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

Optimal. Leaf size=251 \[ -\frac {\sqrt {\frac {\pi }{2}} b^2 \sin \left (a+\frac {b^2}{4 c}\right ) C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}+\frac {\sqrt {\frac {\pi }{2}} \cos \left (a+\frac {b^2}{4 c}\right ) C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} \sin \left (a+\frac {b^2}{4 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} b^2 \cos \left (a+\frac {b^2}{4 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}+\frac {b \cos \left (a+b x-c x^2\right )}{4 c^2}+\frac {x \cos \left (a+b x-c x^2\right )}{2 c} \]

[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 = 0.20, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, 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 = 165, normalized size = 0.66 \[ \frac {-\sqrt {2 \pi } C\left (\frac {2 c x-b}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (2 c \cos \left (a+\frac {b^2}{4 c}\right )-b^2 \sin \left (a+\frac {b^2}{4 c}\right )\right )-\sqrt {2 \pi } S\left (\frac {2 c x-b}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (2 c \sin \left (a+\frac {b^2}{4 c}\right )+b^2 \cos \left (a+\frac {b^2}{4 c}\right )\right )+2 \sqrt {c} (b+2 c x) \cos (a+x (b-c x))}{8 c^{5/2}} \]

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 = 0.43, size = 179, normalized size = 0.71 \[ \frac {\sqrt {2} {\left (\pi b^{2} \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) - 2 \, \pi c \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} {\left (\pi b^{2} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + 2 \, \pi c \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) + 2 \, {\left (2 \, c^{2} x + b c\right )} \cos \left (c x^{2} - b x - a\right )}{8 \, c^{3}} \]

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*sq
rt(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 = 0.18, size = 229, 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 = 0.03, size = 202, normalized size = 0.80 \[ \frac {x \cos \left (-c \,x^{2}+b x +a \right )}{2 c}-\frac {b \left (-\frac {\cos \left (-c \,x^{2}+b x +a \right )}{2 c}+\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}+c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x -\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}+c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x -\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{2 c}-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}+c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x -\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {\frac {b^{2}}{4}+c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x -\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}} \]

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))+si
n((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 = 2.85, size = 1556, normalized size = 6.20 \[ \text {result too large to display} \]

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)*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)*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)*gamma(
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)*sq
rt(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)*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)*s
qrt(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)*gamma(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)*sqrt(pi)*(er
f(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)*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)*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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