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

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

[Out]

1/4*b*cos(c*x^2+b*x+a)/c^2-1/2*x*cos(c*x^2+b*x+a)/c+1/4*cos(a-1/4/c*b^2)*FresnelC(1/2*(2*c*x+b)/c^(1/2)*2^(1/2
)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/c^(3/2)+1/8*b^2*cos(a-1/4/c*b^2)*FresnelS(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))
*2^(1/2)*Pi^(1/2)/c^(5/2)+1/8*b^2*FresnelC(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*sin(a-1/4/c*b^2)*2^(1/2)*Pi
^(1/2)/c^(5/2)-1/4*FresnelS(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*sin(a-1/4/c*b^2)*2^(1/2)*Pi^(1/2)/c^(3/2)

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Rubi [A]  time = 0.24, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.72, size = 157, normalized size = 0.63 \[ \frac {\sqrt {2 \pi } C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (b^2 \sin \left (a-\frac {b^2}{4 c}\right )+2 c \cos \left (a-\frac {b^2}{4 c}\right )\right )+\sqrt {2 \pi } S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (b^2 \cos \left (a-\frac {b^2}{4 c}\right )-2 c \sin \left (a-\frac {b^2}{4 c}\right )\right )+2 \sqrt {c} (b-2 c x) \cos (a+x (b+c x))}{8 c^{5/2}} \]

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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fricas [A]  time = 0.46, size = 172, normalized size = 0.69 \[ \frac {\sqrt {2} {\left (\pi b^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + 2 \, \pi c \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) + \sqrt {2} {\left (\pi b^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - 2 \, \pi c \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - 2 \, {\left (2 \, c^{2} x - b c\right )} \cos \left (c x^{2} + b x + a\right )}{8 \, c^{3}} \]

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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giac [C]  time = 0.20, size = 227, normalized size = 0.91 \[ -\frac {-\frac {i \, \sqrt {2} \sqrt {\pi } {\left (b^{2} + 2 i \, c\right )} \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 i \, {\left (c {\left (2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (i \, c x^{2} + i \, b x + i \, a\right )}}{16 \, c^{2}} - \frac {\frac {i \, \sqrt {2} \sqrt {\pi } {\left (b^{2} - 2 i \, c\right )} \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 i \, {\left (c {\left (2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (-i \, c x^{2} - i \, b x - i \, a\right )}}{16 \, c^{2}} \]

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

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maple [A]  time = 0.02, size = 204, normalized size = 0.82 \[ -\frac {x \cos \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {b \left (-\frac {\cos \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{2 c}+\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}} \]

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 = 1.69, size = 1558, normalized size = 6.26 \[ \text {result too large to display} \]

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]

-1/32*((((-(8*I + 8)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (8*I - 8)*sqr
t(2)*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 + (-(32*I - 32)*sqrt(2)*gamma
(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (32*I + 32)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x
+ I*b^2)/c))*c^4)*cos(-1/4*(b^2 - 4*a*c)/c) + (((8*I - 8)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*
c*x + I*b^2)/c)) - 1) - (8*I + 8)*sqrt(2)*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 + (-(32*I + 32)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (32*I - 32)*sqrt(2)*gamm
a(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*c^4)*sin(-1/4*(b^2 - 4*a*c)/c))*x^3 + (((-(12*I + 12)*sqrt(2
)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (12*I - 12)*sqrt(2)*sqrt(pi)*(erf(1/2*sq
rt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2 + (-(48*I - 48)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 +
4*I*b*c*x + I*b^2)/c) + (48*I + 48)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b*c^3)*cos(-
1/4*(b^2 - 4*a*c)/c) + (((12*I - 12)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1)
- (12*I + 12)*sqrt(2)*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 + (-(48*I +
48)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (48*I - 48)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^
2*x^2 + 4*I*b*c*x + I*b^2)/c))*b*c^3)*sin(-1/4*(b^2 - 4*a*c)/c))*x^2 - (8*b*c^2*(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))*cos(-1/4*(b^2 - 4*a*c)/c) - b*c^2*(-8*I*e^(1/4*
(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + 8*I*e^(-1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*sin(-1/4*(b^2 - 4*a*c
)/c))*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2) + (((-(6*I + 6)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*
b*c*x + I*b^2)/c)) - 1) + (6*I - 6)*sqrt(2)*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 + (-(24*I - 24)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (24*I + 24)*sqrt(2)*gamm
a(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*cos(-1/4*(b^2 - 4*a*c)/c) + (((6*I - 6)*sqrt(2)*sqr
t(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (6*I + 6)*sqrt(2)*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 + (-(24*I + 24)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x
+ I*b^2)/c) + (24*I - 24)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*sin(-1/4*(b^
2 - 4*a*c)/c))*x + ((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (I -
1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^5 + (-(4*I - 4)*sqrt(2)*gamm
a(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (4*I + 4)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x +
I*b^2)/c))*b^3*c)*cos(-1/4*(b^2 - 4*a*c)/c) + (((I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c
*x + I*b^2)/c)) - 1) - (I + 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^5
+ (-(4*I + 4)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (4*I - 4)*sqrt(2)*gamma(3/2, -1/4
*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*sin(-1/4*(b^2 - 4*a*c)/c))/(c^4*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^
(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(a + b*x + c*x^2),x)

[Out]

int(x^2*sin(a + b*x + c*x^2), x)

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

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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