cos((1/4*b^2-c*a)/c
)*FresnelS(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b))-sin((1/4*b^2-c*a)/c)*FresnelC(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1
/2*b))))+1/4/c^(3/2)*2^(1/2)*Pi^(1/2)*(cos((1/4*b^2-c*a)/c)*FresnelC(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b))+sin
((1/4*b^2-c*a)/c)*FresnelS(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b)))

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Maxima [C]  time = 5.03571, size = 4115, normalized size = 16.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*b*c*(e^(1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + e^(-1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(
c)*cos(-1/4*(b^2 - 4*a*c)/c) + b*c*(4*I*e^(1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 4*I*e^(-1/4*(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c))*((4*c^2*x^2 + 4*b*c*x + b^2)/abs(c))^(3/2) + (b^3*
(-4*I*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + 4*I*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b
^2)/c))*abs(c)*cos(-1/4*(b^2 - 4*a*c)/c) - 4*b^3*(gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(
3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c) + (c^3*(-32*I*gamma(3/2, 1/4*
(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + 32*I*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-
1/4*(b^2 - 4*a*c)/c) - 32*c^3*(gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/4*(4*I*c^2*
x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c))*x^3 + (b*c^2*(-48*I*gamma(3/2, 1/4*(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c) + 48*I*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/4*(b^2 - 4
*a*c)/c) - 48*b*c^2*(gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*
b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c))*x^2 + (b^2*c*(-24*I*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c
*x + I*b^2)/c) + 24*I*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/4*(b^2 - 4*a*c)/c) -
24*b^2*c*(gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*
b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c))*x)*cos(3/2*arctan2((4*c^2*x^2 + 4*b*c*x + b^2)/c, 0)) + ((I*sqrt(pi
)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - I*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*
x + I*b^2)/c)) - 1))*b^5*cos(-1/4*(b^2 - 4*a*c)/c) + (sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)
/c)) - 1) + sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^5*sin(-1/4*(b^2 - 4*a*c)/c)
+ ((8*I*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - 8*I*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c
^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3*cos(-1/4*(b^2 - 4*a*c)/c) + 8*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x
^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3
*sin(-1/4*(b^2 - 4*a*c)/c))*x^3 + ((12*I*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - 1
2*I*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2*cos(-1/4*(b^2 - 4*a*c)/c) + 12
*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4
*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2*sin(-1/4*(b^2 - 4*a*c)/c))*x^2 + ((6*I*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c)) - 1) - 6*I*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c
*cos(-1/4*(b^2 - 4*a*c)/c) + 6*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(
erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c*sin(-1/4*(b^2 - 4*a*c)/c))*x)*cos(1/2*arctan2(
(4*c^2*x^2 + 4*b*c*x + b^2)/c, 0)) - (4*b^3*(gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2,
-1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/4*(b^2 - 4*a*c)/c) - b^3*(4*I*gamma(3/2, 1/4*(4*I*c^2
*x^2 + 4*I*b*c*x + I*b^2)/c) - 4*I*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2
- 4*a*c)/c) + (32*c^3*(gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*
I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/4*(b^2 - 4*a*c)/c) - c^3*(32*I*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*
b^2)/c) - 32*I*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c))*x^3 + (
48*b*c^2*(gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b
^2)/c))*abs(c)*cos(-1/4*(b^2 - 4*a*c)/c) - b*c^2*(48*I*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 4
8*I*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c))*x^2 + (24*b^2*c*(g
amma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs
(c)*cos(-1/4*(b^2 - 4*a*c)/c) - b^2*c*(24*I*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 24*I*gamma(3
/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/4*(b^2 - 4*a*c)/c))*x)*sin(3/2*arctan2((4*c^2*x^2
+ 4*b*c*x + b^2)/c, 0)) + ((sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(erf
(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^5*cos(-1/4*(b^2 - 4*a*c)/c) + (-I*sqrt(pi)*(erf(1/2*s
qrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + I*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c
)) - 1))*b^5*sin(-1/4*(b^2 - 4*a*c)/c) + (8*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1)
+ sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3*cos(-1/4*(b^2 - 4*a*c)/c) + (-8
*I*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 8*I*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^
2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3*sin(-1/4*(b^2 - 4*a*c)/c))*x^3 + (12*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2
*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c
^2*cos(-1/4*(b^2 - 4*a*c)/c) + (-12*I*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 12*I
*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2*sin(-1/4*(b^2 - 4*a*c)/c))*x^2 +
(6*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 +
4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c*cos(-1/4*(b^2 - 4*a*c)/c) + (-6*I*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*
I*b*c*x + I*b^2)/c)) - 1) + 6*I*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c*sin(
-1/4*(b^2 - 4*a*c)/c))*x)*sin(1/2*arctan2((4*c^2*x^2 + 4*b*c*x + b^2)/c, 0)))/(c^3*((4*c^2*x^2 + 4*b*c*x + b^2
)/abs(c))^(3/2)*abs(c))

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Fricas [A]  time = 1.55497, size = 448, normalized size = 1.8 \begin{align*} \frac{\sqrt{2}{\left (\pi b^{2} \sin \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right ) + 2 \, \pi c \cos \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt{\frac{c}{\pi }} \operatorname{C}\left (\frac{\sqrt{2}{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + \sqrt{2}{\left (\pi b^{2} \cos \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right ) - 2 \, \pi c \sin \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt{\frac{c}{\pi }} \operatorname{S}\left (\frac{\sqrt{2}{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) - 2 \,{\left (2 \, c^{2} x - b c\right )} \cos \left (c x^{2} + b x + a\right )}{8 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(2)*(pi*b^2*sin(-1/4*(b^2 - 4*a*c)/c) + 2*pi*c*cos(-1/4*(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_cos(1/2*
sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c) + sqrt(2)*(pi*b^2*cos(-1/4*(b^2 - 4*a*c)/c) - 2*pi*c*sin(-1/4*(b^2 - 4*a*c)/
c))*sqrt(c/pi)*fresnel_sin(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c) - 2*(2*c^2*x - b*c)*cos(c*x^2 + b*x + a))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 2.97246, size = 306, normalized size = 1.23 \begin{align*} -\frac{-\frac{i \, \sqrt{2} \sqrt{\pi }{\left (b^{2} + 2 i \, c\right )} \operatorname{erf}\left (-\frac{1}{4} \, \sqrt{2}{\left (2 \, x + \frac{b}{c}\right )}{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}\right ) e^{\left (-\frac{i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} - 2 i \,{\left (c{\left (2 i \, x + \frac{i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (i \, c x^{2} + i \, b x + i \, a\right )}}{16 \, c^{2}} - \frac{\frac{i \, \sqrt{2} \sqrt{\pi }{\left (b^{2} - 2 i \, c\right )} \operatorname{erf}\left (-\frac{1}{4} \, \sqrt{2}{\left (2 \, x + \frac{b}{c}\right )}{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}\right ) e^{\left (-\frac{-i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} - 2 i \,{\left (c{\left (2 i \, x + \frac{i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (-i \, c x^{2} - i \, b x - i \, a\right )}}{16 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-I*sqrt(2)*sqrt(pi)*(b^2 + 2*I*c)*erf(-1/4*sqrt(2)*(2*x + b/c)*(-I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*
(I*b^2 - 4*I*a*c)/c)/((-I*c/abs(c) + 1)*sqrt(abs(c))) - 2*I*(c*(2*I*x + I*b/c) - 2*I*b)*e^(I*c*x^2 + I*b*x + I
*a))/c^2 - 1/16*(I*sqrt(2)*sqrt(pi)*(b^2 - 2*I*c)*erf(-1/4*sqrt(2)*(2*x + b/c)*(I*c/abs(c) + 1)*sqrt(abs(c)))*
e^(-1/4*(-I*b^2 + 4*I*a*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c))) - 2*I*(c*(2*I*x + I*b/c) - 2*I*b)*e^(-I*c*x^2 -
I*b*x - I*a))/c^2