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

Optimal. Leaf size=248 \[ \frac {\sqrt {\pi } \sin \left (2 a+\frac {b^2}{2 c}\right ) C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } b^2 \cos \left (2 a+\frac {b^2}{2 c}\right ) C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } b^2 \sin \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } \cos \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {b \sin \left (2 a+2 b x-2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x-2 c x^2\right )}{8 c}+\frac {x^3}{6} \]

[Out]

1/6*x^3+1/16*b*sin(-2*c*x^2+2*b*x+2*a)/c^2+1/8*x*sin(-2*c*x^2+2*b*x+2*a)/c+1/16*b^2*cos(2*a+1/2/c*b^2)*Fresnel
C((-2*c*x+b)/c^(1/2)/Pi^(1/2))*Pi^(1/2)/c^(5/2)-1/16*cos(2*a+1/2/c*b^2)*FresnelS((-2*c*x+b)/c^(1/2)/Pi^(1/2))*
Pi^(1/2)/c^(3/2)+1/16*FresnelC((-2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a+1/2/c*b^2)*Pi^(1/2)/c^(3/2)+1/16*b^2*Fresn
elS((-2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a+1/2/c*b^2)*Pi^(1/2)/c^(5/2)

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Rubi [A]  time = 0.22, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3467, 3464, 3447, 3351, 3352, 3462, 3448} \[ \frac {\sqrt {\pi } \sin \left (2 a+\frac {b^2}{2 c}\right ) \text {FresnelC}\left (\frac {b-2 c x}{\sqrt {\pi } \sqrt {c}}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } b^2 \cos \left (2 a+\frac {b^2}{2 c}\right ) \text {FresnelC}\left (\frac {b-2 c x}{\sqrt {\pi } \sqrt {c}}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } b^2 \sin \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } \cos \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {b \sin \left (2 a+2 b x-2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x-2 c x^2\right )}{8 c}+\frac {x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[a + b*x - c*x^2]^2,x]

[Out]

x^3/6 + (b^2*Sqrt[Pi]*Cos[2*a + b^2/(2*c)]*FresnelC[(b - 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(16*c^(5/2)) - (Sqrt[Pi]*
Cos[2*a + b^2/(2*c)]*FresnelS[(b - 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(16*c^(3/2)) + (Sqrt[Pi]*FresnelC[(b - 2*c*x)/(
Sqrt[c]*Sqrt[Pi])]*Sin[2*a + b^2/(2*c)])/(16*c^(3/2)) + (b^2*Sqrt[Pi]*FresnelS[(b - 2*c*x)/(Sqrt[c]*Sqrt[Pi])]
*Sin[2*a + b^2/(2*c)])/(16*c^(5/2)) + (b*Sin[2*a + 2*b*x - 2*c*x^2])/(16*c^2) + (x*Sin[2*a + 2*b*x - 2*c*x^2])
/(8*c)

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 3462

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

Rule 3464

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*S
in[a + b*x + c*x^2])/(2*c), x] + (-Dist[(e^2*(m - 1))/(2*c), Int[(d + e*x)^(m - 2)*Sin[a + b*x + c*x^2], x], x
] - Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Cos[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 3467

Int[((d_.) + (e_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]^(n_), x_Symbol] :> Int[ExpandTrigReduce[
(d + e*x)^m, Sin[a + b*x + c*x^2]^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 1]

Rubi steps

\begin {align*} \int x^2 \sin ^2\left (a+b x-c x^2\right ) \, dx &=\int \left (\frac {x^2}{2}-\frac {1}{2} x^2 \cos \left (2 a+2 b x-2 c x^2\right )\right ) \, dx\\ &=\frac {x^3}{6}-\frac {1}{2} \int x^2 \cos \left (2 a+2 b x-2 c x^2\right ) \, dx\\ &=\frac {x^3}{6}+\frac {x \sin \left (2 a+2 b x-2 c x^2\right )}{8 c}-\frac {\int \sin \left (2 a+2 b x-2 c x^2\right ) \, dx}{8 c}-\frac {b \int x \cos \left (2 a+2 b x-2 c x^2\right ) \, dx}{4 c}\\ &=\frac {x^3}{6}+\frac {b \sin \left (2 a+2 b x-2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x-2 c x^2\right )}{8 c}-\frac {b^2 \int \cos \left (2 a+2 b x-2 c x^2\right ) \, dx}{8 c^2}+\frac {\cos \left (2 a+\frac {b^2}{2 c}\right ) \int \sin \left (\frac {(2 b-4 c x)^2}{8 c}\right ) \, dx}{8 c}-\frac {\sin \left (2 a+\frac {b^2}{2 c}\right ) \int \cos \left (\frac {(2 b-4 c x)^2}{8 c}\right ) \, dx}{8 c}\\ &=\frac {x^3}{6}-\frac {\sqrt {\pi } \cos \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a+\frac {b^2}{2 c}\right )}{16 c^{3/2}}+\frac {b \sin \left (2 a+2 b x-2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x-2 c x^2\right )}{8 c}-\frac {\left (b^2 \cos \left (2 a+\frac {b^2}{2 c}\right )\right ) \int \cos \left (\frac {(2 b-4 c x)^2}{8 c}\right ) \, dx}{8 c^2}-\frac {\left (b^2 \sin \left (2 a+\frac {b^2}{2 c}\right )\right ) \int \sin \left (\frac {(2 b-4 c x)^2}{8 c}\right ) \, dx}{8 c^2}\\ &=\frac {x^3}{6}+\frac {b^2 \sqrt {\pi } \cos \left (2 a+\frac {b^2}{2 c}\right ) C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } \cos \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a+\frac {b^2}{2 c}\right )}{16 c^{3/2}}+\frac {b^2 \sqrt {\pi } S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a+\frac {b^2}{2 c}\right )}{16 c^{5/2}}+\frac {b \sin \left (2 a+2 b x-2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x-2 c x^2\right )}{8 c}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 175, normalized size = 0.71 \[ \frac {-3 \sqrt {\pi } C\left (\frac {2 c x-b}{\sqrt {c} \sqrt {\pi }}\right ) \left (c \sin \left (2 a+\frac {b^2}{2 c}\right )+b^2 \cos \left (2 a+\frac {b^2}{2 c}\right )\right )+3 \sqrt {\pi } S\left (\frac {2 c x-b}{\sqrt {c} \sqrt {\pi }}\right ) \left (c \cos \left (2 a+\frac {b^2}{2 c}\right )-b^2 \sin \left (2 a+\frac {b^2}{2 c}\right )\right )+\sqrt {c} \left (3 (b+2 c x) \sin (2 (a+x (b-c x)))+8 c^2 x^3\right )}{48 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[a + b*x - c*x^2]^2,x]

[Out]

(3*Sqrt[Pi]*FresnelS[(-b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*(c*Cos[2*a + b^2/(2*c)] - b^2*Sin[2*a + b^2/(2*c)]) - 3*
Sqrt[Pi]*FresnelC[(-b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*(b^2*Cos[2*a + b^2/(2*c)] + c*Sin[2*a + b^2/(2*c)]) + Sqrt[
c]*(8*c^2*x^3 + 3*(b + 2*c*x)*Sin[2*(a + x*(b - c*x))]))/(48*c^(5/2))

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fricas [A]  time = 0.46, size = 187, normalized size = 0.75 \[ \frac {8 \, c^{3} x^{3} - 6 \, {\left (2 \, c^{2} x + b c\right )} \cos \left (c x^{2} - b x - a\right ) \sin \left (c x^{2} - b x - a\right ) - 3 \, {\left (\pi b^{2} \cos \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) + \pi c \sin \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {{\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) - 3 \, {\left (\pi b^{2} \sin \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) - \pi c \cos \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{c}\right )}{48 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*c^3*x^3 - 6*(2*c^2*x + b*c)*cos(c*x^2 - b*x - a)*sin(c*x^2 - b*x - a) - 3*(pi*b^2*cos(1/2*(b^2 + 4*a*c
)/c) + pi*c*sin(1/2*(b^2 + 4*a*c)/c))*sqrt(c/pi)*fresnel_cos((2*c*x - b)*sqrt(c/pi)/c) - 3*(pi*b^2*sin(1/2*(b^
2 + 4*a*c)/c) - pi*c*cos(1/2*(b^2 + 4*a*c)/c))*sqrt(c/pi)*fresnel_sin((2*c*x - b)*sqrt(c/pi)/c))/c^3

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giac [C]  time = 0.92, size = 216, normalized size = 0.87 \[ \frac {1}{6} \, x^{3} - \frac {{\left (c {\left (-2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (2 i \, c x^{2} - 2 i \, b x - 2 i \, a\right )} - \frac {\sqrt {\pi } {\left (b^{2} + i \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x - \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} - \frac {{\left (c {\left (2 i \, x - \frac {i \, b}{c}\right )} + 2 i \, b\right )} e^{\left (-2 i \, c x^{2} + 2 i \, b x + 2 i \, a\right )} - \frac {\sqrt {\pi } {\left (b^{2} - i \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x - \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/32*((c*(-2*I*x + I*b/c) - 2*I*b)*e^(2*I*c*x^2 - 2*I*b*x - 2*I*a) - sqrt(pi)*(b^2 + I*c)*erf(-1/2*s
qrt(c)*(2*x - b/c)*(-I*c/abs(c) + 1))*e^(-1/2*(I*b^2 + 4*I*a*c)/c)/(sqrt(c)*(-I*c/abs(c) + 1)))/c^2 - 1/32*((c
*(2*I*x - I*b/c) + 2*I*b)*e^(-2*I*c*x^2 + 2*I*b*x + 2*I*a) - sqrt(pi)*(b^2 - I*c)*erf(-1/2*sqrt(c)*(2*x - b/c)
*(I*c/abs(c) + 1))*e^(-1/2*(-I*b^2 - 4*I*a*c)/c)/(sqrt(c)*(I*c/abs(c) + 1)))/c^2

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maple [A]  time = 0.07, size = 199, normalized size = 0.80 \[ \frac {x^{3}}{6}+\frac {x \sin \left (-2 c \,x^{2}+2 b x +2 a \right )}{8 c}-\frac {b \left (-\frac {\sin \left (-2 c \,x^{2}+2 b x +2 a \right )}{4 c}+\frac {b \sqrt {\pi }\, \left (\cos \left (\frac {4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{4 c}+\frac {\sqrt {\pi }\, \left (\cos \left (\frac {4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{16 c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3+1/8*x*sin(-2*c*x^2+2*b*x+2*a)/c-1/4*b/c*(-1/4*sin(-2*c*x^2+2*b*x+2*a)/c+1/4*b/c^(3/2)*Pi^(1/2)*(cos(1/
2*(4*a*c+b^2)/c)*FresnelC(1/Pi^(1/2)/c^(1/2)*(2*c*x-b))+sin(1/2*(4*a*c+b^2)/c)*FresnelS(1/Pi^(1/2)/c^(1/2)*(2*
c*x-b))))+1/16/c^(3/2)*Pi^(1/2)*(cos(1/2*(4*a*c+b^2)/c)*FresnelS(1/Pi^(1/2)/c^(1/2)*(2*c*x-b))-sin(1/2*(4*a*c+
b^2)/c)*FresnelC(1/Pi^(1/2)/c^(1/2)*(2*c*x-b)))

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maxima [C]  time = 2.61, size = 1603, normalized size = 6.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*sqrt(2)*(((((24*I - 24)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1)
- (24*I + 24)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3 + (-(4
8*I + 48)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) + (48*I - 48)*sqrt(2)*gamma(3/2, -1/2*(4
*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*c^4)*cos(1/2*(b^2 + 4*a*c)/c) + ((-(24*I + 24)*sqrt(2)*sqrt(pi)*(erf(sqrt(
1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + (24*I - 24)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*
c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3 + (-(48*I - 48)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*x
+ I*b^2)/c) + (48*I + 48)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*c^4)*sin(1/2*(b^2 + 4
*a*c)/c))*x^3 + (((-(36*I - 36)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1
) + (36*I + 36)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2 + ((
72*I + 72)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) - (72*I - 72)*sqrt(2)*gamma(3/2, -1/2*(
4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*b*c^3)*cos(1/2*(b^2 + 4*a*c)/c) + (((36*I + 36)*sqrt(2)*sqrt(pi)*(erf(sqr
t(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) - (36*I - 36)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*
I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2 + ((72*I - 72)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*
x + I*b^2)/c) - (72*I + 72)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*b*c^3)*sin(1/2*(b^2
+ 4*a*c)/c))*x^2 + 2*sqrt(2)*(16*c^4*x^3 + b*c^2*(6*I*e^(1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) - 6*I*e^(-1/
2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*cos(1/2*(b^2 + 4*a*c)/c) + 6*b*c^2*(e^(1/2*(4*I*c^2*x^2 - 4*I*b*c*x +
I*b^2)/c) + e^(-1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*sin(1/2*(b^2 + 4*a*c)/c))*((4*c^2*x^2 - 4*b*c*x + b^
2)/c)^(3/2) + ((((18*I - 18)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) -
(18*I + 18)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c + (-(36*I
+ 36)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) + (36*I - 36)*sqrt(2)*gamma(3/2, -1/2*(4*I*
c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*cos(1/2*(b^2 + 4*a*c)/c) + ((-(18*I + 18)*sqrt(2)*sqrt(pi)*(erf(sqrt
(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + (18*I - 18)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I
*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c + (-(36*I - 36)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*x
+ I*b^2)/c) + (36*I + 36)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*sin(1/2*(b^2
+ 4*a*c)/c))*x + ((-(3*I - 3)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1)
+ (3*I + 3)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^5 + ((6*I + 6)
*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) - (6*I - 6)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2
- 4*I*b*c*x + I*b^2)/c))*b^3*c)*cos(1/2*(b^2 + 4*a*c)/c) + (((3*I + 3)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4
*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) - (3*I - 3)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*
b*c*x + I*b^2)/c)) - 1))*b^5 + ((6*I - 6)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) - (6*I +
6)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c))*b^3*c)*sin(1/2*(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 \[ \int x^2\,{\sin \left (-c\,x^2+b\,x+a\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(a + b*x - c*x^2)^2,x)

[Out]

int(x^2*sin(a + b*x - c*x^2)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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